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Wave Optics

Question

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The refracting angle of the prism is 60° and the refractive index of the material of the prism is 1.632. Calculate the angle of minimum deviation.

Solution

Given,
Angle of prism, A = 60°
Refractive index, μ = 1.632

Now, using the formula,

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

1.632 = \frac{\sin (\frac{60 ° + δ_{m}}{2})}{\sin \frac{60 °}{2}} = \frac{\sin (\frac{60 + δ_{m}}{2})}{\sin 30 °}

\Rightarrow \sin (\frac{60 ° + δ_{m}}{2}) = 1.632 \sin 30 ° = 1.632 \times . 5

\sin (\frac{60 ° + δ_{m}}{2}) = 0.816 \frac{60 ° + δ_{m}}{2} = 54 ° 42' δ_{m} = 49 ° 24' .

is the required angle of minimum deviation.

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