The speed of light in media 'A' and 'B' are $2.0 \times {10^{10}}$ cm/s and $1.5 \times {10^{10}}$ cm/s respectively. A ray of light enters from the medium B to A at an incident angle '$\theta$'. If the ray suffers total internal reflection, then
Solution
<p>${\mu _A} = {{3 \times {{10}^8}} \over {2 \times {{10}^8}}} = 1.5$</p>
<p>${\mu _B} = {{3 \times {{10}^8}} \over {1.5 \times {{10}^8}}} = 2$</p>
<p>For TIR</p>
<p>$\theta > {i_c}$</p>
<p>$\theta > {\sin ^{ - 1}}\left( {{{1.5} \over 2}} \right)$</p>
<p>$\theta > {\sin ^{ - 1}}\left( {{3 \over 4}} \right)$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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