In a Young's double slit experiment, a light of wavelength 550 nm is used to get fringes on screen 1.2 m away from the slits. The distance between the two slits is 1.5 mm. The fringe width is

0.44 mm

The correct answer is Option (2) → 0.44 mm

Given:

Wavelength, $\lambda = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m}$

Distance between slits, $d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$

Distance between slits and screen, $D = 1.2 \, \text{m}$

Fringe width formula:

$\beta = \frac{\lambda D}{d}$

Substitute values:

$\beta = \frac{550 \times 10^{-9} \times 1.2}{1.5 \times 10^{-3}}$

$\beta = \frac{660 \times 10^{-9}}{1.5 \times 10^{-3}}$

$\beta = 440 \times 10^{-6} \, \text{m}$

$\beta = 0.44 \, \text{mm}$

Final Answer:

$\beta = 0.44 \, \text{mm}$