JEE Main 2025 — 7 April, Morning Shift — Question 60

$\frac{\mu_{1}}{f}-\frac{\mu_{1}}{\alpha}=\left(\mu-\mu_{1}\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$ For air $\mu_{1}=1$ and $\mu=1.6$

So $\frac{1}{f}=0.6\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=\frac{1}{12} \mathrm{~cm}$ For water $\mu_{1}=1.28$ and $\mu=1.6$

So, $\frac{\mu_{1}}{f_{w}}=0.32\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \Rightarrow \frac{1}{f_{w}}=\frac{0.32}{1.28} \times \frac{1}{0.6 \times 12}$

$\Rightarrow \quad f_{w}=\frac{1.28 \times 0.6 \times 12}{0.32}=28.8 \mathrm{~cm} \approx 288 \mathrm{~mm}$