A) \[\left( \frac{\pi }{2}-A \right)\]
B) \[\left( 2\pi -\frac{A}{2} \right)\]
C) \[\left( \frac{\pi -A}{2} \right)\]
D) \[(\pi -2A)\]
Correct Answer: D
Solution :
Key Idea:\[\sin ({{90}^{o}}-\theta )=\cos \theta \] The refractive index \[(\mu )\] of a prism of angle\[A\], and minimum deviation\[{{\delta }_{m}}\], is given by \[\mu =\frac{\left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin A/2}\] Given, \[\mu =\cot \frac{A}{2}\] \[\therefore \] \[\cot \frac{A}{2}=\frac{\sin \frac{(A+{{\delta }_{m}})}{2}}{\sin (A/2)}\] \[\Rightarrow \] \[\frac{\cos A/2}{\sin A/2}=\frac{\sin \frac{(A+{{\delta }_{m}})}{2}}{\sin A/2}\] \[\Rightarrow \] \[\cot \frac{A}{2}=\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)\] \[\therefore \] \[\sin \left( {{90}^{o}}-\frac{A}{2} \right)=\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)\] \[\Rightarrow \] \[{{90}^{o}}-\frac{A}{2}=\frac{A+{{\delta }_{m}}}{2}\] \[\Rightarrow \] \[{{180}^{o}}-A=A+{{\delta }_{m}}\] \[\Rightarrow \] \[{{\delta }_{m}}={{180}^{o}}-2A=\pi -2A\]You need to login to perform this action.
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