Question

In a Young’s double slit experiment, the intensity at a point is \[ \left(\frac{1}{4}\right)^{\text{th}} \] of the maximum intensity, the minimum distance of the point from the central maximum is ____ $\mu$m.
(Given: $\lambda = 600 \, \text{nm}, \, d = 1.0 \, \text{mm}, \, D = 1.0 \, \text{m}$)

Answer 1

Correct Answer: 200

In Young's double slit experiment, the intensity \( I \) at a point is given by the formula: \( I = I_0 \cos^2(\frac{\pi d \sin \theta}{\lambda}) \), where:

• \( I_0 \) is the maximum intensity.
• \( d \) is the slit separation.
• \( \theta \) is the angle of the point from the center.
• \( \lambda \) is the wavelength.

Given \( I = \left(\frac{1}{4}\right) I_0 \), we have: 
\[ \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{4} \] 
\[ \cos\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{2} \] 
Thus: 
\[ \frac{\pi d \sin \theta}{\lambda} = \frac{\pi}{3} \text{ or } \frac{2\pi}{3} \] 
Solving for \( \sin \theta \): 
\[ \sin \theta = \frac{\lambda}{3d} \text{ for } \frac{\pi}{3} \text{, or } -\frac{\lambda}{3d} \text{ for } \frac{2\pi}{3} \] 
Using the small angle approximation, \( \sin \theta \approx \tan \theta \approx \frac{x}{D} \), where \( x \) is the distance from the central maximum: 
\[ \frac{x}{D} = \frac{\lambda}{3d} \] 
Thus: 
\[ x = \frac{\lambda D}{3d} \] 
Substitute \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \), \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \), \( D = 1.0 \, \text{m} \): 
\[ x = \frac{600 \times 10^{-9} \times 1.0}{3 \times 1.0 \times 10^{-3}} \] 
\[ x = 200 \times 10^{-6} \, \text{m} = 200 \, \mu\text{m} \] 
The minimum distance is \( 200 \, \mu\text{m} \), which is within the given range (200,200).

Answer 2

Step 1: Intensity relation The intensity at a point in the interference pattern is given by:

\[ I = I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right). \]

Given:

\[ I = \frac{I_0}{4}. \]

Substitute:

\[ \frac{I_0}{4} = I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right). \]

Simplify:

\[ \cos^2 \left( \frac{\Delta \phi}{2} \right) = \frac{1}{4}. \]

Taking square root:

\[ \cos \left( \frac{\Delta \phi}{2} \right) = \frac{1}{2}. \]

This implies:

\[ \frac{\Delta \phi}{2} = \frac{\pi}{3}. \]

So:

\[ \Delta \phi = \frac{2\pi}{3}. \]

Step 2: Path difference relation The phase difference \(\Delta \phi\) is related to the path difference by:

\[ \Delta \phi = \frac{2\pi}{\lambda} \times \frac{y d}{D}. \]

Substitute \(\Delta \phi = \frac{2\pi}{3}\):

\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda} \times \frac{y d}{D}. \]

Cancel \(2\pi\):

\[ \frac{1}{3} = \frac{y d}{\lambda D}. \]

Rearrange to solve for \(y\):

\[ y = \frac{\lambda D}{3 d}. \]

Step 3: Substitution of values Substitute \(\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}\), \(D = 1.0 \, \text{m}\), and \(d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}\):

\[ y = \frac{600 \times 10^{-9} \times 1.0}{3 \cdot 1.0 \times 10^{-3}}. \]

Simplify:

\[ y = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} = \frac{600}{3} \times 10^{-6}. \]

\[ y = 200 \times 10^{-6} \, \text{m}. \]

Convert to \(\mu \text{m}\):

\[ y = 200 \, \mu \text{m}. \]

Final Answer: \(y = 200 \, \mu \text{m}\).