The width of fringe is 2 mm on the screen in a double slits experiment for the light of wavelength of 400 nm. The width of the fringe for the light of wavelength 600 nm will be:
- 2mm
- 3mm
- 4mm
- 1.33mm
The Correct Option is B
Solution and Explanation
The fringe width (\( \beta \)) in a double-slit experiment is given by: \[ \beta = \frac{\lambda D}{d}, \] where: - \( \lambda \) is the wavelength of the light, - \( D \) is the distance between the slits and the screen, - \( d \) is the separation between the slits.
Step 1: Relation between fringe widths. Since \( D \) and \( d \) are constant, the fringe width \( \beta \) is directly proportional to \( \lambda \): \[ \beta \propto \lambda. \]
Step 2: Calculate the new fringe width. Let the initial fringe width (\( \beta_1 \)) correspond to \( \lambda_1 = 400 \, \text{nm} \) and the new fringe width (\( \beta_2 \)) correspond to \( \lambda_2 = 600 \, \text{nm} \). Then: \[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}. \] Substituting \( \beta_1 = 2 \, \text{mm} \), \( \lambda_1 = 400 \, \text{nm} \), and \( \lambda_2 = 600 \, \text{nm} \): \[ \frac{\beta_2}{2} = \frac{600}{400} \implies \beta_2 = 2 \cdot 1.5 = 3 \, \text{mm}. \]
Final Answer: The fringe width for \( \lambda = 600 \, \text{nm} \) is: \[ \boxed{3 \, \text{mm}}. \]
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