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Question
For any prism, obtain a relation between the angle of the prism (A), the angle of minimum deviation (δm) and the refractive index of its material (μ or n).
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Solution
In the given diagram,
OP is the incident ray, which makes the angle i, with normal, and QR is the angle of emergence, which is represented by i2. A is the prism angle, and it is the refractive index of the prism.
Now, we know that,
A = Prism angle, δ = Angle of deviation
i1 = Angle of incidence
i2 = Angle of emergent
In the case of minimum deviation
∠r1 = ∠r2 = ∠r
A = ∠r1 + ∠r2 = ∠2r
`=> ∠"r" = "A"/2`
Now again
A + δ = i1 + i2 ...(∵ In the case of minimum deviation)
i1 = i2 = i and δ = δm
So, A + δm = i + i = 2i
Now, `angle "i" = ("A" + delta_"m")/2`
For Snell's law:
`mu = (sin i)/(sin r)`
`=> mu = (sin ("A" + delta_m)/2)/(sin "A"/2)`
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