For a thin symmetric prism made of glass (refractive index \(1.5\)), the ratio of incident angle and minimum deviation will be _______.

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: \(3 : 2\)