Question 154: 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 :
![](https://imagex.cdn.examgoal.net/jf762wkjy29lbim/6ebdb95c-3689-4d47-80cd-ae9930c999a6/79e8cae0-a5d4-11e9-902e-23ffedc9804e/file-jf762wkjy29lbin.png)

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} $ ![](https://imagex.cdn.examgoal.net/jf762wkjy2anxiz/dbe2ddaa-78a4-4839-ba22-a39b37c6531f/aba8b4b0-a5d8-11e9-902e-23ffedc9804e/file-jf762wkjy2anxj0.png)

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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