Question:medium

In Young's double slit experiment, we get 15 fringes per cm on the screen, using light of wavelength 5600 \AA. For the same setting, how many fringes per cm will be obtained with light of wavelength 7000 \AA?

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In Young's double slit experiment, fringe separation is inversely proportional to the wavelength of light.
Updated On: Mar 28, 2026
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The Correct Option is B

Solution and Explanation

In Young's double-slit experiment, fringe separation \( \Delta y \) is inversely proportional to wavelength \( \lambda \). The formula for fringe separation is: \[ \Delta y = \frac{\lambda D}{d} \] Where: - \( \lambda \) represents the wavelength of light. - \( D \) is the distance from the slits to the screen. - \( d \) is the separation between the slits. The number of fringes per centimeter is the inverse of the fringe separation: \[ \text{Number of fringes per cm} = \frac{1}{\Delta y} \] For two distinct wavelengths, \( \lambda_1 \) and \( \lambda_2 \), the relationship is: \[ \frac{\text{Number of fringes per cm with } \lambda_2}{\text{Number of fringes per cm with } \lambda_1} = \frac{\lambda_1}{\lambda_2} \] Given \( \lambda_1 = 5600 \, \text{\AA} \) and \( \lambda_2 = 7000 \, \text{\AA} \), the number of fringes with \( \lambda_2 \) is calculated as: \[ \frac{15}{\frac{7000}{5600}} = 12 \] Therefore, there are 12 fringes per cm when the wavelength is \( \lambda_2 = 7000 \, \text{\AA} \).
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