QJEEM25Ph047b: JEE Main 2025 Physics - Class 12 JEEM25Ph047b Solution: A thin plano convex lens made of glass of refractive index 1.5…

$\frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R}$

For a thin plano‑convex lens made of glass (refractive index $n_2 = 1.5$ ) immersed in a liquid (refractive index $n_1 = 1.2$ ), when the plane side is silver‐coated, a ray entering the curved surface from the liquid is refracted into the glass, reflects off the coated plane (thereby reversing its direction) and finally refracts back into the liquid at the same curved surface. In the paraxial approximation the net effect is equivalent to a mirror in the liquid with an effective focal length $f$ .

A ray that is incident parallel to the optical axis (at height $h$ ) will, after this combined refraction–reflection–refraction process, emerge with an angular deviation corresponding to a mirror having a power given by:

$\frac{h}{f} = \frac{2(n_2 - n_1)}{n_1R}\, h\,.$

This gives the relation:

$\frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R}\,.$

Given that the effective focal length is $f = 0.2\,\text{m}$ , substitute the values:

$\frac{1}{0.2} = \frac{2(1.5 - 1.2)}{1.2\,R}\,.$

Simplify the numerator:

$\frac{1}{0.2} = \frac{2(0.3)}{1.2\,R} = \frac{0.6}{1.2\,R}\,.$

Since

$\frac{0.6}{1.2} = 0.5,$

the equation becomes:

$5 = \frac{0.5}{R}\,.$

Solving for $R$ :

$R = \frac{0.5}{5} = 0.10\,\text{m}\,.$

Thus, the radius of curvature of the curved surface of the lens is $0.10\,\text{m}$ .