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KCET · Physics · Ray Optics

The speed of light in media \(M_{1}\) and \(M_{2}\) are \(1.5 \times 10^{8} \mathrm{~ms}^{-1}\) and \(2 \times 10^{8} \mathrm{~ms}^{-1}\) respectively. A ray travels from medium \(M_{1}\) to the medium \(M_{2}\) with an angle of incidence \(\theta\). The ray suffers total internal reflection. Then the value of the angle of incidence \(\theta\) is

  1. A \(>\sin ^{-1}\left(\frac{3}{4}\right)\)
  2. B \( < \sin ^{-1}\left(\frac{3}{4}\right)\)
  3. C \(=\sin ^{-1}\left(\frac{3}{4}\right)\)
  4. D \(\leq \sin ^{-1}\left(\frac{3}{4}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(>\sin ^{-1}\left(\frac{3}{4}\right)\)

Step-by-step Solution

Detailed explanation

For total internal reflection, the angle of incidence of light in denser medium must be greater than the critical angle for the pair of media in contact, i.e.,
We have, \(\frac{1}{\sin C}=\frac{v_{2}}{v_{1}}\)
Given,
\(\begin{aligned}
&v_{1}=M_{1}=1.5 \times 10^{8} \mathrm{~ms}^{-1} \\
&v_{2}=M_{2}=2 \times 10^{8} \mathrm{~ms}^{-1}
\end{aligned}\)
Here,
\(\begin{array}{ll}
\therefore & i>\sin ^{-1}\left(\frac{M_{1}}{M_{2}}\right) \\
\Rightarrow & i>\sin ^{-1}\left(\frac{1.5 \times 10^{8}}{2 \times 10^{8}}\right) \\
\Rightarrow & i>\sin ^{-1}\left(\frac{3}{4}\right)
\end{array}\)