In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength `lamda`. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum.

Explanation

To solve the problem of finding the distances from the central point where the intensity falls to half and one fourth of the maximum in a Young's double slit interference experiment, we can follow these steps:

### Step 1: Understand the Intensity Formula
In a Young's double slit experiment, the intensity \( I \) at a point on the screen is given by the formula:
\[ I = I_0 \cdot \cos^2\left(\frac{\delta}{2}\right) \]
where \( I_0 \) is the maximum intensity and \( \delta \) is the phase difference.

### Step 2: Determine Maximum Intensity
The maximum intensity \( I_{max} \) occurs when \( \cos^2\left(\frac{\delta}{2}\right) = 1 \), which gives:
\[ I_{max} = 4I_0 \]
Thus, the average intensity \( I_0 \) can be expressed as:
\[ I_0 = \frac{I_{max}}{4} \]

### Step 3: Find Distance for Half Maximum Intensity
To find the distance where the intensity is half of the maximum:
\[ I = \frac{I_{max}}{2} \]
Substituting into the intensity formula:
\[ \frac{I_{max}}{2} = 4I_0 \cdot \cos^2\left(\frac{\delta}{2}\right) \]
This simplifies to:
\[ \cos^2\left(\frac{\delta}{2}\right) = \frac{1}{8} \]
Taking the square root:
\[ \cos\left(\frac{\delta}{2}\right) = \frac{1}{2\sqrt{2}} \]
This corresponds to:
\[ \frac{\delta}{2} = \frac{\pi}{4} \quad \Rightarrow \quad \delta = \frac{\pi}{2} \]

### Step 4: Relate Phase Difference to Path Difference
The phase difference \( \delta \) is related to the path difference \( x \) by:
\[ \delta = \frac{2\pi}{\lambda} \cdot x \]
Setting \( \delta = \frac{\pi}{2} \):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \cdot x \quad \Rightarrow \quad x = \frac{\lambda}{4} \]

### Step 5: Calculate the Distance on the Screen
Using the formula for the distance \( y \) from the central maximum:
\[ y = \frac{xD}{d} \]
Substituting \( x = \frac{\lambda}{4} \):
\[ y = \frac{\left(\frac{\lambda}{4}\right)D}{d} = \frac{\lambda D}{4d} \]

### Step 6: Find Distance for One Fourth Maximum Intensity
Now, for the intensity to be one fourth of the maximum:
\[ I = \frac{I_{max}}{4} \]
Substituting into the intensity formula:
\[ \frac{I_{max}}{4} = 4I_0 \cdot \cos^2\left(\frac{\delta}{2}\right) \]
This simplifies to:
\[ \cos^2\left(\frac{\delta}{2}\right) = \frac{1}{16} \]
Taking the square root:
\[ \cos\left(\frac{\delta}{2}\right) = \frac{1}{4} \]
This corresponds to:
\[ \frac{\delta}{2} = \frac{\pi}{3} \quad \Rightarrow \quad \delta = \frac{2\pi}{3} \]

### Step 7: Relate Phase Difference to Path Difference Again
Using the same relationship:
\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda} \cdot x \quad \Rightarrow \quad x = \frac{\lambda}{3} \]

### Step 8: Calculate the Distance for One Fourth Intensity
Using the distance formula again:
\[ y = \frac{xD}{d} \]
Substituting \( x = \frac{\lambda}{3} \):
\[ y = \frac{\left(\frac{\lambda}{3}\right)D}{d} = \frac{\lambda D}{3d} \]

### Final Results
1. The distance from the central point where the intensity falls to half the maximum is:
\[ y_{1/2} = \frac{\lambda D}{4d} \]
2. The distance from the central point where the intensity falls to one fourth of the maximum is:
\[ y_{1/4} = \frac{\lambda D}{3d} \]