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Question
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on the face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Solution
Angle of minimum deviation, δm = 40°
Angle of the prism, A = 60°
Refractive index of water, µ = 1.33
Refractive index of the material of the prism = µ'
The angle of deviation is related to refractive index (µ') as:
µ' = `(sin (("A" + δ_"m"))/2)/(sin "A"/2)`
= `(sin ((60° + 40°))/2)/(sin (60°)/2)`
= `(sin 50°)/(sin 30°)`
= 1.532
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let `δ_"m"^"'"` be the new angle of minimum deviation for the same prism.
The refractive index of glass with respect to water is given by the relation:
`μ_"g"^"w" = (μ"'")/μ = (sin (("A" + δ_"m"^"'"))/2)/(sin "A"/2)`
`sin (("A" + δ_"m"^"'"))/2 = (μ"'")/μ sin "A"/2`
`sin (("A" + δ_"m"^"'"))/2 = 1.532/1.33 xx sin (60°)/2 = 0.5759`
` (("A" + δ_"m"^"'"))/2 = sin^-1 0.5759` = 35.16°
60° + `δ_"m"^"'"` = 70.32°
∴ `δ_"m"^"'"` = 70.32° − 60° = 10.32°
Hence, the new minimum angle of deviation is 10.32°.
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