Skip to content
dronacharya.ai
← JEE Main 2026 · 2 Apr · Shift 2
PhysicsInteger+4 / -1Question 46 of 75

JEE Main 2026 · 2 Apr · Shift 2, Physics question 46

In a Young's double slit experiment, the intensity at some point on the screen is found to be 34\dfrac34 times of the maximum of the interference pattern. The path difference between the interfering waves at this point is λx\dfrac\lambda x where λ\lambda is wavelength of the incident light. The value of xx is ________.

Answer and worked solution

Official answer: 6

Intensity. I=Imax⁡cos⁡2ϕ2=34Imax⁡I=I_{\max}\cos^2\frac\phi2=\frac34I_{\max}, so ϕ2=π6\frac\phi2=\frac\pi6 and ϕ=π3\phi=\frac\pi3.

Path difference. Δ=λ2πϕ=λ6\Delta=\frac\lambda{2\pi}\phi=\frac\lambda6, so x=6x=6.

Answer: 6