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The fringe width at a distance of 50 cm ...

The fringe width at a distance of 50 cm from the slits in young's experiment for light of wavelength `6000 Å` is `0.048cm`. The fringe width at the samedistance for `lambda = 5000 Å` will be

A

`0.04 cm`

B

`0.4 cm`

C

`0.14 cm`

D

`0.45 cm`

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The correct Answer is:
To solve the problem, we need to find the fringe width for a wavelength of \( \lambda = 5000 \, \text{Å} \) given that the fringe width for \( \lambda = 6000 \, \text{Å} \) is \( 0.048 \, \text{cm} \) at a distance of \( 50 \, \text{cm} \) from the slits. ### Step-by-Step Solution: 1. **Understand the relationship between fringe width and wavelength**: The fringe width \( \beta \) in Young's double-slit experiment is given by the formula: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. Since \( D \) and \( d \) are constants for this experiment, we can conclude that the fringe width \( \beta \) is directly proportional to the wavelength \( \lambda \). 2. **Set up the proportionality**: Since \( \beta \) is directly proportional to \( \lambda \), we can write: \[ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} \] where: - \( \beta_1 = 0.048 \, \text{cm} \) (fringe width for \( \lambda_1 = 6000 \, \text{Å} \)) - \( \lambda_1 = 6000 \, \text{Å} \) - \( \lambda_2 = 5000 \, \text{Å} \) 3. **Substituting values**: We want to find \( \beta_2 \): \[ \beta_2 = \beta_1 \cdot \frac{\lambda_2}{\lambda_1} \] Substituting the known values: \[ \beta_2 = 0.048 \, \text{cm} \cdot \frac{5000 \, \text{Å}}{6000 \, \text{Å}} \] 4. **Calculating \( \beta_2 \)**: \[ \beta_2 = 0.048 \, \text{cm} \cdot \frac{5000}{6000} \] Simplifying the fraction: \[ \beta_2 = 0.048 \, \text{cm} \cdot \frac{5}{6} \] Now, calculate \( \beta_2 \): \[ \beta_2 = 0.048 \cdot 0.8333 \approx 0.04 \, \text{cm} \] 5. **Final Answer**: Therefore, the fringe width at the same distance for \( \lambda = 5000 \, \text{Å} \) is: \[ \beta_2 \approx 0.04 \, \text{cm} \]

To solve the problem, we need to find the fringe width for a wavelength of \( \lambda = 5000 \, \text{Å} \) given that the fringe width for \( \lambda = 6000 \, \text{Å} \) is \( 0.048 \, \text{cm} \) at a distance of \( 50 \, \text{cm} \) from the slits. ### Step-by-Step Solution: 1. **Understand the relationship between fringe width and wavelength**: The fringe width \( \beta \) in Young's double-slit experiment is given by the formula: \[ \beta = \frac{\lambda D}{d} ...
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