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

The problem involves calculating the thickness of the lower slab so that the lateral shift of the ray after passing through both slabs is zero. Given the refractive indices and thickness of the upper slab, we use Snell's Law and the lateral shift formula to solve this.

1. Snell's Law at the first interface:

\(n_1 \sin \theta_1 = n_2 \sin \theta_2\)

For the first interface where the light travels from air \((n = \sqrt{2})\) to the slab with \(n = 2\),

\(\sqrt{2} \sin 45^\circ = 2 \sin \theta_2\)

Since \(\sin 45^\circ = \frac{1}{\sqrt{2}}\),

\(1 = 2 \sin \theta_2 \Rightarrow \sin \theta_2 = \frac{1}{2}\)

Therefore, \(\theta_2 = 30^\circ\).

2. Snell's Law at the interface between the two slabs:

\(n_2 \sin \theta_2 = n_3 \sin \theta_3\)

For the interface between slabs with \(n = 2\) and \(n = \frac{2}{\sqrt{3}}\),

\(2 \times \frac{1}{2} = \frac{2}{\sqrt{3}} \sin \theta_3\)

\(\sin \theta_3 = \frac{\sqrt{3}}{2}\)

Therefore, \(\theta_3 = 60^\circ\).

3. Snell's Law at the last interface:

\(n_3 \sin \theta_3 = n_4 \sin \theta_4\)

For the exit where \(n = \frac{2}{\sqrt{3}}\) into the same medium \((\sqrt{2})\),

\(\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \sqrt{2} \sin \theta_4\)

\(1 = \sqrt{2} \sin \theta_4 \Rightarrow \sin \theta_4 = \frac{1}{\sqrt{2}}\)

Therefore, \(\theta_4 = 45^\circ\).

4. Condition for zero lateral shift:

Lateral shift is zero when the path difference is zero:

\(d_1 \tan \theta_2 = d_2 \tan \theta_3\)

Given \(d_1 = 1 \, \text{cm}\),

\(1 \times \tan 30^\circ = d_2 \times \tan 60^\circ\)

\(\tan 30^\circ = \frac{1}{\sqrt{3}}\) and \(\tan 60^\circ = \sqrt{3}\)

Substitute these values:

\(\frac{1}{\sqrt{3}} = d_2 \times \sqrt{3}\)

\(d_2 = \frac{1}{\sqrt{3}} \, \text{cm}\)

Conclusion:

The thickness of the lower slab should be \(\frac{1}{\sqrt{3}} \, \text{cm}\) for the lateral shift to be zero.