A prism of refractive index of √2 has a refracting angle of 60°. At what angle must a ray be incident on it so that it suffers a minimum deviation?

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Solution

Given,
Refractive index of prism =

\sqrt{2}

angle of prism = 60⁰

For minimum deviation, i = e.

Therefore,

A + δ_{m} = i + r

\Rightarrow

i = \frac{A + δ_{m}}{2} for δ minimum i = r

Now,

μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin A / 2} = \frac{\sin i}{\sin A / 2}

i.e.

\sin i = μ \sin A / 2

\Rightarrow

\sin i = μ \sin (\frac{60 °}{2}) = \sqrt{2} \sin 30 ° = \frac{1}{\sqrt{2}}

Hence,

i = 45 ° .

When, angle of incidence is 45^o the ray suffers minimum deviation.

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