Q 12-09-129JEE MainJEE Main 2023 (8 Apr, Shift 2)Medium
Two transparent media having refractive indices $1.0$ and $1.5$ are separated by a spherical refracting surface of radius of curvature $30$ cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principal axis in rarer medium at a distance of $15$ cm from the pole of the surface. The distance of image from the pole of the surface is ______ cm.
Numerical value type. Enter your answer.
Answer: 30
$\dfrac{\mu_2}{v}-\dfrac{\mu_1}{u}=\dfrac{\mu_2-\mu_1}{R}$ with $u=-15$, $R=+30$:
$$\frac{1.5}{v}+\frac1{15}=\frac{0.5}{30}\ \Rightarrow\ \frac{1.5}{v}=-\frac{1}{20}\ \Rightarrow\ v=-30\ \text{cm}$$
A virtual image $30$ cm from the pole, on the object's side.
Solution by Sreeraj P, M.Sc Physics