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(A) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.In the position of minimum deviation,the angle of deviation $\delta_m$ is given a

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$\sqrt{3}$

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(A) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.In the position of minimum deviation,the angle of deviation $\delta_m$ is given as equal to the angle of the prism,so $\delta_m = A = 60^{\circ}$.The formula for the refractive index $\mu$ of the prism material is given by:$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$Substituting the values $A = 60^{\circ}$ and $\delta_m = 60^{\circ}$:$\mu = \frac{\sin\left(\frac{60^{\circ} + 60^{\circ}}{2}\right)}{\sin\left(\frac{60^{\circ}}{2}\right)}$$\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$$\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$

When light of wavelength $\lambda$ is incident on an equilateral prism kept in its minimum deviation position,it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism for the wavelength $\lambda$ is,then

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