Question

In Young's double-slit experiment, the ratio of maximum and minimum intensity on the screen is 9:1. What should be the ratio of the width of the slits?

Hint:

In Young's double-slit experiment, the intensity ratio is influenced by the slit width and the fringe visibility, which are interdependent on the wavelength and the distance between slits.

Answer 1

Step 1: Intensity in Young's Double-Slit Experiment.
The intensity in Young's double-slit experiment is given by: \[ I = I_0 \cos^2 \left( \frac{\pi d \sin \theta}{\lambda} \right) \] where \( I_0 \) is the maximum intensity, \( d \) is the distance between the slits, \( \lambda \) is the wavelength of light, and \( \theta \) is the angle of the screen. The maximum intensity occurs when the phase difference is an integer multiple of \( 2\pi \), and the minimum intensity occurs when the phase difference is an odd multiple of \( \pi \).
Step 2: Intensity Ratio.
The ratio of the maximum intensity (\( I_{\text{max}} \)) to the minimum intensity (\( I_{\text{min}} \)) is given as 9:1. Therefore, \[ \frac{I_{\text{max}}}{I_{\text{min}}} = 9 \] Since the intensity ratio is related to the square of the cosine function of the phase difference, we can use the relation for the maximum and minimum intensities. The ratio of intensities also depends on the slit width, which affects the diffraction pattern and the fringe visibility. To satisfy the given intensity ratio, the slit width must be adjusted accordingly.
Step 3: Conclusion.
The required ratio of the slit width for the given intensity ratio will be determined by specific experimental conditions, which balance the diffraction and interference effects. The detailed calculation would require further information about the wavelength and distance between the slits.