An object is placed at $\frac{f}{2}$ away from first focus of convex lens where f is the focal length of the lens. Its image is formed at a distance $\frac{3 f}{2}$ 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.

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f/2

3f/2

answer is D.

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Detailed Solution

Distance of final image in the slab is 3f/2.
Distance of image formed due to convex lens from faces of the slab facing the lens $= \frac{(3 / 2) f}{μ} = f$

$Let F_{2} S = x; V = 2 f + x;$

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

$\frac{1}{(x + 2 f)} - \frac{1}{(- 3 f / 2)} = \frac{1}{f}$

$\Rightarrow x = f$

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