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Wave Optics question

2021 · 24 Feb · Shift 1 · Q63
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Wave Optics question

2021 · 24 Feb · Shift 1 · Q63

JEE MainPhysicsWave OpticsNumerical+4 / −1
An unpolarised light beam is incident on the polarizer of a polarization experiment and the intensity of light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by 30 ∘^\circ∘ in clockwise direction, the intensity of emerging light will be ‾\underline{\hspace{2cm}}​ Lumens.
Numerical answer
View written solutionFree

Correct answer: 75

  1. Initial setup

    The light is unpolarised and first passes through a polarizer.

    For unpolarised light passing through a polarizer, the transmitted intensity becomes I1=I02I_1=\frac{I_0}{2}I1​=2I0​​ where I0I_0I0​ is the incident intensity.

  2. Intensity through analyzer

    Let the angle between the transmission axes of the polarizer and analyzer initially be θ\thetaθ.

    By Malus' law, intensity after the analyzer is I=I1cos⁡2θI=I_1\cos^2\thetaI=I1​cos2θ

    We are given that initially, I=100 LumensI=100\text{ Lumens}I=100 Lumens

  3. Analyzer is rotated by 30∘30^\circ30∘

    Rotating the analyzer by 30∘30^\circ30∘ changes the angle between the axes from θ\thetaθ to θ+30∘\theta+30^\circθ+30∘.

    New intensity becomes I′=I1cos⁡2(θ+30∘)I'=I_1\cos^2(\theta+30^\circ)I′=I1​cos2(θ+30∘)

  4. Interpreting the usual polarization experiment condition

    In a standard polarization experiment, when the analyzer reading is quoted without any relative angle specified, it is generally taken that the polarizer and analyzer are initially parallel.

    So initially, θ=0∘\theta=0^\circθ=0∘

    Hence, 100=I1cos⁡20∘=I1100=I_1\cos^2 0^\circ=I_1100=I1​cos20∘=I1​

    Therefore, I1=100I_1=100I1​=100

  5. After rotation

    Now the analyzer is rotated by 30∘30^\circ30∘: I′=100cos⁡230∘I'=100\cos^2 30^\circI′=100cos230∘

    Since cos⁡30∘=32\cos 30^\circ=\frac{\sqrt{3}}{2}cos30∘=23​​ so cos⁡230∘=34\cos^2 30^\circ=\frac{3}{4}cos230∘=43​

    Therefore, I′=100×34=75I'=100\times \frac{3}{4}=75I′=100×43​=75

  6. Final answer

    The intensity of emerging light is 75 Lumens\boxed{75\text{ Lumens}}75 Lumens​

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