Wave Optics

JEE Main level, numerical answer, medium

158 of 191

Question 158 of 191

An unpolarized light of intensity IoI_o passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is 0°0° and intensity of light emerged from analyser is 38Io\dfrac{3}{8}I_o, the angle of rotation of the light by the solution with respect to analyser is _______ degrees.

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Answer: 30

Intensity of light after passing through the first polarizer is:

I1=Io2I_1 = \dfrac{I_o}{2}

Let the optically active solution rotate the plane of polarization by an angle θ\theta. Since the angle between the polarizer and the analyser is 0∘0^\circ, the angle between the plane of polarization of the light emerging from the solution and the transmission axis of the analyser is also θ\theta.

According to Malus's law, the intensity of light emerging from the analyser is:

I2=I1cos⁡2θ=Io2cos⁡2θI_2 = I_1 \cos^2 \theta = \dfrac{I_o}{2} \cos^2 \theta

Given that the final intensity is 38Io\dfrac{3}{8}I_o, we have:

Io2cos⁡2θ=38Io\dfrac{I_o}{2} \cos^2 \theta = \dfrac{3}{8}I_o

cos⁡2θ=34\cos^2 \theta = \dfrac{3}{4}

cos⁡θ=32\cos \theta = \dfrac{\sqrt{3}}{2}

θ=30∘\theta = 30^\circ

Answer: 3030

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