Question

In YDSE, light of intensity \( 4I \) and \( 9I \) passes through two slits respectively. Difference of maximum and minimum intensity of interference pattern is:

• A. \( I \)
• B. \( 3I \)
• C. \( 5I \)
• D. \( 4I \)

Hint:

The intensity difference in YDSE interference patterns is calculated using the difference of maximum and minimum intensities.

Answer 1

The Correct Option is A

In Young's Double Slit Experiment (YDSE), the intensity of the maxima and minima can be found using the superposition principle. If two light waves of intensities \( I_1 \) and \( I_2 \) interfere, the maximum and minimum intensities are given by: - Maximum intensity: \( I_{\text{max}} = I_1 + I_2 \) - Minimum intensity: \( I_{\text{min}} = |I_1 - I_2| \) For the given problem: - \( I_1 = 4I \) - \( I_2 = 9I \) Thus, the maximum intensity is \( 4I + 9I = 13I \), and the minimum intensity is \( |4I - 9I| = 5I \). The difference between maximum and minimum intensity is: \[ \Delta I = 13I - 5I = 8I \] Thus, the difference is \( 8I \).