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 shift of the object is 6cm, the shift of screen in cm is
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Solution
Given that,
Case I: When the object is at O and image is formed at I.
Initial magnification of lens m=−3
Let object is at, u=−xthen, m=vu=v−x=−3 ⇒v=3x
By using lens formula, 13x−1−x=1f 43x=1f ...(1)
Case II: When object is at O′ and image is formed at I′
Now when object is shifted by 6cm
Object distance, u=−(6+x)cm
New magnification, m′=vu=−2
Therefore, image distance, v=2(6+x)cm
Applying lens formula, 12(6+x)−1−(6+x)=1f 32(6+x)=1f ....(2)
Equation (1) and (2) gives 43x=32(6+x) ∴9x=48+8x x=48cm
Shift of screen =3x−2(6+x) =3×48−2(6+48) =36cm