Question

In a two-slit setup, each slit is 0.020 mm wide. These apertures are illuminated by plane waves of yellow sodium light (?=589.6 nm). The resulting Fraunhofer fringe pattern consists of 11 narrow bright fringes that gradually decrease in irradiance with distance from the central maximum. Determine the separation between the slits.

Answer 1

Answer:

so separation between slit is 0.12 mm

Explanation:

given data

slit wide = 0.020 mm

wavelength λ = 589.6 nm

fringe pattern n = 11

to find out

separation between the slits

solution

we know that for maxima we can say

d sinθ = m × λ   .........1

and m is mth order maxima

and for central maxima

number of bright fringe is here

N = 2m -1

11 = 2m - 1

m = 6

so here number of maxima on either side of central maxima will be m-1

that is = 5

because we know 6th maxima will be minimized by minima

so

condition for minima is

a sin(θ) = n × λ

now for 1st minima

a sin(θ)  =  λ

and

$\frac{dsin\Theta}{asin\Theta} = \frac{m\lambda}{\lambda}$

$\frac{d}{a}$ = m

d = ma

d = 6 × 0.02

d = 0.12 mm

so separation between slit is 0.12 mm