Q 12-09-040JEE MainJEE Main 2026 (6 Apr, Shift 1)Hard
A spherical interface lens of radius $R$ separates two media of refractive indices $1$ and $1.4$ respectively as shown in the figure below. A point source is placed at a distance of $4R$ in front of spherical interface. The magnitude of the magnification of point source image is ______.
Answer: (A) $1.66$
Refraction at a spherical surface: $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$ with $u = -4R$ and $R$ positive (centre in the second medium):
$$\frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R} \;\Rightarrow\; \frac{1.4}{v} = \frac{0.15}{R} \;\Rightarrow\; v = \frac{28R}{3}$$
$$m = \frac{n_1v}{n_2u} = \frac{28R/3}{1.4 \times (-4R)} = -\frac{5}{3}, \qquad |m| \approx 1.66$$
Solution by Sreeraj P, M.Sc Physics