For the prism given below. Two light rays are incident normally on the surface of the prism. The angle between emergent rays will be

Question

For the prism given below. Two light rays are incident normally on the surface of the prism. The angle between emergent rays will be (A) 27° (B) 32° (C) 37° (D) 42°

Tardigrade Admin · Accepted Answer

The angle between any two lines is equal to the angle between their perpendiculars.

∴ i = 3 0^{\circ}

From Snell's law, we have

\frac{1}{1.5} = \frac{s i n 3 0 ^{\circ}}{s i n r}

\Rightarrow sin r = 0.75 or r = 48. 6^{\circ}

Therefore,

θ = r - i

= 48. 6^{\circ} - 3 0^{\circ} = 18. 6^{\circ}

∴

Required angle between two emergent rays

= 2 \times 18. 6^{\circ}

= 37. 2^{\circ} C \approx 3 7^{\circ}

Q. For the prism given below. Two light rays are incident normally on the surface of the prism. The angle between emergent rays will be

$2 7^{\circ}$

$3 2^{\circ}$

$3 7^{\circ}$

$4 2^{\circ}$

Q. For the prism given below. Two light rays are incident normally on the surface of the prism. The angle between emergent rays will be

27∘

32∘

37∘

42∘

The angle between any two lines is equal to the angle between their perpendiculars. ∴i=30∘ From Snell's law, we have 1.51​=sinrsin30∘​ ⇒sinr=0.75 or r=48.6∘ Therefore, θ=r−i =48.6∘−30∘=18.6∘ ∴ Required angle between two emergent rays =2×18.6∘ =37.2∘C≈37∘

$2 7^{\circ}$

$3 2^{\circ}$

$3 7^{\circ}$

$4 2^{\circ}$