Physics Question 46 – JEE-MAIN 2026
In a Young's double slit experiment, the intensity at some point on the screen is found to be times of the maximum of the interference pattern. The path difference between the interfering waves at this point is where is wavelength of the incident light. The value of is _______.
🧠 Full Solution Path
Step 1: Relate Intensity to Phase Difference✦ Active
The intensity
This simplifies to:
Step 2: Calculate Phase Difference○ Expand
Taking the square root of both sides, we get:
For the smallest non-zero path difference, we consider the positive value. Thus,
💡 Teacher's Secret Hint
Remember that
Step 3: Determine Path Difference and x○ Expand
The phase difference
Solving for
The problem states that the path difference is
Thus, the value of
💡 Teacher's Secret Hint
Ensure units are consistent when relating phase and path difference.
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