JEE Advanced 2017 · Paper 1 · Physics Q7

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PhysicsMultiple correct+4 / -2Question 7 of 54

For an isosceles prism of angle $A$ and refractive index $\mu$ , it is found that the angle of minimum deviation $\delta_m = A$ . Which of the following options is/are correct?

For the angle of incidence

i_1 = A

, the ray inside the prism is parallel to the base of the prism

For this prism, the refractive index

\mu

and the angle of prism

A

are related as

A = \frac{1}{2}\cos^{-1}\left(\frac{\mu}{2}\right)

At minimum deviation, the incident angle

i_1

and the refracting angle

r_1

at the first refracting surface are related by

r_1 = (i_1/2)

For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is

i_1 = \sin^{-1}\left[\sin A\sqrt{4\cos^2\frac{A}{2} - 1} - \cos A\right]

Answer and worked solution

Official answer: A, C, D

Minimum deviation. $r_1=r_2=A/2$ and $i_1=\dfrac{A+\delta_m}2=A$ .

(C) $r_1=\dfrac A2=\dfrac{i_1}2$ . True.
(A) At $i_1=A$ the prism is at minimum deviation, where the ray inside is parallel to the base. True.

Refractive index.
$\mu=\frac{\sin A}{\sin(A/2)}=2\cos\frac A2\implies A=2\cos^{-1}\frac\mu2.$
(B) has the factor $\tfrac12$ instead of 2. False.

Grazing emergence. This needs $r_2=c$ with $\sin c=1/\mu$ , so $r_1=A-c$ :
$\sin i_1=\mu\sin(A-c)=\sin A\sqrt{\mu^2-1}-\cos A=\sin A\sqrt{4\cos^2\tfrac A2-1}-\cos A.$
(D) is true.

(A), (C), (D)

Topics:Prism and dispersion Total internal reflection

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JEE Advanced 2017 · Paper 1, Physics question 7