Question

Upon inserting a parallel-sided slab of glass between the lens and the screen, it was found necessary to move the screen a distance d away from the lens in order for the image to be again sharply focussed. 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

The correct option is C $\frac{\mu d}{(\mu -1)}$

Let the thickness of the screen is $t$
shift after entering the glass slab $=d$

According to formula,

Shift $=(1-\frac{1}{\mu })t$

$\Rightarrow t=\frac{d}{(1-\frac{1}{\mu })}=\frac{\mu d}{(\mu -1)}$