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Question

A convex lens forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of the object. If the increase in distance of the object is 6 cm, the shift of the screen is:

A
36 cm
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B
72 cm
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C
18 cm
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D
9 cm
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Solution

The correct option is A 36 cm
Let object and image distance be u and v respectively and focal length = f.
For the first case when magnification m=±3=vu
v=+3u image is real.
Using lens formula 1v1u=1f
13u+1u=1f
or 43u=1f ….. (1)

For the case when object distance is increased by 6 and m=±2=vu+6
v=2(u+6)
Using lens formula again,
12(6+u)+16+u=1f
or 32(6+u)=1f ….. (2)
Eqs. (1) and (2) gives
43u=32(6+u)
9u=48+8u
u=48 cm
Shift of screen = 3u – 2(6+u)
=3×482(6+48)
= 36 cm

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