A) 2.4
B) 0.8
C) 1.3
D) 1.5
Correct Answer: D
Solution :
For a prism of angle \[A\] and \[\delta \] being angle of minimum deviation, refractive index of the material of prism is given by \[n=\frac{\sin \,\left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] Given, \[A={{60}^{\circ }},\,{{\delta }_{m}}={{30}^{\circ }},\] \[n=\frac{\sin \,\left( \frac{{{60}^{\circ }}+{{38}^{\circ }}}{2} \right)}{\sin \frac{{{60}^{\circ }}}{2}}\] \[=\frac{\sin \,\,{{49}^{\circ }}}{\sin \,\,{{30}^{\circ }}}\] \[n=\frac{0.7547}{0.5}=1.5\]You need to login to perform this action.
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