A) \[{{180}^{\circ }}-2A\]
B) \[{{90}^{\circ }}-A\]
C) \[{{180}^{\circ }}+2A\]
D) \[{{180}^{\circ }}-3A\]
Correct Answer: A
Solution :
In the prism ABC, \[\delta \] is the angle of minimum deviation, A is angle of prism. The refractive index of the material of the prism is given by \[\mu =\frac{\sin \,\left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \,\frac{A}{2}}\] Given, \[\mu =\cot \frac{A}{2},\] \[\cot \frac{A}{2}=\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[\Rightarrow \] \[\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[\Rightarrow \] \[\cos \frac{A}{2}=\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)\] Using \[\sin \left( \frac{\pi }{2}-\theta \right)=\cos \,\,\theta ,\] we have \[\Rightarrow \] \[\sin \left( \frac{\pi }{2}-\frac{A}{2} \right)=\sin \frac{A+{{\delta }_{m}}}{2}\] \[\Rightarrow \] \[{{\delta }_{m}}=\pi -2A={{180}^{\circ }}-2A\] Note: For a prism there is one and only one angle of minimum deviation.You need to login to perform this action.
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