In a Young double slit experiment, the wavelength of incident light is 6000 AA , the separation between slits is 5 cm , the distance between slits plane and screen is 50 cm as shown in the figure. If the resultant intensity at P is equal to the intensity due to individual slits, the path difference between interfering waves is Å.

Physics · Wave Optics

JEE Main 2026 — 6 April, Evening Shift — Question 15

In a Young double slit experiment, the wavelength of incident light is $6000\mathrm{\AA}$ , the separation between slits is $5\mathrm{cm}$ , the distance between slits plane and screen is $50\mathrm{cm}$ as shown in the figure. If the resultant intensity at P is equal to the intensity due to individual slits, the path difference between interfering waves is ______ Å.

Option A:
4000
Option B:
3000
Option C:
2000
Correct
Option D:
1000

Answer: C

Step-by-step solution

Intensity $I = 4I_0 \cos^2(\Delta\phi/2)$ . Given $I = I_0$ , so $\cos^2(\Delta\phi/2)=1/4 \Rightarrow \cos(\Delta\phi/2)=1/2 \Rightarrow \Delta\phi/2 = \pi/3 \Rightarrow \Delta\phi = 2\pi/3$ . Path difference $\Delta x = \frac{\lambda}{2\pi}\Delta\phi = \frac{6000}{2\pi}\times\frac{2\pi}{3} = 2000$ Å.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications