Question

An object is placed at $\frac{f}{2}$ away from first focus of a convex lens where $f$ is the focal length of the lens. Its image is formed at a distance $\frac{3f}{2}$ in a slab of refractive index $\frac{3}{2}$, from the face of the slab facing the lens. Find the distance of this face of the slab from the second focus of the lens.

Answer 1

$u=-\frac{3f}{2}$ ; $f=+f$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{-2}{3f}=\frac{1}{f}$

$v=+3f$

image is formed at a distance $\frac{3f}{2}$ from the face i.e, in air it is a distance $\frac{d}{\mu }=\frac{\frac{3f}{2}}{\frac{3}{2}}=f$ from the face

so the distance between second focus and the face is $3f-f-f=f$

option $D$ is correct