Refraction of Light through Multi-Layered Medium Configuration

To find the number of identical units $n$, we calculate the horizontal displacement (lateral shift) of the light ray as it passes through each layer of one unit.

1. Refraction in the Air Layer: The air layer (layer 0) has a height $h = \frac{1}{3} \text{ cm}$ and refractive index $\mu_0 = 1$. The angle of incidence is $\theta_0 = 60^\circ$. The horizontal displacement in this layer is given by: $\Delta x_{\text{air}} = h \tan \theta_0 = \frac{1}{3} \tan 60^\circ = \frac{1}{3} \times \sqrt{3} = \frac{1}{\sqrt{3}} \text{ cm}$

2. Refraction in the Second Layer ($\mu_1$): The second layer has a thickness $d = \frac{\sqrt{3}-1}{2} \text{ cm}$ and refractive index $\mu_1 = \sqrt{\frac{3}{2}}$. Applying Snell's Law at the interface between air and the second layer: $\mu_0 \sin \theta_0 = \mu_1 \sin \theta_1$ $1 \cdot \sin 60^\circ = \sqrt{\frac{3}{2}} \sin \theta_1$ $\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{2}} \sin \theta_1 \implies \sin \theta_1 = \frac{1}{\sqrt{2}}$ Thus, the angle of refraction is $\theta_1 = 45^\circ$.

The horizontal displacement in this layer is: $\Delta x_1 = d \tan \theta_1 = \left(\frac{\sqrt{3}-1}{2}\right) \tan 45^\circ = \frac{\sqrt{3}-1}{2} \text{ cm}$

3. Refraction in the Third Layer ($\mu_2$): The third layer has a thickness $d = \frac{\sqrt{3}-1}{2} \text{ cm}$ and refractive index $\mu_2 = \sqrt{3}$. Applying Snell's Law: $\mu_0 \sin \theta_0 = \mu_2 \sin \theta_2$ $1 \cdot \sin 60^\circ = \sqrt{3} \sin \theta_2$ $\frac{\sqrt{3}}{2} = \sqrt{3} \sin \theta_2 \implies \sin \theta_2 = \frac{1}{2}$ Thus, the angle of refraction is $\theta_2 = 30^\circ$.

The horizontal displacement in this layer is: $\Delta x_2 = d \tan \theta_2 = \left(\frac{\sqrt{3}-1}{2}\right) \tan 30^\circ = \frac{\sqrt{3}-1}{2\sqrt{3}} \text{ cm}$

4. Total Horizontal Displacement Per Unit: The total horizontal shift per unit $\Delta x_{\text{unit}}$ is the sum of the horizontal shifts in each of its three layers: $\Delta x_{\text{unit}} = \Delta x_{\text{air}} + \Delta x_1 + \Delta x_2$ $\Delta x_{\text{unit}} = \frac{1}{\sqrt{3}} + \frac{\sqrt{3}-1}{2} + \frac{\sqrt{3}-1}{2\sqrt{3}}$ $\Delta x_{\text{unit}} = \frac{1}{\sqrt{3}} + (\sqrt{3}-1) \left(\frac{1}{2} + \frac{1}{2\sqrt{3}}\right)$ $\Delta x_{\text{unit}} = \frac{1}{\sqrt{3}} + (\sqrt{3}-1) \left(\frac{\sqrt{3}+1}{2\sqrt{3}}\right)$ Using the identity $(\sqrt{3}-1)(\sqrt{3}+1) = 3 - 1 = 2$: $\Delta x_{\text{unit}} = \frac{1}{\sqrt{3}} + \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \text{ cm}$

5. Total Displacement for $n$ Units: The total horizontal distance $l$ after emerging from $n$ units is given as $l = \frac{8}{\sqrt{3}} \text{ cm}$: $l = n \cdot \Delta x_{\text{unit}}$ $\frac{8}{\sqrt{3}} = n \cdot \left(\frac{2}{\sqrt{3}}\right)$ $n = \frac{8}{2} = 4$

The value of $n$ is 4.