if angle of minimum deviation of a prism is equal to refracting angle of the prism, then refractive index μ of material of prism satisfies

 When the entrance and departing angles of a prism are the same, the smallest deviation m occurs. The ray of light inside the prism is parallel to the prism’s base in this case. In this case, the angle of refraction within the prism on both sides of the prism is the same. Given the angle of minimal deviation and the angle of the prism, the refractive index of the prism may be calculated as follows:

μ = sin(A+δm/2)/sin(A/2)

Since we’ve been told for the angle of minimum deviation of a prism ( δmin ) to be equal to its refracting angle ( A ), we can write δm=A . On substituting it in the above equation, we can write

μ = sin(A+A/2)/sin(A/2)

μ = sin(A)/sin(A/2)

Now, we know that sin(A)=2sin(A/2)cos(A/2) so we can write,

μ = 2sin(A/2)/Cos(A/2)/sin(A/2

μ = 2cos(A/2)

The angle of the prism A can only lie between 0∘ and 90∘ , so the minimum value of the refractive index will be for 90∘ and will be,

⇒ μmin = 2cos(45)

⇒ μmin = 2/√2 = √2

And the maximum value will be for 0∘ which will be

⇒ μmin = 2cos(0)

⇒ μmin =2

So, √2<μ<2

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