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Question Number 132255 by aurpeyz last updated on 12/Feb/21
A ray of light is incident at 60^0  at an  air−glass plane surface. find the   angle of refraction in the glass(  refractive index for glass=1.5)  (a) 25.3 (b) 35.3 (c) 15.3 (d)30
$${A}\:{ray}\:{of}\:{light}\:{is}\:{incident}\:{at}\:\mathrm{60}^{\mathrm{0}} \:{at}\:{an} \\ $$$${air}−{glass}\:{plane}\:{surface}.\:{find}\:{the}\: \\ $$$${angle}\:{of}\:{refraction}\:{in}\:{the}\:{glass}\left(\right. \\ $$$$\left.{refractive}\:{index}\:{for}\:{glass}=\mathrm{1}.\mathrm{5}\right) \\ $$$$\left({a}\right)\:\mathrm{25}.\mathrm{3}\:\left({b}\right)\:\mathrm{35}.\mathrm{3}\:\left({c}\right)\:\mathrm{15}.\mathrm{3}\:\left({d}\right)\mathrm{30} \\ $$
Commented by Dwaipayan Shikari last updated on 12/Feb/21
sin(θ)=μsin(φ)  ⇒((√3)/2)=(3/2)sin(φ)⇒φ=sin^(−1) ((1/( (√3))))=35.3°
$${sin}\left(\theta\right)=\mu{sin}\left(\phi\right) \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}{sin}\left(\phi\right)\Rightarrow\phi={sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{35}.\mathrm{3}° \\ $$

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