MHT CET · Physics · Ray Optics
If a ray of light in denser medium strikes a rarer medium at angle of incidence \(i\), the angles of reflection and refraction are \(r\) and \(r^{\prime}\) respectively. If the reflected and refracted rays are at right angles to each other, the critical angle for the given pair of media is
- A \(\sin ^{-1}\left(\tan r^{\prime}\right)\)
- B \(\tan ^{-1}(\sin \mathrm{i})\)
- C \(\sin ^{-1}(\tan r)\)
- D \(\cot ^{-1}(\tan \mathrm{i})\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}(\sin \mathrm{i})\)
Step-by-step Solution
Detailed explanation

According to Snell's Law,
\(\begin{array}{ll}
& \frac{\sin i}{\sin r}=\frac{1}{n} \\
& \text { But } i=r, \\
\therefore \quad & \frac{\sin r}{\sin r^{\prime}}=\frac{1}{n} \\
& \text { From figure } r^{\prime}+r+90=180^{\circ} \\
\therefore \quad & r^{\prime}=180^{\circ}-90^{\circ}-r=90^{\circ}-r \\
\therefore \quad & \Rightarrow \frac{\sin r}{\sin \left(90^{\circ}-r\right)}=\frac{1}{n} \\
& \frac{\sin r}{\cos r}=\frac{1}{n}
\end{array}\)
\(\therefore \quad \tan \mathrm{r}=\frac{1}{\mathrm{n}}\)
Critical angle is given by \(\sin i_c=\frac{1}{n}\)
\(\begin{array}{ll}
\therefore & \tan r=\sin i_c \\
\therefore & i_c=\sin ^{-1}(\tan r)
\end{array}\)
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