Q134 / 150 · MHT CET 2026 (11 April Shift 1) · Physics
Two polaroid A and B placed in such a way that the pass-axis of polaroid are perpendicular to each other. Now, another polaroid C is placed between A and B bisecting angle between them. If intensity of unpolarised light is $I_0$ then intensity of transmitted light after passing through polaroid B will be
- A$\dfrac{I_0}{4}$
- B$\dfrac{I_0}{2}$
- Correct answerC$\dfrac{I_0}{8}$
- Dzero
Solution
Intensity of light after passing through polaroid A is $I_A = \dfrac{I_0}{2}$
Angle between the pass-axis of polaroid A and C is $\theta_1 = 45^{\circ}$
Intensity of light after passing through polaroid C is $I_C = I_A \cos^2(45^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{4}$
Angle between the pass-axis of polaroid C and B is $\theta_2 = 45^{\circ}$
Intensity of light after passing through polaroid B is $I_B = I_C \cos^2(45^{\circ}) = \dfrac{I_0}{4} \times \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{8}$
Answer: $\dfrac{I_0}{8}$
Angle between the pass-axis of polaroid A and C is $\theta_1 = 45^{\circ}$
Intensity of light after passing through polaroid C is $I_C = I_A \cos^2(45^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{4}$
Angle between the pass-axis of polaroid C and B is $\theta_2 = 45^{\circ}$
Intensity of light after passing through polaroid B is $I_B = I_C \cos^2(45^{\circ}) = \dfrac{I_0}{4} \times \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{8}$
Answer: $\dfrac{I_0}{8}$
Hence, the correct answer is option: 'C'.