Q 12-09-056JEE MainJEE Main 2026 (28 Jan, Shift 1)Easy
A convex lens of refractive index 1.5 and focal length $f = 18$ cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha\times\text{f}$. The value of $\alpha$ is ______ .
(refractive index of water $= 4/3$ )
Numerical value type. Enter your answer.
Answer: 3
By the lens maker's formula, $\dfrac1f \propto \left(\dfrac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right)$.
$$\frac{f_w}{f} = \frac{1.5 - 1}{\frac{1.5}{4/3} - 1} = \frac{0.5}{0.125} = 4$$
So $f_w = 4f$, and $f_w - f = 3f\Rightarrow\alpha = 3$.
Solution by Sreeraj P, M.Sc Physics