A ray $PQ$ incident on the refracting face $BA$ is refracted in the prism $BAC$ as shown in the figure and emerges from the other refracting face $AC$ as $RS$ such that $AQ=AR$. If the angle of prism $A=60^{\circ}$ and the refractive index of the material of the prism is $\sqrt{3}$, calculate the angle $\theta$.

$\theta=60^{\circ}$ (the angle of deviation).

CBSE Class 12 · 2016 · All India · Geometrical OpticsShort Answer

A ray

P Q

incident on the refracting face

B A

is refracted in the prism

B A C

as shown in the figure and emerges from the other refracting face

A C

R S

such that

A Q = A R

. If the angle of prism

A = 6 0^{\circ}

and the refractive index of the material of the prism is

3

, calculate the angle

θ

Answer: $θ = 6 0^{\circ}$ (the angle of deviation).

Since

A Q = A R

, triangle

A QR

is isosceles, so

∠ A QR = ∠ A R Q = 6 0^{\circ}

and hence

r_{1} = r_{2} = 3 0^{\circ}

, giving

i = e

(symmetric, minimum-deviation path).

μ = \frac{sin i}{sin r _{1}} \Rightarrow 3 = \frac{sin i}{sin 3 0 ^{\circ}} \Rightarrow sin i = \frac{3}{2} \Rightarrow i = 6 0^{\circ} = e

The deviation

θ = i + e - A = 6 0^{\circ} + 6 0^{\circ} - 6 0^{\circ} = 6 0^{\circ}

CBSE Class 12 Board · 2016 · All India · Subjective