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 \[\sqrt{2} sin\,(i) = \sqrt{3} sin\,(r) or r = 45{}^\circ \]You need to login to perform this action.
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