In Young’s double-slit experiment,

Solution:

Distance between the slits and the screen, D = 1.4 m

and the distance between the slits, d = 0.28 mm = 0.28 x 10-3 m

Distance between the central fringe and the fourth (n = 4) fringe,

u = 1.2cm = 1.2 × 10-2 m

For constructive interference, the following is the relation for the distance between the two fringes:

$u=n \lambda \frac{D}{d}$

Where, n = order of fringes

$=4 \lambda=$ Wavelength of light used

Rearranging the formula, we get

$\lambda=\frac{u d}{n D}$

$=\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}$

= 6 × 10-7 m = 600nm

 

600nm is the wavelength of the light.