Question

In Young's double slit experiment, two slits $4\text{mm}$ apart, are illuminated by a light of wavelength $600\text{nm}$. What will be the fringe width on a screen placed $2\text{m}$ from the slits?

• A. $0.3\text{mm}$
• B. $3\text{mm}$
• C. $4\text{mm}$
• D. $0.12\text{mm}$

Answer 1

The correct option is A $0.3\text{mm}$

The fringe width is given by,

$\beta =\frac{\lambda D}{d}=\frac{6\times {10}^{-7}\times 2}{4\times {10}^{-3}}$

$=3\times {10}^{-4}\text{m}$
$=0.3\text{mm}$