JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Geometrical Optics

When the plane face of a plano-convex lens is silvered, light entering through the curved surface refracts once, reflects off the silvered plane, and refracts again through the curved surface — the system behaves as an equivalent mirror.

Refraction at the curved surface (liquid of index 1.2 to glass of index 1.5), parallel rays from infinity ($u=\infty$):
$\frac{n_{glass}}{v}-\frac{n_{liquid}}{u}=\frac{n_{glass}-n_{liquid}}{R}$

$\frac{1.5}{v}=\frac{1.5-1.2}{R}=\frac{0.3}{R}\Rightarrow v=5R$

This image acts as the object for the plane silvered surface, which reflects it back at the same distance; light then refracts through the curved surface a second time, now going from glass to liquid, to form the final image, giving the equivalent focal length $f$ of the mirror system:

$\frac{n_{liquid}}{f}-\frac{n_{glass}}{v}=\frac{n_{liquid}-n_{glass}}{-R}$

$\frac{1.2}{f}-\frac{1.5}{5R}=\frac{0.3}{R}$

$\frac{1.2}{f}=\frac{0.3}{R}+\frac{0.3}{R}=\frac{0.6}{R}$

$f=\frac{1.2R}{0.6}=2R$

With $f=0.2$ m:
$R=\frac{f}{2}=0.10$ m