A monochromatic beam of light is incident at 60 ^ o on one face of...

By Snell's law
$1\sin 60 =nsin {r}_{1} ⇒ sin {r}_{1} = \dfrac{1}{2}{r}_{1} = {30}^{o}$ ...(i)
By differentiating 'w.r.t' n
$O = sin {r}_{1} + ncos {r}_{1}\left(\dfrac{d{r}_{1}}{dn}\right) <br > =\dfrac{1}{2} +$\sqrt{3}\left(\sqrt{\dfrac{3}{2}}\right)\dfrac{d{r}_{1}}{dn}$<br >$= \dfrac{d{r}_{1}}{dn} = \dfrac{1}{3}$...(ii)<br >By applying Snell's law<br >$nsin {r}_{2} = 1sin \theta$<br >$nsin $\left(60 - {r}_{1}\right) = 1sin \theta$
\left[∴ A + {r}_{1} + {r}_{2}\right]$<br >By differentiating 'w.r.t' n<br >$\sin ( 60 - {r}_{1} ) - ncos \left(60 - {r}_{1}\right)\dfrac{d{r}_{1}}{dn} = cos\theta\dfrac{d\theta}{dn}$<br >By substituting value$ of '{r}_{1}$' and$\dfrac{d{r}_{1}}{dn}$from (i) and (ii)<br >$\dfrac{d\theta}{dn} =2