JEE Physics/ Lakshmikanta Satapathy/ Ray optics QA part 2/ Question on Change in Magnification produced by a convex lens due to shifting of object

1. Physics Helpline
L K Satapathy Geometrical Optics 2
Change in Magnification
Shifting of Object
Due to

2. Physics Helpline
L K Satapathy Geometrical Optics 2
When an extended object is placed on
the principal axis of a convex lens of
focal length 30 cm , the magnification
produced is +3. Now the object is
displaced through a distance d without
disturbing the position of the lens such
that the magnification changes to −3.
Then the value of d is
(a) 10 cm (b) 20 cm
(c) 30 cm (d) 40 cm
Question :

3. Physics Helpline
L K Satapathy Geometrical Optics 2
CONCEPTS:
When
image is virtual and erect.
 M is +ve
u f When
image is real and inverted.
 M is − ve
u f
1 1 1
Lens equation
v u f
  Magnificati
I
n
v
Mo
O u
 
OI O
I

4. Physics Helpline
L K Satapathy Geometrical Optics 2
Method
1 1 1
Lens equation
v u f
 
( i ) Define object distance (u)
( ii ) Use the equation for Magnification to find the value of (v)
( iii ) Value of Focal length ( f ) is given in the Question
( iv ) Find the value of (u) using Lens equation
( v ) Repeat the process for the 2nd experiment
( vi ) Difference in the values of u = displacement of object

5. Physics Helpline
L K Satapathy Geometrical Optics 2
Solution
u
v
30f cm
3 3
v
M v u
u
    
3u x v x   
1 1 1 1 1 1 2 1
3 30 3 30v u f x x x
       
 
Case-I
20x cm 

6. Physics Helpline
L K Satapathy Geometrical Optics 2
Solution
u
v
30f cm
3 3
v
M v u
u
     
3u y v y   
1 1 1 4 1
3 30 3 30y y y
   

20y x cm 
 The correct option is (b)
Case-II
40y cm 
 Object is shifted by

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L K Satapathy
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