Question

There is a system of infinite layers of varying refractive indices as shown in the figure. The width of each layer is $b = 15 \, cm$ and refractive index of $n^{th}$ layer from origin is $\mu_n = (\frac{5}{2})^n \pi$. A very narrow laser beam of light incident at an angle of $\theta = \frac{\pi}{100}$ radian at the origin and enters into the system. At very large distance from the origin ($x \rightarrow \infty$) there is a screen placed parallel to the y-axis. Laser beam hit at the screen at $y = k \, mm$. The nearest integer value of $k$ is ____.

Answer 1

8 mm

Answer 2

1. Refractive Indices:

The refractive index of the $n^\text{th}$ layer is

$$\mu_n = \left(\frac{5}{2}\right)^n \pi.$$

Thus, the second layer has

$$\mu_2 = \left(\frac{5}{2}\right)^2 \pi.$$

2. Snell’s Law at a Vertical Interface:

For a vertical boundary the normal is horizontal so if the incident angle in the $n^\text{th}$ layer is $\theta_n$ (with the horizontal), then at the next interface

$$\mu_n \sin\theta_n = \mu_{n+1}\sin\theta_{n+1}.$$

Given:

$$\frac{\mu_n}{\mu_{n+1}} = \frac{1}{\frac{5}{2}} = 0.4,$$

we have:

$$\sin\theta_{n+1} = 0.4\,\sin\theta_n.$$

For small angles ($\theta \approx \sin\theta$),

$$\theta_{n+1} \approx 0.4\,\theta_n.$$

With the initial angle $\theta_1 = \pi/100$, we get:

$$\theta_n \approx \left(0.4\right)^{n-1} \frac{\pi}{100}.$$

3. Vertical Displacement in Each Layer:

In each layer, the beam travels horizontally a distance $b = 15\, \text{cm}$ and gains a vertical shift

$$\Delta y_n = b\,\tan\theta_n \approx b\,\theta_n \quad (\text{for small } \theta_n).$$

4. Total Vertical Displacement:

The total vertical displacement is the sum over all layers:

$$y = \sum_{n=1}^\infty \Delta y_n \approx b\sum_{n=1}^\infty \theta_n = b\,\frac{\pi}{100}\sum_{n=0}^\infty (0.4)^n.$$

The geometric series sum is:

$$\sum_{n=0}^\infty (0.4)^n = \frac{1}{1-0.4} = \frac{1}{0.6} = \frac{5}{3}.$$

Therefore,

$$y \approx 15\,\text{cm} \times \frac{\pi}{100} \times \frac{5}{3} = \frac{15\pi \times 5}{100 \times 3} = \frac{75\pi}{300} = \frac{\pi}{4}\,\text{cm}.$$

5. Converting to Millimeters:

Since $1\,\text{cm} = 10\,\text{mm}$,

$$y = \frac{\pi}{4}\,\text{cm} = \frac{10\pi}{4}\,\text{mm} = \frac{5\pi}{2}\,\text{mm} \approx 7.854\,\text{mm}.$$

The nearest integer value is 8.