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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 34 times of the maximum of the interference pattern. The path difference between the interfering waves at this point is λx where λ is wavelength of the incident light. The value of x is _______.

Recall the formula for intensity in Young's double-slit experiment in terms of maximum intensity and phase difference.

Step 1: Relate Intensity to Phase Difference✦ Active

The intensity I at any point in Young's double-slit experiment is given by I=Imaxcos2(ϕ2), where Imax is the maximum intensity and ϕ is the phase difference. Given that I=34Imax, we can write:

34Imax=Imaxcos2(ϕ2)

This simplifies to:

cos2(ϕ2)=34
Step 2: Calculate Phase Difference○ Expand

Taking the square root of both sides, we get:

cos(ϕ2)=±32

For the smallest non-zero path difference, we consider the positive value. Thus, ϕ2=π6 (or 30). Therefore, the phase difference is:

ϕ=2π6=π3
💡 Teacher's Secret Hint

Remember that cos(θ)=32 for θ=π6.

Step 3: Determine Path Difference and x○ Expand

The phase difference ϕ is related to the path difference Δx by the formula ϕ=2πλΔx. Substituting the calculated phase difference:

π3=2πλΔx

Solving for Δx:

Δx=λ312=λ6

The problem states that the path difference is λx. Comparing this with our result:

λx=λ6

Thus, the value of x is:

x=6
💡 Teacher's Secret Hint

Ensure units are consistent when relating phase and path difference.

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