MHTCET 17 Apr 2026 Shift 2 Paper

Concept:Malus's law describes how the intensity of polarised light changes when it passes through a polariser, varying with $\cos^2$ of the angle between transmission axes.Explanation:When unpolarised light passes through the first polariser, its intensity becomes half:$$I_1 = \frac{I_0}{2} = \frac{64}{2} = 32\ \mathrm{Wm}^{-2}$$Let the angle between the first and second polarisers be $\theta$.Applying Malus's law after the second polariser:$$I_2 = I_1 \cos^2\theta = 32\cos^2\theta$$The third polariser is crossed with the first, so the angle between them is $90^{\circ}$.Thus, the angle between the second and third polarisers is $90^{\circ} - \theta$.After the third polariser:$$I_3 = I_2\cos^2(90^{\circ} - \theta) = 32\cos^2\theta\sin^2\theta$$because $\cos(90^{\circ} - \theta) = \sin\theta$.Using $\sin 2\theta = 2\sin\theta\cos\theta$, we get $\sin^2\theta\cos^2\theta = \frac{1}{4}\sin^2 2\theta$.Therefore, $I_3 = 32 \times \frac{1}{4}\sin^2 2\theta = 8\sin^2 2\theta$.Given the emerging intensity is $6\ \mathrm{Wm}^{-2}$:$$8\sin^2 2\theta = 6 \quad \Rightarrow \quad \sin^2 2\theta = \frac{3}{4}$$$$\sin 2\theta = \frac{\sqrt{3}}{2}$$$$2\theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$$$$\theta = \frac{1}{2}\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$$Answer:The correct option is A: $\frac{1}{2}\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$.