MHT CET · Physics · Ray Optics
The refractive index of material of glass is \(\sqrt{3}\). If the angle of minimum deviation is equal to the angle of prism, the angle of prism is
\(\left(\cos 30^{\circ}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}, \sin 30^{\circ}=\frac{1}{2}=\cos 60^{\circ}\right)\)
- A \(45^{\circ}\)
- B \(60^{\circ}\)
- C \(30^{\circ}\)
- D \(50^{\circ}\)
Answer & Solution
Correct Answer
(B) \(60^{\circ}\)
Step-by-step Solution
Detailed explanation
Given, \(\mu=\sqrt{3}\) and \(\frac{\sin \left(60^{\circ}\right)}{\sin \left(30^{\circ}\right)}=\sqrt{3}\)
When light passes through a prism of refracting angle A, it suffers minimum deviation \(\delta\) given by
\(\mu=\frac{\sin \left(\frac{\mathrm{A}+\delta}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)} \Rightarrow \sqrt{3}=\frac{\sin \left(\frac{\mathrm{A}+\delta}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
\(\therefore \frac{\mathrm{A}}{2}=30^{\circ}\) and \(\frac{\mathrm{A}+\delta}{2}=60^{\circ}\) are the solutions
\(\therefore A=60^{\circ}\)
When light passes through a prism of refracting angle A, it suffers minimum deviation \(\delta\) given by
\(\mu=\frac{\sin \left(\frac{\mathrm{A}+\delta}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)} \Rightarrow \sqrt{3}=\frac{\sin \left(\frac{\mathrm{A}+\delta}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
\(\therefore \frac{\mathrm{A}}{2}=30^{\circ}\) and \(\frac{\mathrm{A}+\delta}{2}=60^{\circ}\) are the solutions
\(\therefore A=60^{\circ}\)
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