JEE Main 2023 — Ray Optics Question with Solution

The focal length of the bi-convex lens $\left(f\right)$ can be calculated as follows:

$\begin{array}{rcl}\frac{1}{f}& =& (\mu -1)\left(\frac{1}{R}-\frac{1}{-R}\right)\\ & =& \frac{2\left(\mu -1\right)}{R} ...\left(1\right)\end{array}$

When the lens is cut into two halves, the focal length $\left(f\text{'}\right)$ of each part can be calculated as follows:

$\begin{array}{rcl}\frac{1}{f^{\text{'}}}& =& (\mu -1)\left(\frac{1}{R}-\frac{1}{\infty }\right)\\ & =& \frac{\left(\mu -1\right)}{R} ...\left(2\right)\end{array}$

Hence, the ratio of the powers for both the lenses can be given by

$\begin{array}{rcl}\frac{P^{\text{'}}}{P}& =& \frac{{\displaystyle \frac{1}{f\text{'}}}}{{\displaystyle \frac{1}{f}}}\\ & =& \frac{f}{f\text{'}}\\ & =& \frac{{\displaystyle \frac{\left(\mu -1\right)}{R}}}{{\displaystyle \frac{2\left(\mu -1\right)}{R}}}\\ & =& \frac{1}{2}\\ & \Rightarrow & P\text{'}= \frac{P}{2} ...\left(3\right)\end{array}$

Substitute the value of the known parameter into equation (3) to calculate the required power.

$\begin{array}{rcl}P\text{'}& =& \frac{10 \mathrm{D}}{2}\\ & =& 5 \mathrm{D}\end{array}$