Question

In Young's double slit experiment, if the light of wavelength \(6000\AA\) is used, the fringe width is \(\beta\). If \(I\) is the intensity of light at a point on the screen where the path difference becomes \(2000\AA\), then the intensity of light at a point on the screen which is located at a distance of \(\frac{\beta}{6}\) from the central maximum is:

• A. \(\frac{I}{2}\)
• B. \(2I\)
• C. \(\frac{I}{3}\)
• D. \(3I\)

Hint:

At a distance \(y\) from the central maximum, the path difference is \(\Delta=\frac{y}{\beta}\lambda\).

Answer 1

The Correct Option is D

Concept: Intensity in YDSE: \[ I=4I_0\cos^2\left(\frac{\phi}{2}\right) \] where \[ \phi=\frac{2\pi\Delta}{\lambda} \]

Step 1: Intensity corresponding to path difference \(2000\AA\). \[ \Delta=2000\AA \] \[ \lambda=6000\AA \] \[ \phi=\frac{2\pi(2000)}{6000} \] \[ =\frac{2\pi}{3} \] Thus, \[ I=4I_0\cos^2\frac{\pi}{3} \] \[ =4I_0\left(\frac12\right)^2 \] \[ =I_0 \]

Step 2: Intensity at \(y=\frac{\beta}{6}\). Path difference \[ \Delta=\frac{\lambda}{6} \] Hence, \[ \phi=\frac{2\pi}{6} \] \[ =\frac{\pi}{3} \] Intensity \[ I'=4I_0\cos^2\frac{\pi}{6} \] \[ =4I_0\left(\frac{\sqrt3}{2}\right)^2 \] \[ =3I_0 \] Since \(I=I_0\), \[ I'=3I \] \[ \boxed{3I} \]