Wave Optics Part 1

Given, intensity of light falling on screen is I. Intensity of light when $P_{1}$ and $P_{2}$ placed between screen and source is I/2. As per Malus’s law, the intensity of transmitted light when passing through a pair of polariser and analyser having angle ϕ between their axis is given as $I=I_{0} \cos^{2} \phi$ Initially let us consider the angle between analyser and polariser is 0°. Then, the intensity of transmitted light is I / 2. Thus, intensity of light become I / 2 after transmitting through 1st polariser as shown in the figure. If we rotate the polarise $P_{2}$ by angle ϕ, then the resulting intensity of light on screen will be $\frac{3}{8}$. Now, for second condition, apply Malus’s law, $\frac{3l}{8}=\frac{l}{2} \cos^{2} \phi$ $\cos^{2} \phi=\frac{3}{4}$ ⇒ $\cos \phi=\frac{\sqrt{3}}{2}$ ϕ = 30°. Thus, $P_{2}$ should be rotated by $30^{\circ}$ to get intensity of light $\frac{31}{8}$ on screen.