Ray Optics

JEE Main level, numerical answer, medium

259 of 295

Question 259 of 295

A ray of light passing through a prism (μ=3)(\mu =\sqrt{3}) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is (in degrees).

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Answer: 60

At minimum deviation, r1=r2=A2{r}_{1}={r}_{2}=\dfrac{A}{2}
Also given i=2r1=Ai=2{r}_{1}=A
Now 1⋅sin⁡i=3sin⁡r11\cdot \sin i=\sqrt{3}\sin {r}_{1}
1sin⁡A=3sin⁡A21\sin A=\sqrt{3}\sin \dfrac{A}{2}
⇒2sin⁡A2cos⁡A2=3sin⁡A2\Rightarrow 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}=\sqrt{3}\sin \dfrac{A}{2}
⇒cos⁡A2=32⇒A2=30∘\Rightarrow \cos \dfrac{A}{2}=\dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{A}{2}=30^{\circ}
⇒A=60∘\Rightarrow A=60^{\circ}

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