Question:
In a Young's double slit experiment two slits are separated by $2 \mathrm{~mm}$ and the screen is placed one meter away. When a light of wavelength $500 \mathrm{~nm}$ is used, the fringe separation will be :
Correct Option: , 4
Solution:
(4)
Fringe width $(\beta)=\frac{\lambda D}{d}$
$d=2 \times 10^{-3} \mathrm{~m}$
$\lambda=500 \times 10^{-9} \mathrm{~m}$
$\mathrm{D}=1 \mathrm{~m}$
Now
$\beta=\frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}}$
$\beta=\frac{5}{2} \times 10^{-4}$
$\beta=2.5 \times 10^{-4}$
$\beta=0.25 \mathrm{~mm}$