A) \[{{30}^{o}}\]
B) \[{{45}^{o}}\]
C) \[{{60}^{o}}\]
D) \[{{75}^{o}}\]
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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