An object is placed at `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 `3f//2` in a slab of refractive index `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.

Options

A. `f//2`
B. `3f//2`
C. 2f
D. f

Answer

d.

Explanation

`u=-(f+(f)/(2))=-(3f)/(2)`
Distance of final image from slabe is `(3)/(2)f`. Distance of image formed due to convex lens from faces to the slab facing the lens `=((3//2)f)/(mu)=f`
Let `F_(2)S=x`
`V=2f+x`
From `(1)/(v)-(1)/(u)=(1)/(f)`
`(1)/((x+2f))-(1)/((-3f//2))=(1)/(f)or x=f`