Determine the angle of deviation experienced by the light ray...

The light ray is incident normally on the first face of the prism, so the angle of incidence $i_1 = 0^\circ$ and the angle of refraction $r_1 = 0^\circ$.

For a prism, the refracting angle $A = r_1 + r_2$.

Given $A = 4^\circ$, we have $r_2 = 4^\circ$.

Applying Snell's law at the second face:

$\mu \sin(r_2) = \sin(e)$

Since the angles are small, $\sin(\theta) \approx \theta$:

$\mu r_2 = e$

$1.5 \times 4^\circ = e \Rightarrow e = 6^\circ$

The angle of deviation $\delta$ is:

$\delta = i_1 + e - A$

$\delta = 0^\circ + 6^\circ - 4^\circ = 2^\circ$

Alternatively, for a thin prism, the deviation is directly given by:

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

$\delta = (1.5 - 1) \times 4^\circ = 2^\circ$