question_answer

The xy plane is the boundary between two transparent media. Medium 1 with \[\operatorname{z}\,\,\ge 0\] has a refractive index of \[\sqrt{2}\] and medium 2 with \[\,\operatorname{z}\,\,\le 0\] has refractive index of\[\sqrt{3}\] . A ray of light in medium 1 given by the vector \[6\sqrt{3}\,\hat{i}+8\sqrt{3}\hat{j}-10\hat{k}\] is incident of the plance of separation. Find the angle of refraction in medium 2 vector in the direction of the refracted ray in medium 2

A) \[30{}^\circ \]  

B)                    \[45{}^\circ \]

C) \[60{}^\circ \]  

D)                    \[75{}^\circ \]

Correct Answer: B

Solution :

Unit vector representing the normal to the plane \[{{\hat{e}}_{n}}=\hat{k}\] Component of the incident ray along the normal is \[-\,1\,\hat{k}\] The unit vector that represents the plane of the incident ray and the normal \[{{\hat{e}}_{p}}\,\,=\,\,\frac{6\sqrt{3}\,\hat{i}+8\sqrt{3}\,\hat{j}}{\sqrt{{{(6\sqrt{3})}^{2}}+\,\,{{(8\sqrt{3})}^{2}}}}\,\,=\,\,0.6\,\hat{i}+0.8\hat{j}\]            Angle between the incident ray and the normal is given by \[\cos \,\theta =\,\,(6\sqrt{3}\,\hat{i}\,\,+\,\,8\sqrt{3}\hat{j}\,-10\hat{k})\,\,\hat{k}\,/{{(6\sqrt{3})}^{2}}\,\,+\,{{(8\sqrt{3})}^{2}}+\,1{{0}^{2}}\]or \[\cos  \theta  =-\,0.5\] Therefore, the angle \[\theta  = 120{}^\circ \] [b] The angle of incidence is \[\operatorname{i} = 180{}^\circ  -120{}^\circ  = 60{}^\circ \] The angle of the refracted beam is given by \[\sqrt{2} sin\,(i) = \sqrt{3} sin\,(r) or r = 45{}^\circ \]