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 calculated using the formula: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) is the wavelength of light, \( D \) is the distance between the slits and the screen, and \( d \) is the separation between the slits. Given an initial screen distance \( D_1 \), the new distance is \( D_2 = D_1 - 30 \, \text{cm} \). The fringe width decreases by 0.09 mm, representing the difference between the initial and new fringe widths. The initial fringe width is: \[ \beta_1 = \frac{\lambda D_1}{d} \] The new fringe width is: \[ \beta_2 = \frac{\lambda D_2}{d} \] The change in fringe width is: \[ \Delta \beta = \beta_1 - \beta_2 = 0.09 \, \text{mm} = 9 \times 10^{-5} \, \text{m} \] (Note: The problem statement implies \( \beta_1 - \beta_2 \) as the fringe width decreases. If it was \( \beta_2 - \beta_1 \), the difference would be negative, but the magnitude is given as positive.) From the difference in fringe widths: \[ \Delta \beta = \frac{\lambda D_1}{d} - \frac{\lambda D_2}{d} = \frac{\lambda (D_1 - D_2)}{d} \] Given \( D_1 - D_2 = 30 \, \text{cm} = 0.3 \, \text{m} \) and \( d = 2 \times 10^{-3} \, \text{m} \): \[ 9 \times 10^{-5} = \frac{\lambda (0.3)}{2 \times 10^{-3}} \] Solving for \( \lambda \): \[ \lambda = \frac{9 \times 10^{-5} \times 2 \times 10^{-3}}{0.3} = 6 \times 10^{-7} \, \text{m} \] The wavelength of light used is: \[ \boxed{6 \times 10^{-7} \, \text{m} = 600 \, \text{nm}} \]