JEE Main 2025 Apr 7 Shift 1, Physics Q8: Thin Lens and Mirrors

Use the lens-maker's equation:

$\frac{1}{f}=\left(\frac{{\mu }_L}{{\mu }_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

In air (${\mu }_m=1$):

$\frac{1}{12}=(1.6-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=0.6\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\Rightarrow \left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{12\cdot 0.6}=\frac{1}{7.2}=\frac{10}{72}$

In water (${\mu }_m=1.28$):

$\frac{1}{f_w}=\left(\frac{1.6}{1.28}-1\right)\left(\frac{10}{72}\right)$

$\frac{1.6}{1.28}-1=\frac{0.32}{1.28}=\frac{1}{4}$

$\frac{1}{f_w}=\frac{1}{4}\cdot \frac{10}{72}=\frac{10}{288}$

$f_w=28.8$ cm $=288$ mm.

$\therefore$ Option (2).