The diagram shows part of the path of a ray of light through a right-angled prism - AQA - A-Level Physics - Question 19 - 2018 - Paper 1

To determine the path of the ray after it is incident on face XY, we can apply Snell's Law, which states:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

Here, the incident ray from the prism is parallel to face XY, and since the refractive index of air is approximately 1, the light will refract at face XY according to the angles involved.

Considering the angle of incidence, which is the same as the angle inside the prism (45 degrees), we can calculate:

• Let $n_1 = 1.5$ (for glass), $\theta_1 = 45^{\circ}$, and $n_2 = 1$ (for air).
• Therefore, from Snell's law:

$1.5 \sin(45^{\circ}) = 1 \sin(\theta_2)$

Calculating for $\sin(\theta_2)$ gives:

$\sin(\theta_2) = 1.5 \cdot \frac{\sqrt{2}}{2}$

Since the light is transitioning from a denser medium to a less dense medium, it refracts away from the normal.

Based on the provided options (A, B, C, D), the correct path of the ray after it exits face XY will correspond with the option that shows the ray continuing in a straight line at an angle away from the normal.

Thus, the correct answer is Option D.