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$\sf{Refractive \: Index \: of \: water \: with \: respect \: to \: air \: is \: \sqrt{2}. }$
$\sf{A \: light \: ray \: is \: incident \: on \: the \: surface \: at \: an \: angle \: of \: 60° }$
$\sf{travelling \: through \: water. The \:angle \: of \: deviation \: of \: light \: ray \: is}$

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Answer 1

$\mathsf{We\:know\:Refractive\:index=\frac{sini}{sinr}}$

Angle of incidence ( i ) = 60° .  

$\mathsf{\sqrt{3}=\frac{sin60^\circ}{sinr}}\\\\\mathsf{=\:>\sqrt{3}=\frac{\frac{\sqrt{3}}{2}}{sinr}}\\\\\mathsf{=\:>\sqrt{3}=\frac{\sqrt{3}}{2sinr}}\\\\\mathsf{=\:>1=\frac{1}{2sinr}}$

 

$\mathsf{=\:>sinr=\frac{1}{2}}\\\\\mathsf{\implies sinr=sin30^\circ}\\\\\mathsf{\implies r=30^\circ}$

$\textsf{If angle of incidence is more :}$

$\textsf{Angle of deviation = angle of incidence - angle of refraction}$

$\mathsf{\implies60^\circ-30^\circ}\\\\\mathsf{\implies30^\circ}$

$\mathfrak{\underline{ANSWER}}$

$\boxed{\mathsf{\huge{30^\circ}}}$

Answer 2

$\mathsf{We\:know\:Refractive\:index=\frac{sini}{sinr}}$

Angle of incidence ( i ) = 60

$\begin{lgathered}\mathsf{\sqrt{3}=\frac{sin60^\circ}{sinr}}\\\\\mathsf{=\:>\sqrt{3}=\frac{\frac{\sqrt{3}}{2}}{sinr}}\\\\\mathsf{=\:>\sqrt{3}=\frac{\sqrt{3}}{2sinr}}\\\\\mathsf{=\:>1=\frac{1}{2sinr}}\end{lgathered} \\\\</p><p></p><p> </p><p>\begin{lgathered}\mathsf{=\:>sinr=\frac{1}{2}}\\\\\mathsf{\implies sinr=sin30^\circ}\\\\\mathsf{\implies r=30^\circ}\end{lgathered} \\\\</p><p> </p><p> </p><p> \textsf{If angle of incidence is more :}\\\\</p><p></p><p></p><p>\textsf{Angle of deviation = angle of incidence - angle of refraction}\\\\</p><p></p><p></p><p>\begin{lgathered}\mathsf{\implies60^\circ-30^\circ}\\\\\mathsf{\implies30^\circ}\end{lgathered} \\\\</p><p></p><p> </p><p>\mathfrak{\underline{ANSWER}} \\\\</p><p> </p><p> </p><p></p><p>\boxed{\mathsf{\huge{30^\circ}}}$