Refractive index of an equilateral prism is 2 . Then

From μ=sin⁡A+δm2sin⁡A2We can see thatδm=300 for μ=2 and A=60∘Further, at minimum deviation,r1=r2=A2=300sin⁡i1=μsin⁡r1=(2)sin⁡300=12⇒i1=450

Refractive index of an equilateral prism is 2 . Then

From μ=sin⁡A+δm2sin⁡A2We can see thatδm=300 for μ=2 and A=60∘Further, at minimum deviation,r1=r2=A2=300sin⁡i1=μsin⁡r1=(2)sin⁡300=12⇒i1=450

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Refractive index of an equilateral prism is $\sqrt{2}$ . Then

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at angle of incidence 60⁰, deviation is minimum

at angle of incidence 45⁰, deviation is minimum

minimum deviation from this prism can be 30⁰

minimum deviation from this prism can be 45⁰

answer is A, C.

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Detailed Solution

From $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$

We can see that

$δ_{m} = 30^{0} for (μ = \sqrt{2} and A = 60^{\circ})$

Further, at minimum deviation,

$r_{1} = r_{2} = \frac{A}{2} = 30^{0}$

$\sin i_{1} = μsin r_{1} = (\sqrt{2}) \sin 30^{0} = \frac{1}{\sqrt{2}} \Rightarrow i_{1} = 45^{0}$

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Refractive index of an equilateral prism is $\sqrt{2}$ . Then