Refraction of Light Through Stack of Variable Refractive Index Layers

According to Snell's Law for a system of parallel planar interfaces, the product of the refractive index of a medium $n_i$ and the sine of the angle of incidence/refraction $\theta_i$ with the normal to the interface remains constant throughout the propagation:

$n_i \sin \theta_i = \text{constant}$

For the light ray entering from the initial medium with refractive index $n = 1.6$ at an angle of incidence $\theta = 30^\circ$ to the normal:

$\text{constant} = n \sin \theta = 1.6 \times \sin 30^\circ = 1.6 \times \frac{1}{2} = 0.8$

The ray travels through the successive parallel layers and emerges parallel to the interface separating the $(m-1)^{\text{th}}$ and $m^{\text{th}}$ layers. This implies that the angle of refraction inside the $m^{\text{th}}$ layer is $\theta_m = 90^\circ$.

Applying Snell's Law to the $m^{\text{th}}$ layer:

$n_m \sin 90^\circ = n \sin \theta$

Since $\sin 90^\circ = 1$:

$n_m = 0.8$

We are given that the refractive index of the $m^{\text{th}}$ layer is defined as:

$n_m = n - m\Delta n$

Substituting $n = 1.6$, $\Delta n = 0.1$, and $n_m = 0.8$ into the equation:

$0.8 = 1.6 - m(0.1)$

Rearranging the terms to solve for $m$:

$0.1 m = 1.6 - 0.8 = 0.8$

$m = \frac{0.8}{0.1} = 8$