Question

A polaroid sheet \(C\) is rotated between two crossed polaroids \(A\) and \(B\). If the light emerged from the first polaroid \(A\) is plane polarized, then the angle between the pass axes of polaroids \(A\) and \(C\) for which the intensity of transmitted light from polaroid \(B\) becomes maximum is

• A. \(45^\circ\)
• B. \(30^\circ\)
• C. \(60^\circ\)
• D. \(37^\circ\)

Hint:

For three polaroids with the first and last crossed, \[ \boxed{ I=I_0\cos^2\theta\sin^2\theta =\frac{I_0}{4}\sin^22\theta } \] Maximum transmission occurs at \[ \boxed{\theta=45^\circ.} \]

Answer 1

The Correct Option is A

Step 1: Apply Malus' law. Let the angle between the pass axes of polaroids \(A\) and \(C\) be \(\theta\). Since \(A\) and \(B\) are crossed, \[ \angle(C,B)=90^\circ-\theta. \] The intensity after polaroid \(C\) is \[ I_1=I_0\cos^2\theta. \] The intensity after polaroid \(B\) is \[ I = I_1\cos^2(90^\circ-\theta) = I_0\cos^2\theta\sin^2\theta. \]

Step 2: Maximize the intensity. Using \[ \sin2\theta = 2\sin\theta\cos\theta, \] \[ I = \frac{I_0}{4}\sin^22\theta. \] Maximum intensity occurs when \[ \sin2\theta=1, \] i.e., \[ 2\theta=90^\circ. \] Hence, \[ \boxed{\theta=45^\circ.} \] Therefore, \[ \boxed{(A)} \] is the correct answer.