Question

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?

• A. 10
• B. 12
• C. 15
• D. 18

Hint:

In Young's double slit experiment, fringe separation is inversely proportional to the wavelength of light.

Answer 1

The Correct Option is B

In Young's double slit experiment, the fringe separation \( \Delta y \) is inversely proportional to the wavelength \( \lambda \). Since the fringe separation is given by: \[ \Delta y = \frac{\lambda D}{d} \] Where: - \( \lambda \) is the wavelength of light, - \( D \) is the distance from the slits to the screen, - \( d \) is the distance between the slits. The number of fringes per cm is the reciprocal of the fringe separation, so: \[ \text{Number of fringes per cm} = \frac{1}{\Delta y} \] For two different wavelengths \( \lambda_1 \) and \( \lambda_2 \), we have the relationship: \[ \frac{\text{Number of fringes per cm with } \lambda_2}{\text{Number of fringes per cm with } \lambda_1} = \frac{\lambda_1}{\lambda_2} \] Given that \( \lambda_1 = 5600 \, \text{\AA} \) and \( \lambda_2 = 7000 \, \text{\AA} \), the number of fringes with \( \lambda_2 \) is: \[ \frac{15}{\frac{7000}{5600}} = 12 \] Thus, the number of fringes per cm with \( \lambda_2 = 7000 \, \text{\AA} \) is 12.