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JEE MAIN_2026
06-04-26_S2
Question
In a Young double slit experiment, the wavelength of incident light is 6000 A , the separation between slits $S_1$ and $S_2$ is 5 cm and the distance between slits plane and screen is 50 cm , as shown in the figure below. If the resultant intensity at $P$ is equal to the intensity due to individual slits, the path difference between interfering waves is $\_\_\_\_$ A.
Question Image
Select the correct option:
A
4000
B
3000
C
2000
D
1000
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{array}{ll}\mathrm{I}_{\mathrm{p}}=\mathrm{I}_0=4 \mathrm{I}_0 \cos ^2 \frac{\Delta \phi}{2} & \\ \cos \frac{\Delta \phi}{2}=\frac{1}{2} \text { or }-\frac{1}{2} & \\ \frac{\Delta \phi}{2}=\frac{\pi}{3} \text { or } \frac{2 \pi}{3} & \\ \frac{2 \pi}{\lambda}=\Delta \mathrm{x}=\Delta \phi=\frac{2 \pi}{3} & \text { or } \\ \frac{2 \pi}{\lambda}=\Delta \mathrm{x}=\Delta \phi=\frac{4 \pi}{3} \\ \Delta \mathrm{x}=\frac{\lambda}{3}=2000 \mathrm{~A}^{\circ} & \text { or } \\ \Delta \mathrm{x}=\frac{2 \lambda}{3}=4000 \mathrm{~A}^{\circ}\end{array}$
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