For a glass prism (mu = sqrt{3}),the angle of | vedclass

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(C) The refractive index of the material of the prism is given by the formula:$\mu = \frac{\sin \left(\frac{A+D_{m}}{2}\right)}{\sin \left(\frac{A}{2}

Vedclass · Accepted Answer

$60$

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(C) The refractive index of the material of the prism is given by the formula:$\mu = \frac{\sin \left(\frac{A+D_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$Given that the angle of minimum deviation $D_{m}$ is equal to the angle of the prism $A$,i.e.,$D_{m} = A$.Substituting this into the formula:$\mu = \frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)} = \frac{\sin A}{\sin \frac{A}{2}}$Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$:$\mu = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$Given $\mu = \sqrt{3}$,we have:$\sqrt{3} = 2 \cos \frac{A}{2}$$\cos \frac{A}{2} = \frac{\sqrt{3}}{2}$Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we get:$\frac{A}{2} = 30^{\circ}$$A = 60^{\circ}$

For a glass prism $(\mu = \sqrt{3})$,the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism. (in $^{\circ}$)

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