Question

Find the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.

Hint:

For fringe coincidence in double slit interference, the condition is that the fringe widths for both wavelengths must be the same.

Answer 1

Least Distance from the Central Maximum Where Bright Fringes Coincide

Given:

• Two wavelengths: \( \lambda_1 \) and \( \lambda_2 \)
• Distance from slits to screen: \( D \)
• Separation between slits: \( d \)

Formula for Bright Fringe Position:

The position of the \( m \)-th bright fringe for wavelength \( \lambda \) is given by: \[ y_m = \frac{m \lambda D}{d} \]

Condition for Coincidence of Bright Fringes:

For the bright fringes due to both wavelengths to coincide, their positions must be the same: \[ \frac{m \lambda_1 D}{d} = \frac{n \lambda_2 D}{d} \] Simplifying: \[ m \lambda_1 = n \lambda_2 \] \[ \frac{m}{n} = \frac{\lambda_2}{\lambda_1} \]

Solution Example:

• Let \( \lambda_1 = 600 \, \text{nm} \) (for light 1)
• Let \( \lambda_2 = 450 \, \text{nm} \) (for light 2)
• Let \( D = 1 \, \text{m} \)
• Let \( d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \)

Step 1: Calculate the Ratio of \( m \) and \( n \):

\[ \frac{m}{n} = \frac{\lambda_2}{\lambda_1} = \frac{450 \times 10^{-9}}{600 \times 10^{-9}} = \frac{3}{4} \]

Thus, the smallest values of \( m = 3 \) and \( n = 4 \).

Step 2: Calculate the Coincidence Distance:

The least distance \( y \) where the fringes coincide is: \[ y = \frac{m \lambda_1 D}{d} = \frac{3 \times 600 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 3.6 \, \text{mm} \]

Conclusion:

The least distance from the central maximum where the bright fringes due to both wavelengths coincide is **3.6 mm**.