Ray Optics

JEE Main level, single correct, medium

235 of 295

Question 235 of 295

For a thin symmetric prism made of glass (refractive index 1.51.5), the ratio of incident angle and minimum deviation will be _______.

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Answer: (B) 3:23 : 2

For a thin prism, the angle of minimum deviation is given by δm=(μ−1)A\delta_m = (\mu - 1)A.

At the position of minimum deviation, the angle of incidence ii and angle of emergence ee are equal, and the angle of refraction is r=A2r = \dfrac{A}{2}.

Using Snell's law for small angles, i=μr=μ(A2)i = \mu r = \mu \left(\dfrac{A}{2}\right).

The ratio of the incident angle to the minimum deviation is:

iδm=μA2(μ−1)A=μ2(μ−1)\dfrac{i}{\delta_m} = \dfrac{\dfrac{\mu A}{2}}{(\mu - 1)A} = \dfrac{\mu}{2(\mu - 1)}

Given μ=1.5\mu = 1.5, substituting the value:

iδm=1.52(1.5−1)=1.52(0.5)=1.51=32\dfrac{i}{\delta_m} = \dfrac{1.5}{2(1.5 - 1)} = \dfrac{1.5}{2(0.5)} = \dfrac{1.5}{1} = \dfrac{3}{2}

Answer: 3:23 : 2

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