KCET · Physics · Ray Optics
A prism having refractive index \(1.414\) and refracting angle \(30^{\circ}\) has one of the refracting surfaces silvered. A beam of light incident on the other refracting surface will retrace its path, if the angle of incidence is
- A \(0^{\circ}\)
- B \(30^{\circ}\)
- C \(60^{\circ}\)
- D \(45^{\circ}\)
Answer & Solution
Correct Answer
(D) \(45^{\circ}\)
Step-by-step Solution
Detailed explanation
At second surface, there is no refraction so
\[
\therefore \quad r_{1}=A=30^{\circ}
\]
From Snell's law
\(\begin{aligned} n=& \frac{\sin i_{1}}{\sin r_{1}} \\ n=& \frac{\sin i_{1}}{\sin 30^{\circ}} \\ \sin i_{1} &=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}} \\ \sin i_{1} &=\frac{1}{\sqrt{2}} \\ i_{1} &=45^{\circ} \end{aligned}\)
\[
\therefore \quad r_{1}=A=30^{\circ}
\]
From Snell's law
\(\begin{aligned} n=& \frac{\sin i_{1}}{\sin r_{1}} \\ n=& \frac{\sin i_{1}}{\sin 30^{\circ}} \\ \sin i_{1} &=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}} \\ \sin i_{1} &=\frac{1}{\sqrt{2}} \\ i_{1} &=45^{\circ} \end{aligned}\)
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