Question

What are coherent sources? In a Young’s double slit experiment, the distance between two coherent sources is 2 mm, and the distance of the screen is 1.5 m. If monochromatic light of wavelength 6000 \(\text{Å}\) is used, then find the fringe width and the distance of the third dark fringe from the centre.

Hint:

The fringe width is directly proportional to the wavelength of light and the distance to the screen, and inversely proportional to the distance between the slits.

Answer 1

Step 1: Formula for Fringe Width.
In Young’s double slit experiment, the fringe width (\( \beta \)) is given by: \[ \beta = \frac{\lambda D}{d} \] Where:
- \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \) is the wavelength of light,
- \( D = 1.5 \, \text{m} \) is the distance of the screen from the slits,
- \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \) is the distance between the two slits.
Substitute the values into the formula: \[ \beta = \frac{6000 \times 10^{-10} \times 1.5}{2 \times 10^{-3}} = \frac{9000 \times 10^{-10}}{2 \times 10^{-3}} = 4.5 \times 10^{-4} \, \text{m} = 0.45 \, \text{mm} \]
Step 2: Distance of the Third Dark Fringe.
The position of the \( n \)-th dark fringe is given by: \[ y_n = (n - \frac{1}{2}) \beta \] For the third dark fringe (\( n = 3 \)): \[ y_3 = \left( 3 - \frac{1}{2} \right) \times 0.45 = 5 \times 0.45 = 1.125 \, \text{mm} \]
Final Answer:
The fringe width is \( \boxed{0.45 \, \text{mm}} \), and the distance of the third dark fringe from the centre is \( \boxed{1.125 \, \text{mm}} \).