JEE Main 2023 — Wave Optics Question with Solution

When an unpolarized light passes through a polariser, the intensity of the emergent light from the polariser depends on the square of the cosine of the angle between the plane of the polariser and the direction of the incident light.

The formula to calculate the intensity $\left(I_1\right)$ of the emergent light from a polariser which makes an angle $\theta$ with the direction of the incident light is given by

$I_1= I{\cos }^2\theta ..................\left(1\right)$

where, $I$ is the intensity of the incident light.

Substitute $45°$ for $\theta$ into equation (1) to obtain the intensity of the first emergent light.

$\begin{array}{rcl}{I}_1& =& I{\cos }^{2}45°\\ & =& \frac{I}{2}\end{array}$

In a similar manner, the intensity $\left(I_2\right)$ of the second emergent light can be written as

$\begin{array}{rcl}{I}_2& =& I_1{\cos }^{2}45°\\ & =& \frac{I_1}{2}\\ & =& \frac{I}{4}\\ & =& \frac{I}{2^2}\end{array}$

Thus, for $n$ number of such polarisers, the net intensity $\left(I_n\right)$ of the emergent light can be written as

$I_n= \frac{I}{2^n}......................\left(1\right)$

Comparing equation (1) with the given expression for the intensity, it can be concluded that

$n= 6$