Question

Axes of polariser and analyser are inclined at an angle $60^{\circ}$. Intensity of emerging light from analyser is $I$. Intensity of light on the polariser is

• A. $8 \, I$
• B. $4 \, I$
• C. $2 \, I$
• D. $I$

Answer 1

Answer: (A)

$8 \, I$

Answer 2

Let intensity of light on the polariser be $I_0$.
First, light passes through polariser and its intensity becomes $I_o / 2$.
Angle between polariser and analyser is $60^{\circ}$.
Using Malus law, intensity of transmitted light,
$I =\frac{I_{0}}{2} \cos^{2} 60^{\circ} =\frac{I_{0}}{2} \times \frac{1}{4}$
$\therefore \, I_{0} =8I$