Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is :

Solution

Physics sir · 4 steps

Treat each glass surface on its own, with light going towards it, then compare where the two images sit.

1. Set up distances. The figure is symmetric, so PB = PA = R. Then OB = R/2 and OA = R + R/2 = 3R/2. Each surface curves towards the air, so its centre of curvature lies on the air side, against the incoming light. That makes the radius negative for both surfaces.

   μ₂/v - μ₁/u = (μ₂ - μ₁)/Rₛ, μ₁ = 1, μ₂ = 1.5

2. Surface B. With u = -R/2 and Rₛ = -R: 1.5/v + 2/R = -0.5/R, so v = -0.6R. The image is 0.6R from B on the air side, at a distance 2R - 0.6R = 1.4R from A.
3. Surface A. With u = -3R/2 and Rₛ = -R: 1.5/v + 2/3R = -0.5/R, so 1.5/v = -7/6R and v = -9R/7 ≈ -1.286R. The image is 1.286R from A on the air side.
4. So. Both images lie between A and B. Their separation is 1.4R - 9R/7 = 4R/35 ≈ 0.114R, which is option (b).

The idea to keep. For each refracting surface, measure distances from its own pole and take the light's direction as positive. A concave surface facing the object always gives a virtual image on the object's side.

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