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
- A \(>\sin ^{-1}\left(\frac{3}{4}\right)\)
- B \( < \sin ^{-1}\left(\frac{3}{4}\right)\)
- C \(=\sin ^{-1}\left(\frac{3}{4}\right)\)
- D \(\leq \sin ^{-1}\left(\frac{3}{4}\right)\)
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,
\(\begin{array}{ll}
\therefore & i>\sin ^{-1}\left(\frac{M_{1}}{M_{2}}\right) \\
\Rightarrow & i>\sin ^{-1}\left(\frac{1.5 \times 10^{8}}{2 \times 10^{8}}\right) \\
\Rightarrow & i>\sin ^{-1}\left(\frac{3}{4}\right)
\end{array}\)
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,
\(\begin{array}{ll}
\therefore & i>\sin ^{-1}\left(\frac{M_{1}}{M_{2}}\right) \\
\Rightarrow & i>\sin ^{-1}\left(\frac{1.5 \times 10^{8}}{2 \times 10^{8}}\right) \\
\Rightarrow & i>\sin ^{-1}\left(\frac{3}{4}\right)
\end{array}\)
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