A) \[25{}^\circ \]
B) \[60{}^\circ \]
C) \[45{}^\circ \]
D) \[30{}^\circ \]
Correct Answer: B
Solution :
Refractive index of the material\[\mu =\sqrt{3}\] The relation for refractive index is \[\mu =\frac{\sin i}{\sin r}\] \[\sqrt{3}=\frac{\sin i}{\sin r}\] \[\sqrt{3}=\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\sin \frac{60{}^\circ }{2}}\] \[\upsilon =\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\frac{1}{2}}\] Or \[\sin \frac{A+{{\delta }_{m}}}{2}=\frac{\sqrt{3}}{2}\] \[=\sin 60{}^\circ \] (where A is the angle of equilateral prism) Hence, \[\frac{A+{{\delta }_{m}}}{2}=60{}^\circ \] Thus, \[A+{{\delta }_{m}}=120\] \[{{\delta }_{m}}=120{}^\circ -60{}^\circ \] \[{{\delta }_{m}}=60{}^\circ \]You need to login to perform this action.
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