For the refractive index of the slab material. We are given the angle of incidence (i) and the angle of refraction (r). We assume the light is entering the slab from air, so the refractive index o

The problem asks for the refractive index of the slab material. We are given the angle of incidence ($i$) and the angle of refraction ($r$). We assume the light is entering the slab from air, so the refractive index of the first medium (air) is $n_1 = 1$.

Given:
Angle of incidence, $i = 40^\circ$
Angle of refraction, $r = 26^\circ$
Refractive index of air, $n_1 = 1$

We use Snell's Law:

$$n_1 \sin(i) = n_2 \sin(r)$$

where $n_2$ is the refractive index of the slab material.

$$\sin(40^\circ) \approx 0.6428$$ $$\sin(26^\circ) \approx 0.4384$$

$$0.6428 = n_2 \cdot 0.4384$$ $$n_2 = \frac{0.6428}{0.4384}$$

$$n_2 \approx 1.47$$

The refractive index of the slab material is $\text{1.47}$.