Q 12-09-050JEE MainJEE Main 2026 (2 Apr, Shift 1)Medium
Refer the figure given below. $\mu_1$ and $\mu_2$ are refractive indices of air and lens material. The height of image will be ______ cm.
Answer: (A) $1$
Take light travelling left to right from P. The object O is $40$ cm to the left, so $u=-40$ cm. The centre of curvature C is on the object's side, $20$ cm from P, so $R=-20$ cm.
$$\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}\ \Rightarrow\ \frac{1.54}{v}+\frac{1}{40}=\frac{0.54}{-20}$$
$\dfrac{1.54}{v}=-0.027-0.025=-0.052\Rightarrow v\approx-29.6$ cm (virtual image on the object's side).
Magnification: $m=\dfrac{\mu_1v}{\mu_2u}=\dfrac{1\times(-29.6)}{1.54\times(-40)}\approx0.48$
Image height $=0.48\times2\approx0.96$ cm $\approx1$ cm (erect).
Solution by Sreeraj P, M.Sc Physics