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Wave Optics

Question

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How does the angle of minimum deviation of a glass prism of refractive index 1.5 change, if it is immersed in a liquid of refractive index 1.3?

Solution

Given,
Refractive index of glass w.r.to air, ^aμ_g = 1.5
Refractive index of water w.r.to air, ^aμ_w= 1.3

As,

δ = (μ - 1) A

For deviation in air,

μ = \frac{μ_{g}}{μ_{a}} = \frac{1.5}{1} = 1.50

∴

Angle of deviation,

δ = (1.15 - 1) \times 60 ° = 30 °

For deviation in water,

μ = \frac{μ_{g}}{μ_{w}} = \frac{1.5}{1.3} = 1.15

∴

Angle of deviation,

δ = (1.15 - 1) \times 60 ° = 9 °

Therefore, the angle of deviation of glass prism is decreased when it is immersed in aliquid of refractive index of 1.3.

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