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प्रश्न
A ray of light is refracted by a glass prism. Obtain an expression for the refractive index of the glass in terms of the angle of prism A and the angle of minimum deviation δm.
उत्तर
In the given diagram,
OP is the incident ray, which makes the angle i1 normal, and ∠N'QR is the angle of emergence, which is represented by i2. A is the prism angle, and µ is the refractive index of the prism.
A = Prism angle, δ = Angle of deviation, i1 = Angle of incidence, i2 = Angle of emergence.
In the case of minimum deviation,
`∠r_1 = ∠r_2 = ∠r`
A = `∠r_1 + ∠r_2`
So, A = ∠r + ∠r = ∠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, `i = ((A + δ_m))/2`
Now, from Snell's rule,
`mu = (sini)/(sinr)`
`mu = (sin((A + delta_m)/2))/(sin A/2)`
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