Question

The wavelength of the monochromatic light used in Young's double slit experiment is 550 nm and the screen is placed at a distance of 120 cm from the plane of the slits. If third dark fringe is formed on the screen at a distance of 1.5 mm from the central bright fringe, then the distance of separation between the two slits is

• A. 5.5 mm
• B. 1.1 mm
• C. 2.2 mm
• D. 3.3 mm

Hint:

- Use correct fringe formula for dark/bright fringes: $y = m \lambda L/d$ (bright), $y = (m+1/2)\lambda L/d$ (dark). - Convert all lengths to meters before calculation. - Always verify m-th order for the given fringe.

Answer 1

The Correct Option is B

1. Dark fringe formula: $y = (m + \frac{1}{2}) \frac{\lambda L}{d}$, $m = 2$ for third dark fringe (m = 0,1,2)
2. $1.5 \times 10^{-3} = (2 + 0.5) \cdot \frac{550 \times 10^{-9} \cdot 1.2}{d}$
3. $d = \frac{2.5 \cdot 550 \cdot 10^{-9} \cdot 1.2}{1.5 \times 10^{-3}} \approx 1.1 \times 10^{-3}~\text{m} = 1.1~\text{mm}$