Question 208: Ray Optics

Question

A transparent cube of side, made of a material of refractive index $$\mu $$2, is immersed in a liquid of refractive
index $$\mu $$1($$\mu $$1 < $$\mu $$2). A ray is incident on the face AB at an angle $$	heta $$(shown in the figure). Total internal
reflection takes place at point E on the face BC.

Then $$	heta $$ must satisfy :

Accepted Answer

By SnellÃ¢ÂÂs law at AB $$\mu $$ 1 sin $$ heta $$ = $$\mu $$ 2 sin $$\alpha $$ sin $$\alpha $$ = $${{{\mu _1}} \over {{\mu _2}}}\sin heta $$ ....... (1) Critical angle at BC, $$\mu $$ 2 sin i c = $$\mu $$ 1 sin 90 o $$ \Rightarrow $$ sin i c = $${{{\mu _1}} \over {{\mu _2}}}$$ ..... (2) For Total internal reflection to occur, 90 o - $$\alpha $$ > i c Taking sin both side, sin(90 o - $$\alpha $$) > sin i c $$ \Rightarrow $$ cos $$\alpha $$ > $${{{\mu _1}} \over {{\mu _2}}}$$ (from equation 2) $$ \Rightarrow $$ $$\sqrt {1 - {{\mu _1^2} \over {\mu _2^2}}{{\sin }^2} heta } $$ > $${{{\mu _1}} \over {{\mu _2}}}$$ (from equation 1) Squaring both sides, we get $${1 - {{\mu _1^2} \over {\mu _2^2}}{{\sin }^2} heta }$$ > $${{{\mu _1^2} \over {\mu _2^2}}}$$ $$ \Rightarrow $$ $${1 - {{\mu _1^2} \over {\mu _2^2}}}$$ > $${{{\mu _1^2} \over {\mu _2^2}}{{\sin }^2} heta }$$ $$ \Rightarrow $$ $${{\mu _2^2} \over {\mu _1^2}} - 1$$ > $${{{\sin }^2} heta }$$ $$ \Rightarrow $$ sin $$ heta $$ < $$\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $$ $$ \Rightarrow $$ $$ heta < {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $$

Answer

$$	heta  > {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $$

Answer

$$	heta  < {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $$

Answer

$$	heta  < {\sin ^{ - 1}}{{{\mu _1}} \over {{\mu _2}}}$$

Answer

$$	heta  > {\sin ^{ - 1}}{{{\mu _1}} \over {{\mu _2}}}$$

Ray Optics

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