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

$3 f / 2$

$2 f$

$f$

answer is D.

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Detailed Solution

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

Let F₂S = x ; V = 2F + x ; from $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{(x + 2 f)} - \frac{1}{(- 3 f / 2)} = \frac{1}{f}$ or x = f

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