Question

A lens form a sharp image on a screen. On inserting a parallel-sides slab of glass between the lens and the screen it is found necessary to move the screen a distance $d$ away from the lens in order for the image to be again sharply focused. If the refractive index of glass relative to air is $\mu$, then the thickness of the slab is

• A. $\frac{d}{\mu }$
• B. $\frac{\mu }{d}$
• C. $\frac{\mu d}{\mu -1}$
• D. $\frac{(\mu -1)d}{\mu }$

Answer 1

Answer: (C)

$\frac{\mu d}{\mu -1}$

When a glass slab of thickness t is inserted in the path of the light rays, the point of intersection of ray shifts by $t(1-\frac{1}{\mu })$ in the direction in which the ray is moving.

As per the question $d=t(1-\frac{1}{\mu })$

Solving we get, $t=\frac{d\mu }{(\mu -1)}$