Applied physics sem 2 polytechnic

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Six
Heat and thermodynamics
Heat and Thermodynamics Temperature and zeroth law of thermodynamics, heat
and work, specific heat capacity and latent heat, first and second laws
of thermodynamics, heat engine and its efficiency, entropy and its non-
conservation
6.1 Introduction
Thermodynamics is a word that is a combination of thermal and dynamics. Thermal is related
to heat energy and dynamics is related to motion. History of thermodynamics is involved
with the conversion of heat energy into useful mechanical work and vice versa. Historically
heat was treated as a completely different physical quantity in comparison to other forms of
energy. In fact it was believed that hot objects possessed invisible fluid called caloric. Thus
when one hot object came in contact with a cold object this invisible fluid caloric flows to the
cold object. But today we know that heat is just another form of energy that is transferred
due to the random motion of the constituent particles of the object. Temperature on the other
hand is a measure of the hotness or coldness of any object.
Thermodynamics is concerned with energy relationships involving heat, mechanical energy
and other aspects of energy and energy transfer. In thermodynamics we usually have a well
defined system which is simply a specified quantity of matter. A thermodynamic system is one
that can interact with its surroundings in at least two ways one of which must be transfer of
heat. A familiar example is a quantity of a gas confined in a cylinder with a piston. Energy can
be added to the system by conduction of heat and it is also possible to do Mechanical work
on the system since the Piston exerts a force that can move fluid and produce displacement.
When a hot body is kept in contact with a cold body, the cold body warms up and the
71

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72 6. Heat and thermodynamics
hot body cools down. The internal energy of the hot body decreases and the internal energy
of the cold body increases. The energy is transferred when the bodies are in contact. The
transfer of energy is a non mechanical process i.e. no actual displacements are involved. The
energy that is transferred from one body to another without any mechanical work involved is
called heat.
6.2 Temperature and zeroth law of thermodynamics
If no heat transfer takes place between two bodies kept in contact we say that the bodies are
in thermal equilibrium with each other. The zeroth law of thermodynamics states that
If two bodies X and Y are in thermal equilibrium and X and Z are also in thermal equilibrium
then Y and Z are also in thermal equilibrium.
This law allows us to introduce the idea of temperature to measure the relative hotness or
coldness of objects. All bodies in thermal equilibrium for which no heat transfer takes place
are defined to have equal temperatures. A hotter body has a higher temperature and a colder
body has a lower temperature by definition. It is known that the heat flows from a hotter
body i.e. one at higher temperature to the colder one at lower temperature. Thus temperature
determines the direction of the heat flow.
To actually compare the temperatures of two bodies, we need to measure it. Direct mea-
surement of temperature is difficult. So instead we use a convenient physical property of an
object that changes with temperature. For example the volume of mercury changes with tem-
perature. Thus volume is one of the convenient properties. Other properties that can be used
are electrical resistance, magnetization of objects, pressure of a gas etc. These properties that
depend on temperature are known as thermometric properties. As an example we take some
mercury in a glass bulb with a long narrow capillary. The length of the mercury column then
changes with the temperature. To assign numerical values we need to chose two fixed points
of temperature which can easily be reproduced in laboratory. The temperature of melting ice
at 1 atmosphere called the ice point and the temperature of boiling water at 1 atmosphere
called the steam point are often chosen as the fixed points. The thermometer is first brought
in contact with melting Ice and the resulting level of mercury is noted then the same is made
to contact with the boiling water and the resulting level of mercury is noted. The distance
between these two points is divided into equal parts. In centigrade scale traditionally the ice
point is taken to be at zero degree Celsius and the boiling point is taken to be 100 degree
Celsius. Another popular system known as Fahrenheit systems assumes 32 degree Fahrenheit

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6.3. IDEAL GAS AS A THERMODYNAMIC SYSTEM 73
for the ice point and 212 degree Fahrenheit for the steam point.
F = 32 +
9
5
C. (6.2.1)
It is also possible to define an absolute temperature scale which does not depend on any
property of any substance. The unit of this temperature is called Kelvin and is abbreviated as
K, It is also called the Absolute Temperature scale. The temperature of the ice point on this
scale is 273.15 K and of the steam point it is 373.15 K. If C denotes the Celsius temperature
and T denotes the Kelvin temperature then,
C = T − 273.15 K. (6.2.2)
6.3 Ideal Gas as a thermodynamic system
Gases are most extensively studied systems in thermodynamics. The state of a gas is specified
by specifying the values of its Pressure (P), Volume (V), Temperature (T), its amount (in moles
n or in mass m) and its type (oxygen, hydrogen etc or its combination). All gases behave
somewhat differently when subjected to similar conditions. However under very low pressures
and high temperatures the behavior of gases is very similar. We may study the behavior of a
gas by means of a cylinder with a movable piston with a pressure gauge and a thermometer.
We may vary the pressure, volume and temperature and pump any desired mass of any gas
into the cylinder in order to explore the relationship between pressure, volume, temperature
and quantity of substance. It is often convenient to measure the amount of gas in terms of the
number of moles n.
The values of P, V , T, and n specify the state of the gas. Whenever a change in the
values of these variables take place, we say that the gas is undergoing a process. Thus going
from initial state P1, V1, T1, n1 to P2, V2, T2, n2 is a process. These state variables are not
independent of one another but are connected by an equation known as the equation of state.
Observations made on gases at very low pressures and high temperatures can be summarized
as below.
When we take a gas of fixed quantity n from initial state P1, V1 to final state P2, V2 by
keeping its temperature T = constant, then the process is said to be isothermal. In such a
process we find that the Pressure is inversely proportional to the Volume. This is known as
Boyle’s law i.e.
P ∝
1
V
or PV = const, or P1V1 = P2V2. (6.3.1)

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When we take an ideal gas of fixed quantity n from initial state V1, T1 to final state V2, T2
by keeping its pressure P = constant, then the process is said to be isobaric. In such a
process we find that the volume is directly proportional to its (absolute) temperature . This
is known as Charles’s law
V ∝ T or
V
T
= const, or
V1
T1
=
V2
T2
. (6.3.2)
When we take an ideal gas of fixed quantity n from initial state P1, T1 to final state P2, T2
by keeping its volume V = constant, then the process is said to be isochoric. In such a process
we find that the pressure is directly proportional to its (absolute) temperature. This is known
as Gay-lussac’s law
P ∝ T or
P
T
= const, or
P1
T1
=
P2
T2
. (6.3.3)
All of these conclusions can be summarized neatly in a single equation, known as the
equation of state;
PV = nRT, (6.3.4)
where R is a constant whose value is found to be same for all gases (at low pressures and
high temperatures). The constant R = 8.314 J ·mol−1
·K−1
is therefore known as the universal
gas constant. Thus we can at most fix three values of the state variables (say P,V and n) then
the fourth (T in this case) gets fixed by equation (6.3.4). We now define an ideal gas as a
gas for which equation (6.3.4) holds true for all pressures and temperatures. Ideal gas is an
idealized model which represents the behavior of gases very well in certain circumstances.
Generally gas behavior approximates the ideal gas model most closely at very low pressures
when the gas molecules are far apart.
Numerical Problems
Example 1: Find the volume of 1 mole of any ideal gas at standard temperature and pressure
STP that is 1 atmosphere pressure and zero degree Celsius.
Example 2: A tank attached to an air compressor contains 20 liters of air at a temperature of 30
degree Celsius and gauge pressure of 4 × 105
Pa. What is the number of moles of air and what
volume would it occupy at normal atmospheric pressure and zero degree Celsius. (gauge pressure
is pressure in excess of atmospheric pressure).
Example 3: Convert 27o
C into Fahrenheit and kelvin.

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6.4. WORK 75
6.4 Work
If a system undergoes a displacement under the action of a force work is said to be done. The
amount of work done is equal to the product of the force and the component of the displacement
parallel to the force. In Thermodynamics only the work that involves an interaction between
a system and its surroundings, known as external work is analyzed. Work done by one
part of a system on another part is called internal work and it is not considered. Thus a
gas confined in a cylinder and at uniform pressure while expanding and imparting motion
to a piston does external work on its surroundings. When a system does external work it
changes the state of the system. There is also a corresponding change in the surroundings.
For example in the surroundings there may be raising or lowering of a suspended weight, the
winding or unwinding of a spring or in general the alteration of the position or configuration of
some external mechanical device. Usually work done by the system (expansion) is considered
positive whereas work done onto the system (compression) is considered negative.
6.4.1 Work in a volume change
Imagine a system such as a gas contained in a closed cylinder equipped with a frictionless
movable piston as shown in fig (6.4.1) on which the system and the surroundings may act.
Suppose that the cylinder has a cross section area A and that the pressure exerted by the
system at the piston face is P.
A
PA
dx
Figure 6.4.1: Work in a volume change
The force on to the Piston face is therefore PA. If under these conditions the Piston moves

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in an infinitesimal distance dx during expansion as shown in the figure then the system
performs an infinitesimal amount of positive work dW ,
dW = Fdx = PAdx. (6.4.1)
But, during expansion, the volume of the system is increasing, so
Adx = dV ,
and hence
dW = PdV .
In a finite (very slow) process if the volume changes from Vi to Vf , the amount of work
W done by the system is
W =
Vf
ˆ
Vi
PdV . (6.4.2)
The amount of work done in general will depend on the kind of process involved e.g. isothermal,
isobaric, etc.
6.4.2 Path dependence of work
Work performed on or by the system can also be calculated from a PV diagram. The values of
Pressure and Volume are recorded throughout the process from initial state P1, V1 to a final
state P2, V2. Then the area enclosed by the curve with volume axis gives the work done in the
process as shown in the diagram in fig. (6.4.2a) . On the PV diagram in the fig (6.4.2b) an
initial state 1 (characterized by pressure P1 and volume V1) and a final state 2 (characterized
by pressure P2 and volume V2) are represented by the two points 1 and 2. There are many
ways in which the system may be taken from state 1 to state 2. For example the pressure
may be kept constant from 1 → 3 and then the volume kept constant from 3 → 2 . In which
case the work done is equal to the area under the line 1 → 3. Another possibility is the path
1 → 4 → 2 in which case the work is the area under the line 4 → 2. Smooth curve from
1 → 2 represents another possibility and the work done is different in each case. We can
therefore say that the work done depends not only on the initial and final States but also on
the intermediate states, that is on the path that the system takes.

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6.5. HEAT 77
V1 V2
1
2
V
P
(a) Work as area
4 2
3
1
V1
P2
V2
P1
V
P
(b) Path dependece of work
Figure 6.4.2: Work on PV plane
6.5 Heat
The energy being transferred between two bodies or between adjacent parts of a body as a
result of temperature difference is called heat. Once it is transferred it becomes the internal
energy of the receiving body. The SI unit of heat is Joule. Another unit of heat Calorie is
also used.
The amount of heat needed to increase the temperature of 1 gram of water from 14.5
degree Celsius to 15.5 degree Celsius at a pressure of 1 atmosphere is called 1 calorie. The
calorie defined in terms of Joule is 1 calorie = 4.186 Joule.
Just as work performed depends on the kind of process that is done, heat transferred to or
by the system is different for different processes, i.e. heat is not a property of a system and
therefore is not a state variable.
6.5.1 Heat transfer
Conduction
It is the primary mode of transfer of heat in solids. The flow of energy between adjacent
parts of a body due to the temperature difference between them is called conduction.
Consider a solid slab having a cross sectional area A and of thickness d as shown in figure
(6.5.1). Let the opposite parallel faces be at temperature T1 and T2with T2 > T1. Now heat
will start flowing from face at a higher temperature to the face at lower temperature and the
direction of heat flow will be normal to the two parallel faces of the slab.

