Question

In Young’s double slit experiment, light of wavelength $6000\,\mathring{A} $ is used to get an interference pattern on a screen. The fringe width changes by 1.5 mm, when the screen is brought towards the double slit by 50 cm. The distance between the two slits is

• A. 0.2 mm
• B. 0.4 mm
• C. 0.6 mm
• D. 0.8 mm

Answer 1

Answer: (A)

Fringe width, $\beta=\frac{D\lambda}{d}$
As $\lambda$ and $d$ are constant, so the change in fringe width is given by
$\Delta\beta=\frac{\lambda\,\Delta D}{d}$
$\therefore $Distance between the two slits,
$d=\frac{\lambda \,\Delta D}{\Delta\beta}=\frac{6000\times10^{-10}\times50\times10^{-2}}{1.5\times10^{-3}}$
$=0.2\times10^{-3}\,m=0.2\,mm.$