For an isosceles prism of angle A and refractive index , it is found that the angle of minimum deviation δ_m=A which of the following is/are correct [JEE Adv 2017 P1]

At minimum deviation, the incident angle i₁ and the refracting angle r₁ at the first refracting surface are realted by r₁=i₁/2

For this prism, the refractive index μ and the angle of prism A are related as A=1/2^{-1}≤ft(/2)

For the angle of incidence i₁=A,the ray inside the prism is parallel to the base of the prism

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 i₁= ^{-1}A{4²≤ft(A/2)-1- A}

Test Index

Geometrical Optics

Section: Physics

Question : 26 of 56

Marks: +1, -0

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 is/are correct

[JEE Adv 2017 P1]

At minimum deviation, the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are realted by $r_1=\frac{i_1}{2}$
For this prism, the refractive index μ and the angle of prism A are related as $A=\frac{1}{2}\cos^{-1}\left(\frac{\mu}{2}\right)$
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 emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1=$ $\sin^{-1}A\sqrt{4\cos^2\left(\frac{A}{2}\right)-1-\cos A}$

Validate

Solution:

Angular deviation :

\delta=i_1+i_2-A

For min deviation

:i_1=1_2

so,

i_1=A

...(i)

Also

r_1+r_2=A

i.e.,

r_1=\frac{A}{2}

.....(ii)

\therefore r_1=\frac{i_1}{2}

(A is correct)

By Snell's law,

\sin i_1=\mu r_1

i.e

\sin A=\mu\sin\left(\frac{A}{2}\right)

\Rightarrow 2\cos\left(\frac{A}{2}\right)=\mu

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

Hence (B) is correct

For tangential emergence,

i_2\rightarrow 90^{\circ}

So,

\mu\sin r_2=1

\cos r_2=\sqrt{1-\frac{1}{\mu^2}}

Also

r_1=A-r_2

\sin r_1=\sin A\cos r_2-\cos A\sin r_2

=\frac{(\sin A)\sqrt{\mu^2-1}}{\mu}-\frac{(\cos A)1}{\mu}

By Snell's law on 1st surface,

\sin i_2=\mu\sin r_1

\Rightarrow \sin i_1=\sin A(\mu^2-1)^{1/2}-\cos A

\Rightarrow i_1=\sin^{-1}\left(\sin A\sqrt{4\cos^2\left(\frac{A}{2}\right)-1}-\cos A\right)

i.e. (D) is correct

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