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?

(A) For the angle of incidence $i_{1}=A$, the ray inside the prism is parallel to the base of the prism; (C) At minimum deviation, the incident angle $i_{1}$ and the refracting angle $r_{1}$ at the first refracting surface are related by $r_{1}=\left(i_{1} / 2\right)$; (D) 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 $i_{1}=\sin ^{-1}\left[\sin A \sqrt{4 \cos ^{2} \frac{A}{2}-1}-\cos A\right]$

(B) For this prism, the refractive index $\mu$ and the angle of prism $A$ are related as $A=\frac{1}{2} \cos ^{-1}\left(\frac{\mu}{2}\right)$

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Geometrical Optics

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For an isosceles prism of angle

A

and refractive index

μ

, it is found that the angle of minimum deviation

δ_{m} = A

. Which of the following options is/are correct?

(A)For the angle of incidence $i_{1} = A$ , the ray inside the prism is parallel to the base of the prism✓ Correct
(B)For this prism, the refractive index $μ$ and the angle of prism $A$ are related as $A = \frac{1}{2} cos^{- 1} (\frac{μ}{2})$
(C)At minimum deviation, the incident angle $i_{1}$ and the refracting angle $r_{1}$ at the first refracting surface are related by $r_{1} = (i_{1} /2)$ ✓ Correct
(D)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 $i_{1} = sin^{- 1} [sin A 4 cos^{2} \frac{A}{2} - 1 - cos A]$ ✓ Correct

Answer: (A), (C), (D)

Solution

At minimum deviation for a prism,

δ_{m} = 2 i - A

. Given

δ_{m} = A

, we have

2 i = 2 A

, so

i_{1} = A

. Also,

r_{1} + r_{2} = A

and by symmetry at minimum deviation,

r_{1} = r_{2} = A /2

. Thus

r_{1} = i_{1} /2 = A /2

, confirming option (C). At minimum deviation, the ray inside the prism is parallel to the base, confirming option (A). Applying Snell's law:

sin i_{1} = μ sin r_{1}

, so

sin A = μ sin (A /2)

. Using

sin A = 2 sin (A /2) cos (A /2)

, we get

μ = 2 cos (A /2)

, which gives

A = cos^{- 1} (μ /2)

, not

A = \frac{1}{2} cos^{- 1} (μ /2)

, so option (B) is incorrect. For the emergent ray to be tangential,

r_{2} = θ_{c}

where

sin θ_{c} = 1/ μ

. With

r_{1} = A - θ_{c}

and applying Snell's law at the first surface, we can derive the expression in option (D), confirming it.

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · One or more correct options

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Geometrical Optics

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