In Young's experiment the slits, separated by `d = 0.8 mm`, are illuminated with light of wavelength 7200 Å, Interference pattern is obtained on a screen `D = 2 m` from the slits. Find minimum distance from central maximum at which the a average intensity is 50% if of maximum?

Explanation

To solve the problem step by step, we will use the principles of interference and the formula for intensity in Young's double-slit experiment.

### Step 1: Understand the given parameters
- Slit separation, \( d = 0.8 \, \text{mm} = 0.8 \times 10^{-3} \, \text{m} \)
- Wavelength of light, \( \lambda = 7200 \, \text{Å} = 7200 \times 10^{-10} \, \text{m} \)
- Distance from slits to screen, \( D = 2 \, \text{m} \)

### Step 2: Write the formula for intensity
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