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The amount of heat Q flowing from one face to another is
1. Directly proportional to the face area A
2. Directly proportional to the time t for which conduction takes place
3. directly proportional to the temperature difference (T2 − T1) between the faces
4. inversely proportional to the thickness d of the slab
From this it is clear that
Q ∝
A (T2 − T1) t
d
(6.5.1)
Q = K
A (T2 − T1) t
d
(6.5.2)
Here K the proportionality constant called the coeffient of thermal conductivity (or simply
thermal conductivity) of the nature of the material
K =
Qd
A (T2 − T1)
(6.5.3)
If A = 1 m2
,(T2 − T1) = 1 K, t = 1 s, and d = 1 m then K = Q numerically. Thus, coeficient
of thermal conductivity is equal to the quantity of heat which flows in one second through
a unit cube of material when it’s opposite faces are maintained at a temperature difference
of 1 ◦
C. Units of thermal conductivity are W
mK
and cal
cm ◦Cs
or kcal
mKs
.
Figure 6.5.1: Heat conduction

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6.5. HEAT 79
Numerical Problems
Example 4: Find the quantity of heat conducted in 10 min across a silver sheet of size 40×30 cm2
,
thickness 6 mm, if its two faces are at temperatures of 40 ◦
C and 30 ◦
C, Ksilver = 0.1 kcal/m ◦
Cs.
Example 5: A nickel plate of thickness 4 mm has a temperature difference of 32 ◦
C between its
faces. It transmits 200 kcal per hour through an area of 5 cm2
. Calculate the thermal conductivity
of nickel.
Example 6: Heat is conducted through a composite slab of two different metals having thermal
conductivity 0.2 and 0.3 kcal/m ◦
Cs and equal thickness. The outer faces are at 100 ◦
C and 0
◦
C. Find the temperature of interface.
6.5.2 Specific heat capacity and latent heat
Usually when an object is heated there are two changes that take place. Either the temper-
ature of object changes or the state of the object changes; i.e. it goes from solid to liquid
say.
Lets first take the case of change in temperature. It is found that the amount of heat
required to change the temperature of a given mass of object depends on the amount of
temperature change we want to bring about and also onto the mass of the object itself. That
is,
Q ∝ 4T (6.5.4)
Q ∝ m (6.5.5)
and if we remove the proportionality and write the combined equation then
Q = mS4T; (6.5.6)
where S; the proportionality constant is known as the specific heat capacity. S is the property
of the material, that is it is different for different materials. The specific heat of water is
S = 4.186 kJ/kg. Formula for specific heat capacity becomes
S =
Q
m4T
(6.5.7)
Specific heat capacity may be defined as the amount of heat required to change the temper-
ature of a unit mass of substance by 1 o
C. Its units are Jkg−1
K−1
or cal · g−1
·o
C.
Now instead of a rise in temperature, we also find that the state of a substance can change
when heat is given to or removed from it. During the process of change of state it is found

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that the temperature of the object does not change at all, for as long as complete state change
takes place. For example, ice melts at 0o
C, so if we supply heat to ice at 0o
C we do not see
any rise in temperature but the ice starts melting. So the ice does gain heat but we don’t
observe it as a rise in temperature, thus heat given gets used in changing the state of the
ice into water. When the ice gets completely converted in to water at 0o
C, supplying further
heat results in the increase of temperature of water. Thus the change of state/phase is an
isothermal process.
Let m be the mass of the object. If supplying Q amount of heat changes its phase completely
at constant temperature then through experiments it is found that
Q ∝ m. (6.5.8)
That is, the amount of heat required to change the state depends on the mass. Removing
proportionality
Q = mL; (6.5.9)
where L is called the latent heat.
L =
Q
m
(6.5.10)
denotes the heat required per unit mass to change the state and its SI unit is J/kg. If the
process is conversion of solid to liquid known as melting or fusion, the heat required per unit
mass is known as Latent heat of melting or fusion. The reverse process is solidification. If
the process is of vaporization, the heat required per unit mass is known as latent heat of
vaporization. The process of direct conversion from solid to liquid is known as sublimation.
6.5.3 Specific heat capacity of gases
As heat is given to gases, we see large expansion in gases i.e. their volume changes. But if
we take a gas in a fixed volume container then on heating; instead of volume, the pressure
changes. It is found that the heat required to raise the temperature of a gas depends on the
kind of process undertaken i.e. isobaric, isochoric, iso-thermal or some other combination.
Depending on the process the specific heat capacities of gases are different. Two of the most
important specific heat capacities are for the constant pressure and constant volume processes.
6.5.3.1 Specific heat capacity at constant pressure CP
The specific heat capacity of a gas at constant pressure is defined as the amount of heat
required to change the temperature of a unit mass of a gas by 1 ◦
C at constant pressure,

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6.5. HEAT 81
denoted by CP. If instead of unit mass the specific heat capacity is defined for unit moles,
then it is known as molar specific heat capacity. Thus
CP =
Q
m4T
or CP =
Q
n4T
(6.5.11)
6.5.3.2 Specific heat capacity at constant volume CV
The specific heat capacity of a gas at constant volume is defined as the amount of heat required
to change the temperature of a unit mass of a gas by 1 ◦
C at constant volume, denoted by
CV .
CV =
Q
m4T
or CV =
Q
n4T
(6.5.12)
6.5.3.3 Relation between CP and CV for an ideal gas
When an ideal gas is heated its temperature increases. But the increase of temperature is
different for the isobaric and isochoric processes. In an isochoric process since the volume
of the gas is kept constant, the gas is not allowed to expand and hence no work is done by
the gas. Hence all of the heat goes on to increase the temperature of the gas. Whereas in
the case of isobaric process the gas is allowed to expand in order to keep pressure constant,
thereby the gas performs work against surroundings. So part of the heat given during the
isobaric process goes to perform work and so the temperature change is not as large as that
in an isochoric process. Thus we find that
4TV > 4TP. (6.5.13)
Since specific heat ∝ 1
4T
, we must have
CP > CV .
For an ideal gas it is easy to establish the difference of the specific heat capacities. We
have seen that the difference between the isochoric and isobaric process is the performance
of work done by the gas during the isobaric process. If we want the same rise in temperature
during both these processes, then the isobaric process will require more heat to be given,
since it must also contribute to perform external work. Let QP be the heat given for the
isobaric process and QV be the heat given for the isochoric process to n moles of ideal gas
and let 4T = T2 − T1 be the resulting change in temperature. During the isobaric process
let the volume change from V1 to V2 and the pressure of the gas be P. Thus we may write

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the difference of heat as
QP − QV = work done during isobaric expansion (6.5.14)
QP − QV = P(V2 − V1) (6.5.15)
Since during an isobaric process Gay lussac’s law holds; we have from the ideal gas equation
of state
P(V2 − V1) = nR(T2 − T1) = nR4T. (6.5.16)
So eq (6.5.15) can be written as
QP − QV = nR4T (6.5.17)
QP
n4T
−
QV
n4T
= R (6.5.18)
CP − CV = R. (6.5.19)
Thus the difference between molar specific heat capacities of an ideal gas is equal to the
universal gas constant.
6.5.4 Principle of Calorimetry
Calorimetry is measurement of heat. When substances are mixed together or kept in contact,
heat flow takes place and eventually all substances will acquire same temperature. The
principle of Calorimetry says, ’heat gained (by some substances) = heat lost (by some other
substances)’. As it is formulated the principle is just the conservation of energy principle.
Usually heat lost to the environment is neglected if it is small.
Numerical Problems
Example 7: A sphere of aluminium of 0.047 kg at 100 o
C is transferred to 0.14 kg copper
calorimeter containing 0.25 kg of water at 20 o
C. The temperature of water rises and attains a
steady state at 23 o
C. Calculate the specific heat capacity of aluminium. Specific heat capacities
of water and copper are 4186 J · kg−1
· K−1
and 386.4 J · kg−1
· K−1
.
Example 8: When 0.15 kg of ice at 0 o
C is mixed with 0.30 kg of water at 50 o
C, the resulting
temperature is 6.7 o
C. Calculate the latent heat of fusion of ice. (Swater = 4186 J · kg−1
· K−1
).
Example 9: Calculate the heat required to convert 3 kg of ice at -12 o
C into steam at 100
o
C. Given specific heat capacity of ice = 2100 J · kg−1
· K−1
, specific heat capacity of water
= 4186 J ·kg−1
·K−1
, latent heat of fusion of ice = 3.35×105
J ·kg−1
and latent heat of steam
(vaporization) = 2.256 × 106
J · kg−1
.

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6.6. INTERNAL ENERGY AND THE FIRST LAW OF THERMODYNAMICS 83
6.6 Internal energy and the first law of thermodynamics
Transfer of heat and doing mechanical work are two different ways of transferring energy to
and from the system. Once the transfer of energy takes place we say that the internal energy
of the system has changed.
Suppose a system changes from state 1 to state 2 along a definite path and that the
heat Q absorbed by the system and the work W done by it are measured. We may then
calculate the difference Q − W . If now we do the same thing over again for many different
paths between the same states 1 and 2, we find experimentally that Q − W is the same for
all paths connecting 1 and 2. But since Q is the energy added to the system and W is the
performance of work then the difference Q − W must represent the internal energy change
of the system. It follows that the internal energy change of a system is independent of the
path, and is therefore equal to the energy of the system in state 2 minus the energy in state
1. Internal energy is usually denoted by the symbol U. Thus if U1 is the internal energy in
state 1 and U2, the internal energy in state 2, then
U2 − U1 = 4U = Q − W = Q − P4V . (6.6.1)
This equation is known as the first law of thermodynamics. While applying this equation, it
must be remembered that, (1) all quantities must be expressed in the same units, (2) Q is
positive when the heat goes into the system, and (3) W is positive when the system expands
and does positive work on its surroundings.
We may rewrite eq (6.6.1) as
Q = 4U + W (6.6.2)
which simply means heat Q enters a system during a process, some of it (4U) remains in the
system as increased internal energy, while the remainder (W) leaves the system again in the
form of work done by the system against its surroundings. Internal energy can be interpreted
in terms of microscopic mechanical energy, that is, kinetic and potential energies of individual
molecules in a material.
If a system is carried through a process that eventually returns it to its initial state (known
as a cyclic process), then U2 = U1 and we have Q = W . Whatever heat given to the system
is used up in performing external work. An isolated system is a system that does no external
work and cannot exchange heat from surroundings. In which case Q = W = 0 and we have
4U = 0 =⇒ U2 = U1. (6.6.3)

