Question 202: Wave Optics

Question

Unpolarised light of intensity 32 Wm$$^{-2}$$ passes through the combination of three polaroids such that the pass axis of the last polaroid is perpendicular to that of the pass axis of first polaroid. If intensity of emerging light is 3 Wm$$^{-2}$$, then the angle between pass axis of first two polaroids is ______________ $$^\circ$$.

Accepted Answer

When dealing with three polaroids, the intensity after the second polaroid will be given by Malus' law as $I_0 \cos^2 	heta$, where $	heta$ is the angle between the pass axes of the first two polaroids. However, as the third polaroid is orthogonal to the first, no light from the first polaroid passes through, only light from the second polaroid. So the final intensity is also modulated by a $\sin^2 	heta$ term (as the second and third polaroids are orthogonal).
So if we set up the equation for the final intensity $I_{	ext{net}}$:
$$I_{	ext{net}} = I_0 \cos^2 	heta \sin^2 	heta$$
And we substitute the given values $I_{	ext{net}} = 3 \, 	ext{W/m}^2$ and $I_0 = \frac{32 \, 	ext{W/m}^2}{2} = 16 \, 	ext{W/m}^2$:
$$3 = 16 \cos^2 	heta \sin^2 	heta$$
This simplifies to:
$$\frac{3}{16} = \sin^2 	heta \cos^2 	heta = \left(\frac{1}{2} \sin 2	hetaight)^2$$
Taking the square root of both sides gives:
$$\frac{\sqrt{3}}{2} = \left|\sin 2	hetaight|$$
The solutions for this are $	heta = 30^\circ$ and $	heta = 60^\circ$.
So, the angle between the pass axes of the first two polaroids is either $30^\circ$ or $60^\circ$.

Wave Optics

Integer / Numeric Answer Type