Question

The refractive indices (\(n\)) of two transparent slabs are \(2\) and \(2/\sqrt{3}\). They are attached together and placed in a third transparent medium of refractive index \(\sqrt{2}\), as shown in the figure. The thickness of the upper slab is \(1\ \text{cm}\). A monochromatic light ray is incident on the upper slab at \(45^\circ\). What would be the thickness in \(\text{cm}\) of the lower slab such that the lateral shift of the ray after passing through both the slabs is zero?

• A. \(1/\sqrt{3}\)
• B. \(1/\sqrt{2}\)
• C. \(1/2\)
• D. \(\sqrt{3}/2\)

Hint:

Using Snell's law, we can easily find the refraction angles as \(30^\circ\) and \(60^\circ\).
Setting the actual horizontal shift equal to the straight line shift \((1+t_2)\tan 45^\circ\) gives a simple linear equation in \(t_2\) which yields the answer in one step.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a light ray passing through two parallel-faced slabs of different refractive indices.
The assembly is immersed in a medium of refractive index \(\sqrt{2}\).
We need to find the thickness of the second slab so that the exiting ray aligns exactly with the straight-line extension of the incident ray (i.e., zero net lateral shift).

Step 2: Key Formula or Approach:

• Snell's law of refraction:
\[ n_0 \sin i = n \sin r \]
• Horizontal displacement inside slab \(i\) of thickness \(t_i\) and refraction angle \(r_i\).:
\[ \Delta x_i = t_i \tan r_i \]
• For zero net lateral displacement relative to the undeflected ray, the total actual horizontal displacement must equal the hypothetical undeflected horizontal displacement:
\[ \sum t_i \tan r_i = \left(\sum t_i\right) \tan i \]

Step 3: Detailed Explanation:

• Apply Snell's law at the first interface:
\[ \sqrt{2} \sin(45^\circ) = 2 \sin(r_1) \implies \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \sin(r_1) \] \[ \sin(r_1) = \frac{1}{2} \implies r_1 = 30^\circ \]
• Apply Snell's law for the second slab:
\[ \sqrt{2} \sin(45^\circ) = \frac{2}{\sqrt{3}} \sin(r_2) \implies 1 = \frac{2}{\sqrt{3}} \sin(r_2) \] \[ \sin(r_2) = \frac{\sqrt{3}}{2} \implies r_2 = 60^\circ \]
• The horizontal shift inside the first slab (\(t_1 = 1\ \text{cm}\)) is:
\[ \Delta x_1 = t_1 \tan(30^\circ) = 1 \cdot \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}\ \text{cm} \]
• The horizontal shift inside the second slab of thickness \(t_2\) is:
\[ \Delta x_2 = t_2 \tan(60^\circ) = t_2 \sqrt{3} \]
• The hypothetical horizontal shift if the ray travelled undeflected at \(45^\circ\) is:
\[ \Delta x_{\text{undeflected}} = (t_1 + t_2) \tan(45^\circ) = (1 + t_2) \cdot 1 = 1 + t_2 \]
• Setting actual shift equal to undeflected shift:
\[ \frac{1}{\sqrt{3}} + t_2 \sqrt{3} = 1 + t_2 \] \[ t_2 (\sqrt{3} - 1) = 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \] \[ t_2 = \frac{1}{\sqrt{3}}\ \text{cm} \]

Step 4: Final Answer:
The thickness of the lower slab is \(1/\sqrt{3}\ \text{cm}\).