KCET · Physics · Ray Optics
If the angle of minimum deviation is equal to angle of a prism for an equilateral prism, then the speed of light inside the prism is .....
- A \(3 \times 10^8 \mathrm{~ms}^{-1}\)
- B \(2 \sqrt{3} \times 10^8 \mathrm{~ms}^{-1}\)
- C \(\sqrt{3} \times 10^8 \mathrm{~ms}^{-1}\)
- D \(\frac{\sqrt{3}}{2} \times 10^8 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{3} \times 10^8 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(\delta_m=A=60^{\circ}\)
We know that, \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\)
\(=\frac{\sin \left(\frac{60^{\circ}+60^{\circ}}{2}\right)}{\sin \frac{60^{\circ}}{2}}=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\sqrt{3} / 2}{1 / 2}\)
\(\Rightarrow \mu=\sqrt{3}\)
\(\therefore\) If \(v\) be the speed light inside the prism, then
\(\mu=\frac{c}{v}\)
\(\Rightarrow v=\frac{c}{\mu}=\frac{3 \times 10^8}{\sqrt{3}}=\sqrt{3} \times 10^8 \mathrm{~m} / \mathrm{s}\)
We know that, \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\)
\(=\frac{\sin \left(\frac{60^{\circ}+60^{\circ}}{2}\right)}{\sin \frac{60^{\circ}}{2}}=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\sqrt{3} / 2}{1 / 2}\)
\(\Rightarrow \mu=\sqrt{3}\)
\(\therefore\) If \(v\) be the speed light inside the prism, then
\(\mu=\frac{c}{v}\)
\(\Rightarrow v=\frac{c}{\mu}=\frac{3 \times 10^8}{\sqrt{3}}=\sqrt{3} \times 10^8 \mathrm{~m} / \mathrm{s}\)
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