The refractive index of the material of a gla | vedclass

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(B) The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin \left(\frac{A + \delta_{min}}{2}\right)}{\sin \frac{A}{2}}$.Gi

Vedclass · Accepted Answer

$60$

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(B) The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin \left(\frac{A + \delta_{min}}{2}\right)}{\sin \frac{A}{2}}$.Given that the angle of minimum deviation $\delta_{min}$ is equal to the angle of the prism $A$,we substitute $\delta_{min} = A$ into the formula.$\mu = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin \frac{A}{2}} = \frac{\sin A}{\sin \frac{A}{2}}$.Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we get $\mu = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$.Given $\mu = \sqrt{3}$,we have $\sqrt{3} = 2 \cos \frac{A}{2}$,which implies $\cos \frac{A}{2} = \frac{\sqrt{3}}{2}$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $\frac{A}{2} = 30^{\circ}$,so $A = 60^{\circ}$.

The refractive index of the material of a glass prism is $\sqrt{3}$. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism (in $^{\circ}$)?

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