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 between the maxima is given by the formula: \[ y_m = \frac{m \lambda D}{d} \] Where:
- \( y_m \) is the distance between the \( m^{\text{th}} \) maxima and the central maximum,
- \( m \) is the order of the maxima (for first maxima, \( m = 1 \)),
- \( \lambda \) is the wavelength of the light,
- \( D \) is the distance between the slits and the screen,
- \( d \) is the separation between the slits. For the 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 between the first maxima for the two wavelengths: 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 distance between the first maxima for the two wavelengths is: \[ \Delta y = y_2
- y_1 = 0.6 \, \text{mm}
- 0.5 \, \text{mm} = 0.1 \, \text{mm} \] Thus, the distance between the first maxima for the two wavelengths is \( 0.1 \, \text{mm} \). (b) The least distance from the central maximum, where the bright fringes due to both the wavelengths coincide.

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

For the bright fringes of two different wavelengths to coincide, the path difference must be an integer multiple of both wavelengths. This happens when the path difference is equal to the least common multiple (LCM) of the two wavelengths.

The condition for constructive interference is given by:

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

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 find the position \( y \) on the screen where both bright fringes coincide, we use the fringe position formula:

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

Substitute 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 where the bright fringes due to both wavelengths coincide is 3.0 mm.