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This is nothing but conservation of energy principle. The internal energy of an isolated system
can not change on its own by any process (electrical, mechanical, chemical etc). The energy
of a system can only be changed by a flow of heat through its boundaries or performance of
work or both.
6.6.1 Isothermal process
A process taking place at constant temperature is said to be isothermal. In this process the
expansion or compression of a gas is infinitesimally slow. Hence the internal energy of the
system remains constant. In general, none of the quantities Q, W, or 4U is zero. But for
an ideal gas the internal energy only depends on its temperature and since in an isothermal
process temperature remains constant we have
4U = 0
for an ideal gas, which means
Q = W .
Here the formula for work done for an ideal gas is obtained as below..
6.6.1.1 Work in an isothermal process for an ideal gas
For an ideal gas we have
PV = nRT,
and hence
P =
nRT
V
. (6.6.4)
Using this equation for pressure in eq (6.4.2), we can write for work done as
W =
V2
ˆ
V1
nRT
V
dV . (6.6.5)
Since in an isothermal process T = constant, we have
W = nRT
V2
ˆ
V1
dV
V
W = nRT ln
V2
V1
. (6.6.6)
Since P1V1 = P2V2 we also have
W = nRT ln
P1
P2
. (6.6.7)

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6.6.2 Adiabatic process
A process in which no heat enters or leaves a system is called an adiabatic process. The wall
or boundary of the system that does not allow flow of heat is known as an adiabatic wall or
boundary. Also any process that takes place very quickly will be adiabatic since flow of heat
takes finite time. For every adiabatic process, Q = 0 and from the first law we have
U2 − U1 = 4U = −W (6.6.8)
Thus, the change in the internal energy of a system, in an adiabatic process, is equal in
magnitude to the work done by the system. If the work W is negative, as when a system is
compressed, then - W is positive, U2, is greater than U1 and the internal energy of the system
increases which is usually seen as a rise in temperature.
6.6.2.1 Work done in adiabatic process for an ideal gas
In an adiabatic process the system is insulated from its surroundings and heat absorbed or
released is zero. In this process compression or expansion of a gas takes place suddenly;
which causes change in internal energy of the system. We have seen that in an isothermal
process
P ∝
1
V
but in an adiabatic process it is found that for an ideal gas
P ∝
1
V γ
(6.6.9)
where γ = 1.4 for ideal monatomic gas and 1.66 for air. More precisely we can write
PV γ
= constant. (6.6.10)
It is found that γ is numerically equal to the ratio of specific heat capacities at constant
pressure to volume (ordinary or molar) of the ideal gas. That is,
γ =
CP
CV
. (6.6.11)
If we use equation (6.6.10) to calculate the work done using equation (6.4.2), we get
W =
1
1 − γ
[P2V2 − P1V1] =
nR(T1 − T2)
γ − 1
. (6.6.12)

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6.6.3 Isochoric process
When a substance undergoes a process in which the volume remains unchanged (4V = 0),
the process is called isochoric. When the volume does not change, no work is done; W = 0
and therefore, from the first law,
U2 − U1 = 4U = Q (6.6.13)
All the added heat has served to increase the internal energy.
6.6.4 Isobaric process
A process that takes place at constant pressure is known as an isobaric process.
6.6.4.1 Work in an isobaric process
In an isobaric process pressure remains constant while the volume changes. If the volume
changes from V1 to V2 then from eq (6.4.2),
W =
V2
ˆ
V1
PdV = P
V2
ˆ
V!
dV = P(V2 − V1). (6.6.14)
If the system is compressed the work is negative and if it expands work is positive.
Thus the first law can be written as
4U = Q − P (V2 − V1)
A simple example is the vaporization of a mass m of liquid at constant pressure and
temperature. If VL is the volume of liquid and VV is the volume of vapor, the work done by
the system in expanding from VL to VV at constant pressure P is
W = P(VV − VL). (6.6.15)
The heat absorbed per unit mass is the heat of vaporization L. Hence,
Q = mL. (6.6.16)
From the first law
UV − UL = mL − P(VV − VL). (6.6.17)

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6.6.5 Alternative derivation of CP − CV = R
The internal energy of an ideal gas only depends on its temperature. Consider an ideal
gas which is taken from some intial state 1 to a final state 2 resulting into same temperature
change 4T = T2−T1; by two different processes: (one) isochoric, V = constant (two) isobaric,
P = constant. The heat given for both the processes is different. For the isochoric process
W = 0 since the gas is not allowed to expand then from the first law of thermodynamics
4U = QV = nCV 4T. (6.6.18)
During the isobaric process both heat and work are non-zero, further W = P(V2 − V1), hence
from the first law of thermodynamics
4U = QP − P(V2 − V1) = nCP4T − P(V2 − V1). (6.6.19)
Now since in both the processes the change in temperature is same the change in internal
energy in both isochoric and isobaric process must be same, thus from eq 6.6.18 and eq 6.6.19
we have
nCV 4T = nCP4T − P(V2 − V1)
Further for an ideal gas in isobaric process we have, PV2 = nRT2 and PV1 = nRT1 from the
equation of state. Thus the above equation reduces to
nCV 4T = nCP4T − nR(T2 − T1) (6.6.20)
CP − CV = R. (6.6.21)

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Numerical Problems
Example 10: One gram of water (1 cm3
) becomes 1671 cm3
of steam when boiled at a pressure
of 1 atm. The heat of vaporization at this pressure is 2256 J · g−1
. Compute the external work
and the increase in internal energy.
Example 11: A thermodynamic process is shown in figure (6.6.1). In process ab, 600 J of heat
are added, and in process bd 200 J of heat are added. Find (a) the internal energy change in
process ab; (b) the internal energy change in process abd; and (c) the total heat added in process
acd.
a c
d
b
2 × 10−3
m3
3 × 104
Pa
5 × 10−3
m3
8 × 104
Pa
0 V
P
Figure 6.6.1: Figure for example 8.
6.7 Heat engine and refrigerator (heat pump)
Most our energy comes from the burning of fossil fuels and from nuclear reactions, both of
these supply energy that is transferred to heat. Very little of heat is directly used, such
as for cooking and keeping warm, but to operate a machine or propel a vehicle we require
mechanical energy. Any device for converting heat to mechanical energy is called a heat
engine.
In a heat engine a certain quantity of matter (system) undergoes various thermal and
mechanical processes such as addition or subtraction of heat, expansion, compression, and

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6.7. HEAT ENGINE AND REFRIGERATOR (HEAT PUMP) 89
change of phase. This material is called the working substance of the engine. For simplicity
we will consider an engine in which the working substance is carried through a cyclic process,
that is, a sequence of processes in which it eventually returns to its original state. In a
condensing type of steam engine, as used in marine propulsion, the working substance is
pure water, which is actually used over and over again. Water is evaporated in boilers at
high pressure and temperature, does work in expanding against a piston or in a turbine, is
condensed by cooling water from the ocean, and pumped back into the boilers. The refrigerant
in a household refrigerator also undergoes a cyclic process. Internal combustion engines and
steam locomotives do not carry a system through a closed cycle.
All these substances absorb heat from a source at high temperature known as a hot
reservoir (such as boiler), perform some mechanical work and discard some heat at a lower
temperature source known as cold reservoir (such as running cold water). When a system is
carried through a cyclic process its initial and final internal energies are same, hence from
the first law
U2 − U1 = 0 = Q − W (6.7.1)
Q = W . (6.7.2)
That is, the net heat flowing into the engine in a cyclic process equals the net work done
by the engine. On a PV diagram, a cyclic process is represented by a closed curve and that
the area enclosed by the curve corresponds to the net work done by the system during one
cycle. If curve is traveled in clockwise direction the net work is positive, if counterclockwise,
negative.
Let QH represent the heat absorbed by the working substance from hot reservoir and QC
represent the heat discarded by the working substance to the cold reservoir. The energy
transformations in a heat engine are conveniently represented schematically by the flow
diagram fig (6.7.1a). The engine itself is represented by the ellipse.
The net heat absorbed by the engine in one cycle is
Q = QH − QC . (6.7.3)
The useful output is the net work W done by the working substance, and from the first law,
W = Q = QH − QC . (6.7.4)
The thermal efficiency of a cycle, denoted by η is defined as the ratio of the useful work to
the heat absorbed,
η =
W
QH
=
QH − QC
QH
= 1 −
QC
QH
. (6.7.5)

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Hot reservoir
at temperature
TH
Cold reservoir
at temperature
TC
Heat engine
QH
W
QC
(a) Heat engine
Hot reservoir
at temperature
TH
Cold reservoir
at temperature
TC
Refrigerator
QH
W
QC
(b) Refrigerator
Figure 6.7.1: Heat engine and refrigerator
The actual efficiency of a cycle is even less than the thermal efficiency due to friction and
other losses. For 100 % efficiency means QC = 0, i.e. no heat be rejected to the cold reservoir.
6.7.1 The refrigerator
A refrigerator is nothing but a heat engine that is run in reverse. A refrigerator takes in heat
from a cold reservoir, the compressor supplies mechanical work input, and heat is rejected to
a hot reservoir. With respect to the home refrigerator, the food is the cold reservoir, work is
done by the electric motor, and the hot reservoir is the air in the kitchen. The flow diagram
of a refrigerator is as shown in fig (6.7.1b).
In one cycle, heat QC enters the refrigerator at a low temperature TC , work W is done on
the refrigerator and heat QH leaves at a higher temperature TH. It follows from the first law
that
QC + W = QH. (6.7.6)
The heat rejected to the hot reservoir is the sum of the heat taken from the cold reservoir
and the heat equivalent of the work done by the motor. The best refrigerator is the one that
removes greatest heat QC with the least expenditure of mechanical work W . We therefore

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6.8. THE SECOND LAW OF THERMODYNAMICS 91
define the coefficient of performance (c.o.p.) of a refrigerator κ as,
κ =
QC
W
=
QC
QH − QC
. (6.7.7)
6.8 The second law of thermodynamics
Any engine that we build cannot have a thermal efficiency of 100%. That is, none of them
absorb heat and converts it completely into mechanical work. This is because of the difference
between the nature of internal energy and that of mechanical work. The internal energy is
the energy of random molecular motion, while the work represents ordered molecular motion.
The molecules of a moving body have an ordered motion in the direction of the motion of the
body; which is over an above their random motion within the body itself. The total molecular
kinetic energy associated with the ordered motion is what we call in mechanics the kinetic
energy of the moving body. The kinetic and potential energy associated with the random
motion constitutes the internal energy. When the moving body makes an inelastic collision
and comes to rest, the ordered portion of the molecular kinetic energy gets converted to
random motion. Since we cannot control the motion of individual molecules, it is impossible to
reconvert the random motion completely to ordered motion. We can however convert a portion
of it, this is what is accomplished by a heat engine.
The impossibility of converting heat completely into mechanical energy forms the basis of
the second law of thermodynamics. One statement of the second law of thermodynamics is as
follows:
It is impossible for any system to undergo a process in which it absorbs heat from
a reservoir at a single temperature and converts it completely into mechanical
work, while ending in the same state in which it begins.
Two other statements are also equivalent
Kelvin-Planck statement: No process is possible whose sole result is the ab-
sorption of heat from a reservoir and the complete conversion of the heat into
work.
Clausius statement: No process is possible whose sole result is the transfer of
heat from a colder object to a hotter object.

