Question

A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of glass layers from the bottom side at an angle $\theta$ = 30°. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $n_m$ = n - m$\Delta$n, where n – m is the refractive index of the $m^{th}$ slab and $\Delta$n = 0.1 (see the figure). The ray is refracted out parallel to the interface between the (m-1)$^{th}$ and $m^{th}$ slabs from the right side of the stack. What is the value of m?

• A. 8
• B. 9
• C. 10
• D. 11

Answer 1

Answer: (A)

8

Answer 2

The problem describes a light ray passing through a stack of glass layers with decreasing refractive indices. We use Snell's law at each interface.

Let $n$ be the refractive index of the initial medium and $\theta$ be the angle of incidence. The refractive index of the $k^{th}$ slab from the bottom is given by $n_k = n - k\Delta n$.

The condition "the ray is refracted out parallel to the interface between the $(m-1)^{th}$ and $m^{th}$ slabs" can be interpreted as the ray being refracted into the $m^{th}$ slab at an angle of $90^\circ$ with respect to the normal. This means the ray travels along the interface between the $m^{th}$ and $(m+1)^{th}$ slabs.

Applying Snell's law at the interface between the initial medium and the first slab, and considering the refraction into the $m^{th}$ slab where the angle of refraction is $90^\circ$: $n \sin \theta = n_m \sin 90^\circ$ $n \sin \theta = n_m$

We are given: $n = 1.6$ $\theta = 30^\circ$, so $\sin \theta = 0.5$ $\Delta n = 0.1$ The refractive index of the $m^{th}$ slab is $n_m = n - m\Delta n$.

Substitute the values into the equation: $1.6 \times 0.5 = 1.6 - m \times 0.1$ $0.8 = 1.6 - 0.1m$

Now, solve for $m$: $0.1m = 1.6 - 0.8$ $0.1m = 0.8$ $m = \frac{0.8}{0.1}$ $m = 8$

This interpretation aligns with one of the provided options. The ray is refracted into the $8^{th}$ slab at an angle of $90^\circ$, meaning it travels along the interface between the $8^{th}$ and $9^{th}$ slabs. The phrase "refracted out parallel to the interface between the $(m-1)^{th}$ and $m^{th}$ slabs" suggests that the ray's path within the $m^{th}$ slab is parallel to the interfaces, which is consistent with an angle of refraction of $90^\circ$ into that slab.

Therefore, the value of m is 8.