Question

In Young's double slit experiment, the screen is moved 30 cm towards the slits. As a consequence, the fringe width of the pattern changes by 0.09 mm. If the slits separation used is 2 mm, calculate the wavelength of light used in the experiment.

Hint:

The fringe width in Young's double slit experiment is directly proportional to the wavelength and the distance from the slits to the screen. The fringe width decreases when the screen is moved closer to the slits.

Answer 1

In Young's double slit experiment, the fringe width \( \beta \) is given by the formula: \[ \beta = \frac{\lambda D}{d} \] where: - \( \lambda \) is the wavelength of light, - \( D \) is the distance between the slits and the screen, - \( d \) is the separation between the slits. Let the initial distance between the screen and the slits be \( D_1 \), and the new distance be \( D_2 = D_1 - 30 \, \text{cm} \). The fringe width changes by 0.09 mm, which is the difference in fringe width for the initial and new screen distances. Thus, the initial fringe width is: \[ \beta_1 = \frac{\lambda D_1}{d} \] and the new fringe width is: \[ \beta_2 = \frac{\lambda D_2}{d} \] The change in fringe width is: \[ \Delta \beta = \beta_2 - \beta_1 = 0.09 \, \text{mm} = 9 \times 10^{-5} \, \text{m} \] From the difference in fringe widths: \[ \Delta \beta = \frac{\lambda (D_2 - D_1)}{d} \] Substitute the given values: \[ 9 \times 10^{-5} = \frac{\lambda (30)}{2 \times 10^{-3}} \] Solving for \( \lambda \): \[ \lambda = \frac{9 \times 10^{-5} \times 2 \times 10^{-3}}{30} = 6 \times 10^{-7} \, \text{m} \] Thus, the wavelength of light used in the experiment is: \[ \boxed{6 \times 10^{-7} \, \text{m} = 600 \, \text{nm}} \]