Wave Optics

JEE Main level, single correct, medium

159 of 191

Question 159 of 191

An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at 30°30° and 90°90°, respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at 60°60° with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is ______.

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Answer: (C) 9/49/4

Let the intensity of the incident unpolarized light be I0I_0.

After passing through the first polarizer, the intensity becomes I1=I02I_1 = \dfrac{I_0}{2}.

Without the third polarizer, the light passes directly from the first polarizer (at 30∘30^{\circ}) to the second polarizer (at 90∘90^{\circ}). The angle between their transmission axes is 90∘−30∘=60∘90^{\circ} - 30^{\circ} = 60^{\circ}.

Using Malus's law, the output intensity without the third polarizer is:
Iwithout=I1cos⁡2(60∘)=I02×(12)2=I08I_{without} = I_1 \cos^2(60^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{1}{2}\right)^2 = \dfrac{I_0}{8}

When the third polarizer (at 60∘60^{\circ}) is placed between them, the light first passes from the first polarizer to the third polarizer. The angle between their axes is 60∘−30∘=30∘60^{\circ} - 30^{\circ} = 30^{\circ}.

The intensity after the third polarizer is:
I3=I1cos⁡2(30∘)=I02×(32)2=3I08I_3 = I_1 \cos^2(30^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_0}{8}

The light then passes from the third polarizer to the second polarizer. The angle between their axes is 90∘−60∘=30∘90^{\circ} - 60^{\circ} = 30^{\circ}.

The final output intensity with the third polarizer is:
Iwith=I3cos⁡2(30∘)=3I08×(32)2=9I032I_{with} = I_3 \cos^2(30^{\circ}) = \dfrac{3I_0}{8} \times \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{9I_0}{32}

The ratio of the output intensities with and without the third polarizer is:
Ratio=IwithIwithout=9I032I08=94\text{Ratio} = \dfrac{I_{with}}{I_{without}} = \dfrac{\dfrac{9I_0}{32}}{\dfrac{I_0}{8}} = \dfrac{9}{4}

Answer: 9/49/4

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