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 1v−1u=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=48cm Shift of screen = 3u – 2(6+u) =3×48–2(6+48) = 36 cm