In YDSE, light of intensity \( 4I \) and \( 9I \) passes through two slits respectively. Difference of maximum and minimum intensity of interference pattern is:
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The intensity difference in YDSE interference patterns is calculated using the difference of maximum and minimum intensities.
In Young's Double Slit Experiment (YDSE), the superposition principle determines the intensity of maxima and minima. When two light waves with intensities \( I_1 \) and \( I_2 \) interfere, the maximum and minimum intensities are calculated as follows:
- Maximum intensity: \( I_{\text{max}} = I_1 + I_2 \)
- Minimum intensity: \( I_{\text{min}} = |I_1 - I_2| \)
For the specified problem:
- \( I_1 = 4I \)
- \( I_2 = 9I \)
Consequently, the maximum intensity is \( 4I + 9I = 13I \), and the minimum intensity is \( |4I - 9I| = 5I \).
The difference between the maximum and minimum intensity is:
\[
\Delta I = 13I - 5I = 8I
\]
Therefore, the difference is \( 8I \).
