For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

Solution

Physics sir · 4 steps

The slab pushes the image back by 1/3 cm, so the lens must now focus 1/3 cm short of the screen.

1. Find the focal length. Take u = -240 cm and v = +12 cm. The lens formula gives 1/f = 1/12 + 1/240 = 21/240, so 1/f = 7/80 cm⁻¹.
2. Shift by the slab. A glass plate moves the image away from the lens by t(1 - 1/μ) = 1(1 - 1/1.5) = 1/3 cm. For the final image to land on the screen at 12 cm, the lens alone must form its image at 12 - 1/3 = 35/3 cm.

   1/u = 1/v - 1/f = 3/35 - 7/80 = (240 - 245)/2800 = -1/560

3. New object distance. So u = -560 cm. The object must now sit 5.6 m from the lens. That is farther away, because a nearer image needs a more distant object.
4. So. The object moves from 2.4 m to 5.6 m, a shift of 5.6 - 2.4 = 3.2 m away from the lens. The answer is (b).

The idea to keep. A slab of thickness t shifts the image by t(1 - 1/μ) away from the lens. Work backwards: find where the lens itself must focus, then use the lens formula for the new u.

Answer from the NTA final answer key. Solution by PrepWiser.