For a thin symmetric prism made of glass (refractive index 1.5 ),...

For a thin prism, the angle of minimum deviation is given by $\delta_m = (\mu - 1)A$.

At the position of minimum deviation, the angle of incidence $i$ and angle of emergence $e$ are equal, and the angle of refraction is $r = \dfrac{A}{2}$.

Using Snell's law for small angles, $i = \mu r = \mu \left(\dfrac{A}{2}\right)$.

The ratio of the incident angle to the minimum deviation is:

$\dfrac{i}{\delta_m} = \dfrac{\dfrac{\mu A}{2}}{(\mu - 1)A} = \dfrac{\mu}{2(\mu - 1)}$

Given $\mu = 1.5$, substituting the value:

$\dfrac{i}{\delta_m} = \dfrac{1.5}{2(1.5 - 1)} = \dfrac{1.5}{2(0.5)} = \dfrac{1.5}{1} = \dfrac{3}{2}$

Answer: $3 : 2$