Question

80. The slab of a material of refractive index 2 shown in figure has a curved surface
APB of radius of curvature 10 em and a plane surface CD. on the left of APB is air
and on the right of CD is water with retractive indices as given in the figure. An
object O is placed at distance of 15 cm from the pole P as shown. The distance of
the final image of o from P, as viewed from the left is.
A
13
200m
(A) 36 cm​

Answer 1

Answer:

In case of refraction from a curved surface, we have μ 2/v − μ 1/u = μ 2 −μ 1/R ⇒ 1/v -2/(−15)= (1−2)/-10⇒v=−30 cm.

i.e. the curved surface will form virtual image I at distance of 30 cm from P. Since the image is virtual there will be no refraction at the plane surface CD ( as the rays are not actually passing through the boundary), the distance of final image I from P will remain 30 cm.