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1. ⇒ (MHT CET 2023 12th May Morning Shift )

A beam of light is incident on a glass plate at an angle of $60^{\circ}$ . The reflected ray is polarized. If angle of incidence is $45^{\circ}$ then angle of refraction is

A. $\sin^{- 1} (\frac{1}{\sqrt{6}})$

B. $\sin^{- 1} (\frac{1}{\sqrt{3}})$

C. $\sin^{- 1} (\sqrt{\frac{3}{2}})$

D. $\cos^{- 1} (\sqrt{\frac{3}{2}})$

Correct Option is (A)

According to Brewster's law,

$\begin{aligned} \tan θ_{B} = n \\ ∴ & \tan 60^{\circ} = n \end{aligned}$

$\begin{aligned} \begin{array}{ll} ∴ & n = \sqrt{3} \\ Now, \frac{\sin i}{\sin r} = n \\ ∴ & \sin r = \frac{\sin i}{n} \\ ∴ & \sin r = \frac{\sin 45^{\circ}}{\sqrt{3}} \\ ∴ & \sin r = \frac{1}{\sqrt{6}} \dots . (∵ \sin 45^{\circ} = \frac{1}{\sqrt{2}}) \\ ∴ & r = \sin^{- 1} (\frac{1}{\sqrt{6}}) \end{array} \end{aligned}$

2. ⇒ (MHT CET 2023 9th May Morning Shift )

A beam of unpolarized light passes through a tourmaline crystal A and then it passes through a second tourmaline crystal B oriented so that its principal plane is parallel to that of A. The intensity of emergent light is $I_{0}$ . Now B is rotated by $45^{\circ}$ about the ray. The emergent light will have intensity $(\cos 45^{\circ} = \frac{1}{\sqrt{2}})$

A. $\frac{I_{0}}{2}$

B. $\frac{I_{0}}{\sqrt{2}}$

C. $\frac{\sqrt{2}}{I_{0}}$

D. $\frac{2}{I_{0}}$

Correct Option is (A)

Using law of Malus,

$\begin{aligned} I & = I_{0} \cos^{2} θ \\ = I_{0} (\cos^{2} 45) \\ = I_{0} {(\frac{1}{\sqrt{2}})}^{2} \\ = \frac{I_{0}}{2} \end{aligned}$

⇒Wave Optics