Question

In YDSE, when light of wavelength 700nm is used, a fringe width of 0.5mm is obtained. What happens when light of wavelength 500nm is used?

Hint:

Remember: The fringe width in YDSE is directly proportional to the wavelength of the light used.

Answer 1

Step 1: Use the formula for fringe width 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 the light, \( D \) is the distance between the screen and the slits, and \( d \) is the distance between the slits.
Step 2: Calculate the new fringe width.
Let \( \beta_1 = 0.5 \, \text{mm} \) be the fringe width when \( \lambda_1 = 700 \, \text{nm} \) is used. When the wavelength is changed to \( \lambda_2 = 500 \, \text{nm} \), the new fringe width \( \beta_2 \) is related to the initial fringe width by: \[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \] Substitute the known values: \[ \frac{\beta_2}{0.5} = \frac{500}{700} \quad \Rightarrow \quad \beta_2 = 0.5 \times \frac{500}{700} = 0.357 \, \text{mm} \] Thus, when light of wavelength 500nm is used, the fringe width becomes: \[ \boxed{0.357 \, \text{mm}} \]