JEE Main 2022 · 28 Jun · Shift 1 · Physics Q42

← JEE Main 2022 · 28 Jun · Shift 1

PhysicsSingle correct+4 / -1Question 42 of 90

The refracting angle of a prism is A and refractive index of the material of the prism is $\cot(A/2)$ . Then the angle of minimum deviation will be -

180 − 2A

90 − A

180 + 2A

180 − 3A

Answer and worked solution

Official answer: A

Minimum deviation. $\dfrac{\sin\frac{A+\delta}2}{\sin\frac A2}=\cot\frac A2$ , so $\sin\frac{A+\delta}2=\cos\frac A2=\sin\left(90°-\frac A2\right)$ .

So $\frac{A+\delta}2=90°-\frac A2$ , $\delta=180°-2A$ .

(A)

Topics:Prism and dispersion

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JEE Main 2022 · 28 Jun · Shift 1, Physics question 42