Question

In a Young's double slit experiment distance between the slits is 1 mm. The fringe width is found to be 0.6 mm. When the screen is moved through a distance of 0.25 m away from the plane of the slit, the fringe width becomes 0.75 mm. Find the wavelength of light used.

Answer 1

Distance between the slits $d=1mm={10}^{-3}m$

Change in position of screen $\Delta D=0.25m$

Change in fringe width $\Delta \beta =(0.75-0.6)mm$

$\therefore$ $\Delta \beta =0.15mm=0.15\times {10}^{-3}m$

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

$0.15\times {10}^{-3}=\frac{\lambda }{{10}^{-3}}\times 0.25$

$⟹$ $\lambda =600nm$