Angle of minimum deviation is equal to the angle of prism A of an equilateral glass prism. The angle of incident at which minimum deviation will be obtained is

60^{\circ}

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30^{\circ}

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45^{\circ}

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s i n^{- 1} (2 / 3)

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Solution

The correct option is A $60^{\circ}$
$A = δ_{m} = 60^{\circ}$
At minimum deviation $i = (\frac{A + δ_{m}}{2}) = 60^{\circ}$ .

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Angle of minimum deviation is equal to the angle of prism A of an equilateral glass prism. The angle of incident at which minimum deviation will be obtained is

The correct option is A 60∘A=δm=60∘At minimum deviation i=(A+δm2)=60∘.

The correct option is A $60^{\circ}$
$A = δ_{m} = 60^{\circ}$
At minimum deviation $i = (\frac{A + δ_{m}}{2}) = 60^{\circ}$ .