Question

In a Young's double slit experiment, the intensity at some point on the screen is found to be \( \frac{3}{4} \) times of the maximum of the interference pattern. The path difference between the interfering waves at this point is \( \frac{\lambda}{x} \), where \( \lambda \) is the wavelength of the incident light. The value of \( x \) is _______.}

Answer 1

Correct Answer: 6

Step 1: Understanding the Concept:
In an interference pattern, the resultant intensity at any point on the screen depends on the phase difference between the two interfering waves.
Once the phase difference is found, it can be directly converted into the corresponding path difference.
Step 2: Key Formula or Approach:
The intensity formula is \(I = I_{max} \cos^2\left(\frac{\phi}{2}\right)\), where \(\phi\) is the phase difference.
The relation between phase difference and path difference is \(\phi = \frac{2\pi}{\lambda} \Delta x\).
Step 3: Detailed Explanation:
Given that the intensity is \(I = \frac{3}{4} I_{max}\).
Substitute this into the intensity formula:
\[ \frac{3}{4} I_{max} = I_{max} \cos^2\left(\frac{\phi}{2}\right) \] \[ \cos^2\left(\frac{\phi}{2}\right) = \frac{3}{4} \] Taking the square root of both sides:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{\sqrt{3}}{2} \] The principal angle whose cosine is \(\frac{\sqrt{3}}{2}\) is \(30^\circ\) or \(\frac{\pi}{6}\) radians.
\[ \frac{\phi}{2} = \frac{\pi}{6} \implies \phi = \frac{\pi}{3} \text{ radians} \] Now, convert the phase difference to a path difference \(\Delta x\):
\[ \phi = \frac{2\pi}{\lambda} \Delta x \] \[ \frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x \] \[ \Delta x = \frac{\lambda}{3 \times 2} = \frac{\lambda}{6} \] Comparing this to the given format \(\frac{\lambda}{x}\), we find that \(x = 6\).
Step 4: Final Answer:
The value of \(x\) is \(6\).