Physics · Ch 6 — Optics

Interference in Thin Films

6.10.6

Interference in Thin Films

A thin transparent film of refractive index $\mu$ and thickness $d$ , struck by a parallel beam at (near-)normal incidence, splits the light at its upper surface into a reflected part and a refracted part; the refracted part is further partly reflected at the lower surface and re-emerges through the top -- so both the reflected and the transmitted light consist of multiple coherent contributions that can interfere, producing colour in soap films and oil slicks. For the extra optical path $2\mu d$ travelled inside the film (near-normal incidence), the transmitted light interferes constructively when $2\mu d=n\lambda$ and destructively when $2\mu d=(2n-1)\lambda/2$ . The reflected light picks up an extra phase shift of $\pi$ (equivalent to an extra path $\lambda/2$ ) whenever reflection happens while travelling from a rarer into a denser medium, so its conditions are reversed relative to transmission: constructive reflection requires $2\mu d+\lambda/2=n\lambda$ , i.e. $2\mu d=(2n-1)\lambda/2$ , and destructive reflection requires $2\mu d=n\lambda$ . For a general (non-normal) angle of incidence with internal refraction angle $r$ , every occurrence of $2\mu d$ above is replaced by $2\mu d\cos r$ . T …

Figure 6.62Interference in thin films

What this figure shows. A thin transparent film, thickness d and refractive index mu, is struck by a ray at angle i; at the upper surface it splits into a reflected ray and a refracted ray that enters the film. That refracted ray reaches the lower surface and there splits again -- part transmitted onward out of the film, part reflected back up through the film and out through the top surface a second time. The two rays leaving the top surface (the original direct reflection, and the twice-refracted-once-internally-reflected ray) are coherent and interfere, as are the two rays eventually leaving through the bottom of the film; multiplying reflections inside a real film add further, pro …

Misc Example 6.30Minimum thickness of a film for strong reflection and for anti-reflection

Worked out. A film of refractive index 1.25 is to strongly reflect light of wavelength 589 nm. For strong (constructive) reflection, the least optical path difference of lambda/2 requires 2(mu)d = lambda/2, giving d = lambda/(4 mu) = 589e-9/(4 times 1.25) = 117.8e-9 m = 117.8 nm, the minimum thickness for a highly-reflective coating. For the same film to instead be anti-reflecting, destructive reflection requires the larger path difference 2(mu)d = lambda, giving d = lambda/(2 mu) = 589e-9/(2 times 1.25) = 235.6e-9 m = 235.6 nm -- exactly double the strong-reflection thickness, since anti-reflection coatings must instead cancel, rather than reinf …

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