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NEET PHYSICSEasy

A linearly polarized monochromatic light of intensity 10 lumen is incident on a polarizer. The angle between the direction of polarization of the light and that of the polarizer such that the intensity of output light is 2.5 lumen is:

A

60°

B

75°

C

30°

D

45°

Step-by-Step Solution

  1. Identify Given Values:
  • Incident Intensity (I0I_0) = 10 lumen (Correcting typo '101010').
  • Transmitted/Output Intensity (II) = 2.5 lumen (Correcting typo '2.52.52.5').
  1. Formula: According to Malus' Law, when linearly polarized light is incident on a polarizer, the transmitted intensity is given by: I=I0cos⁡2θI = I_0 \cos^2 \theta where θ\theta is the angle between the direction of polarization of the incident light and the pass axis of the polarizer.
  2. Calculation: 2.5=10cos⁡2θ2.5 = 10 \cos^2 \theta cos⁡2θ=2.510=14=0.25\cos^2 \theta = \frac{2.5}{10} = \frac{1}{4} = 0.25 Taking the square root: cos⁡θ=0.5\cos \theta = 0.5 θ=cos⁡−1(0.5)=60∘\theta = \cos^{-1}(0.5) = 60^\circ
  3. Conclusion: The angle must be 60∘60^\circ.
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