Temporal coherence and Young slits ; displacement and interference fringes 

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A detailed solution is then proposed to you.

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Consider the interference device of Young slits, with observation in the focal plane of a convergent thin lens \((L)\).

We give :

\(F_1 F_2 = a = 1\;mm\;;\;\lambda _0 = 600\;nm\;;\;f = 50\;cm\)

Question

Describing the figure of interference observed and the intensity distribution \(I(x)\).

Solution

The result is classical (see the review of the lecture). The particular interfringe is :

\(i = \frac{{{\lambda _0}f}}{a}\)

The intensity is given by the Fresnel's formula :

\(I(x) = 2{I_0}(1 + \cos (\frac{{2\pi }}{{{\lambda _0}}}\frac{{ax}}{f}))\)

Question

We interpose in the path of one of the beams the blade of parallel faces of thickness of \(e\) and of index \(n\).

The faces are perpendicular to the axis of symmetry.

Determine the number \(N\) fringes which defiled in \(F_i\).

We take :

\(n_{air}=1\;\;  ;\;\;n=1,5\;\; ;\;\;e=0,5\;mm\)

Solution

The new path difference is :

\(\delta = \frac{{ax}}{f} - (n - 1)e\)

The position of the central fringe becomes (it is obtained for a path difference of zero) :

\({x_0} = (n - 1)\frac{{ef}}{a}\)

The fringe has not changed and the number of fringes that defiled :

\(N = \frac{{{x_0}}}{i} = (n - 1)\frac{e}{{{\lambda _0}}} = 416\)

Question

The blade is removed with parallel faces.

The slots are now illuminated by the yellow doublet of sodium formed by two radiation assumed monochromatic and same intensity, wavelength :

\(\lambda _1 = 589,0\;nm\) and \(\lambda _2 = 589,6\;nm\)

How far from the central fringe do the fringes disappear for the \(1st\) time ?

Solution

Qualitatively, the first interference appears when a dark fringe corresponding to \(\lambda_1\) is superimposed on a bright fringe associated with \(\lambda_2\) or for a verified interference of order \(p\) :

\(x = (p + \frac{1}{2}){i_1} = p{i_2}\)

Or :

\(i_1 = \frac{{{\lambda _1}f}}{a}\) and \(i_2 = \frac{{{\lambda _2}f}}{a}\)

We can deduce :

\(p = \frac{{{i_1}}}{{2({i_2} - {i_1})}} = \frac{{{\lambda _1}}}{{2({\lambda _2} - {\lambda _1})}}\)

Then the value of the abscissa where the interference occurs :

\(x = \frac{{{\lambda _1}}}{{2({\lambda _2} - {\lambda _1})}}{i_2} = \frac{{{\lambda _1}{\lambda _2}}}{{2({\lambda _2} - {\lambda _1})}}\frac{f}{a} = \frac{{\lambda _0^2}}{{2\Delta \lambda }}\frac{f}{a} = 14,5\;cm\)

Where were asked :

\({\lambda _0} = \frac{{{\lambda _1} + {\lambda _2}}}{2}\)

The average value of the two wavelengths.