Young experiment

In the Young slit experiment, the slits are spaced by distance of \(a= 0.2~\mbox{mm}\) and the screen is placed at a distance of \(d_0 = 1.5~\mbox{m}\).The distance between the fifth minima situated on both sides of the fringe of the order of \(0\) is equal to \(34.73~\mbox{mm}\).

Question

Determine the wavelength of the used source.

Solution

The bright maxima corresponding to the orders of \(k = \{ -4,~-3,~-2,~-1,~0,~+1,~+2,~+2,~+3,~+4 \}\) are located between the fifth minima situated on both sides of the fringe to the order of \(0\). We can count 9 fringes on the screen.

By assuming that \(\Delta x = 34.73~\mbox{mm}\), we can write the relation:

Δ x = 9 i = 9 λ d 0 a %DELTA x = 9 i = 9 {%lambda d_0}over{a}

from which:

λ = 0.2 × 34.73 9 × 1500 = 514 × 10 − 4 mm = 514 nm %lambda = {0.2 times 34.73} over {9 times 1500} = 514 times 10^{-4} "mm" = 514 "nm"