Question:
In Young's experiment, using red light ($\lambda = 6600$ \AA), 60 fringes are seen in the field of view. How many fringes will be seen by using violet light ($\lambda = 4400$ A)?
In Young's experiment, using red light ($\lambda = 6600$ \AA), 60 fringes are seen in the field of view. How many fringes will be seen by using violet light ($\lambda = 4400$ A)?
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Number of fringes \(\propto \dfrac{1}{\lambda}\). A smaller wavelength (violet) gives more fringes in the same field of view.
Updated On: May 2, 2026
- 10
- 20
- 45
- 90
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
In Young's double slit experiment, the fringe width is given by \(\beta = \dfrac{\lambda D}{d}\). The number of fringes visible in a given field of view is inversely proportional to the fringe width, hence inversely proportional to the wavelength.
Step 2: Detailed Explanation:
If \(n_1\) fringes are seen with wavelength \(\lambda_1\) and \(n_2\) fringes with \(\lambda_2\), then: \[ n_1 \lambda_1 = n_2 \lambda_2 \] \[ 60 \times 6600 = n_2 \times 4400 \] \[ n_2 = \frac{60 \times 6600}{4400} = 90 \]
Step 3: Final Answer:
The number of fringes seen with violet light is 90.
In Young's double slit experiment, the fringe width is given by \(\beta = \dfrac{\lambda D}{d}\). The number of fringes visible in a given field of view is inversely proportional to the fringe width, hence inversely proportional to the wavelength.
Step 2: Detailed Explanation:
If \(n_1\) fringes are seen with wavelength \(\lambda_1\) and \(n_2\) fringes with \(\lambda_2\), then: \[ n_1 \lambda_1 = n_2 \lambda_2 \] \[ 60 \times 6600 = n_2 \times 4400 \] \[ n_2 = \frac{60 \times 6600}{4400} = 90 \]
Step 3: Final Answer:
The number of fringes seen with violet light is 90.
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