KCET · Physics · Ray Optics
The angle of minimum deviation for an incident light ray on an equilateral prism is equal to its refracting angle. The refractive index of its material is
- A \(\frac{1}{\sqrt{2}}\)
- B \(\sqrt{3}\)
- C \(\frac{\sqrt{3}}{2}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
For an equilateral prism, angle of prism of refracting angle \(\mathrm{A}=60^{\circ}\)
Here, \(\quad \delta_{\mathrm{m}}=\mathrm{A}=60^{\circ}\)
\(\therefore\) Refractive index,
\(\begin{aligned} \mu=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \frac{\mathrm{A}}{2}}=\frac{\sin \left(\frac{60^{\circ}+60^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)} \\
=& \frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\sin 60^{\circ}}{\cos 60^{\circ}}=\tan 60^{\circ}=\sqrt{3} \end{aligned}\)
Here, \(\quad \delta_{\mathrm{m}}=\mathrm{A}=60^{\circ}\)
\(\therefore\) Refractive index,
\(\begin{aligned} \mu=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \frac{\mathrm{A}}{2}}=\frac{\sin \left(\frac{60^{\circ}+60^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)} \\
=& \frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\sin 60^{\circ}}{\cos 60^{\circ}}=\tan 60^{\circ}=\sqrt{3} \end{aligned}\)
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