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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).
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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