Question #7abf9

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Question #7abf9

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Answer 1 · 2017-12-28T21:53:30

$Critical angle = \frac{π}{6} = 30^{\circ}$ $"Critical angle"=pi/6=30^circ$

Explanation:

The equation for critical angle is:
$sin C = \frac{n_{1}}{n_{2}}$ $sinC=n_1/n_2$ , where:

$C$ $C$ = critical angle
$n_{1}$ $n_1$ = boundary medium
$n_{2}$ $n_2$ = initial medium

In this case $n_{1}$ $n_1$ is the vacuum as it is the medium acting as a boundary, while $n_{2}$ $n_2$ is the other medium, as the light is travelling through that medium.

$n_{1} = 1$ $n_1=1$

$sin C = \frac{1}{2}$ $sinC=1/2$

$C = {sin}^{- 1} (\frac{1}{2}) = \frac{π}{6} = 30^{\circ}$ $C=sin^(-1)(1/2)=pi/6=30^circ$

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