Light is incident on a prism with an equilateral triangular cross-section, making an angle of $θ$ w.r.t the normal as shown in the figure. The index of refraction of the prism is $\frac{2}{\sqrt{3}}$ . If the angle $θ$ starts at $60^{\circ}$ and is gradually made smaller until no light exist from side BC, the final value of $θ$ in degrees is

Question

Light is incident on a prism with an equilateral triangular cross-section, making an angle of $θ$ w.r.t the normal as shown in the figure. The index of refraction of the prism is $\frac{2}{\sqrt{3}}$ . If the angle $θ$ starts at $60^{\circ}$ and is gradually made smaller until no light exist from side BC, the final value of $θ$ in degrees is

Toppr Admin · Accepted Answer

A-60-x2218-r-x2032-r-ASo- sin90-x2218-x221A-2sinr-x21D2-r-x2032-45-x2218-45-r-A-x21D2-r-15-x2218-Since-1-sin-x3B8-x221A-2sin15-x2218-sin-x3B8-x221A-2sin15-x2218-x3B8-sin-x2212-1-x221A-2sin15-x2218-

A=60∘r′+r=ASo, sin90∘=√2sinr⇒r′=45∘45+r=A⇒r=15∘Since(1)sinθ=√2sin15∘sinθ=√2sin15∘θ=sin−1(√2sin15∘)

$A = 60^{\circ}$

$r^{'} + r = A$

So, $sin 90^{\circ} = \sqrt{2} sin r$

$\Rightarrow r^{'} = 45^{\circ}$

$45 + r = A$

$\Rightarrow r = 15^{\circ}$

Since
$(1) sin θ = \sqrt{2} sin 15^{\circ}$

$sin θ = \sqrt{2} sin 15^{\circ}$

$θ = {sin}^{- 1} (\sqrt{2} sin 15^{\circ})$