17-Phys-A7 Optics · December 2018

Question 10 of 10: Thin-film interference in a glass plate

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 10: Thin-film interference in a glass plate (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — which reflections carry a $180^\circ$ phase shift? A reflection carries a $180^\circ$ phase shift only when reflecting off a higher-index medium (low-to-high); reflection off a lower-index medium (high-to-low) has no phase shift. Ray (1)'s reflection is ray (2), at the top (air, $n=1$ $\to$ glass, $n_g=1.45$) surface — low to high, so ray (2) has a $180^\circ$ phase shift. Ray (3)'s reflection is ray (4), at the bottom (glass, $n_g=1.45$ $\to$ air, $n=1$) surface — high to low, so ray (4) has no phase shift ($0^\circ$).

Part (b) — angles of ray (2) and ray (3).

Given. Angle of incidence $\theta_i=30^\circ$ (ray 1, in air), $n_g=1.45$.

1. Angle of reflection (ray 2). By the law of reflection, $\theta_{r,2}=\theta_i=\boxed{30.0^\circ}$ (measured from the normal, same side of it as the incident ray's mirror image).
2. Angle of refraction (ray 3). Snell's law: $\sin\theta_i=n_g\sin\theta_{t,3}\Rightarrow\sin\theta_{t,3}=\dfrac{\sin30^\circ}{1.45}=\dfrac{0.5}{1.45}=0.3448\Rightarrow\theta_{t,3}=\boxed{20.17^\circ}.$

Part (c) — amplitude reflection/transmission coefficients ($p$-polarization).

Given. $E$-field polarized in the plane of incidence ($p$-pol); top surface: $n_i=1$ (air), $n_t=1.45$ (glass), $\theta_i=30^\circ$, $\theta_t=20.17^\circ$ (part b); bottom surface (ray 3 $\to$ rays 4,5): $n_i=1.45$, $n_t=1$, angle inside the glass $=20.17^\circ$, and by the parallel-surface geometry the ray exits at $30^\circ$ — the mirror of the entry angle.

Approach. Apply the given Fresnel formulas at each interface with its own $(n_i,n_t,\theta_i,\theta_t)$ pair.

1. Top surface (air $\to$ glass). $$\begin{aligned}r_{\parallel,\text{top}}&=\frac{n_t\cos\theta_i-n_i\cos\theta_t}{n_t\cos\theta_i+n_i\cos\theta_t}=\frac{1.45\cos30^\circ-\cos20.17^\circ}{1.45\cos30^\circ+\cos20.17^\circ}\\&=\frac{1.2557-0.9385}{1.2557+0.9385}=\boxed{0.1445}\end{aligned}$$ $$t_{\parallel,\text{top}}=\frac{2\cos30^\circ}{1.45\cos30^\circ+\cos20.17^\circ}=\frac{1.7321}{2.1942}=\boxed{0.7893}$$
2. Bottom surface (glass $\to$ air). Here $n_i=1.45$, $n_t=1$, incidence angle (in glass) $=20.17^\circ$, transmission angle (in air) $=30^\circ$: $$\begin{aligned}r_{\parallel,\text{bot}}&=\frac{\cos20.17^\circ-1.45\cos30^\circ}{\cos20.17^\circ+1.45\cos30^\circ}=\frac{0.9385-1.2557}{0.9385+1.2557}\\&=\boxed{-0.1445}\end{aligned}$$ $$t_{\parallel,\text{bot}}=\frac{2(1.45)\cos20.17^\circ}{\cos20.17^\circ+1.45\cos30^\circ}=\frac{2.7217}{2.1942}=\boxed{1.2405}$$
3. Consistency check (Stokes relations). $r_{\parallel,\text{bot}}=-r_{\parallel,\text{top}}$ exactly, as required for a lossless, non-magnetic interface traversed in opposite directions — a useful self-check on any two-surface Fresnel calculation.

Part (d) — minimum thickness for destructive interference at normal incidence.

Given. Normal incidence ($\theta_i=0$), $n_g=1.45$, $\lambda=1\ \mu\text{m}$; ray (2) carries a $180^\circ$ reflection phase shift (part a), ray (5) does not (its one reflection, ray 4, is high-to-low, no shift; transmission never introduces this kind of phase jump).

Find. Minimum non-zero $D$ for rays (2) and (5) to interfere destructively.

Approach. Ray (5) travels an extra optical round-trip path $2n_gD$ inside the glass relative to ray (2), and carries zero net reflection phase shift, while ray (2) carries a fixed $180^\circ$ shift and zero path. Because exactly one of the two interfering rays picks up the extra $180^\circ$, the roles of the "usual" thin-film formulas invert: the film behaves, for interference purposes, as if there were no net phase-shift difference beyond the path term, so destructive (dark) reflection occurs at $2n_gD=m\lambda$ ($m=1,2,\dots$; $m=0$ is the trivial zero-thickness case).

1. Phase bookkeeping. Total phase of ray 2 (reference, zero path here) $=\pi$. Total phase of ray 5 $=\dfrac{2\pi}{\lambda}(2n_gD)+0$. Destructive interference requires the phase difference to be an odd multiple of $\pi$: $\dfrac{4\pi n_gD}{\lambda}-\pi=(2m+1)\pi\ \Rightarrow\ 2n_gD=(m+1)\lambda$.
2. Minimum non-zero thickness ($m=0$). $$D_{\min}=\frac{\lambda}{2n_g}=\frac{1\times10^{-6}}{2(1.45)}=\boxed{3.448\times10^{-7}\ \text{m}=344.8\ \text{nm}=0.3448\ \mu\text{m}}.$$

Practical applications. Yes — engineering a thin dielectric film so that reflection destructively interferes at a design wavelength is exactly the working principle of anti-reflection (AR) coatings on camera lenses, eyeglasses, and solar-cell cover glass (reducing unwanted reflected light and increasing transmitted/collected light), and of thin-film interference (dichroic) filters that selectively pass or reject narrow wavelength bands by stacking multiple layers, each tuned via its own destructive/constructive condition. The same physics, viewed for the transmitted rather than reflected beam, also underlies anti-reflective etalons and the structural colours seen in soap films and oxide layers on metals.