Question:
In interference with coherent sources, the fringe width varies as:
In interference with coherent sources, the fringe width varies as:
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In Young’s Double Slit Experiment, increasing the wavelength increases the fringe width.
Updated On: Mar 26, 2025
- Directly as wavelength
- Inversely as wavelength
- Directly as the separation between slits
- Inversely as the distance between the slits and screen
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The Correct Option is A
Solution and Explanation
For Young’s Double Slit Experiment, the fringe width \( \beta \) is given by:
\[ \beta = \frac{\lambda D}{d} \] where:
- \( \lambda \) = wavelength of light
- \( D \) = distance between slits and screen
- \( d \) = separation between slits
Since \( \beta \propto \lambda \), the fringe width increases with wavelength.
\[ \beta = \frac{\lambda D}{d} \] where:
- \( \lambda \) = wavelength of light
- \( D \) = distance between slits and screen
- \( d \) = separation between slits
Since \( \beta \propto \lambda \), the fringe width increases with wavelength.
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