Question 6 of 6: Thin-film antireflection coating, normal and oblique incidence
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A7, Optics — National Exams, December 2016. 3 hours; closed book (formula sheet supplied on pages 6–8). Notes 1–7 on the cover page require Q1 and Q2 (mandatory) plus any two of Q3–Q6 for a complete paper; every question is solved below as a full study resource, including all four of Q3–Q6.
Approach. Because $n_f$ is intermediate between $n_{air}$ and $n_s$, BOTH the air–film and the film–substrate reflections are low-to-high-index reflections, so both pick up the same $\pi$ phase shift on reflection — the two shifts cancel in the relative phase between the two reflected beams, leaving the round-trip OPTICAL PATH inside the film, $2n_ft$, as the only phase difference. Minimum reflectance (destructive interference between the two reflected beams) then requires a half-integer number of wavelengths: $2n_ft=(m+\tfrac12)\lambda_0$.
Fig. Q6(a) — both reflections (air–film, film–substrate) are low-to-high index, so both add a π phase shift; the only NET phase difference is the extra round-trip path $2n_ft$ through the film.
Minimum-thickness solution ($m=0$). $t=\dfrac{\lambda_0}{4n_f}=\dfrac{500\ \text{nm}}{4(1.38)}=\dfrac{500}{5.52}=\boxed{90.6\ \text{nm}}$ (the classic "quarter-wave" AR coating).
Quantity
Value
Film thickness, $t$
$90.6$ nm
Part (b) — minimally-reflected wavelength at oblique incidence $\theta$.
Given. Same film ($n_f=1.38$, $t=90.6$ nm from part (a)) and substrate ($n_s=1.50$); angle of incidence in air is now $\theta$ (general).
Find. $\lambda(\theta)$, the wavelength minimally reflected at incidence angle $\theta$.
Approach. Both reflections remain low-to-high index at any angle (the index ordering $n_{air}\lt n_f\lt n_s$ doesn't depend on $\theta$), so the $\pi$-shift cancellation of part (a) still holds; only the round-trip path length inside the film changes, because the ray now crosses the film at the REFRACTED angle $\theta_f$ (from Snell's law $\sin\theta=n_f\sin\theta_f$) rather than straight through. The round-trip physical path becomes $2t/\cos\theta_f$, but the relevant OPTICAL path difference between the two reflected rays (accounting for the extra lateral offset between them, a standard thin-film result) reduces to $2n_ft\cos\theta_f$.
Fig. Q6(b) — at oblique incidence θ the ray refracts to θf inside the film ($\sin\theta=n_f\sin\theta_f$); the round-trip optical path difference between the two reflected rays is $2n_ft\cos\theta_f$.
Minimum-reflection condition ($m=0$, same design order as part (a)). $2n_ft\cos\theta_f=\tfrac12\lambda(\theta)\ \Rightarrow\ \lambda(\theta)=4n_ft\cos\theta_f$. Since $4n_ft=\lambda_0$ (the normal-incidence design result), this simplifies to $\boxed{\lambda(\theta)=\lambda_0\cos\theta_f=\lambda_0\sqrt{1-\left(\dfrac{\sin\theta}{n_f}\right)^2}}$, using $\cos\theta_f=\sqrt{1-\sin^2\theta_f}$ and Snell's law $\sin\theta_f=\sin\theta/n_f$.
Illustrative evaluation, $\theta=45^\circ$. $\sin\theta_f=\sin45^\circ/1.38=0.5124$, $\cos\theta_f=0.8587$, so $\lambda(45^\circ)=500(0.8587)=\boxed{429.4\ \text{nm}}$ — the coating's reflection minimum blue-shifts as the angle of incidence increases, since the effective optical path through the film shortens.