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6.8.1 Irreversibility
The fact that work maybe dissipated completely into heat where as heat may not be converted
entirely into work expresses an essential one-sidedness of nature. Natural spontaneous
processes may be studied in the light of the second law and in all such cases this peculiar
one sidedness is found. Thus heat always flows spontaneously from a hotter to a colder
body gases always seep through an opening spontaneously from region of high pressure to
a region of low pressure, gases and liquids left by themselves always tend to mix not to un-
mix, salt dissolves in water but a salt solution does not separate by itself into pure salt and
pure water, rocks weather and crumble, iron rusts, people grow old. These are all examples
of irreversible processes that take place naturally in only one direction and by their one
sidedness, express the second law of thermodynamics. We also note that in the thermodynamic
sense an irreversible process is always a non equilibrium process. Irreversibility arises mainly
from two causes (one) many processes like a free expansion or an explosive chemical reaction
take the system to non equilibrium states (two) most processes involve friction, viscosity and
other dissipative effects. A thermodynamic process is reversible if the process can be turned
back such that both the system and the surroundings return to their original states with no
other change anywhere else in the universe. Reversible process is an idealized notion. A
process is reversible only if it is Quasi-static that is the process is carried out so slowly
so that the system is in equilibrium with the surrounding at every stage and there are no
dissipative effects
6.8.2 Carnot cycle and engine
According to the second law no heat engine can have 100 % efficiency, but what is the maximum
possible efficiency of an engine given two heat reservoirs at temperatures TH and TC ? This
question was answered in 1824 by the French engineer Sadi Carnot. His engine is called
the Carnot engine and its cycle is called the Carnot cycle.
The Carnot cycle involves only reversible processes since irreversibility leads to dissipation
and therefore minimizes efficiency. The Carnot cycle consists of two isothermal and two
adiabatic processes. The Carnot cycle using an ideal gas as the working substance is shown
in the PV diagram in the fig. (6.8.1).
The Carnot cycle consists of the following steps.
• (a) The gas expands isothermally at temperature TH, absorbing heat QH.
• (b) It expands adiabatically until its temperature drops to TC .

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6.8. THE SECOND LAW OF THERMODYNAMICS 93
Work W
(a) absorb QH
(c) release QC
(
b
)
(
d
)
0 V
P
Figure 6.8.1: The Carnot cycle.
• (c) It is compressed isothermally at TC , rejecting heat QC .
• (d) It is compressed adiabatically back to its initial state.
The work achieved from the Carnot cycle is the shaded portion. The engine cycle is repeated
again to extract more work. The heat and work in each of these steps can be calculated in
terms of the volumes and temperatures, and from these it is found that
QC
QH
=
TC
TH
. (6.8.1)
Then the efficiency of the Carnot engine can be written as
η = 1 −
TC
TH
. (6.8.2)
This surprising result says that the efficiency of a Carnot engine depends only on the tem-
peratures of the two heat reservoirs. When the difference between the temperatures of the
two reservoirs is large, the efficiency is nearly unity.
Because each step in the Carnot cycle is reversible, the entire cycle may be reversed, con-
verting the engine into a refrigerator. The coefficient of performance of the Carnot refrigerator
is then
K =
TC
TH − TC
. (6.8.3)
Numerical Problems

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Example 12: A refrigerator is to maintain eatables kept inside at 9 o
C. If room temperature is
36 o
C, calculated the coefficient of performance.
Exercises
Q:1 State the zeroth law of thermodynamics.
Q:2 State Boyle’s law, Charles’ law and Gay Lussac’s law. Also state the equation of state
of ideal gas.
Q:3 Define work in a volume change. Draw the necessary diagram and prove that W =
´ V2
V1
PdV .
Q:4 Prove that in an isobaric volume change, the work done is W = P(V2 − V1).
Q:5 Obtain the formula for work done in the case of isothermal expansion of an ideal gas.
Q:6 Describe the work done by an ideal gas in an adiabatic expansion.
Q:7 Define isothermal, isochoric, adiabatic and isobaric process.
Q:8 Prove that the amount of work done depends on the path followed in the PV plane.
Q:9 Define heat and state its SI unit.
Q:10 Define specific heat capacity and latent heat and give their SI units.
Q:11 Define molar CP and molar CV in case of gases. Also derive the relation CP −CV = R.
Q:12 Write the general formula for the first law of thermodynamics. Also obtain the first law
of thermodynamics for isochoric, isothermal and adiabatic process by taking suitable example.
Q:13 Define heat engine. Draw the flow diagram and obtain a formula for the efficiency of
heat engine.
Q:14 Define a refrigerator. Draw the flow diagram and obtain a formula for the coefficient
of performance of refrigerator.
Q:15 Give any of the statements of the second law of thermodynamics.
Q:16 Explain irrersibility in brief.
Q:17 Describe the Carnot cycle by drawing necessary diagram in case of ideal gas as
working substance. Write the formula for efficiency of a Carnot cycle with ideal gas as a
working substance.

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Seven
Properties of Matter
Properties of Matter Surface tension, molecular force, cohesive and adhesive force,
molecular range, definition of surface tension, angle of contact, surface ten-
sion through capillary rise method, Viscosity; Fluid friction, viscous force,
definition of viscosity, coefficient of viscosity, stokes; law, Fluid flow, equa-
tion of continuity, Bernoulli theorem and its applications
7.1 Thrust and pressure
Consider a liquid of density “d” in a beaker of cross-section area “a”, up to a height “h”. The
force acting on the surface of solid by liquid is called thrust. Thrust has the same unit as
force, i.e: Newton.
a
h
ρ
Figure 7.1.1: Pressure
95

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96 7. Properties of Matter
Total thrust produced by the liquid = weight of the liquid
= mass of liquid × g
= volume of liquid × density of liquid × g
Thrust = ahdg (7.1.1)
Thrust per unit area is called pressure of the liquid. Its unit is N/m2
= Pa.
Pressure =
thrust
area
P =
ahdg
a
P = hdg (7.1.2)
Density:
• It is the mass per unit volume of substance.
• Its unit is g/cm3
or kg/m3
• Its value is different in the two different system of measurement.
Specific gravity: -
Specific gravity =
density of object
density of water
(7.1.3)
also
Specific gravity =
weight of object in air
weight lost in water
=
W1
W1 − W2
; (7.1.4)
where W1 = weight in air, W2 = weight in water.
• It is a ratio of the density of substance to the density of water.
• It has no unit.
• Its value remain unchanged whatever be the system of units is used.
Laws of liquid pressure:
• The pressure at any point in the liquid is proportional to the density of the liquid.
• The pressure at any point in the liquid is proportional to the depth of the point below
the free surface of the liquid.

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7.2. SURFACE TENSION 97
• Liquid transmit pressure equally in all directions. (Pascal’s Law)
• Pressure at equal depth is the same.
• Pressure dose not depend on the shape of the containing vessel.
Upthrust:
• When we submerge a solid into liquid, the liquid exerts and upward force. This force is
known as upthrust or force of buoyancy.
• Archimedes’ principle:
upthrust = weight of liquid displaced by the body (7.1.5)
upthrust = mass of liquid displaced × g (7.1.6)
7.2 Surface tension
Surface tension is the property of liquid by virtue of which, the free surface of liquid at rest
behave like elastic stretched membrane with a tendency to contract so as to occupy minimum
surface area.
7.2.1 molecular force: cohesive and adhesive force, molecular range
• The forces of attraction between different molecules are known as molecular forces.
• There are two types of molecular forces:
1. Adhesive force: - Force of attraction between molecules of different substances.
2. Cohesive force: - Force of attraction between molecules of same substances.
Molecular range:
• The maximum distance up to which the molecular force can act.
Sphere of influence:
• An imaginary sphere drawn with radius equal to molecular range and having a liquid
molecule at its center is called sphere of influence.

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7.2.2 Molecular theory of surface tension
A
A
A
A
A
A
A
A
A
A
A
A
C
B
C
B
C
B
C
B
C
B
C
B
C
B
C
B
C
B
C
B
C
B
C
B
Figure 7.2.1: Sphere of influence
• Consider a container filled up with some liquid.
• Consider three molecule A, B, and C, with its sphere of influence; such that A is com-
pletely inside the liquid, B is just below the surface of the liquid and C is at the surface
of the liquid.
• In case of molecule A, the sphere of influence is completely inside the liquid. Thus it is
attracted equally in all directions by the other molecules lying within its sphere. Hence
there is no resultant cohesive force acting on molecule A.
• In case of molecule B, the sphere of influence is partly outside the liquid. There are
more molecules in lower hemisphere in comparison of upper hemisphere. Hence there
is net resultant cohesive force acting on molecule B is in downward direction.
• In case of molecule C, the sphere of influence is exactly half outside and half inside
the liquid. There are all molecules in lower hemisphere. Hence there is net resultant
cohesive force acting on molecule C is in downward direction and its magnitude is
maximum
• As a result, the molecules at the surface of the liquid are attracted sideways and the
tension or pull in the surface is called surface tension. The layer of the liquid at surface
of thickness equal to the molecular range is known as surface film.
• If ‘F’ is the total tangential force acting on either side of an imaginary line AB of length
‘L’. Surface tension

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A B
F
F
Figure 7.2.2: surface tension
σ =
force
length
σ =
F
L
(7.2.1)
• Unit of surface tension is N/m
7.2.3 Angle of contact
θ
θ
Figure 7.2.3: angle of contact
• It is the angle between the tangent to the liquid surface and the solid surface inside
the liquid at the point contact.
• It is denoted by θ and measured in degrees.
• For water and glass surface; mercury and glass surface the angle of contact is shown
in figure.
• The angle of contact between water and glass is very small and taken to be zero.

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• The angle of contact between mercury and glass is 140◦
.
7.2.4 Surface Energy and surface tension
The energy possessed by the molecules of surface film of unit are compared to the molecules
in the interior is called surface energy.
Surface energy =
work done
increase in surface area
(7.2.2)
7.2.4.1 The relation between surface energy and surface tension
D
C B
A
B0
A0
x
F
2σL
L
Soap film
Figure 7.2.4: Work done in surface film
Consider a rectangular fram ABCD in which wire AB is movable. Dip the frame in soap
solution. A film is formed which pulls the wire AB inward due to surface tension with a force
F = 2σL (7.2.3)
( since there are two free surfaces of the soap film)
Suppose AB is moved out through distance x to the position A0
B0
. Then
work done = force × displacement
W = 2σLx (7.2.4)
Increase in surface area of the film = 2lx (7.2.5)
Therefore,

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surface energy per unit area =
work done
increase in surface area
(7.2.6)
=
2σlx
2lx
(7.2.7)
surface energy per unit area = σ (7.2.8)
7.2.5 surface tension through capillary rise method
σ
σ sin θ
R = σ σ cos θ
r
h
θ
liquid of density d
Figure 7.2.5: Capillary rise
• If capillary tube is immersed in a liquid, the liquid rises or depress in the capillary tube.
• This phenomenon is due to the surface tension and is known as capillarity
• Consider a capillary tube of radius ‘r’ dipped in to a liquid of density ‘d’.
• Let,
h = height to which liquid rises in tube
θ = angle of contact
σ = surface tension of liquid
R = reaction force due to σ
r = radius of the tube

