In Young's double-slit experiment, light of wavelength 600 nm is used. When the slits are kept at a distance of 1mm, interference pattern is observed at a diatance of 1m. What will be the distance between central maxima and fringe with intensity equal to one-forth of maximum intensity.

Answer

To solve the problem of finding the distance between the central maxima and the fringe with intensity equal to one-fourth of the maximum intensity in Young's double-slit experiment, we can follow these steps:

### Step 1: Understand the Maximum Intensity
In Young's double-slit experiment, the maximum intensity \( I_{\text{max}} \) at the central maxima occurs when the path difference between the two waves is zero. The intensity at the central maxima is given by:
\[ I_{\text{max}} = 4I \]
where \( I \) is the intensity from each slit.

### Step 2: Determine the Intensity Condition
We need to find the position where the intensity \( I \) is one-fourth of the maximum intensity:
\[ I = \frac{1}{4} I_{\text{max}} = \frac{1}{4} \times 4I = I \]
This means we are looking for the position where the resultant intensity is equal to \( I \).

### Step 3: Use the Intensity Formula
The net intensity \( I_{\text{net}} \) at a point in the interference pattern is given by:
\[ I_{\text{net}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
Assuming \( I_1 = I_2 = I \), we have:
\[ I_{\text{net}} = I + I + 2I \cos \phi = 2I + 2I \cos \phi \]
Setting \( I_{\text{net}} = I \):
\[ I = 2I + 2I \cos \phi \]
This simplifies to:
\[ 0 = I + 2I \cos \phi \quad \Rightarrow \quad \cos \phi = -\frac{1}{2} \]

### Step 4: Find the Phase Difference
The phase difference \( \phi \) can be expressed as:
\[ \phi = 2\pi \frac{\Delta x}{\lambda} \]
where \( \Delta x \) is the path difference. From \( \cos \phi = -\frac{1}{2} \), we find:
\[ \phi = \frac{2\pi}{3} \text{ or } \frac{4\pi}{3} \]

### Step 5: Relate Path Difference to Geometry
The path difference \( \Delta x \) can be expressed as:
\[ \Delta x = d \sin \theta \]
where \( d \) is the distance between the slits and \( \theta \) is the angle from the central maxima to the point of interest.

### Step 6: Use Small Angle Approximation
For small angles, \( \sin \theta \approx \tan \theta \approx \frac{y}{D} \), where \( D \) is the distance from the slits to the screen and \( y \) is the distance from the central maxima to the point of interest:
\[ \Delta x = d \frac{y}{D} \]

### Step 7: Substitute and Solve for \( y \)
Substituting \( \Delta x \) into the phase difference equation:
\[ \frac{2\pi}{3} = 2\pi \frac{d \frac{y}{D}}{\lambda} \]
This simplifies to:
\[ \frac{1}{3} = \frac{d y}{\lambda D} \]
Rearranging gives:
\[ y = \frac{\lambda D}{3d} \]

### Step 8: Plug in the Values
Given:
- \( \lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m} \)
- \( D = 1 \text{ m} \)
- \( d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)

Substituting these values:
\[ y = \frac{600 \times 10^{-9} \times 1}{3 \times 1 \times 10^{-3}} = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} = \frac{600}{3} \times 10^{-6} = 200 \times 10^{-6} \text{ m} = 0.2 \text{ mm} \]

### Final Answer
The distance between the central maxima and the fringe with intensity equal to one-fourth of the maximum intensity is:
\[ y = 0.2 \text{ mm} \]