Question Number 132255 by aurpeyz last updated on 12/Feb/21
$${A}\:{ray}\:{of}\:{light}\:{is}\:{incident}\:{at}\:\mathrm{60}^{\mathrm{0}} \:{at}\:{an} \\ $$$${air}−{glass}\:{plane}\:{surface}.\:{find}\:{the}\: \\ $$$${angle}\:{of}\:{refraction}\:{in}\:{the}\:{glass}\left(\right. \\ $$$$\left.{refractive}\:{index}\:{for}\:{glass}=\mathrm{1}.\mathrm{5}\right) \\ $$$$\left({a}\right)\:\mathrm{25}.\mathrm{3}\:\left({b}\right)\:\mathrm{35}.\mathrm{3}\:\left({c}\right)\:\mathrm{15}.\mathrm{3}\:\left({d}\right)\mathrm{30} \\ $$
Commented by Dwaipayan Shikari last updated on 12/Feb/21
$${sin}\left(\theta\right)=\mu{sin}\left(\phi\right) \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}{sin}\left(\phi\right)\Rightarrow\phi={sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{35}.\mathrm{3}° \\ $$