A) \[{{30}^{\text{o}}}\]
B) \[{{45}^{\text{o}}}\]
C) \[{{60}^{\text{o}}}\]
D) \[{{75}^{\text{o}}}\]
Correct Answer: C
Solution :
The refractive index of material of prism (from Snell's law) is \[\mu =\frac{\sin i}{\sin r}\] Here, \[i=\frac{A+{{\delta }_{m}}}{2}\] and \[r=\frac{A}{2}\] where A is the angle of prism and \[{{\delta }_{m}}\] the angle of minimum deviation. Hence, \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[Given,\,\mu =\sqrt{3},A={{60}^{o}}\,(for\,prism)\] \[Thus,\,\sqrt{3}=\frac{\sin \left( \frac{60+{{\delta }_{m}}}{2} \right)}{\sin {{30}^{o}}}\] \[or\sin \left( \frac{60+{{\delta }_{m}}}{2} \right)=\frac{1}{2}\times \sqrt{3}\] \[or\sin \left( \frac{60+{{\delta }_{m}}}{2} \right)=\sin {{60}^{o}}\] \[or\frac{60+{{\delta }_{m}}}{2}=60\] \[or{{\delta }_{m}}=2\times 60-60={{60}^{o}}\]
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