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The $\mathrm{X-Y}$ plane be taken as the boundary between two transparent media $\mathrm{M_{1}\: and \: M_{2} . \: M_{1} \: in\: \mathrm{Z} \geqslant 0}$ has a refractive index of $\sqrt{2}$ and $\mathrm{M}_{2}$ with $\mathrm{Z}<0$ has a refractive index of $\sqrt{3}$ . A ray of light travelling in $\mathrm{M}_{1}$ along the direction given by the vector $\overrightarrow{\mathrm{P}}=4 \sqrt{3} \hat{i}-3 \sqrt{3} \hat{j}-5 \hat{k}$ , is incident on the plane of separation. The value of difference between the angle of incident in $\mathrm{M}_{1}$ and the angle of refraction in $\mathrm{M}_{2}$ will be___________ degree.
Option: 1
15

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

unit vector along incident ray ,

$\mathrm{ \hat{e}_1=\frac{\bar{p}}{|\bar{p} \mid}=\frac{4 \sqrt{3}}{10} \hat{i}-\frac{3 \sqrt{3}}{10} \hat{\jmath}-\frac{\hat{k}}{2}}$

$\mathrm{\hat{n}=\hat{k} (unit \: vector \: along\: normal)}$

By law of refraction (vector form)

$\mathrm{ \mu_1\left|\left(\hat{e_1} \times \hat{n}\right)\right|=\mu_2\left|\left(\hat{e_2} \times \hat{n}\right)\right| }$

$\mathrm{\left|\hat{e}_1 \times \hat{n}\right|=\sin i }$

$\mathrm{\left|\frac{4}{10} \sqrt{3}(-\hat{\jmath})-\frac{3 \sqrt{3}}{10} \hat{i}\right|=\sin i }$

$\mathrm{\sin \hat{i}=\frac{3}{4}=\frac{i}{1}=53^{\circ} }\rightarrow (1)$

$\mathrm{\left|\hat{e}_2 \times \hat{n}\right|=\frac{\mu_1}{\mu_2}\left|\hat{e}_1 \times \hat{n}\right| }$

$\mathrm{\sin r=\sqrt{\frac{2}{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2 \sqrt{2}} }$

$\mathrm{\sin r=38^{\circ} }$

$\mathrm{i-r=15^{\circ}}$

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HARSH KANKARIA

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