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: Understand the interference pattern equation.
In Young's double slit experiment, the intensity at a point on the screen is given by the equation: \[ I = I_{\text{max}} \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right) \] where \( I_{\text{max}} \) is the maximum intensity, \( \Delta x \) is the path difference, and \( \lambda \) is the wavelength of the light.
Step 2: Use the given intensity ratio.
The problem states that the intensity is \( \frac{3}{4} \) of the maximum intensity, so: \[ \frac{I}{I_{\text{max}}} = \frac{3}{4} \] Thus, we have the equation: \[ \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right) = \frac{3}{4} \]
Step 3: Solve for the path difference.
Taking the square root of both sides: \[ \cos \left( \frac{\pi \Delta x}{\lambda} \right) = \frac{\sqrt{3}}{2} \] The cosine value of \( \frac{\sqrt{3}}{2} \) corresponds to \( \frac{\pi}{6} \), so: \[ \frac{\pi \Delta x}{\lambda} = \frac{\pi}{6} \]
Step 4: Find \( x \).
Solving for \( \Delta x \): \[ \Delta x = \frac{\lambda}{6} \] Since \( \Delta x = \frac{\lambda}{x} \), we conclude: \[ \frac{\lambda}{x} = \frac{\lambda}{6} \] Thus, the value of \( x \) is: \[ x = 6 \]