Question 2 of 8intermediate🔧 ApplyNumerical3 marks

Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?

Correct Answer

Let us consider two crossed polaroids P₁ and P₃ with a polaroid sheet P₂ rotated between them. Let I₀ be the intensity of polarised light after passing through the first polariser P₁.

When this light passes through the second polariser P₂, which makes an angle θ with the pass axis of P₁, by Malus' law, the intensity becomes: I = I₀ cos²θ

Since P₁ and P₃ are crossed (their pass axes are perpendicular), the angle between the pass axes of P₂ and P₃ will be (π/2 - θ). Applying Malus' law again for P₃: I = I₀ cos²θ cos²(π/2 - θ)

Using the identity cos(π/2 - θ) = sin θ: I = I₀ cos²θ sin²θ = (I₀/4) sin²2θ

Therefore, the transmitted intensity varies as (I₀/4) sin²2θ. The intensity is maximum when θ = π/4 (45°) and zero when θ = 0° or 90°.

Exercise: Example 10.2 | Q: Example 10.2 | (Page Page 17)
For More Understanding

Explanation

This example demonstrates how inserting a third polaroid between two crossed polaroids allows light transmission, which seems counterintuitive. The key insight is that the intermediate polaroid P₂ partially aligns the light's polarization, enabling some component to pass through P₃. The mathematical derivation uses Malus' law twice and trigonometric identities to arrive at the final expression showing that maximum transmission occurs at 45°.

Solution Steps

  1. Step 1: Apply Malus' law for P₂: I = I₀ cos²θ

  2. Step 2: Determine angle between P₂ and P₃: (π/2 - θ) since P₁ and P₃ are crossed

  3. Step 3: Apply Malus' law for P₃: I = I₀ cos²θ cos²(π/2 - θ)

  4. Step 4: Simplify using cos(π/2 - θ) = sin θ: I = I₀ cos²θ sin²θ

  5. Step 5: Apply identity sin²2θ = 4 sin²θ cos²θ: I = (I₀/4) sin²2θ

  6. Step 6: Maximum occurs when sin²2θ = 1, i.e., θ = π/4

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