Derive the condition for maxima due to interference of light reflected from thin film of uniform thickness.

Set-up: A thin transparent film of uniform thickness \(t\) and refractive index \(\mu\) is in air. A ray AB is incident at angle \(i\). Part of it is reflected at B (ray BR1), and part is refracted at angle \(r\), reflected at C on the lower surface, and comes out at D (ray DR2). The two reflected rays are coherent and interfere.

        A     R1        R2
         \    /         /
   air    \ i/         /
 ----------B----------D-------- upper surface
   film   \  r      / \
   (μ, t)  \      /
 -----------\----/-------------- lower surface
             \  /
              C

Geometric path difference: \(\Delta=\mu(BC+CD)-BN\), where DN is drawn perpendicular to BR1. Using the geometry (\(BC=CD=t/\cos r\), \(BN=BD\sin i\), \(BD=2t\tan r\), and \(\sin i=\mu\sin r\)):

\[ \Delta=\frac{2\mu t}{\cos r}-2\mu t\tan r\sin r=\frac{2\mu t}{\cos r}(1-\sin^2 r)=2\mu t\cos r \]

Stokes' phase change: Reflection at B (rarer to denser) adds a phase change of \(\pi\), equivalent to an extra path of \(\lambda/2\). Reflection at C (denser to rarer) adds none.

\[ \text{Effective path difference } \Delta=2\mu t\cos r-\frac{\lambda}{2} \]

Maxima (bright): \(\Delta=n\lambda\Rightarrow\ \boxed{2\mu t\cos r=(2n+1)\dfrac{\lambda}{2}},\ n=0,1,2,\dots\)

Minima (dark): \(\Delta=(2n+1)\dfrac{\lambda}{2}\Rightarrow\ \boxed{2\mu t\cos r=n\lambda},\ n=0,1,2,\dots\)

The conditions in transmitted light are the reverse: \(2\mu t\cos r=n\lambda\) gives a bright fringe, so the reflected and transmitted patterns are complementary. With white light, different wavelengths satisfy the condition at different places, which is why thin films look coloured.