To find the displacement $d$ of the emergent ray as it exits the flat slab, we can use geometric optics principles. Here's how to approach the problem:

Step 1: Refraction at the first surface

When a light ray passes from air (with refractive index $n_1 = 1$) into a medium with refractive index $n_2 = 1.49$, it bends according to Snell's law:

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

where:

• $n_1 = 1$ is the refractive index of air,
• $n_2 = 1.49$ is the refractive index of the slab,
• $\theta_1 = 60^\circ$ is the angle of incidence,
• $\theta_2$ is the angle of refraction in the slab.

Let's solve for $\theta_2$:

$\sin \theta_2 = \frac{n_1 \sin \theta_1}{n_2} = \frac{1 \times \sin 60^\circ}{1.49}$

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$:

$\sin \theta_2 = \frac{\frac{\sqrt{3}}{2}}{1.49}$

Now calculate $\theta_2$.