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: 1
Solution:
$\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}}$
$\beta=\frac{5}{2} \times 10^{-4} \mathrm{~m}=2.5 \times 10^{-1} \mathrm{~mm}$
$\mathrm{b}=0.25 \mathrm{~mm}$