Question

Two polaroide $A$ and $B$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other Now, another polaroid $C$ is placed between $A$ and $B$ bisecting angle between them If intensity of unpolarized light is $I _0$ then intensity of transmitted light after passing through polaroid $B$ will be:

• A. $\frac{I_0}{8}$
• B. $Z erO$
• C. $\frac{I_0}{2}$
• D. $\frac{I_0}{4}$

Hint:

The intensity of light passing through successive polarising filters is given by the product of intensity and the cosine square of the angle between the polarising axes.

Answer 1

The Correct Option is A

To solve this problem, we first need to understand the behavior of light as it passes through multiple polaroids. Here, we have three polaroids: A, B, and C.

1. Polaroids A and B are oriented such that their pass axes are perpendicular to each other. Therefore, without polaroid C, no light would pass from A to B, as the light polarized by A will be completely absorbed by B.
2. Polaroid C is inserted between A and B and is oriented to bisect the angle between A and B. Since A and B are perpendicular (90° apart), polaroid C will make an angle of 45° with each of them.
3. The intensity of light after passing through a polaroid can be calculated using Malus' Law, which states that the intensity of polarized light passing through a polaroid is given by: \(I = I_0 \cdot \cos^2(\theta)\), where \(\theta\) is the angle between the light's polarization direction and the polaroid's pass axis.
4. The initial unpolarized light has intensity \(I_0\). After passing through polaroid A, it becomes polarized with an intensity of \(\frac{I_0}{2}\) because unpolarized light loses half its intensity when first polarized.
5. When the light passes through polaroid C:
   • \(\theta = 45^\circ\)
   • The intensity becomes: \(I_C = \frac{I_0}{2} \cdot \cos^2(45^\circ) = \frac{I_0}{2} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4}\)
6. Finally, light passes through polaroid B:
   • Again, \(\theta = 45^\circ\) between polaroid C and B.
   • The intensity now becomes: \(I_B = \frac{I_0}{4} \cdot \cos^2(45^\circ) = \frac{I_0}{4} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{4} \cdot \frac{1}{2} = \frac{I_0}{8}\)
7. Therefore, the intensity of the transmitted light after passing through all three polaroids is \(\frac{I_0}{8}\).

Thus, the correct option is \(\frac{I_0}{8}\).