Answer: Path difference \(=2\mu t\cos r-\lambda/2\). Bright (maxima): \(2\mu t\cos r=(2n-1)\dfrac\lambda2\); dark (minima): \(2\mu t\cos r=n\lambda\), \(n=1,2,3\dots\)
Figure (described): a film of thickness \(t\) and refractive index \(\mu\) lies in air between two parallel surfaces. A ray AB strikes the top surface at B at angle of incidence \(i\). Part is reflected along BR₁; part is refracted into the film at angle \(r\), reflected from the bottom surface at C, and comes out at D along DR₂, parallel to BR₁. DN is drawn perpendicular to BR₁ (N on BR₁).
Geometrical path difference between the two reflected rays BR₁ and DR₂:
\[\Delta=\mu(BC+CD)-BN\]
From the geometry: \(BC=CD=\dfrac{t}{\cos r}\), and \(BD=2t\tan r\).
In \(\triangle BND\): \(BN=BD\sin i=2t\tan r\sin i\). By Snell's law \(\sin i=\mu\sin r\), so \(BN=2\mu t\tan r\sin r=\dfrac{2\mu t\sin^2r}{\cos r}\).
\[\Delta=\frac{2\mu t}{\cos r}-\frac{2\mu t\sin^2r}{\cos r}=\frac{2\mu t(1-\sin^2r)}{\cos r}=2\mu t\cos r\]
Stokes' phase change. Ray BR₁ is reflected at the surface of a denser medium (air to film), so it suffers an extra phase change of \(\pi\), i.e. an extra path of \(\lambda/2\). The ray reflected at C (film to air) does not. Effective path difference:
\[\Delta=2\mu t\cos r-\frac\lambda2\]
Maxima (bright): \(\Delta=n\lambda\Rightarrow2\mu t\cos r=(2n+1)\dfrac\lambda2\), \(n=0,1,2\dots\) (equivalently \((2n-1)\lambda/2\) with \(n=1,2,\dots\)).
Minima (dark): \(\Delta=(2n+1)\dfrac\lambda2\Rightarrow2\mu t\cos r=(n+1)\lambda\), i.e. \(2\mu t\cos r=n\lambda\), \(n=0,1,2\dots\)
Since the condition depends on \(\lambda\), a film in white light shows colours: different wavelengths are reinforced at different thicknesses or angles.