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• Surface tension force ‘σ’ acts along the circumference of the tube and is resolved into
two components; Horizontal component σ sin θ and Vertical component σ cos θ as shown
in figure.
• All the horizontal components σ sin θ are equal and opposite. Therefore they cancel
out effect of each other whereas all vertical components σ cos θ add up. The vertical
components σ cos θ are only effective.
• The total upward force due to the surface tension, acts along the circumference of the
tube is given by
total upward force = vertical component σ cos θ × circumference of the tube (7.2.9)
F = σ cos θ × 2πr (7.2.10)
• This total upward force balances the weight of liquid in the capillary tube that rises
above the liquid surface in beaker.
total upward force = weight of liquid that rises in capillary
σ cos θ × 2πr = mass of liquid that rises × g
σ cos θ × 2πr = density of liquid × volume of liquid that rises × g
σ cos θ × 2πr = d

πr2
h

g
σ =
rhdg
2 cos θ
. (7.2.11)
7.2.6 Application/Examples of surface tension:
• The plants get their water supply by the capillary action at roots
• The rise of oil in the oil lamp and rise of melted wax in a candle.
• Blotting paper sucks the ink.
• In ink-pen a pen is split at the tip to provide the narrow capillary and the ink is drawn
up to the point continuously.
• The pores in the earth act as capillary and because of this rain water is soaked by the
earth. Numericals:

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7.3. VISCOSITY 103
Numerical Problems
Example 1: Calculate the height to which water will rise in capillary tube of 1.5 mm diameter.
Surface tension of water is 7.4 × 10−3
N/m.
Example 2: Calculate the diameter of a capillary tube in which mercury is depressed by 1.21
cm. Given the surface tension of mercury is 540 × 10−3
N/m, the angle of contact with glass
is 140◦
and density of mercury is 13.6 × 103
kg/m3
..
7.3 Viscosity
A B
C D
v + dv
v
x x + dx
Maximum velocity layer
Stationary layer along fixed surface
Figure 7.3.1: Laminar flow
Viscosity is the property of liquids by virtue of which it opposes relative motion between the
adjacent layers of the liquid. It is also known as fluid friction.
• Figure shows a liquid flowing steadily over a horizontal surface. •
• The liquid column is made up of different horizontal layers. The layer which is in contact
with the solid surface is at rest due to adhesive forces.
• The velocities of layers go on increasing as we move up from the solid surface. The top
most layer of the liquid has maximum velocity.
• Thus different layers of the liquid move with different velocities. Any two adjacent layers
are therefore in relative motion.
• The upper layer tries to accelerate on the lower layer.

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• Due to this a force tangential to the surface of the layer acts in such a way that the
relative motion between different layers is reduced.
• This is called viscous force and the property of the liquid is called as viscosity.
7.3.1 Examples of viscosity
1. When we swim in a pool of water, we experience some resistance or viscous drag force
to our motion.
2. Cough syrup have a high viscosity so that it gets coated in the sore throat
3. If we pour water and oil into separate funnels, then water comes out quickly then oil.
This is because the oil is more viscous then water.
4. The oil used in the engines are less viscous, so that it works smoothly without any wear
and tear of the machine. We can say that viscosity is the oil resistance to the flow.
5. The phenomena of viscosity plays an important role in the circulation of blood through
veins and arteries of human body.
7.3.2 Velocity gradient:
• Consider a layer CD of a liquid moving with a velocity v at a distance x and a parallel
layer AB, moving with a velocity (v+dv), at a distance of (x+dx) from fixed surface.
• Velocity gradient is defined as the rate of change of velocity (dv) with respect to the the
distance (dx) measured perpendicular to the layer of the flow of liquid. It is denoted by
(dv/dx). Its unit is second-1.
7.3.3 Newton’s law of viscosity:
• Newton found that the viscous force F acting tangentially on a layer of a liquid is
directly proportional to:
1. Velocity gradient (dv/dx); and
2. The surface area ‘A’ of the layer.

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7.4. STOKES’ LAW AND TERMINAL VELOCITY 105
F ∝ A
dv
dx
F = ηA
dv
dx
(7.3.1)
hence
η =
F
Adv
dx
. (7.3.2)
• η is a constant, known as co-efficient of viscosity, it depends upon the nature and
temperature of liquid.
• -ve sign indicate that ‘F’ is in backward dragging force.
• Unit of η in M.K.S. is N · s/m2
.
• Unit of η in C.G.S. is dyne · s/cm2
= Poise.
• The co-efficient of viscosity of a liquid is defined as the tangential viscous force (F) per
unit area (A) per unit velocity gradient.
7.3.4 Effect of temperature and pressure on Viscosity of Liquid
• When a liquid is heated, the kinetic energy of its molecules increases and the in-
termolecular attractions become weaker. Hence the viscosity of liquid decreases with
increase in temperature.
• Except water the viscosity of liquids increases with increase in pressure
7.4 Stokes’ law and terminal velocity
• According to stoke’s law, when a small spherical body moves in a viscous liquid, the
viscous force ‘F’ acting on a small spherical body is directly proportional to
1. radius r of spherical body
2. coefficient of viscosity η
3. terminal velocity v
F = 6πηrv (7.4.1)
• When a small spherical body moves in a viscous liquid, liquid offers resistance to its
motion, which increases with the increase in velocity of the spherical body.

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• A stage is reached when the resultant downward force due to weight of the body acts
vertically in downward direction, become equal to the viscous force acts vertically up-
ward. The resultant force at this stage becomes zero and the body moves with a constant
velocity called the terminal velocity
7.4.1 Determination of η by falling sphere method (Stoke’s law method)
F1
F2
F3
Figure 7.4.1: Measurement of η
• Suppose spherical ball of radius ‘r’ and density ‘ρ’ is falling freely in a liquid of density
σ and coefficient of viscosity ‘η’.
• After covering small distance, the sphere attains a constant terminal velocity ‘v’.
• Three forces acting on the sphere are
1. F1 = Weight of the sphere in the downward direction,
2. F2 = Upthrust acting in upward direction and
3. F3 = Force of viscosity in upward direction
• Now
F1 = volume of sphere × density of sphere × g
F1 =
4
3
πr3
ρg (7.4.2)

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7.5. FLUID FLOW 107
F2 = upthrust
F2 = weight of liquid displaced by sphere
F2 = volume of liquid displaced by sphere × density of liquid × g
F2 =
4
3
πr3
σg (7.4.3)
and
viscose force F3 = 6πηrv (7.4.4)
When sphere attains constant terminal velocity v,
downward force = resultant upward force
F1 = F2 + F3
4
3
πr3
ρg =
4
3
πr3
σg + 6πηrv
η =
2
9
r2
(ρ − σ)
v
. (7.4.5)
Numerical Problems
Example 3: A metal plate of area 5 cm2
is placed on a 0.5 mm thick castor oil layer. If a force
of 22,500 dyne is needed to move the plate with a velocity of 3 cm/s. calculate the coefficient
of viscosity of castor oil.
Example 4: A spherical glass ball of mass 1.34 × 10−4
kg and diameter 4.4 × 10−3
m takes
6.4 s to fall steadily through a height of 0.381 m inside large volume of oil of specific gravity
0.943. calculate the viscosity of oil.
7.5 Fluid flow
7.5.1 Streamline and turbulent flow
• In streamline flow, a liquid flows steadily, such that each particle passing a certain point
follows exactly the same path and has the same velocity as its preceding particle. The
tangent at any point on the streamline gives the direction of velocity of fluid particle at
that point.
• Fluid velocity remains constant at any point of streamline, but it may be different at
different points of the same streamline.

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• In turbulent flow, a liquid flows with large velocity and there is a free mixing of the fluid
particles of the various shells.
• The velocity at which the steady or streamline flow changes into turbulent flow is called
critical velocity.
7.5.2 Reynold’s number
• Reynold studied the motion of fluids in detail and observed that the critical velocity
VC of a liquid is directly proportional to its Co-efficient of viscosity ‘η’ and inversely
proportional to density ‘ρ’ of the liquid and radius ‘r’ of the tube.
• Mathematically,
VC ∝
η
ρr
(7.5.1)
VC =
Rη
ρr
(7.5.2)
• Where ‘R’ is constant and known as Reynolds number.
• On the basis of experiments it is observed that for tube of radius 1 cm flow is streamline
if R 2000, whereas flow is turbulent if R 3000
• If the value of Reynolds number lies between 2000 and 3000 then the flow of liquid is
of intermediate type.
Steady flow: In steady flow, the fluid velocity at each point does not change with time, either
in magnitude or direction.
Incompressible flow: In this flow, the density of the fluid remains constant.
Non-viscous flow: Any object moving the fluid does not experience any viscous force due to
fluid.
Irrotational flow: There is no angular momentum of the fluid about any point.
7.5.3 Equation of continuity
Principle: It states that when an incompressible, nonviscous fluid and flows steadily through
a tube of non-uniform cross-section, then the product of the area of cross-section and
the velocity of flow will remain constant at every point in the tube.

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7.5. FLUID FLOW 109
A1 A2
v14t
v24t
4x1
4x2
Figure 7.5.1: Continuity equation
At point 1
Area of cross-section = A1, velocity of fluid = v1, density of fluid = ρ1, therefore mass of
fluid that flows through in time 4t,
m1 = ρ1A1v14t (7.5.3)
At point 2
Area of cross-section = A2, velocity of fluid = v2, density of fluid = ρ2, therefore mass of
fluid that flows through in time 4t,
m2 = ρ2A2v24t (7.5.4)
Since the liquid is incompressible and assuming no changes in temperature, the mass must
remain conserved, we must have
m1 = m2
ρ1A1v14t = ρ2A2v24t
A1v1 = A2v2 (7.5.5)
or in general
Av = constant (7.5.6)

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7.5.4 Bernoulli’s theorem and its applications
Figure 7.5.2: Bernouli’s principle
Let us consider a streamline,steady, irrotational, incompressible and non-viscous flow of liquid
through a pipe as shown in figure. The portion of pipe with smaller diameter has cross-section
of area A1 at height of h1 from the surface. The velocity of fluid is v1. The portion of pipe with
large diameter has cross-section of area A2 at a height of h2 from the surface. The velocity
of fluid is v2.
Now, the work done due to pressure P1 is
W1 = F1s1 (7.5.7)
W1 = P1A1s1 (7.5.8)
and the work done by the system against the pressure P2
W2 = −P2A2s2 (7.5.9)
In time 4t it can be thought that the mass m was transferred from lower point to the upper
point. In this transfer the work done by the gravitational force on the fluid is
W3 = −mg(h2 − h1). (7.5.10)
Therefore, the work done by the resultant force on the system is
W = W1 + W2 + W3. (7.5.11)

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7.5. FLUID FLOW 111
Now according to the work energy theorem, Work
W = change in kinetic energy of the system
W = final kinetic energy − initial kinetic energy
W1 + W2 + W3 =
1
2
mv2
2 −
1
2
mv2
1
P1A1s1 − P2A2s2 − mg(h2 − h1) =
1
2
mv2
2 −
1
2
mv2
1 (7.5.12)
But
A1s1 = A2s2 = V =
m
ρ
(7.5.13)
∴ P1

m
ρ

− P2

m
ρ

− mg(h2 − h1) =
1
2
mv2
2 −
1
2
mv2
1
P1 +
1
2
ρv2
1 + ρgh1 = P2 +
1
2
ρv2
2 + ρgh2
Thus
P +
1
2
ρv2
+ ρgh = constant (7.5.14)
This equation is known as Bernoulli’s equation.
7.5.4.1 Application of Bernoulli’s principle
Venturimeter: A venturimeter is used to measure the flow speed of a fluid in pipe. The meter
is connected between two sections of pipe; the cross-sectional area A of the entrance and
exit of the meter matches the pipe’s cross-sectional area. Between the entrance and exit, the
fluid with density ρ flows from the pipe with speed V and then through a narrow ‘throat’ of
cross section area ‘a’ and speed v. A manometer connects the wider portion of the meter
to the narrower portion. The change in fluid speed is accompanied by a change in the fluid
pressure, which causes a height difference h of the liquid in two arms of manometer. Let P1
and P2 be the pressure at A and a.

