Question

In a Young's double slit experiment using monochromatic light of wavelength \(\lambda\), the intensity of light at a point on the screen is \(I_0\), where the path difference between two interfering waves is \(\lambda\). The path difference between the interfering waves at a point on the screen where the intensity is \(\dfrac{I_0}{4}\) will be

• A. \(\dfrac{\lambda}{4}\)
• B. \(\dfrac{\lambda}{3}\)
• C. \(\dfrac{3\lambda}{2}\)
• D. \(2\lambda\)

Hint:

For equal-intensity coherent sources: \[ I=4I\cos^2\left(\frac{\phi}{2}\right) \] and \[ \phi=\frac{2\pi}{\lambda}\Delta \] Always convert the given intensity ratio into a value of \(\cos^2(\phi/2)\) and then find the path difference.

Answer 1

The Correct Option is B

Concept: For two coherent sources of equal intensity, the intensity at any point is \[ I=4I\cos^2\left(\frac{\phi}{2}\right) \] where \(\phi\) is the phase difference. Also, \[ \phi=\frac{2\pi}{\lambda}\Delta \] where \(\Delta\) is the path difference.

Step 1: Determine the maximum intensity. Given path difference \[ \Delta=\lambda \] Therefore, \[ \phi=\frac{2\pi}{\lambda}\lambda=2\pi \] Hence, \[ I_0=4I\cos^2\pi \] \[ I_0=4I \]

Step 2: Use the condition for intensity \(\dfrac{I_0}{4}\). Given, \[ I=\frac{I_0}{4} \] Since \[ I_0=4I \] we get \[ \frac{I}{I_0}=\frac14 \] Thus, \[ \cos^2\left(\frac{\phi}{2}\right)=\frac14 \] \[ \cos\left(\frac{\phi}{2}\right)=\pm\frac12 \] Taking the smallest positive value, \[ \frac{\phi}{2}=\frac{\pi}{3} \] \[ \phi=\frac{2\pi}{3} \]

Step 3: Calculate the corresponding path difference. \[ \phi=\frac{2\pi}{\lambda}\Delta \] \[ \frac{2\pi}{3} = \frac{2\pi}{\lambda}\Delta \] \[ \Delta=\frac{\lambda}{3} \]

Step 4: State the answer. \[ \boxed{ \Delta=\frac{\lambda}{3} } \] Hence, the correct option is \[ \boxed{(B)} \]