Question

In Young's double slit experiment, fringes of certain width are produced on the screen kept at a certain distance from the slits. When the screen is moved away from the slits by $0.1$m, fringe width increase by $6\times {10}^{-5}m$. The separation between the slits is 1 mm. Calculate the wavelength of the light used.

Answer 1

Fringe width of the fringes formed $\beta =\frac{\lambda D}{d}$

Thus change in fringe width $\Delta \beta =\frac{\lambda \Delta D}{d}$

Change in position of screen $\Delta D=0.1m$

Slit width $d=1mm=0.001m$

Increase in fringe width $\beta =6\times {10}^{-5}m$

$\therefore$ $6\times {10}^{-5}=\frac{\lambda \times 0.1}{0.001}$

$⟹$ $\lambda =600nm$