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Figure 7.5.3: venturi meter
P1 +
1
2
ρV 2
+ ρgh1 = P2 +
1
2
ρv2
+ ρgh2
But h1 = h2
P1 +
1
2
ρV 2
= P2 +
1
2
ρv2
P1 − P2 =
1
2
ρv2
−
1
2
ρV 2
(7.5.15)
Now from the continuity equation,
AV = av (7.5.16)
v =
A
a
V (7.5.17)
Using this in equation 7.5.15
P1 − P2 =
1
2
ρ

A
a
2
V 2
− V 2
!
P1 − P2 =
1
2
ρV 2

A2
a2
− 1

2(P1 − P2)
ρ

A2
a2 − 1
= V 2

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7.5. FLUID FLOW 113
Thus the flow speed is
V =
v
u
u
t
2(P1 − P2)
ρ

A2
a2 − 1
(7.5.18)
The difference of pressure can easily be read from mercury manometer as
P1 − P2 = ρmgh (7.5.19)
where ρm is the density of mercury, thus
V =
v
u
u
t
2ρmgh
ρ

A2
a2 − 1
. (7.5.20)
Numerical Problems
Example 5: Water flows through a horizontal pipe whose internal diameter is 2 cm at speed
of 1 m/s. what should be the diameter of the nozzle, if the water is to emerge at a speed of 4
m/s ?
Example 6: The cross sectional area of water pipe entering the basement is 4×10−4
m2
. The
pressure at this point is 3 × 105
N/m2
and the speed of water is 2 m/s. This pipe tapers to
a cross sectional of 2 × 10−4
m2
, when it reaches the second floor 8 m above. Calculate the
speed and pressure at the second floor.

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Exercises
Q:1 What is thrust? state its unit.
Q:2 Define pressure? Give its unit.
Q:3 State laws of pressure.
Q:4 Define density and specific gravity of materials. Explain why the value of specific grav-
ity is independent of choice of units.
Q:5 What do you mean by surface tension of liquid?
Q:6 Write formula for surface tension and give its unit.
Q:7 Explain the surface tension based on molecular theory.
Q:8 Define: cohesive force, adhesive force, molecular range and sphere of influence.
Q:9 Define angle of contact.
Q:10 Prove that surface energy per unit area equals surface tension.
Q:11 What is capillarity? Derive formula for surface tension through capillary rise method.
Q:12 Give some examples or applications of surface tension.
Q:13 What do you mean by viscosity?
Q:14 What is velocity gradient or speed gradient? State its unit.
Q:15 Write Newton’s law of viscosity. Obtain the SI unit of coefficient of viscosity from
Newtons law.
Q:16 Write some applications of viscosity.
Q:17 State Stoke’s law and define terminal velocity.
Q:18 Obtain a formula for the coefficient of viscosity based on the falling sphere method
and Stoke’s law.
Q:19 Define: Streamline flow and turbulent flow.
Q:20 Define critical velocity in fluid flow.
Q:21 Write formula for Reynold’s number. What is the range of Reynold’s number for stream-
line and turbulent flow.
Q:22 Write equation of continuity and derive its formula, draw the necessary diagram.
Q:23 Write Bernoulli’s theorem. Derive the Bernoulli’s eqation for fluid flow. Draw the
necessary diagram.
Q:24 What is Venturimeter? Draw its diagram and obtain the formula for flow speed.

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7.5. FLUID FLOW 115
Additional Problems
1 Calculate the height of a column of water that gives a pressure of 1 atmosphere at the
bottom. 1 atmospheric pressure = 1.01 × 10−5
N/m2
.
2 A swimming pool has dimensions 24m × 9m × 2m. When it is filled with water, what is
the force on the bottom ?
3 Weight of a body in air is 100N. What is its weight in water if it displaces 400cc of water?
4 A fish weigh 348gm in air and 23gm in pure water. Calculate the relative density of fish.
5 A capillary tube of diameter 1.5 mm is dipped in a liquid of density 8 gm/cc. The liquid
rises through 7.2 mm. Calculate the surface tension of liquid . The angle of contact is 28 ◦
.
6 A capillary tube of inner radius 0.5 mm is dipped in water of surface tension 75 dyne/cm.
To what height will water rise? Calculate the weight of water raised.
7 A liquid rises to a height of 2.8 cm in a capillary tube of diameter 0.25 mm. How far will
it rise in a tube of diameter 0.55 mm ?
8 A capillary tube of inner diameter 0.5 mm is dipped in a liquid of specific gravity 13.6,
surface tension 545 dyne/cm and angle of contact 130◦
. Find the depression or elevation of
the liquid in the tube.
9 Calculate the horizontal force required to move a metal plate of area 200cm2
with a
velocity of 4.5 cm/s when it rests on a layer of oil 1.5 mm thick.
10 Calculate the viscous force on a rain drop of diameter 4 mm falling with constant velocity
of 4 m/s through air. Coefficient of viscosity of air is 1.8 × 10−5
N · s/m2
and density of air
is 1.21 kg/m3
.
11 Calculate the radius of drop of water falling through air if the terminal velocity of drop
is 1.2 cm/sec, coefficient of viscosity for air is 1.8 × 10−5
N · s/m2
, density of water is 1000
kg/m3
and the density of air is 1.21 kg/m3
.
12 Calculate the terminal velocity of an air bubble of radius 5 × 10−4
m rising in a liquid of
viscosity 0.15 N · s/m2
and density 900 kg/m3
.
13 Two drops of water of the same size are falling through air with terminal velocity 10cm/sec.
If two drops combine to form a single drop, what will be the new terminal velocity?
14 Find the difference in height of mercury columns in two communicating vertical capillaries
whose diameters are 0.50 mm and 1.00 mm respectively, if the angle of contact is 138◦
.
15 Diameters of the arms of a U-tube are 10 mm and 1 mm respectively. It is partially filled
with water and it is held in a vertical plane. Find the difference in heights of water in both

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arms. Surface tension of water is 70 dyne/cm and angle of contact is zero.

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Eight
Optics
Optics Light as a wave and its characteristics, equation of a plane progressive
wave, Interference of light, principle of superposition, young’s double-slit
experiment, constructive and destructive interference, diffraction of light,
types of diffraction, diffraction through grating, polarization, types of po-
larization, Brewster’s law and Malus’ law, applications
8.1 Wave and its equation
Whenever a system is disturbed from equilibrium and the disturbance can travel or propagate,
from one region of the system to another a wave can occur. It can carry energy much like the
sun warms the surface of earth. Most waves are mechanical waves-waves that travel within
some material called medium.
To understand waves we take the simplest case of waves on a string. Such waves are
important in music. A mechanical wave is a disturbance that travels through some material
or substance called the medium. As the wave propagates the medium particles are also
put in motion and they undergo displacement. If the displacement of the medium particles
are perpendicular or transverse to the direction of propagation, then the waves are called
transverse waves. Exampels are waves on a string, water ripples etc. If motion of the
particles are back and forth along (parallel to) the direction of propagation of the wave , then
the waves are called longitudinal waves. Examples are sound waves and waves on a slinky
spring. The speed at which the disturbance or turbulance travels is called the wave speed. Its
value is determined by the mechanical properties of the medium. This wave speed is not the
same as the speed of the particles of the medium when they move as the wave passes through.
The medium itself does not travel i.e. the particles of the medium undergo up and down or
117

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118 8. Optics
back and forth motion around their mean position. The overall pattern of the disturbance only
travels. The waves are generated by applying force i.e. by doing work and a wave simply
transports this energy from one place to another without actual motion of matter.
Figure 8.1.1: Pulse on rope
We can produce a disturbance on a rope that is fixed at one end by giving it an upward
jerk (force) as shown in the figure. This kind of disturbance (of finite length) is knwown as a
wave pulse. We are more interested in continuous waves also known as periodic waves. This
kind of wave can be produced by repeating the up and down motion periodically resulting
into periodic waves.
In order to describe the wave consider the following diagram.
Figure 8.1.2: Wave properties
Amplitude: The maximum displacement that the medium particles (in this case points on rope)
undergo in either direction is called the amplitude. The symbol used for amplitude is
usuallay A and its unit is m. In the diagram 8.1.2 y0 is used for the amplitude.

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8.1. WAVE AND ITS EQUATION 119
Time period: The time taken by any point of the wave to complete one oscillation is called
the time period of the wave. The symbol used is T and the unit is s. Time period of
wave is also called one cycle.
Frequency: Frequency is the number of oscillations performed by any particle of the medium
in 1 s. The frequency is f = 1/T. Some times symbol ν is also used. Its unit is Hz.
Wavelength: The distance travelled by the wave/disturbance in one periodic time T is called
the wavelength. The symbol used is λ and its unit is m.
The region of half wavelength where the displacement is positive is called positive cycle and
the one with negative displacement is called a negative cycle. Together they constitute a
wavelength. The point with maximum positive displacement is called a crest and the one
with maximum displacement in negative direction is called a trough. There exists a simple
relationship between properties of a wave. Since in one time period T the wave moves a
distance of one wave length λ, the speed of a wave can be written as
speed =
distance
time
v =
λ
T
(8.1.1)
v = fλ. (8.1.2)
Periodic waves can be produced on a string/rope that is attached to a spring that undergoes
simple harmonic oscillations in the vertical direction as shown in the following diagram.
Figure 8.1.3: Periodic waves
Each point on the rope also oscillates with the same period and amplitude but the dis-
turbance takes finite amount of time to reach as it moves forward. We know that the simple

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120 8. Optics
harmonic oscillation is represented by the equation
y = y0 sin ωt (8.1.3)
where ω = 2π/T is the angular frequency. Therefore this must also be the equation that
represents the motion of any individual particle of the rope. But all points of the rope are not
in the same state of oscillation (displacement and speed).
String
x = 0
x = vt
Figure 8.1.4: Wave propagation
The equation 8.1.3 represents the state of motion of the point of string at x = 0. Suppose
the disturbance starts at x = 0 at time t = 0 and propagtes in +x direction with a speed v.
Then as seen from the diagram 8.1.4 the wave will take a time of t = x/v to reach to a point
located at distance x from the spring. Thus the section of rope at location x starts vibrating
exactly as the one at x = 0 (spring) only after a time of x/v. Thus the equation 8.1.3 will be
modified as
y = y0 sin ω

t −
x
v

(8.1.4)
in order to correctly represent the motion of any section located at any position x and at any
instant t. Thus this represents the equation of a sinusoidal harmonic wave travelling in the
+x direction. If instead the wave moves in −x direction the equation gets modified to
y = y0 sin ω

t +
x
v

(8.1.5)
Since sine is an odd function we may also write equation 8.1.4 as
y = y0 sin ω

t −
x
v

= −y0 sin ω
x
v
− t

(8.1.6)
this also represents a wave going in the positive x direction, but the amplitude is reversed.
Thus we can also use the equation
y = y0 sin ω
x
v
− t

(8.1.7)

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to represent the wave going in the +x direction. Since v = λ/T and ω = 2π/T, the equation
8.1.4 can be written as
y = y0 sin

2π

x
λ
−
t
T

(8.1.8)
y = y0 sin

2π
λ
(x − vt)

. (8.1.9)
If we define
k =
2π
λ
(8.1.10)
then the wave equation can be written as
y = y0 sin (kx − ωt) . (8.1.11)
We will use this form of the wave equation henceforth. In the wave function 8.1.11, the
quantity (kx − ωt) = φ is called the phase. It plays the role of an angular quantity which is
always measured in radians. The quantity k = 2π
λ
is called the wave number, physically it
represents the number of waves (wavelengths) that make up 2π. Since ω = 2π/T, we have
v =
λ
T
(8.1.12)
v =
λ
2π
×
2π
T
(8.1.13)
v =
ω
k
. (8.1.14)
8.1.1 Phase difference and path difference
Consider the two waves shown in the following diagram.

