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JEE Main 2019 · Physics · Waves and Sound · medium

Unpolarised light of intensity I0 passes through three ideal polaroids in succession. The pass axis of the second makes an angle of 30^° with that of…

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Unpolarised light of intensity I0I_0 passes through three ideal polaroids in succession. The pass axis of the second makes an angle of 30∘30^\circ with that of the first, and the pass axis of the third makes an angle of 60∘60^\circ with that of the second, rotated in the same sense. The intensity of the light emerging from the third polaroid is
  1. A. 3I016\dfrac{3I_0}{16}
  2. B. I08\dfrac{I_0}{8}
  3. C. 3I08\dfrac{3I_0}{8}
  4. D. 3I032\dfrac{3I_0}{32}
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Answer: (D) 3I032\dfrac{3I_0}{32}

The first polaroid transmits half of the unpolarised intensity: I1=I0/2I_1=I_0/2. Malus's law applies at each later polaroid with the angle between consecutive pass axes: I2=I1cos⁡230∘=I02⋅34=3I08I_2=I_1\cos^2 30^\circ=\dfrac{I_0}{2}\cdot\dfrac34=\dfrac{3I_0}{8}, and then I3=I2cos⁡260∘=3I08⋅14=3I032I_3=I_2\cos^260^\circ=\dfrac{3I_0}{8}\cdot\dfrac14=\dfrac{3I_0}{32}. The last angle to use is between the second and third axes (60∘60^\circ), not between the first and third (90∘90^\circ, which would wrongly give zero). Omitting the initial factor of 12\tfrac12 gives 3I0/163I_0/16.
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