Question

Light of wavelength 478 nm in air falls on two slits 6.53 mm apart. The slits are immersed in water, as is a viewing screen 30.8 cm away. How far apart are the fringes on the screen

Answer 1

The distance between the two fringes is 0.0225 mm.

Explanation:

As per Young's double slit experiment, the distance between the fringes can be observed for the bright constructive interference fringes. So if we know the distance between the separation of slits. the distance of the mirror placed away from the slit, then the distance between two fringes can be observed using the below formula.

$y=\frac{m\times wavelength\times D}{d}$

Here y is the distance between the fringes on screen, D is the distance of the screen from the slits and d is the distance between the slits. Also m is the order of the interference which is generally considered as 1.

So, in the present case, D = 30.8 cm = 0.308 m, d = 6.53×10⁻³ m, wavelength = 478 nm and m = 1.

So, $y = \frac{1\times 478\times 10^{-9} \times 0.308}{6.53\times 10^{-3} }$

$y =22.546\times 10^{-9+3}=22.546\times 10^{-6} m=0.0225\ mm.$

So, the distance between the two fringes is 0.0225 mm.