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122 8. Optics
λ
λ/2
2π
π
λ/4
π/2
x
x
y
1 2 3
1 2 3
-2
2
-2
2
(a) Graph of y1 = 2 sin 2π
1 x

(b) Graph of y2 = 2 sin 2π
1 x + π
2

Figure 8.1.5: Phase and path difference
The first (a) is a (frozen t = 0) wave with wavelength 1 m. The equation of the wave
is y1 = 2 sin 2π
1
x

= 2 sin (φ1). The points at positions 0 m, 0.5 m (λ/2), 1 m (2λ/2), 1.5
m (3λ/2) ... etc are at zero displacements i.e. y1 = 0. These are the points for which
φ1 = 0, π, 2π, 3π, . . . These wave points at which the displacements of medium particles are
zero are known as nodes. Similarly the points at which the displacemnet is maximum (in both
directions) are known as anti-nodes. These occur at positions 0.25 m (λ/4), 0.75 m (3λ/4),
1.25 m (5λ/4) .. etc for which the phase φ1 = π/2, 3π/2, 5π/2, . . .. The distance between two
successive nodes or antinodes is λ/2. The distance between successive node and an antinode
is λ/4. In a travelling wave the positions of nodes and antinodes are not fixed. The second
(frozen) wave (b) is also exactly similar to the first one (a) only difference is that the positions
of nodes and anti-nodes are interchanged. The equation of this wave (b) can therefore be
written as y2 = 2 sin 2π
1
x + π
2

= 2 sin (φ2).
The phase difference is simply the differene of phases of the wave.
4φ = φ2 − φ1 (8.1.15)
In the present case 4φ = π/2. The phase difference when represented in terms of length is
known as the path difference. The path difference in the present case is 0.25 m. It is convenient
to write path difference in terms of wavelength λ. A phase difference of 2π corresponds to
path difference of λ,thus in general, the path difference
4x =
λ · 4φ
2π
. (8.1.16)

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8.1.2 Principle of superposition
When two waves propagate in the same medium and if they cross at any perticular point,
then interference can occur and the shape of resultant wave is obtained by the principle of
superpositon. If two or more traveling waves are moving through a medium, the resultant
value of the wave function at any point is the algebraic sum of the values of the wave
functions of the individual waves. In other words, the wave function y (x, t) for the resulting
motion is obtained by adding the two wave functions for the two separate waves:
y (x, t) = y1 (x, t) + y2 (x, t) . (8.1.17)
thus
wave function of combined wave = sum of individual wave functions. (8.1.18)
A B
A B
x
A B
x
x
x
x
A
B
y
1 2 3
2
2
4
2
2
Figure 8.1.6: Superposition of pulse
Numerical Problems
Example 1: The equation of a wave is given by y = 2 sin 3x − t
4

. Calculate the amplitude,
wave number, wavelength, time period, frequency and speed of wave.
Example 2: Two waves are represented by y1 = 2 sin(4x − 8t) and y2 = 3 sin 4x − 8t + π
3

.
Calculate (a) the amplitude, wave number, wavelength, time period, frequency and speed of
wave for both the waves, (b) the phase difference and path difference between the waves and
(c) the resultant displacement of a particle located at x = 2 m at time t = 3 s.

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124 8. Optics
8.1.3 Interference in waves
The combination of separate waves in the same region of space to produce a resultant wave
is called interference.
When the displacements caused by the two waves are in the same direction, we refer to
their superposition as constructive interference. This occurs whenever the phase difference
between the two waves is 0, 2π, 4π, . . . or the path difference between the two waves is
0, λ, 2λ, . . . The total displacement of the combined wave is y = y1 + y2. Figure 8.1.7 (a)
shows the constructive interference. If the waves have same displacement then the resulting
displacement is twice the individuals.
When the displacements caused by the two waves are in the opposite direction, we refer
to their superposition as destructive interference. This occurs whenever the phase difference
between the two waves is π, 3π, 5π, . . . or the path difference between the two waves is
0, λ/2, 3λ/2, . . . The total displacement of the combined wave is y = y1 − y2. Figure 8.1.7
(b) shows the destructive interference. If the two waves have same displacement then the
resulting displacement is zero.
(a) constructive interference (a) destructive interference
Figure 8.1.7: Interference in waves
For any other phase difference the resulting displacement varies between max and zero.
In figure 8.1.8b the two waves represented by dashed and dotted curves are out of phase by
π/2 (at instant t = 0) and the resulting wave pattern is given by the solid line. Even though
the individual waves have amplitude of 1, the maximum resultant displacement is about 1.4.
In figure 8.1.8a the two waves represented by dashed and dotted curves are out of phase by

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8.2. LIGHT AS A WAVE AND ITS CHARACTERISTICS 125
(a) Phase diff of 3π/4 (b) Phase diff of π/2
Figure 8.1.8: Interference for various phase differences
3π/4 (at instant t = 0) and the resulting wave pattern is given by the solid line. Even though
the individual waves have amplitude of 1, the maximum resultant displacement is about 0.78.
8.2 Light as a wave and its characteristics
Until Newton everyone thought that light conists of particles known as corpuscles emitted by
light sources. Around 1665 evindence for wave nature of light began to appear. In 1873 James
Clerk Maxwell predicted the existence of electromagnetic waves and calcuated the speed of
its propagation. This along with experiements by Hertz in 1887 and subsequently showed
that light indeed is an electromagnetic wave. When it comes to light we often write the speed
with symbol c and its value in vacuum is c = 3 × 108
m/s. The frequency of light waves is
represented by ν and we also have the fundamental relation
c = νλ. (8.2.1)
In an electromagnetic wave the value of electric and magnetic fields oscillate similar to the
particles of a rope. EM waves have following properties.

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126 8. Optics
x
y
z
c
E
B
Figure 8.2.1: EM wave
1. EM waves are transverse waves.
2. EM waves do not require medium to propagate: non-mechanical waves.
3. They move with the speed c = 3 × 108
m/s in vacuum.
4. The direction of oscillation (variation) of E and B fields is mutually perpendicuar and
also perpendicular to the direction of propagation (see fig.)
5. EM waves can be polarized i.e. it is possible to restrict oscillatioin of E or B field in a
perticular direction.
6. According to Planck the energy of an EM wave is proportional to its frequency.
7. When an EM wave travels from one medium to another its wavelength and the speed
changes however the frequency (and therefore energy) remains unchanged.
When a light wave travels from one transparent medium to another it bends. This is called
refraction and is due to the fact that the speed of light is different in different media. The
amount of bending depends upon the speeds of the light in the two media. We define refractive
index
refractive index of medium 2 with respect to medium 1 =
speed of light in medium 1
speed of light in medium 2
(8.2.2)
1µ2 =
v1
v2
(8.2.3)

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8.2. LIGHT AS A WAVE AND ITS CHARACTERISTICS 127
In perticular the refractive index of glass with respect to air is
µa
g =
vair
vglass
. (8.2.4)
When a light wave travels from medium 1 to 2 the wavelength increases as
λ2 = µ × λ1. (8.2.5)
In perticular the wavelength of light in glass is
λg = λa × µa
g. (8.2.6)
8.2.1 Huygen’s principle
We often use the concept of a wave front to describe wave propagation. Wave front can be
thought to describe the leading edge of a wave. More generally, a wave front is the locus
of all adjacent points at which the phase of vibration of a physical quantity (displacement)
associated with the wave is the same. That is, at any instant, all points on a wave front are
at the same part of the cycle of their variation.
When we drop a small stone into a calm pool, the expanding circles formed by the wave
crests, as well as the circles formed by the wave troughs between them, are wave fronts. Simi-
larly, when sound waves spread out in still air from a pointlike source, or when electromagnetic
radiation spreads out from a pointlike emitter, any spherical surface that is concentric with
the source is a wave front. As a spherical or cylndrical wavefront advances, its curvature
decreases progressively. So a small portion of such a wavefront at a large distance (ideally at
infinity) from the source will be a plane wavefront. Thus a spherical or cylindrical wavefront
at infinity becomes a plane wavefront.
Figure 8.2.2: Wavefronts

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128 8. Optics
8.3 Interference of light
Interference in light waves from two sources was first demonstrated by Thomas Young in 1801.
8.3.1 Young’s double-slit experiment
A schematic diagram of the apparatus Young used is shown in Figure 8.3.1.
S
S2
S1
Screen
bright fringe
dark fringe
Figure 8.3.1: Young’s double slit experiment
Light waves arrive at a barrier that contains two slits S1 and S2. The light from S1 and
S2 produces on a viewing screen a visible pattern of bright and dark parallel bands called
fringes as shown in fig 8.3.1. When the light from S1 and that from S2 both arrive at a point on
the screen such that constructive interference occurs at that location, a bright fringe appears.
When the light from the two slits combines destructively at any location on the screen, a dark
fringe results.

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8.3. INTERFERENCE OF LIGHT 129
Figure 8.3.2: Constructive and destructive interference in two slits
In Figure 8.3.2 (a), the two waves, which leave the two slits in phase, strike the screen at
the central point. Because both waves travel the same distance, they arrive at in phase. As
a result, constructive interference occurs at this location and a bright fringe is observed. In
Figure 8.3.2 (b), the two waves also start in phase, but here the lower wave has to travel one
wavelength farther than the upper wave to reach point. Because the lower wave falls behind
the upper one by exactly one wavelength, they still arrive in phase at P and a second bright
fringe appears at this location. At point R in Figure 8.3.2 (c), however, between points O and
P, the lower wave has fallen half a wavelength behind the upper wave and a crest of the upper
wave overlaps a trough of the lower wave, giving rise to destructive interference at point R. A
dark fringe is therefore observed at this location.
No interference effects are observed with two ordinary bulbs because the light waves
from bulbs are emitted independently. The emissions from the light bulbs do not maintain a
constant phase relationship; because light emtted from ordinary bulbs undergo random phase
changes in time intervals of less than a nanosecond and the eye cannot follow such rapid
changes. Such light sources that do not maintain a constant phase relatonship are said to
be incoherent.
To observe interference of waves from two sources, the following conditions must be met:
• The sources must be coherent; that is, they must maintain a constant phase with respect
to each other.
• The sources should be monochromatic; that is, they should be of a single wavelength.

