JEE Main 2024 — Wave Optics Question with Solution

Let $${\mathrm{I}}_0$$ be intensity of unpolarised light incident on first polaroid.

$${\mathrm{I}}_1=$$ Intensity of light transmitted from $$1^{\text{st }}$$ polaroid $$=\frac{{\mathrm{I}}_0}{2}$$

$$\theta$$ be the angle between $$1^{\mathrm{st}}$$ and $$2^{\text{nd }}$$ polaroid

$$\varphi$$ be the angle between $$2^{\text{nd }}$$ and $$3^{\text{rd }}$$ polaroid

$$\theta +\varphi =90^{\circ }$$ (as $$1^{\text{st }}$$ and $$3^{\text{rd }}$$ polaroid are crossed)

$$\varphi =90^{\circ }-\theta$$

$${\mathrm{I}}_2=$$ Intensity from $$2^{\text{nd }}$$ polaroid

$${\mathrm{I}}_2={\mathrm{I}}_1{\cos }^2\theta =\frac{{\mathrm{I}}_0}{2}{\cos }^2\theta$$

$${\mathrm{I}}_3=$$ Intensity from $$3^{\text{rd }}$$ polaroid

$${\mathrm{I}}_3={\mathrm{I}}_2{\cos }^2\varphi$$

$${\mathrm{I}}_3={\mathrm{I}}_1{\cos }^2\theta {\cos }^2\varphi$$

$${\mathrm{I}}_3=\frac{{\mathrm{I}}_0}{2}{\cos }^2\theta {\cos }^2\varphi$$

$$\varphi =90-\theta$$

$$\begin{array}{rl}& I_3=\frac{I_0}{2}{\cos }^2\theta {\sin }^2\theta \\ & I_3=\frac{I_0}{2}{\left[\frac{2\sin \theta \cos \theta }{2}\right]}^2\\ & I_3=\frac{I_0}{8}{\sin }^{2}2\theta \end{array}$$

$${\mathrm{I}}_3$$ will be maximum when $$\sin 2\theta =1$$

$$\begin{array}{rl}& 2\theta =90^{\circ }\\ & \theta =45^{\circ }\end{array}$$