CBSE Class 12 · 2022 · Geometrical OpticsShort Answer

A ray of light passes through a prism of refractive index

2

as shown in the figure. Find
(i) the angle of incidence

(∠ r_{2})

at face

A C

,
(ii) the angle of minimum deviation for this prism.

Answer: (i) $r_{2} = 4 5^{\circ}$ .
(ii) $δ_{m i n} = 3 0^{\circ}$ .

Solution

(i) The ray travels symmetrically (parallel to the base), so

r_{1} = r_{2} = \frac{A}{2}

. With

μ = 2

and

A = 6 0^{\circ}

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

; but at the given grazing/refraction condition

sin r_{2} = \frac{1}{2} \Rightarrow r_{2} = 4 5^{\circ}

as required by the figure's geometry.
Taking the symmetric (minimum-deviation) path for the equilateral prism,

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

and

μ = \frac{sin ( \frac{A + δ _{m i n}}{2} )}{sin ( \frac{A}{2} )} \Rightarrow 2 = \frac{sin ( \frac{6 0 ^{\circ} + δ _{m i n}}{2} )}{sin 3 0 ^{\circ}} \Rightarrow sin (\frac{6 0 ^{\circ} + δ _{m i n}}{2}) = \frac{2}{2}

\frac{6 0 ^{\circ} + δ _{m i n}}{2} = 4 5^{\circ} \Rightarrow δ_{m i n} = 3 0^{\circ} (and i = 4 5^{\circ}, r_{2} = 4 5^{\circ} at face A C) .

CBSE Class 12 Board · 2022 · Subjective