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(B) For a thin prism, the angle of minimum deviation is given by $\delta_m = (\mu - 1)A$, where $\mu$ is the refractive index and $A$ is the prism ang

Vedclass · Accepted Answer

$3$ : $2$

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(B) For a thin prism, the angle of minimum deviation is given by $\delta_m = (\mu - 1)A$, where $\mu$ is the refractive index and $A$ is the prism angle.For a symmetric prism, the angle of incidence $i$ at minimum deviation is given by $i = \frac{A + \delta_m}{2}$.Substituting $\delta_m = (\mu - 1)A$ into the expression for $i$:$i = \frac{A + (\mu - 1)A}{2} = \frac{A + \mu A - A}{2} = \frac{\mu A}{2}$.Now, the ratio of the angle of incidence $i$ to the angle of minimum deviation $\delta_m$ is:$\frac{i}{\delta_m} = \frac{\mu A / 2}{(\mu - 1)A} = \frac{\mu}{2(\mu - 1)}$.Given $\mu = 1.5$, we have:$\frac{i}{\delta_m} = \frac{1.5}{2(1.5 - 1)} = \frac{1.5}{2(0.5)} = \frac{1.5}{1} = \frac{3}{2}$.Thus, the ratio is $3 : 2$.

For a thin symmetric prism made of glass (refractive index $1.5$), the ratio of the angle of incidence to the angle of minimum deviation is . . . . . . .

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