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MHT CET 202622 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
22 April
Shift
Evening Shift
Subject
Physics
Chapter
Wave Optics

MHT CET 2026 · 22 April · Evening Shift · Physics · Wave Optics

Unpolarized light of intensity I_0 is made to pass through three...

Unpolarized light of intensity I0I_0 is made to pass through three polarizers P1P_1, P2andP3P_2 and P_3 successively. The polarization axis of P2P_2 makes an angle of 60^{}^{\circ} with that of P1P_1, while that of P3P_3 makes an angle of 45^{}^{\circ} with that of the P2P_2. If I0=xI3I_0 = xI_3, the value of 'xx' is \left(\cos 45^{}^{\circ} = \dfrac{1}{\sqrt{2}}, \cos 60^{}^{\circ} = \dfrac{1}{2}\right)

Options

  1. A88
  2. B1616
  3. C3232
  4. D6464

Correct answer

B. 1616

Step-by-step solution

Intensity of unpolarized light is I0I_0.

Intensity of light after passing through the first polarizer P1P_1 is:

I1=I02I_1 = \dfrac{I_0}{2}

The polarization axis of P2P_2 makes an angle of 60^{}^{\circ} with P1P_1. According to Malus's law, the intensity after passing through P2P_2 is:

I_2 = I_1 \cos^2(60^{}^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{1}{2}\right)^2 = \dfrac{I_0}{8}

The polarization axis of P3P_3 makes an angle of 45^{}^{\circ} with P2P_2. The intensity after passing through P3P_3 is:

I_3 = I_2 \cos^2(45^{}^{\circ}) = \dfrac{I_0}{8} \times \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{16}

Given I0=xI3I_0 = x I_3, substituting the value of I3I_3 gives:

I0=x(I016)<br><br>I_0 = x \left(\dfrac{I_0}{16}\right)<br><br>x = 16<br><br>Answer:16<br><br>Answer: 16
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