Ray Optics

JEE Main level, single correct, medium

234 of 295

Question 234 of 295

A ray of light passing through an equilateral prism is having velocity 2.12×1082.12 \times 10^8 m/s in the prism material, then the minimum angle of deviation is _________ degrees.

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Answer: (B) 3030

Velocity of light in vacuum, c=3×108c = 3 \times 10^8 m/s

Velocity of light in the prism, v=2.12×108v = 2.12 \times 10^8 m/s

Refractive index of the prism material, μ=cv=3×1082.12×108≈1.414≈2\mu = \dfrac{c}{v} = \dfrac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 \approx \sqrt{2}

For an equilateral prism, the angle of the prism A=60∘A = 60^\circ.

Using the prism formula:

μ=sin⁡(A+δm2)sin⁡(A2)\mu = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}

2=sin⁡(60∘+δm2)sin⁡(30∘)\sqrt{2} = \dfrac{\sin\left(\dfrac{60^\circ + \delta_m}{2}\right)}{\sin(30^\circ)}

2=sin⁡(30∘+δm2)12\sqrt{2} = \dfrac{\sin\left(30^\circ + \dfrac{\delta_m}{2}\right)}{\dfrac{1}{2}}

sin⁡(30∘+δm2)=22=12\sin\left(30^\circ + \dfrac{\delta_m}{2}\right) = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}

30∘+δm2=45∘30^\circ + \dfrac{\delta_m}{2} = 45^\circ

δm2=15∘\dfrac{\delta_m}{2} = 15^\circ

δm=30∘\delta_m = 30^\circ

Answer: 3030

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