Ratio of Incident Angle and Minimum Deviation for Thin Prism

To find the ratio of the angle of incidence ($i$) to the angle of minimum deviation ($\delta_m$) for a thin symmetric prism, we proceed as follows:

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

For a symmetric prism at minimum deviation, the angle of incidence $i$ and the angle of emergence $e$ are equal ($i = e$), and the angle of refraction inside the prism is: $r = \frac{A}{2}$

Using the relation for minimum deviation: $\delta_m = 2i - A$

Rearranging the equation to find the angle of incidence ($i$): $i = \frac{A + \delta_m}{2}$

Substitute $\delta_m = (\mu - 1)A$ into the equation: $i = \frac{A + (\mu - 1)A}{2} = \frac{\mu A}{2}$

Now, we take the ratio of the incident angle ($i$) to the minimum deviation ($\delta_m$): $\frac{i}{\delta_m} = \frac{\frac{\mu A}{2}}{(\mu - 1)A} = \frac{\mu}{2(\mu - 1)}$

Given that the refractive index of glass is $\mu = 1.5$: $\frac{i}{\delta_m} = \frac{1.5}{2(1.5 - 1)} = \frac{1.5}{2 \times 0.5} = \frac{1.5}{1} = \frac{3}{2}$

Thus, the ratio of the incident angle to the minimum deviation is $3 : 2$.