Question

Unpolarized light with intensity \(I_{0}\) passes through three polaroids such that the transmission axis of the last polaroid is crossed with the first. If the intensity of emerging light is \(I_{1}\left(I_{1}=\frac{3}{32}I_{0}\right)\). The angle between the first two polaroids and the angle for which the transmitted intensity is maximum are respectively \(\{\)given, \(\cos (90)=0\}\)

• A. \(\frac{1}{2}\sin ^{-1}\frac{\sqrt{3}}{2}\) and \(45^{\circ }\)
• B. \(\sin ^{-1}\frac{\sqrt{3}}{2}\) and \(90^{\circ }\)
• C. \(\frac{1}{2}\cos ^{-1}\frac{\sqrt{3}}{2}\) and \(45^{\circ }\)
• D. \(\cos ^{-1}\frac{\sqrt{3}}{2}\) and \(90^{\circ }\)

Hint:

Apply Malus's law twice with angles theta and 90 degrees minus theta, then find the angle at which the final intensity is largest.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Unpolarised light loses half its intensity at the first polaroid, giving \(\dfrac{I_0}{2}\).
Each further polaroid follows Malus's law \(I=I_{in}\cos^{2}\theta\).

Step 2: Angles
Let \(\theta\) be the angle between the first two polaroids.
The last polaroid is crossed with the first, so the angle between the second and third is \(90^{\circ}-\theta\).

Step 3: Emerging Intensity
\(I_1=\dfrac{I_0}{2}\cos^{2}\theta\cos^{2}(90^{\circ}-\theta)=\dfrac{I_0}{2}\cos^{2}\theta\sin^{2}\theta=\dfrac{I_0}{8}\sin^{2}2\theta\)
Given \(I_1=\dfrac{3}{32}I_0\), we get \(\sin^{2}2\theta=\dfrac{3}{4}\), so \(\sin 2\theta=\dfrac{\sqrt{3}}{2}\).

Step 4: Angles Found
\(\theta=\dfrac{1}{2}\sin^{-1}\dfrac{\sqrt{3}}{2}\).
The intensity is maximum when \(\sin 2\theta=1\), that is \(\theta=45^{\circ}\). This matches option (A).

Final Answer:
The first angle is half of arcsin of root 3 over 2, and the intensity is maximum at 45 degrees. \[ \boxed{\text{(A) }\dfrac{1}{2}\sin^{-1}\dfrac{\sqrt{3}}{2}\text{ and }45^{\circ}} \]