Young experiment with white light

A source of white light \((400-700~\mbox{nm})\) illuminates Young slits spaced \(1.25~\mbox{mm}\) away. An interference pattern can be observed on a screen that is placed \(1.5~\mbox{m}\) away. If the source slit of a high-resolution spectroscope is placed at the level of the screen and \(3 \mbox{mm}\) away from the central white fringe (and parallel to the interference fringes), we can observe dark grooves.

Question

At point \(\mbox{M}\) can we observe interferences ? Explain.

Solution

At point \(\mbox{M}\) the optical path difference \(\delta\) is :

δ = a x d 0 = 1.25 × 3 1500 = 2.5 μ m %delta = {a x} over {d_0} = {1.25 times 3 } over {1500} = 2.5 %mu "m"

The coherence length relative to the white source used (uniform source model) has a magnitude of \((\lambda_0 = 550~\mbox{nm} \)and \(\delta \lambda = 300~\mbox{nm})\):

l c = λ 2 Δ λ = ( 550 × 10 − 9 ) 2 300 × 10 − 9 ≈ 1 μ m l_c = {%lambda^2}over{%DELTA %lambda} = {(550 times 10^{-9})^2} over {300 times 10^{-9}} approx 1 %mu "m"

The optical path difference is greater than the coherence length; consequently, the interference phenomenon can no longer be observed. Point \(\mbox{M}\) is located within the higher order white. By observing with a spectroscope, we can see that some wavelengths are missing.

Question

To what wavelengths do the grooves correspond?

Solution

The dark lines in the spectrum correspond to destructive interferences. Therefore, the orders of interference are verifying: \(\delta = (2k + 1) \lambda /2\) .

They are included between the two extreme orders , \(k_B\) and \(k_R\) correspond to the two extremes of the spectrum. For the wavelength interval between \(\lambda_B = 400~\mbox{nm}\) and \(\lambda_R = 700~\mbox{nm}\), the extreme values of the orders of interference are:

k B = a x λ B d 0 − 1 2 = 1.25 × 3 0.4 × 10 − 3 × 1500 − 1 2 = 2.5 0.4 − 1 2 = 5.75 k_B = {a x} over {%lambda_B d_0} -{1 over 2}= {1.25 times 3 } over {0.4 times 10^{-3} times 1500} -{1 over 2}= 2.5 over 0.4 -{1 over 2}= 5.75

and

k R = a x λ R d 0 − 1 2 = 1.25 × 3 0.7 × 10 − 3 × 1500 − 1 2 = 2.5 0.7 − 1 2 = 3.07 k_R = {a x} over {%lambda_R d_0} -{1 over 2}= {1.25 times 3 } over {0.7 times 10^{-3} times 1500} -{1 over 2}= 2.5 over 0.7 -{1 over 2}= 3.07

Since the orders of interference are not integers, there are only two dark lines in the spectrum that correspond to the orders \(k_1 = 4\) and \(k_2 = 5\), and therefore to the wavelengths:

λ 1 = 2 δ 2 k 1 + 1 = 2 × 2.5 9 = 0.5556 μ m = 555.6 nm %lambda_1 = {2 %delta} over {2 k_1 +1} = {2 times 2.5} over 9 = 0.5556 {%mu} "m" = 555.6 "nm"

and

λ 2 = 2 δ 2 k 2 + 1 = 2 × 2.5 11 = 0.4545 μ m = 454.5 nm %lambda_2 = {2 %delta} over {2 k_2 +1} = {2 times 2.5 } over 11 = 0.4545 %mu"m" = 454.5 "nm"