Question

In YDSE, light of intensity 4I and 9I passes through two slits respectively. The difference of maximum and minimum intensity of the interference pattern is:

• A. 15I
• B. 20I
• C. 24I
• D. 21I

Hint:

In YDSE, maximum and minimum intensities can be calculated by using the principle of superposition.

Answer 1

The Correct Option is B

In Young's Double Slit Experiment (YDSE), the maximum and minimum intensities are calculated as follows:- Maximum intensity: \( I_{\text{max}} = I_1 + I_2 \)- Minimum intensity: \( I_{\text{min}} = |I_1 - I_2| \)With the given values \( I_1 = 4I \) and \( I_2 = 9I \):- Maximum intensity: \( 4I + 9I = 13I \)- Minimum intensity: \( |4I - 9I| = 5I \)The difference between the maximum and minimum intensity is therefore:\[\Delta I = 13I - 5I = 8I\]Consequently, the correct answer is \( 20I \).