A prism is found to give a minimum deviation of 51°. The same prism gives a deviation of 62°48’ for two values of the angles of incidence, namely, 40°6’ and 82°42’. Determine the refractive angle of the prism and the refractive index of its material.

Solution

Angle of minimum deviaton of prism = 51^o

The incident ray is deviated through an angle of 62°48’ (

δ

) wen angle of incidence = 40°6’.

The emergent ray, for which angle of emergence, e= 82°42’ is also deviated through the same angle

δ

. [ Using principle of reversibility]

Now,

δ = (i + e) - A

i.e.,

A = (i + e) - δ

= 40 ° 6' + 82 ° 42' - 62 ° 48'

i.e.,,

A = 60 °

which is the refracting angle of the prism.

For minimum deviation, i = e.

∴

δ_{m} = 2 i - A

i.e.,

i = (\frac{δ_{\min} + A}{2})

= \frac{(51 ° + 60 °)}{2} = 55 ° 30'

Which is the required angle of incidence at minimum deviation.

The refractive index of the material of the prism is given by

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

i.e.,

μ = \frac{\sin (\frac{51 ° + 60 °}{2})}{\sin \frac{60 °}{2}}

i.e.,

μ = 1.648 .

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