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130 8. Optics
A common method for producing two coherent light sources is to use a monochromatic source
to illuminate a barrier containing two small openings, usually in the shape of slits, as in the
case of Young’s experiment illustrated in Figure 8.3.1. Thus a single light beam is split into
two beams. Any random change in the light emitted by the source occurs in both beams at
the same time. As a result, interference effects can be observed when the light from the two
slits arrives at a viewing screen.
8.3.2 Conditions for constructive and destructive interference
Let the distance between the two slits be d and the screen be located at a perpendicular
distance of L. A point P is located at a vertical distance of y on the screen as shown in
figure 8.3.3. The light travels a distance of r1 from slit S1and a distance of r2 from slit S2 as
it reaches the point P. Usually the distance d L and also y L, so that one can consider
the pahts r1 and r2 to be parallel as shown in fig 8.3.3. In that case the path difference
between the two rays is
δ = r2 − r1 = d sin θ (8.3.1)
Viewing screen
Q
O
P
S2
S1
r2
r1
L
d
y
θ
θ
δ
d
r2
r1
δ = r2 − r1 = d sin θ
θ
When we assume r1 is
parallel to r2, the path
difference between the two
rays is r2 − r1 = d sin θ
Figure 8.3.3: Geometric construction for Young’s double slit experiment
The value of δ determines whether the two waves are in phase when they arrive at point
P. If δ is either zero or some integer multiple of the wavelength, the two waves are in phase
at point P and constructive interference results. Therefore, the condition for bright fringes, or
constructive interference, at point P is
d sin θbright = mλ ; m = 0, ±1, ±2, . . . (8.3.2)

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8.3. INTERFERENCE OF LIGHT 131
The number m is called the order number. For constructive interference, the order number is
the same as the number of wavelengths that represents the path difference between the waves
from the two slits. The central bright fringe at θbright = 0 is called the zeroth-order maximum.
The first maximum on either side, where m = ±1, is called the first-order maximum, and so
forth.
When δ is an odd multiple of λ/2, the two waves arriving at point P are 180° (π) out of
phase and give rise to destructive interference. Therefore, the condition for dark fringes, or
destructive interference, at point P is
d sin θdark =

m +
1
2

λ ; m = 0, ±1, ±2, . . . (8.3.3)
The angles θbright and θdark give the angular positions of bright and dark fringes. To find the
linear distance of a perticular fringe y we note that from the geometry of diagram
tan θ =
y
L
(8.3.4)
and hence
ybright = L tan θbright (8.3.5)
ydark = L tan θdark. (8.3.6)
Often the angle θ is quite small since the screen distance L is typically a few meters whereas
the distance d is about a millimeter. In this small angle approximation
sinθ ' tan θ.
Hence from set of equations 8.3.2 , 8.3.3 and 8.3.5, 8.3.6, we get
ybright = L
mλ
d
, (8.3.7)
ydark = L
m + 1
2

λ
d
. (8.3.8)
Fringe width: It is the separation between two successive bright or dark fringes, for
example
width of bright fringe = distance between consecutive dark fringes (8.3.9)
α = ym − ym−1 (8.3.10)
α =
Lλ
d

m +
1
2

−

m −
1
2

(8.3.11)
α =
λL
d
. (8.3.12)

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132 8. Optics
Numerical Problems
Example 3: In a Young’s double slit experiment, the two parallel slits are made 1 mm apart and
a screen is placed 1 m away. What is the fringe separation when blue-green light of 500 nm is
used?
Example 4: Laser light of wavelength 630 nm incident on a pair of slits produces an interfer-
ence pattern in which the bright fringes are separted by 8.1 mm. A second light produces an
interference pattern in which the fringes are separted by 7.2 mm. Calculate the wavelength
of second light.
Example 5: The two parallel slits used for Young’s interference expt are 0.5 mm apart The
screen on which fringes are projected is 1.5 m from the slits. How far is the third dark fringe
from the central bright one. Wavelength of light used is 600 nm.
8.4 Diffraction of light
Light travels in straight line. However, when light passes through a small hole, there is a
certain amount of spreading of light. Similarly, when light encounters an obstacle, it appears
to bend round the edges of the obstacle and enters its geometrical shadow.
The phenomenon of bending of light around the corners of small obstacles or apertures
and its consequent spreading into the regions of geometrical shadow is called diffraction
of light.
Soure
Geometrical
shadow
obstacle
Figure 8.4.1: Diffraction at an obstacle
In fig 8.4.1 light from a source of light is blocked by an obstacle and its geometrical
shadow is cast on the screen placed behind it. It is found that the geometrical shadow is
not completely dark, but appears to be somewhat bright. This occurs due to diffraction: the
light bends around the edges of the obstacle and is able to reach the portion of shadow. For

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8.4. DIFFRACTION OF LIGHT 133
diffraction to be observable the size of the object should be very small comparable to the size
of the wavelength of light.
Figure 8.4.2: Diffraction pattern of an opaque object
When an opaque object is placed in front of a light source, we dont see a sharp shadow
of the object. But a pattern of alternative bright and dark lines is observed which is similar
to the interference pattern. These patterns are called diffraction patterns. Similar diffraction
pattern is also observed in the case of a single slit when it is illuminated by a monochromatic
light such as laser.
8.4.1 Types of diffraction
8.4.1.1 Fresnel’s diffraction
In Fresnel’s diffraction, the source and screen are placed close to the aperture or the obstacle
and light after diffraction appears to be converging towards the screen and hence no lens is
required to observe it. The incident wavefronts are either spherical or cylindrical.
S
P
P
P
P
P
(a) Fresnel Diffraction (b) Fraunhofer Diffraction
Figure 8.4.3: Two types of diffractoin

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134 8. Optics
8.4.1.2 Fraunhoffer diffraction
In Fraunhoffer diffraction, the source and screen are placed at large distances (effectively at
infinity) from the aperture or the obstacle and converging lens is used to observe the diffraction
pattern. The incident wave front is a plane wave front.
8.4.2 Diffraction through a single slit
A slit of finite width a is placed in the path of a source kept very far. A screen is placed
on the other side of the slit and a converging lens is used to focus the diffraction on to the
screen (Fraunhoffer diffraction). The arrangement is as shown in the diagram above. Figure
8.4.4aa shows light entering a single slit from the left and diffracting as it propagates toward
a screen. Figure 8.4.4ab shows the fringe structure of a Fraunhofer diffraction pattern. A
bright fringe is observed along the axis at θ = 0 known as central maxima, with alternating
dark and bright fringes on each side of the central bright fringe.
According to Huygens’s principle, each portion of the slit acts as a source of light waves.
Hence, light from one portion of the slit can interfere with light from another portion, and the
resultant light intensity on a viewing screen depends on the direction θ. Thus a diffraction
pattern is actually an interference pattern in which the different sources of light are different
portions of the single slit!
To analyze the diffraction pattern, let’s divide the slit into two halves as shown in Figure
8.4.4b. Keeping in mind that all the waves are in phase as they leave the slit, consider rays
1 and 3. As these two rays travel toward a viewing screen far to the right of the figure, ray 1
travels farther than ray 3 by an amount equal to the path difference (a/2) sin θ, where a is the
width of the slit. Similarly, the path difference between rays 2 and 4 is also (a/2) sin θ, as is
that between rays 3 and 5. If this path difference is exactly half a wavelength (corresponding
to a phase difference of 180°), the pairs of waves cancel each other and destructive interference
results. This cancellation occurs for any two rays that originate at points separated by half
the slit width because the phase difference between two such points is 180°. Therefore, waves
from the upper half of the slit interfere destructively with waves from the lower half when
a
2
sin θ =
λ
2
(8.4.1)
or if we consider both the above and below parts of the screen
a
2
sin θ = ±
λ
2
(8.4.2)

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8.4. DIFFRACTION OF LIGHT 135
(a) Expt arrangement and pattern (b) Single slit analysis
Figure 8.4.4: Single slit diffraction
or
sin θ = ±
λ
a
(8.4.3)
Dividing the slit into four equal parts and using similar reasoning, we find that the viewing
screen is also dark when
sin θ = ±2
λ
a
. (8.4.4)
Similarly, dividing the slit into six equal parts shows that darkness occurs on the screen when
sin θ = ±3
λ
a
. (8.4.5)
Therefore, the general condition for destructive interference is
sin θdark = m
λ
2
; m = ±1, ±2, ±3, . . . . (8.4.6)
8.4.3 Diffraction grating
A large number of parallel, closely spaced slits constitutes a diffraction grating. A transmission
grating can be made by cutting parallel grooves on a glass plate with a precision ruling
machine. The spaces between the grooves are transparent to the light and hence act as

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136 8. Optics
separate slits. A reflection grating can be made by cutting parallel grooves on the surface
of a reflective material. Current technology can produce gratings that have very small slit
spacings. For example, a typical grating ruled with 5000 grooves/cm has a slit spacing
d = (1/5000)cm = 2×10−4
cm. A section of a diffraction grating is illustrated in Figure 8.4.5.
The pattern observed on the screen far to the right of the grating is the result of the com-
bined effects of interference and diffraction. Each slit produces diffraction, and the diffracted
beams interfere with one another to produce the final pattern.
Figure 8.4.5: Diffraction grating
The path difference δ between rays from any two adjacent slits is equal to d sin θ. If this
path difference equals one wavelength or some integral multiple of a wavelength, waves from
all slits are in phase at the screen and a bright fringe is observed. Therefore, the condition
for maxima in the interference pattern at the angle θbright is
d sin θbright = mλ ; m = 0, ±1, ±2, . . . . (8.4.7)
We can use this expression to calculate the wavelength if we know the grating spacing d
and the angle θbright. If the incident radiation contains several wavelengths, the mth
-order
maximum for each wavelength occurs at a specific angle.

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8.5. DIFFERENCE BETWEEN INTERFERENCE AND DIFFRACTION 137
8.5 Difference between interference and diffraction
Interference Diffraction
1 Interference is the result of
superposition of secondary waves
starting from two different
wavefronts originating from two
coherent sources.
Diffraction is the result of
superposition of secondary waves
starting from different parts of
the same wavefront.
2 All bright and dark fringes are of
equal width.
The width of central bright fringe
is twice the width of any
secondary maximum.
3 All bright fringes are of same
intensity.
Intensity of bright fringes
decreases as we move away from
central bright fringe on either
side.
4 Regions of dark fringes are
perfectly dark.
Regions of dark fringes are not
perfectly dark.
5
2 1 0 1 2
Position of mth
order maxima
Intensity distribution in interference
2 1 0 1 2
Position of mth
order maxima
Intensity distribution in diffraction
Table 8.5.1: Interference vs diffraction
Exercises
Q:1 Define a wave.
Q:2 What is a mechanical and non-mechanical wave?
Q:3 Differentiate between transverse and longitudinal wave.
Q:4 Define properties of a wave; amplitude, wavelength, time period, frequency and speed
of a wave.
Q:5 Derive the wave equation for a plane progressive wave from the simple harmonic oscil-

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