Question

A light ray is incident on a glass slab of thickness \(4\sqrt{3}\) cm and refractive index \(\sqrt{2}\). The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of the ray after passing through the glass slab is _____ cm.
(Given \(\sin 15^\circ = 0.25\))

Answer 1

Correct Answer: 2

1. Calculate Total Mass of Reactants (\(m_r\)):
\[ m_r = 235.0439 \, \text{u}. \] 
2. Calculate Total Mass of Products (\(m_p\)): 
\[ m_p = 139.9054 + 93.9063 + 1.0086 = 234.8203 \, \text{u}. \] 
3. Calculate Disintegration Energy (\(Q\)): 
The disintegration energy \(Q\) is given by: 
\[ Q = (m_r - m_p)c^2. \] 
Substitute the values: 
\[ Q = (235.0439 - 234.8203) \times 931. \] 
Simplify: 
\[ Q = 0.2236 \times 931 = 208.1716 \, \text{MeV}. \] 
Answer: \(208 \, \text{MeV}\)

Answer 2

Step 1: Identify given data and the setup
A monochromatic light ray is incident from air onto a parallel-sided glass slab of thickness t = 4√3 cm and refractive index μ = √2. The angle of incidence i equals the critical angle of the glass–air interface. We are to find the lateral displacement (also called the lateral shift) δ after the ray emerges from the second face of the slab back into air.

Step 2: Recall the critical angle relation
For a medium of refractive index μ surrounded by air (n = 1), the critical angle C satisfies: \[ \sin C = \frac{1}{\mu}. \] With \(\mu = \sqrt{2}\), we get: \[ \sin C = \frac{1}{\sqrt{2}} \Rightarrow C = 45^\circ. \] Given that the ray is incident at the critical angle, we have: \[ i = C = 45^\circ. \]

Step 3: Find the refraction angle inside the slab
Use Snell’s law at the first interface (air to glass): \[ n_1 \sin i = n_2 \sin r \quad \Rightarrow \quad 1 \cdot \sin 45^\circ = \sqrt{2} \cdot \sin r. \] Since \(\sin 45^\circ = \frac{\sqrt{2}}{2}\), this gives: \[ \frac{\sqrt{2}}{2} = \sqrt{2} \sin r \quad \Rightarrow \quad \sin r = \frac{1}{2} \Rightarrow r = 30^\circ. \]

Step 4: Use the lateral displacement formula for a parallel slab
For a slab of thickness t with incident angle i and refraction angle r, the lateral shift δ is: \[ \delta = \frac{t \sin(i - r)}{\cos r}. \] Substitute \(t = 4\sqrt{3}\,\text{cm}\), \(i = 45^\circ\), \(r = 30^\circ\). We are given \(\sin 15^\circ = 0.25\) and we know \(\cos 30^\circ = \frac{\sqrt{3}}{2}\): \[ \delta = \frac{4\sqrt{3} \cdot \sin(45^\circ - 30^\circ)}{\cos 30^\circ} = \frac{4\sqrt{3} \cdot \sin 15^\circ}{\frac{\sqrt{3}}{2}} = \frac{4\sqrt{3} \cdot 0.25}{\frac{\sqrt{3}}{2}}. \] Compute stepwise: \[ 4 \times 0.25 = 1 \Rightarrow \text{numerator} = \sqrt{3}, \quad \delta = \sqrt{3} \times \frac{2}{\sqrt{3}} = 2\,\text{cm}. \]

Step 5: Quick consistency checks
1) Angle trend: with \(i = 45^\circ\) and \(r = 30^\circ\), the ray bends toward the normal inside glass, so the shift is positive and moderate.
2) Dimensional check: \(\delta\) has units of length; the formula scales with \(t\) as expected.
3) Numerical check: using the exact values \(\sin 15^\circ = 0.25\) and \(\cos 30^\circ = \sqrt{3}/2\) gives an exact cancellation of \(\sqrt{3}\), yielding a clean integer result.

Final answer

2