A monochromatic beam of light is incident at `60^@` on one face of an equilateral prism of refractive inder `n` and emerges from the opposite face making an angle `theta` with the normal. For `n = sqrt(3)`, the value of `theta` is `60^@ and (d theta)/(dn) = m`. The value of `m` is.

Answer

B

Explanation

Here`angleMPQ+angleMQP=60^circ` .If `angleMPQ=rthenangleMQP=60-r`
Applying Snell's law at P
`sin60^circ=nsinr`….(i)
differentiating w.r.t.`n` we get
`O=sinr+ncosrxx(dr)/(dn)`…(ii)
Applying Snell's law at Q
`sintheta=nsin(60^circ-r)`...(iii)
Differentiating the above equation w.r.t `n` we get
`costheta(d theta)/(dn)=sin(60^circ-r)+ncos(60^circ-r)[-(dr)/(dn)]`
`costheta=(d theta)/(dn)=sin(60^circ-r)-ncos(60^circ-r)[-(tanr)/(n)]` [from (ii)]
`(d theta)/(dn)=(1)/(costheta)[sin(60^circ-r)+cos(60^circ-r)tanr]` ...(iv)
From eq. (i) substituting `n=sqrt3` we get `r=30^circ`
From eq (iii) substituting `n=sqrt3`,`r=30^circ`we get `theta=60^circ`
On substituting the values of r and `theta` in eq (iv) we get
`(d theta)/(dn)=(1)/(cos60^circ)[sin30^circ+cos30^circtan30^circ]=2`