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Physics Question on Ray optics and optical instruments

A ray incident at a point at an angle of incidence of 60∘60^{\circ} enters a glass sphere of refractive index 3\sqrt{3} and is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is

A

50∘50^{\circ}

B

60∘60^{\circ}

C

90∘90^{\circ}

D

40∘40^{\circ}

Answer

90∘90^{\circ}

Explanation

Solution

Refraction at PP, sin 60∘sin r1=3\frac{sin\,60^{\circ}}{sin\,r_1} = \sqrt{3} sin r1=12sin\,r_1 = \frac{1}{2} or r1=30∘r_1 = 30^{\circ} Since, r2=r1r_2 = r_1; ∴r2=30∘\therefore r_2 = 30^{\circ} Refraction at Q,sin r2sin i2=13Q, \frac{sin\,r_2}{sin\,i_2} = \frac{1}{\sqrt{3}} or sin 30∘sin i2=13\frac{sin\,30^{\circ}}{sin\,i_2} = \frac{1}{\sqrt{3}} or i2=60∘i_2 = 60^{\circ} At point Q,r2′=r2=30∘Q, r'_2 = r_2 = 30^{\circ} ∴α=180∘−(r2′+i2)\therefore \alpha = 180^{\circ} - (r'_2 +i_2) =180∘−(30∘+60∘)=90∘= 180^{\circ} -(30^{\circ} + 60^{\circ} ) = 90^{\circ}