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Chapter 26
The Refraction of
Light: Lenses and
Optical Instruments
26.1 The Index of Refraction
sm1000.3 8
×=cLight travels through a vacuum at a speed
Light travels through materials at a speed less than its speed
in a vacuum.
DEFINITION OF THE INDEX OF REFRACTION
The index of refraction of a material is the ratio of the speed
of light in a vacuum to the speed of light in the material:
v
c
n ==
materialin thelightofSpeed
in vacuumlightofSpeed
26.1 The Index of Refraction
Page 808
26.2 Snell’s Law and the Refraction of Light
SNELL’S LAW OF REFRACTION
When light travels from a material with
one index of refraction to a material with
a different index of refraction, the angle
of incidence is related to the angle of
refraction by
2211 sinsin θθ nn =
SNELL’S LAW
26.2 Snell’s Law and the Refraction of Light
Example 1 Determining the Angle of Refraction
A light ray strikes an air/water surface at an
angle of 46 degrees with respect to the
normal. Find the angle of refraction when
the direction of the ray is (a) from air to
water and (b) from water to air.
26.2 Snell’s Law and the Refraction of Light
( ) 54.0
33.1
46sin00.1sin
sin
2
11
2 ===

n
n θ
θ(a)
(b)

332 =θ
( ) 96.0
00.1
46sin33.1sin
sin
2
11
2 ===

n
n θ
θ

742 =θ
26.2 Snell’s Law and the Refraction of Light
APPARENT DEPTH
Example 2 Finding a Sunken Chest
The searchlight on a yacht is being used to illuminate a sunken
chest. At what angle of incidence should the light be aimed?
26.2 Snell’s Law and the Refraction of Light
( ) 69.0
00.1
31sin33.1sin
sin
1
22
1 ===

n
n θ
θ

441 =θ
( ) 
313.30.2tan 1
2 == −
θ
26.2 Snell’s Law and the Refraction of Light






=′
1
2
n
n
dd
Apparent depth,
observer directly
above object
26.2 Snell’s Law and the Refraction of Light
Conceptual Example 4 On the Inside Looking Out
A swimmer is under water and looking up at the surface. Someone
holds a coin in the air, directly above the swimmer’s eyes. To the
swimmer, the coin appears to be at a certain height above the
water. Is the apparent height of the coin greater, less than, or the
same as its actual height?
Light rays are refracted AWAY from the normal
when going from a higher index of refraction to
a lower index of refraction.
When it is the opposite, the light bends
TOWARD the normal
26.3 Total Internal Reflection
When the angle of incidence exceeds the critical angle, there is no refracted
light. All the incident light is reflected back into the medium
Critical angle – the angle of
incidence at which the
refracted ray is = 90○ 21
1
2
sin nn
n
n
c >=θ
26.3 Total Internal Reflection
Example 5 Total Internal Reflection
A beam of light is propagating through diamond and strikes the diamond-air
interface at an angle of incidence of 28 degrees. (a) Will part of the beam
enter the air or will there be total internal reflection? (b) Repeat part (a)
assuming that the diamond is surrounded by water.
26.3 Total Internal Reflection

