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A monochromatic beam of light is incided...

A monochromatic beam of light is incided at `60^(@)` in one face of an equilateral prism of refractive index `n` and emerges from the opposite face making an angle `theta(n)` with the normal (see the figure). For `n=sqrt3` the value of `theta` is `60^(@)` and `d(theta)/(dn)=m`. The value of `m` is

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The correct Answer is:
2
By snell's law
`1 "sin" 60 = "n sin"r_(1)`
`implies "sin"r_(1) = (1)/(2) r_(1) = 30^(@)` …..(i)
By differentiating 'w.r.t'n
`O = "sin"r_(1) + "n cos" r_(1)(dr_(1))/(dn)`
`= (1)/(2) + sqrt3 (sqrt((3)/(2))) (dr_(1))/(dn)`
`(dr_(1))/(dn) = (1)/(3)` .... (ii)
By applying snells's law
`"n sin" r_(2) = 1 "sin" theta`
`"n sin"(60 - r_(1)) = 1 "sin" theta [therefore A = r_(1) + r_(2)]`
By differentiating 'w.r.t' n
`"sin"(60- r_(1)) - "n cos"(60 -r_(1)) (dr_(1))/(dn) = "cos"theta (dtheta)/(dn)`
By substituting value of `'r_(1)`' and `(dr_(1))/(dn)`
from (1) and (2)
`(d theta)/(dn) =2`
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