Question

In young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths 1 = 12000 å and 2 = 10000 å. at what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Answer 1

Answer:

The minimum distance will be 6 mm

Explanation:

According to problem the wavelengths of the photons

let a1= 12000 å and a2 = 10000 å

Therefore,

p1a1 = p2a2

where p1, p2 are the number of  fringe position.

Therefore,

p1 / p2 = a2 / a1

=> p1 / p2 = 10000 å  / 12000 å

=>p1/p2 = 5/6

let the minimum distance is s,

s = p1a1R /r

we have calculated that n1 = 5

r = 2mm = 2 x 10^(-3) m

R= 2m

s = [ 5 x 12000 x 10^(-10) x 2 ] / 2 x 10^(-3)

s= 6 mm