4.24
42.2
00.1
sinsin 1
1
21
=





=





= −−
n
n
cθ(a)
(b) 
3.33
42.2
33.1
sinsin 1
1
21
=





=





= −−
n
n
cθ
26.3 Total Internal Reflection
Conceptual Example 6 The Sparkle of a Diamond
The diamond is famous for its sparkle because the light coming from
it glitters as the diamond is moved about. Why does a diamond
exhibit such brilliance? Why does it lose much of its brilliance when
placed under water?
26.4 Polarization and the Reflection and Refraction of Light
1
2
tan
n
n
B =θ
Brewster’s angle – the one
angle of incidence at which the
reflected light is completely polarized
parallel to the surface.
26.5 The Dispersion of Light: Prisms and Rainbows
The net effect of a prism is to change the direction of a light ray.
Light rays corresponding to different colors bend by different amounts.
26.5 The Dispersion of Light: Prisms and Rainbows
26.5 The Dispersion of Light: Prisms and Rainbows
Conceptual Example 7 The Refraction of Light Depends on Two
Refractive Indices
It is possible for a prism to bend light upward,
downward, or not at all. How can the situations
depicted in the figure arise?
26.6 Lenses
Lenses refract light in such a way that an image of the light source is
formed.
With a converging lens, paraxial rays that are parallel to the principal
axis converge to the focal point.
26.6 Lenses
With a diverging lens, paraxial rays that are parallel to the principal
axis appear to originate from the focal point.
26.6 Lenses
26.7 The Formation of Images by Lenses
RAY DIAGRAMS
26.7 The Formation of Images by Lenses
IMAGE FORMATION BY A CONVERGING LENS
In this example, when the object is placed further than
twice the focal length from the lens, the real image is
inverted and smaller than the object.
26.7 The Formation of Images by Lenses
When the object is placed between F and 2F, the real image is
inverted and larger than the object.
26.7 The Formation of Images by Lenses
When the object is placed between F and the lens, the virtual image is
upright and larger than the object.
26.7 The Formation of Images by Lenses
IMAGE FORMATION BY A DIVERGING LENS
A diverging lens always forms an upright, virtual, diminished image.
26.8 The Thin-Lens Equation and the Magnification Equation
fdd io
111
=+
o
i
o
i
d
d
h
h
m −==
26.8 The Thin-Lens Equation and the Magnification Equation
Summary of Sign Conventions for Lenses
lens.convergingaforis +f
lens.divergingaforis −f
lens.theofleftthetoisobjecttheifis +od
lens.theofrightthetoisobjecttheifis −od
image).(reallenstheofrightthetoformedimageanforis +id
image).(virtuallenstheofleftthetoformedimageanforis −id
image.uprightanforis +m
image.invertedanforis −m
26.8 The Thin-Lens Equation and the Magnification Equation
Example 9 The Real Image Formed by a Camera Lens
A 1.70-m tall person is standing 2.50 m in front of a camera. The
camera uses a converging lens whose focal length is 0.0500 m.
(a)Find the image distance and determine whether the image is
real or virtual. (b) Find the magnification and height of the image
on the film.
1
m6.19
m50.2
1
m0500.0
1111 −
=−=−=
oi dfd
(a)
m0510.0=id real image
(b) 0204.0
m50.2
m0510.0
−=−=−=
o
i
d
d
m
( )( ) m0347.0m50.20204.0 −=−== oi mhh
26.9 Lenses in Combination
The image produced
by one lens serves as
the object for the next
lens.
26.10 The Human Eye
OPTICS
The lens only contributes about 20-25% of the refraction, but its function
is important.
26.10 The Human Eye
NEARSIGNTEDNESS
The lens creates an image of the distance object at the far point
of the nearsighted eye.
26.10 The Human Eye
Example 12 Eyeglasses for the Nearsighted Person
A nearsighted person has a far point located only 521 cm from the
eye. Assuming that eyeglasses are to be worn 2 cm in front of the
eye, find the focal length needed for the diverging lens of the glasses
so the person can see distant objects.
26.10 The Human Eye
cm519
11111
−
∞
=+=
io ddf
cm519−=f
26.10 The Human Eye
FARSIGNTEDNESS
The lens creates an image of the close object at the near point
of the farsighted eye.
26.10 The Human Eye
THE REFRACTIVE POWER OF A LENS – THE DIOPTER
( )metersin
1
diopters)(inpowerRefractive
f
=
Optometrists who prescribe correctional lenses and the opticians
who make the lenses do not specify the focal length. Instead
they use the concept of refractive power.
26.11 Angular Magnification and the Magnifying Glass
The size of the image on the retina determines how large
an object appears to be.
26.11 Angular Magnification and the Magnifying Glass
( )
o
o
d
h
≈= sizeAngularradiansinθ
26.11 Angular Magnification and the Magnifying Glass
Example 14 A Penny and the Moon
Compare the angular size of a penny held at arms length with that of
the moon.
rad027.0
cm71
cm9.1
==≈
o
o
d
h
θPenny
Moon rad0090.0
m103.9
m105.3
8
6
=
×
×
=≈
o
o
d
h
θ
26.11 Angular Magnification and the Magnifying Glass
Angular magnification
θ
θ′
=M
N
df
M
i






−≈
11
Angular magnification
of a magnifying glass
26.12 The Compound Microscope
To increase the angular magnification
beyond that possible with a magnifying
glass, an additional converging lens
can be included to “premagnify” the
object.
Angular magnification of
a compound microscope
( )
eo
e
ff
NfL
M
−
−≈
26.13 The Telescope
Angular magnification of
an astronomical telescope e
o
f
f
M −≈
26.14 Lens Aberrations
In a converging lens, spherical
aberration prevents light rays
parallel to the principal axis from
converging at a single point.
Spherical aberration can be reduced
by using a variable-aperture diaphragm.
26.14 Lens Aberrations
Chromatic aberration arises when different colors are focused at
different points along the principal axis.

