KCET · Physics · Wave Optics
When the angle of incidence is \(60^{\circ}\) on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is
- A \(\sqrt{2} \times 10^{8} \mathrm{~ms}^{-1}\)
- B \(\sqrt{3} \times 10^{8} \mathrm{~ms}^{-1}\)
- C \(2 \times 10^{8} \mathrm{~ms}^{-1}\)
- D \(3 \times 10^{8} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3} \times 10^{8} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Refractive index of glass,
\(\begin{aligned} \mu_{g} &=\tan \theta_{p} \\ \text { where } \theta_{P} &=\text { polarising angle } \\ \Rightarrow \quad \mu_{g} &=\tan 60^{\circ}=\sqrt{3} \end{aligned}\)
Now, \(\mu_{g}=\frac{c}{v_{g}}\)
\[
\begin{aligned}
\therefore \quad \frac{c}{v_{g}} &=\sqrt{3} \Rightarrow v_{g}=\frac{3 \times 10^{8}}{\sqrt{3}} \\
&=\sqrt{3} \times 10^{8} \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
\(\begin{aligned} \mu_{g} &=\tan \theta_{p} \\ \text { where } \theta_{P} &=\text { polarising angle } \\ \Rightarrow \quad \mu_{g} &=\tan 60^{\circ}=\sqrt{3} \end{aligned}\)
Now, \(\mu_{g}=\frac{c}{v_{g}}\)
\[
\begin{aligned}
\therefore \quad \frac{c}{v_{g}} &=\sqrt{3} \Rightarrow v_{g}=\frac{3 \times 10^{8}}{\sqrt{3}} \\
&=\sqrt{3} \times 10^{8} \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
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