QJEEAdv24Ph8980: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph8980 Solution: A point source emits unpolarized light uniformly in all…

To solve this problem, we start by understanding the information given and how the polarization of light affects light intensity.

Initially, we have the ratio of the intensities at points A and B given as:

$r = \frac{I_A}{I_B} = 2$

This means:

$I_A = 2I_B$

Now, we place two polaroids with their pass-axes at an angle of $45^\circ$ between them before point B. When unpolarized light passes through the first polaroid, it gets polarized, and its intensity is reduced to half of its original value:

$I_{B1} = \frac{1}{2} I_B$

Here, $I_{B1}$ is the intensity of light after passing through the first polaroid.

Next, this polarized light passes through the second polaroid, which is at an angle of $45^\circ$ to the first one. According to Malus's law, the intensity of light after passing through the second polaroid is given by:

$I_{B2} = I_{B1} \cos^2(45^\circ)$

We know that:

$\cos(45^\circ) = \frac{1}{\sqrt{2}}$

Thus,

$I_{B2} = \frac{1}{2} I_B \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} I_B \cdot \frac{1}{2} = \frac{1}{4} I_B$

Now, the new intensity at point B is $I_{B2}$ . We need to calculate the new ratio $r'$ of the intensities at points A and B:

$r' = \frac{I_A}{I_{B2}}$

Substituting the values $I_A = 2I_B$ and $I_{B2} = \frac{1}{4} I_B$ , we get:

$r' = \frac{2 I_B}{\frac{1}{4} I_B} = 2 \times 4 = 8$

Therefore, the new value of $r$ will be:

$r' = 8$