Malus' theorem, superposition of a plane wave and of a spherical wave
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A converging lens is used \((L)\), pierced at its center, as two-wave interference system.
A point source \(S\), monochromatic wavelength \(\lambda_0\), is placed at the focal object \(F\) of the lens.
It follows that the wave emerging from the lens is plane and that directly transmitted through the hole is spherical.
The hole has a diameter \(2R = 10 \;mm\) on the output face \((L)\) and a depth \(e = 3 \;mm\) along the axis.
Question
Give the analytical expressions of waves that overlap ; be adopted as the origin of phase waves in \(S\) and it is assumed that these two waves have the same amplitude.
Solution
The source is the focus of the lens : it emits a spherical wave.
The lens transforms into plane wave.
The optical path \(\delta_1\) to go from \(S\) to \(M\) through the lens is equal to that to go from \(S\) to \(F'\) (use the theorem of Malus) through \(O\). Therefore :
\(\delta _1 = 2f' + (n - 1)e\)
Expression of the plane wave in \(M\) is :
\(a_1 (M) = a_0 e^{j\omega t} e^{ - j\frac{{2\pi }}{{\lambda _0 }}(2f' + (n - 1)e)}\)
For the spherical wave that goes directly from \(F\) to \(M\), the optical path is (with \(r << f'\)) :
\(\delta _2 = FM = \sqrt {4f'^2 + r^2 } \approx 2f'\left( {1 + \frac{{r^2 }}{{4f'^2 }}} \right)^{1/2} \approx 2f'\left( {1 + \frac{{r^2 }}{{8f'^2 }}} \right) \approx 2f' + \frac{{r^2 }}{{4f'}}\)
The amplitude of the spherical wave is :
\(a_2(M)=\frac{a(F)}{FM}e^{j\omega t} e^{-j\frac{2\pi}{\lambda_0}\delta_2}\)
Whether, assuming that this wave has the same amplitude as the plane wave :
\(a_2(M)=a_0 e^{j\omega t} e^{-j\frac{2\pi}{\lambda_0}(2f'+r^2/4f')}\)
The resulting amplitude in \(M\) is :
\(a(M) = a_1 (M) + a_2 (M) = a_0 e^{j\omega t} e^{ - j\frac{{2\pi }}{{\lambda _0 }}(2f' + (n - 1)e)} \left[ {1 + e^{ - j\frac{2\pi }{\lambda _0 }(\frac{r^2 }{4f'} - (n - 1)e)} } \right]\)
Question
What is the intensity in the plane \((P)\) located at the same distance \(f'\) of the output face of \((L)\) as \((S)\), as a function of the cylindrical coordinate \(r\) ?
Deduce the nature of the interference fringes.
Solution
The light intensity is :
\(I(M) = 2I_0 \left[ {1 + \cos \left( {\frac{{2\pi }}{{\lambda _0 }}(\frac{{r^2 }}{{4f'}} - (n - 1)e)} \right)} \right]\)
The fringes are here rings (corresponding to \(r = constant\)) centered on \(F'\). A bright ray is given by :
\(\frac{{2\pi }}{{\lambda _0 }}(\frac{{r^2 }}{{4f'}} - (n - 1)e) = 2p\pi\) (with \(p\) an integer)
So :
\(r_p^{} = \sqrt {4f'\left( {(n - 1)e + p\lambda _0 } \right)}\)
Question
Calculate the radius extreme bright fringes knowing that :
\(\lambda_0=546\;nm\) ; \(f'=20\;cm\) ; \(n=1,52\) (index of the glass of \((L)\))
Solution
The point \(M\) is in the interference field if \(R \le r \le 2R\).
\(p\) is calculated by taking \(r=R\) then taking \(r=2R\).
We find \(p=-2\;800\) and \(p=-2\;628,2\).
In the latter case, take \(p = - 2 \;629\) and recalculate \(r\). We find \( r = 9,98\; mm\).