JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 18)
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 $$\theta $$(shown in the figure). Total internal
reflection takes place at point E on the face BC.
_12th_April_Evening_Slot_en_18_1.png)
Then $$\theta $$ must satisfy :
Then $$\theta $$ must satisfy :
$$\theta > {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $$
$$\theta < {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $$
$$\theta < {\sin ^{ - 1}}{{{\mu _1}} \over {{\mu _2}}}$$
$$\theta > {\sin ^{ - 1}}{{{\mu _1}} \over {{\mu _2}}}$$
Explanation
_12th_April_Evening_Slot_en_18_2.png)
By Snell’s law at AB
$$\mu $$1 sin $$\theta $$ = $$\mu $$2 sin $$\alpha $$
sin $$\alpha $$ = $${{{\mu _1}} \over {{\mu _2}}}\sin \theta $$ ....... (1)
Critical angle at BC,
$$\mu $$2 sin ic = $$\mu $$1 sin 90o
$$ \Rightarrow $$ sin ic = $${{{\mu _1}} \over {{\mu _2}}}$$ ..... (2)
For Total internal reflection to occur,
90o - $$\alpha $$ > ic
Taking sin both side,
sin(90o - $$\alpha $$) > sin ic
$$ \Rightarrow $$ cos $$\alpha $$ > $${{{\mu _1}} \over {{\mu _2}}}$$ (from equation 2)
$$ \Rightarrow $$ $$\sqrt {1 - {{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta } $$ > $${{{\mu _1}} \over {{\mu _2}}}$$ (from equation 1)
Squaring both sides, we get
$${1 - {{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta }$$ > $${{{\mu _1^2} \over {\mu _2^2}}}$$
$$ \Rightarrow $$ $${1 - {{\mu _1^2} \over {\mu _2^2}}}$$ > $${{{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta }$$
$$ \Rightarrow $$ $${{\mu _2^2} \over {\mu _1^2}} - 1$$ > $${{{\sin }^2}\theta }$$
$$ \Rightarrow $$ sin $$\theta $$ < $$\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $$
$$ \Rightarrow $$ $$\theta < {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $$
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