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NEET PHYSICSWave OpticsMedium

Question

Two polaroids P1P_1 and P2P_2 are placed with their axis perpendicular to each other. Unpolarised light IoI_o is incident on P1P_1. A third polaroid P3P_3 is kept in between P1P_1 and P2P_2 such that its axis makes an angle 4545^\circ with that of P1P_1. The intensity of transmitted light through P2P_2 is

A

Io2\frac{I_o}{2}

B

Io4\frac{I_o}{4}

C

Io8\frac{I_o}{8}

D

Io16\frac{I_o}{16}

Step-by-Step Solution

Let the intensity of unpolarised light be IoI_o. When unpolarised light passes through the first polaroid P1P_1, its intensity becomes half: I1=Io2I_1 = \frac{I_o}{2} The transmitted light is now plane-polarised. The third polaroid P3P_3 is placed between P1P_1 and P2P_2 with its pass axis at an angle of 4545^\circ with respect to P1P_1. According to Malus's Law, the intensity of light transmitted through P3P_3 is: I3=I1cos2(45)=Io2×(12)2=Io2×12=Io4I_3 = I_1 \cos^2(45^\circ) = \frac{I_o}{2} \times \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_o}{2} \times \frac{1}{2} = \frac{I_o}{4} The polaroids P1P_1 and P2P_2 are crossed (their axes are perpendicular to each other). Since P3P_3 makes an angle of 4545^\circ with P1P_1, the angle between the transmission axes of P3P_3 and P2P_2 is 9045=4590^\circ - 45^\circ = 45^\circ. Again, applying Malus's Law, the intensity of light transmitted through P2P_2 is: I2=I3cos2(45)=Io4×(12)2=Io4×12=Io8I_2 = I_3 \cos^2(45^\circ) = \frac{I_o}{4} \times \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_o}{4} \times \frac{1}{2} = \frac{I_o}{8}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Wave Optics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWave Opticspolaroidsplacedperpendicularunpolarisedincident

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