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Chapter26powerpoint 090317003440-phpapp01

  • 1. Chapter 26 The Refraction of Light: Lenses and Optical Instruments
  • 2. 26.1 The Index of Refraction sm1000.3 8 ×=cLight travels through a vacuum at a speed Light travels through materials at a speed less than its speed in a vacuum. DEFINITION OF THE INDEX OF REFRACTION The index of refraction of a material is the ratio of the speed of light in a vacuum to the speed of light in the material: v c n == materialin thelightofSpeed in vacuumlightofSpeed
  • 3. 26.1 The Index of Refraction Page 808
  • 4. 26.2 Snell’s Law and the Refraction of Light SNELL’S LAW OF REFRACTION When light travels from a material with one index of refraction to a material with a different index of refraction, the angle of incidence is related to the angle of refraction by 2211 sinsin θθ nn = SNELL’S LAW
  • 5. 26.2 Snell’s Law and the Refraction of Light Example 1 Determining the Angle of Refraction A light ray strikes an air/water surface at an angle of 46 degrees with respect to the normal. Find the angle of refraction when the direction of the ray is (a) from air to water and (b) from water to air.
  • 6. 26.2 Snell’s Law and the Refraction of Light ( ) 54.0 33.1 46sin00.1sin sin 2 11 2 ===  n n θ θ(a) (b)  332 =θ ( ) 96.0 00.1 46sin33.1sin sin 2 11 2 ===  n n θ θ  742 =θ
  • 7. 26.2 Snell’s Law and the Refraction of Light APPARENT DEPTH Example 2 Finding a Sunken Chest The searchlight on a yacht is being used to illuminate a sunken chest. At what angle of incidence should the light be aimed?
  • 8. 26.2 Snell’s Law and the Refraction of Light ( ) 69.0 00.1 31sin33.1sin sin 1 22 1 ===  n n θ θ  441 =θ ( )  313.30.2tan 1 2 == − θ
  • 9. 26.2 Snell’s Law and the Refraction of Light       =′ 1 2 n n dd Apparent depth, observer directly above object
  • 10. 26.2 Snell’s Law and the Refraction of Light Conceptual Example 4 On the Inside Looking Out A swimmer is under water and looking up at the surface. Someone holds a coin in the air, directly above the swimmer’s eyes. To the swimmer, the coin appears to be at a certain height above the water. Is the apparent height of the coin greater, less than, or the same as its actual height? Light rays are refracted AWAY from the normal when going from a higher index of refraction to a lower index of refraction. When it is the opposite, the light bends TOWARD the normal
  • 11. 26.3 Total Internal Reflection When the angle of incidence exceeds the critical angle, there is no refracted light. All the incident light is reflected back into the medium Critical angle – the angle of incidence at which the refracted ray is = 90○ 21 1 2 sin nn n n c >=θ
  • 12. 26.3 Total Internal Reflection Example 5 Total Internal Reflection A beam of light is propagating through diamond and strikes the diamond-air interface at an angle of incidence of 28 degrees. (a) Will part of the beam enter the air or will there be total internal reflection? (b) Repeat part (a) assuming that the diamond is surrounded by water.
  • 13. 26.3 Total Internal Reflection  4.24 42.2 00.1 sinsin 1 1 21 =      =      = −− n n cθ(a) (b)  3.33 42.2 33.1 sinsin 1 1 21 =      =      = −− n n cθ
  • 14. 26.3 Total Internal Reflection Conceptual Example 6 The Sparkle of a Diamond The diamond is famous for its sparkle because the light coming from it glitters as the diamond is moved about. Why does a diamond exhibit such brilliance? Why does it lose much of its brilliance when placed under water?
  • 15. 26.4 Polarization and the Reflection and Refraction of Light 1 2 tan n n B =θ Brewster’s angle – the one angle of incidence at which the reflected light is completely polarized parallel to the surface.
  • 16. 26.5 The Dispersion of Light: Prisms and Rainbows The net effect of a prism is to change the direction of a light ray. Light rays corresponding to different colors bend by different amounts.
  • 17. 26.5 The Dispersion of Light: Prisms and Rainbows
  • 18. 26.5 The Dispersion of Light: Prisms and Rainbows Conceptual Example 7 The Refraction of Light Depends on Two Refractive Indices It is possible for a prism to bend light upward, downward, or not at all. How can the situations depicted in the figure arise?
  • 19. 26.6 Lenses Lenses refract light in such a way that an image of the light source is formed. With a converging lens, paraxial rays that are parallel to the principal axis converge to the focal point.
