JEE Main 2023 — Ray Optics Question with Solution

Let's consider the following diagram-

The relation between the object distance, the image distance and the radius of curvature of the spherical surface can be written as

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R} ...\left(1\right)$

Substitute the values of the known parameters into equation (1) and solve to calculate the required image distance.

$\begin{array}{rcl}\frac{1.5}{v}-\frac{1}{-15}& =& \frac{1.5-1.0}{30}\\ & \Rightarrow & \frac{1.5}{v}=\frac{1}{60}-\frac{1}{15}\\ & =& -\frac{1}{20}\\ & \Rightarrow & v=-30\end{array}$

Hence, the image is formed at a distance of $30 \mathrm{cm}$ from the pole on the same direction as the object is placed.