Q 12-09-159JEE MainJEE Main 2021 (20 Jul, Shift 1)Medium
Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is:
Answer: (D) 37.58 cm
Light travels from region I into region II. With the pole as origin and the incident direction positive: $u = -40$ cm, and the centre C is on the incident side, so $R = -25$ cm.
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \Rightarrow \frac{1.4}{v} + \frac{1.25}{40} = \frac{0.15}{-25}$$
$$\frac{1.4}{v} = -0.006 - 0.03125 = -0.03725 \Rightarrow v = -37.58\ \text{cm}$$
The image is virtual, 37.58 cm from the surface in region I.
Solution by Sreeraj P, M.Sc Physics