A) \[45{}^\circ \]
B) \[54.74{}^\circ \]
C) \[60{}^\circ \]
D) \[90{}^\circ \]
E) \[30{}^\circ \]
Correct Answer: E
Solution :
By Brewsters law, \[\mu =\tan {{\theta }_{p}}\] or \[\mu =\tan 54.74{}^\circ \] or \[\mu =1.414\] For an equilateral prism, \[\angle A=60{}^\circ \] \[\therefore \] \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \left( \frac{A}{2} \right)}\] Or \[1.414=\frac{\sin \left( \frac{60{}^\circ +{{\delta }_{m}}}{2} \right)}{\sin \left( \frac{60{}^\circ }{2} \right)}\] Or \[\frac{1.414\times 1}{2}=\sin \left( \frac{60{}^\circ +{{\delta }_{m}}}{2} \right)\] \[[\because 1.414=\sqrt{2}]\] Or \[\frac{\sqrt{2}}{2}=\sin \left( \frac{60{}^\circ +{{\delta }_{m}}}{2} \right)\] Or \[\frac{1}{\sqrt{2}}=\sin \left( \frac{60{}^\circ +{{\delta }_{m}}}{2} \right)\] Or \[45{}^\circ =\left( \frac{60{}^\circ +{{\delta }_{m}}}{2} \right)\] Or \[\delta =30{}^\circ \]
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