  • 20. 26.6 Lenses With a diverging lens, paraxial rays that are parallel to the principal axis appear to originate from the focal point.
  • 22. 26.7 The Formation of Images by Lenses RAY DIAGRAMS
  • 23. 26.7 The Formation of Images by Lenses IMAGE FORMATION BY A CONVERGING LENS In this example, when the object is placed further than twice the focal length from the lens, the real image is inverted and smaller than the object.
  • 24. 26.7 The Formation of Images by Lenses When the object is placed between F and 2F, the real image is inverted and larger than the object.
  • 25. 26.7 The Formation of Images by Lenses When the object is placed between F and the lens, the virtual image is upright and larger than the object.
  • 26. 26.7 The Formation of Images by Lenses IMAGE FORMATION BY A DIVERGING LENS A diverging lens always forms an upright, virtual, diminished image.
  • 27. 26.8 The Thin-Lens Equation and the Magnification Equation fdd io 111 =+ o i o i d d h h m −==
  • 28. 26.8 The Thin-Lens Equation and the Magnification Equation Summary of Sign Conventions for Lenses lens.convergingaforis +f lens.divergingaforis −f lens.theofleftthetoisobjecttheifis +od lens.theofrightthetoisobjecttheifis −od image).(reallenstheofrightthetoformedimageanforis +id image).(virtuallenstheofleftthetoformedimageanforis −id image.uprightanforis +m image.invertedanforis −m
  • 29. 26.8 The Thin-Lens Equation and the Magnification Equation Example 9 The Real Image Formed by a Camera Lens A 1.70-m tall person is standing 2.50 m in front of a camera. The camera uses a converging lens whose focal length is 0.0500 m. (a)Find the image distance and determine whether the image is real or virtual. (b) Find the magnification and height of the image on the film. 1 m6.19 m50.2 1 m0500.0 1111 − =−=−= oi dfd (a) m0510.0=id real image (b) 0204.0 m50.2 m0510.0 −=−=−= o i d d m ( )( ) m0347.0m50.20204.0 −=−== oi mhh
  • 30. 26.9 Lenses in Combination The image produced by one lens serves as the object for the next lens.
  • 31. 26.10 The Human Eye OPTICS The lens only contributes about 20-25% of the refraction, but its function is important.
  • 32. 26.10 The Human Eye NEARSIGNTEDNESS The lens creates an image of the distance object at the far point of the nearsighted eye.
  • 33. 26.10 The Human Eye Example 12 Eyeglasses for the Nearsighted Person A nearsighted person has a far point located only 521 cm from the eye. Assuming that eyeglasses are to be worn 2 cm in front of the eye, find the focal length needed for the diverging lens of the glasses so the person can see distant objects.
  • 34. 26.10 The Human Eye cm519 11111 − ∞ =+= io ddf cm519−=f
  • 35. 26.10 The Human Eye FARSIGNTEDNESS The lens creates an image of the close object at the near point of the farsighted eye.
  • 36. 26.10 The Human Eye THE REFRACTIVE POWER OF A LENS – THE DIOPTER ( )metersin 1 diopters)(inpowerRefractive f = Optometrists who prescribe correctional lenses and the opticians who make the lenses do not specify the focal length. Instead they use the concept of refractive power.
  • 37. 26.11 Angular Magnification and the Magnifying Glass The size of the image on the retina determines how large an object appears to be.
  • 38. 26.11 Angular Magnification and the Magnifying Glass ( ) o o d h ≈= sizeAngularradiansinθ
  • 39. 26.11 Angular Magnification and the Magnifying Glass Example 14 A Penny and the Moon Compare the angular size of a penny held at arms length with that of the moon. rad027.0 cm71 cm9.1 ==≈ o o d h θPenny Moon rad0090.0 m103.9 m105.3 8 6 = × × =≈ o o d h θ
  • 40. 26.11 Angular Magnification and the Magnifying Glass Angular magnification θ θ′ =M N df M i       −≈ 11 Angular magnification of a magnifying glass
  • 41. 26.12 The Compound Microscope To increase the angular magnification beyond that possible with a magnifying glass, an additional converging lens can be included to “premagnify” the object. Angular magnification of a compound microscope ( ) eo e ff NfL M − −≈
  • 42. 26.13 The Telescope Angular magnification of an astronomical telescope e o f f M −≈
  • 43. 26.14 Lens Aberrations In a converging lens, spherical aberration prevents light rays parallel to the principal axis from converging at a single point. Spherical aberration can be reduced by using a variable-aperture diaphragm.
  • 44. 26.14 Lens Aberrations Chromatic aberration arises when different colors are focused at different points along the principal axis.