Question

A ray of light is incident at an angle of 60° on a prism of refracting angle 60°. If refractive index of the material of the prism is 3^½, then angle between incident ray and emergent ray will be

(2) 60°
(1) 45°
(3) 37°
(4) 30°​

Answer 1

Answer:

2.60°

Explanation:

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Answer 2

Answer:

The angle between the incident ray and emergent ray will be 60°i.e.option(2).

Explanation:

From the Snell's law, we have,

$\mu_1sini=\mu_2sinr_1$        (1)

Where,

μ₁=refractive index of the medium in which light is present

i=angle of incidence

μ₂=refractive index of the prism

r₁=refracting angle at the first surface of the prism

From the question we have,

i=60°

The refracting angle of the prism(A)=60°

μ₁=1      (air)

μ₂=$\sqrt{3}$

By substituting the values in equation (1) we get;

$1\times sin60\textdegree=\sqrt{3} \times sinr_1$

$\frac{\sqrt{3} }{2} =\sqrt{3} \times sinr_1$

$sinr_1=\frac{1}{2}$

$r_1=30\textdegree$     (2)

And,

$A=r_1+r_2$

$60\textdegree=30\textdegree+r_2$

$r_2=30\textdegree$       (3)

By applying Snell's law at the second surface of the prism we get;

$\mu_2sinr_2=\mu_1sine$    (4)

By substituting the required values in equation (4) we get;

$\sqrt{3} \times sinr30\textdegree=1\times sine$

$sine=\frac{\sqrt{3} }{2}$

$e=60\textdegree$     (5)

The angle between the incident and emergent ray(δ),

$\delta=i+e-A$       (6)

By substituting the required values in equation (6) we get;

$\delta=60\textdegree+60\textdegree-60\textdegree$

$\delta=60\textdegree$

Hence, the angle between the incident ray and emergent ray will be 60°i.e.option(2).