Question

In Young's double-slit experiment interference bands are produced on the screen placed at 1.5 m from two slits 0.15 mm apart and illuminated by the light of wavelength $6000 \, Å$ . If the screen is now taken away from the slits by 50 cm then find the change in fringe width (in mm)?

Answer 1

$D_{2}-D_{1}=50 \, cm$ ; $D_{1}=1.5 \, m=150 \, cm$
$\lambda =6000 \, Å=6000\times 10^{- 8} \, cm$ ; $d=0.15 \, mm$
Change in fringe width $\Delta \beta=\left(\mathrm{D}_2-\mathrm{D}_1\right) \frac{\lambda}{\mathrm{d}}=\frac{50 \times 6000 \times 10^{-8}}{0.15 \times 10^{-1}}$
$\Delta \beta =\frac{20 \times 10^{- 4}}{10^{- 3}}=2\times 10^{- 1} \, cm=2mm$
$\Delta \beta =2 \, mm$