A) 1.531
B) 1.414
C) 1.593
D) 1.825
Correct Answer: B
Solution :
The refractive index [a] of prism is given by \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] where A is angle of prism and \[{{\delta }_{m}}\] is angle of minimum deviation. Given, \[A={{60}^{o}},\,\,{{\delta }_{m}}={{30}^{o}}\] \[\therefore \] \[\mu =\frac{\sin \left( \frac{{{60}^{o}}+{{30}^{o}}}{2} \right)}{\sin \frac{{{60}^{o}}}{2}}=\frac{\sin \,{{45}^{o}}}{\sin \,{{30}^{o}}}\] \[=\frac{0.707}{0.5}=1.414\] Note: For a specific prism, there is one and only one angle of minimum deviation.
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