Question

In a two slit experiment with monochromatic light fringes are obtained on a screen placed a some distance from the plane of slits . If the screen is moves by 53×10^-3 m. If the distance between slits is 10^-3 m, the wavelength of light will be :

• A. 3000Å
• B. 4000Å
• C. 6000Å
• D. 7000Å

Answer 1

Answer: (C)

6000Å

Answer 2

: Fringe width $\beta = \frac{\lambda D}{d}$

Where $\lambda$is the wavelength of light, D is the distance between screen and the slits and d is the distance between two slits

$\therefore\Delta\beta = \frac{\lambda}{d}\Delta D.$ (As $\lambda$and d are constant)

Or $\lambda = \frac{\Delta\beta d}{\Delta D}$

Substituting the given values , we get

$\lambda = \frac{(3 \times 10^{- 5}m)(10^{- 3}m)}{(5 \times 10^{- 2}m)} = 6 \times 10^{- 7}m = 6000Å$