Question

In Young's double slit experiment, the separation between the two slits is 1.0 mm and the screen is 1.0 m away from the slits. A beam of light consisting of two wavelengths, 500 nm and 600 nm, is used to obtain interference fringes. Calculate:
(a) The distance between the first maxima for the two wavelengths.
(b) The least distance from the central maximum, where the bright fringes due to both the wavelengths coincide.

Hint:

In Young's double slit experiment, the separation between maxima for different wavelengths is directly proportional to the wavelength. The least distance where the bright fringes coincide is obtained when the path difference is a multiple of both wavelengths.

Answer 1

In Young's double slit experiment, the distance from the central maximum to the \( m^{\text{th}} \) maxima is calculated using the formula: \[y_m = \frac{m \lambda D}{d}\] where \( y_m \) denotes this distance, \( m \) represents the order of the maxima (e.g., \( m=1 \) for the first maxima), \( \lambda \) is the wavelength of the light, \( D \) is the distance from the slits to the screen, and \( d \) is the separation between the slits. For two wavelengths:
- \( \lambda_1 = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- \( \lambda_2 = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- \( D = 1.0 \, \text{m} \)
- \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \) The distance of the first maxima from the central maximum for each wavelength is: For \( \lambda_1 = 500 \, \text{nm} \): \[y_1 = \frac{1 \times 500 \times 10^{-9} \times 1.0}{1.0 \times 10^{-3}} = 5.0 \times 10^{-4} \, \text{m} = 0.5 \, \text{mm}\] For \( \lambda_2 = 600 \, \text{nm} \): \[y_2 = \frac{1 \times 600 \times 10^{-9} \times 1.0}{1.0 \times 10^{-3}} = 6.0 \times 10^{-4} \, \text{m} = 0.6 \, \text{mm}\] The difference in distance between the first maxima for the two wavelengths is: \[\Delta y = y_2 - y_1 = 0.6 \, \text{mm} - 0.5 \, \text{mm} = 0.1 \, \text{mm}\] Therefore, the distance between the first maxima of the two wavelengths is \( 0.1 \, \text{mm} \). (b) The least distance from the central maximum where the bright fringes due to both wavelengths coincide.

Condition for Coincidence of Bright Fringes in Young’s Double Slit Experiment

Bright fringes from two different wavelengths coincide when their path differences correspond to an integer multiple of both wavelengths. This occurs when the path difference is equal to the least common multiple (LCM) of the two wavelengths.

The condition for constructive interference is:

\[ \Delta y = m \lambda_1 = n \lambda_2 \]

where \( m \) and \( n \) are integers.

Given:
Wavelengths: \( \lambda_1 = 500 \, \text{nm}, \lambda_2 = 600 \, \text{nm} \)
LCM of \( 500 \, \text{nm} \) and \( 600 \, \text{nm} \) is \( 3000 \, \text{nm} = 3.0 \times 10^{-6} \, \text{m} \)

To determine the position \( y \) on the screen where both bright fringes coincide, the fringe position formula is applied:

\[ y = \frac{m \lambda D}{d} \]

Substituting values:
\( \lambda = 3.0 \times 10^{-6} \, \text{m} \)
\( D = 1.0 \, \text{m} \)
\( d = 1.0 \times 10^{-3} \, \text{m} \)

\[ y = \frac{1 \cdot 3.0 \times 10^{-6} \cdot 1.0}{1.0 \times 10^{-3}} = 3.0 \times 10^{-3} \, \text{m} = 3.0 \, \text{mm} \]

Final Answer:
The least distance from the central maximum at which the bright fringes of both wavelengths coincide is 3.0 mm.