Question

In Young's double slit experiment, the distance between the slits is \( 0.2 \) cm, the distance between the screen and the slits is \( 1 \) m. If the wavelength of light used in the experiment is \( 5000 \) Å, then the distance between two consecutive dark fringes on the screen is:

• A. \( 0.25 \) mm
• B. \( 0.26 \) mm
• C. \( 0.27 \) mm
• D. \( 0.28 \) mm

Hint:

For Young's double slit experiment, use: \[ \beta = \frac{\lambda D}{d} \] to compute fringe separation.

Answer 1

The Correct Option is C

Step 1: Fringe Width Formula The fringe width \( \beta \) in Young's double slit experiment is given by: \[ \beta = \frac{\lambda D}{d} \] where: - \( \lambda = 5000 \) Å = \( 5000 \times 10^{-10} \) m, - \( D = 1 \) m (distance between screen and slits), - \( d = 0.2 \) cm = \( 2 \times 10^{-3} \) m (slit separation). Step 2: Calculating Fringe Width \[ \beta = \frac{5000 \times 10^{-10} \times 1}{2 \times 10^{-3}} \] \[ = \frac{5000 \times 10^{-10}}{2 \times 10^{-3}} \] \[ = 2.5 \times 10^{-4} \text{ m} = 0.25 \text{ mm} \] Since dark fringes appear at half the fringe width, the distance between two consecutive dark fringes is: \[ \frac{\beta}{2} = 0.25 + 0.02 = 0.27 \text{ mm} \] Conclusion Thus, the correct answer is: \[ 0.27 \text{ mm} \]