Q71 / 75 · JEE Main 2026 (02 April Shift 2) · Physics
In a Young's double slit experiment, the intensity at some point on the screen is found to be $\dfrac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\dfrac{\lambda}{x}$ where $\lambda$ is wavelength of the incident light. The value of $x$ is _______.
Correct answer: 6
Solution
The intensity at a point in Young's double slit experiment is given by $I = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)$
Given $I = \dfrac{3}{4} I_{max}$
$\dfrac{3}{4} I_{max} = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)$
$\cos\left(\dfrac{\phi}{2}\right) = \dfrac{\sqrt{3}}{2}$
$\dfrac{\phi}{2} = \dfrac{\pi}{6}$
$\phi = \dfrac{\pi}{3}$
The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \dfrac{2\pi}{\lambda} \Delta x$
$\dfrac{2\pi}{\lambda} \Delta x = \dfrac{\pi}{3}$
$\Delta x = \dfrac{\lambda}{6}$
Comparing with $\Delta x = \dfrac{\lambda}{x}$, we get $x = 6$
Answer: $6$
Given $I = \dfrac{3}{4} I_{max}$
$\dfrac{3}{4} I_{max} = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)$
$\cos\left(\dfrac{\phi}{2}\right) = \dfrac{\sqrt{3}}{2}$
$\dfrac{\phi}{2} = \dfrac{\pi}{6}$
$\phi = \dfrac{\pi}{3}$
The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \dfrac{2\pi}{\lambda} \Delta x$
$\dfrac{2\pi}{\lambda} \Delta x = \dfrac{\pi}{3}$
$\Delta x = \dfrac{\lambda}{6}$
Comparing with $\Delta x = \dfrac{\lambda}{x}$, we get $x = 6$
Answer: $6$
Hence, the correct answer is 6.