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 relationship between the refractive index $$\mu$$, the prism angle $$A$$, and the minimum deviation $$\delta_m$$ is given by:

$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

Since the prism is thin (small angle $$A$$), we can use the small angle approximation $$\sin\theta \approx \theta$$:

$$\mu \approx \frac{\frac{A + \delta_m}{2}}{\frac{A}{2}} = \frac{A + \delta_m}{A}$$

So we get:

$$\mu A = A + \delta_m$$

$$\delta_m = (\mu - 1)A$$

Given $$\mu = 1.5$$:

$$\delta_m = (1.5 - 1)A = 0.5A$$

At minimum deviation, the angle of incidence $$i$$ is related to the prism angle and minimum deviation by:

$$i = \frac{A + \delta_m}{2}$$

Substituting $$\delta_m = 0.5A$$:

$$i = \frac{A + 0.5A}{2} = \frac{1.5A}{2} = 0.75A$$

Now we find the ratio of the incident angle to the minimum deviation:

$$\frac{i}{\delta_m} = \frac{0.75A}{0.5A} = \frac{0.75}{0.5} = \frac{3}{2}$$

So the ratio $$i : \delta_m = 3 : 2$$.

Hence, the correct answer is Option B.