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(A) The formula for the refractive index of a prism in the case of minimum deviation is given by: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\fra

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$1.5$

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(A) The formula for the refractive index of a prism in the case of minimum deviation is given by: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$Given: Refracting angle $A = 60^{\circ}$ and minimum deviation $\delta_m = 38^{\circ}$.Substituting these values into the formula:$\mu = \frac{\sin(\frac{60^{\circ} + 38^{\circ}}{2})}{\sin(\frac{60^{\circ}}{2})}$$\mu = \frac{\sin(\frac{98^{\circ}}{2})}{\sin(30^{\circ})}$$\mu = \frac{\sin(49^{\circ})}{\sin(30^{\circ})}$Using the value $\sin(49^{\circ}) \approx 0.7547$ and $\sin(30^{\circ}) = 0.5$:$\mu = \frac{0.7547}{0.5} = 1.5094 \approx 1.5$.

$A$ ray of monochromatic light suffers a minimum deviation of $38^{\circ}$ while passing through a prism of refracting angle $60^{\circ}$. The refractive index of the prism material is:

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