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Physics Question on Wave optics

The Brewster angle for the glass-air interface is 54.74∘54.74^{\circ}. If a ray of light going from air to glass strikes at an angle of incidence 45∘45^{\circ}, then the angle of refraction is (given, tan⁡54.74∘=2\tan 54.74^{\circ}=\sqrt{2} )

A

60∘60^{\circ}

B

30∘30^{\circ}

C

25∘25^{\circ}

D

54.74∘54.74^{\circ}

Answer

30∘30^{\circ}

Explanation

Solution

Refractive index μ=tan⁡θ=2\mu=\tan \theta=\sqrt{2} Now according to snell's law 2=sin⁡(54.74)sin⁡r\sqrt{2}=\frac{\sin (54.74)}{\sin r} sin⁡r=0.8162sin⁡r=0.57\sin r=\frac{0.816}{\sqrt{2}}\sin r=0.57 r=35∘≈30∘r=35^{\circ} \approx 30^{\circ}