Question

In a Young's double slit experiment two slits are separated by $2\text{mm}$ and the screen is placed one meter away. When a light of wavelength $500\text{nm}$ is used, the fringe separation will be:

• A. $0.75\text{mm}$
• B. $0.50\text{mm}$
• C. $1\text{mm}$
• D. $0.25\text{mm}$

Answer 1

The correct option is D $0.25\text{mm}$

Fringe width $\left(\beta \right)=\frac{\lambda D}{d}$
$d$ $=$ $2\times {10}^{-3}$ $\text{m}$

$\lambda =500\times {10}^{-9}$ $\text{m}$

$D=1\text{m}$

Now,

$\beta =\frac{500\times {10}^{-9}\times 1}{2\times {10}^{-3}}$

$\beta =\frac{5}{2}\times {10}^{-4}$

$\beta =2.5\times {10}^{-4}$

$\beta$ = $0.25\text{mm}$