17-Phys-A7 Optics · Undated paper
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
17-Phys-A7, Optics — National Exams, May 2019. 3 hours; closed book (approved Casio/Sharp calculator only). Each question value is as indicated; exam is out of 67. Questions 1–6 are mandatory; the paper then offers a choice of Question 7 or 8, and a choice of Question 9 or 10. Every question is solved in full below as a complete study resource, including both members of each either/or pair.
Reference texts. Hecht, Optics, 5th ed.; Pedrotti, Pedrotti & Pedrotti, Introduction to Optics, 3rd ed.; Griffiths, Introduction to Electrodynamics, 4th ed. (Ch. 7–9, Maxwell’s equations and EM waves).
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.
a) Interference at the contact point. Two rays combine at any point of the air gap between the lens and the flat plate: ray 1 reflects off the lower (curved) surface of the plano-convex lens — a glass→air reflection, with no phase shift (reflection off a lower-index medium) — and ray 2 reflects off the top surface of the flat glass plate — an air→glass reflection, off a higher-index medium, which imposes a $\pi$ phase shift. Exactly at the contact point the two rays travel the same physical path (air-gap thickness $t=0$), so the only difference between them is that single $\pi$ shift; they arrive exactly out of phase and interfere destructively — the observer sees a dark spot at the point of contact, which is the well-known central dark spot of the Newton’s-rings pattern.
b) Radius of the first dark ring.
Given. Lens radius of curvature $R=1$ m; $\lambda=0.5\ \mu\text{m}=5\times10^{-7}$ m.
Find. $X=r_1$, the radius of the first dark ring away from the (also dark) central contact point.
Approach. With the single $\pi$ phase shift established in (a), dark (destructive) rings occur where the round-trip air-gap path $2t$ is an integer number of wavelengths, $2t=m\lambda$ ($m=0,1,2,\dots$; $m=0$ is the central spot itself, not a resolvable ring). For a lens of large radius of curvature, the gap thickness at radial distance $r$ from the contact point is $t\approx r^2/(2R)$ (from the sagitta of a shallow spherical surface).
| Quantity | Value |
|---|---|
| Distance to first dark ring, $X$ | $0.707$ mm |
c) Rainbow-like colours in an oil film on water. The oil layer is a thin film of (locally) varying thickness $t$, with the sequence air ($n=1$) → oil ($n\approx1.4$–$1.5$) → water ($n\approx1.33$). White (broadband) ambient light reflects off both the air–oil and oil–water interfaces, and the two reflected beams interfere; the condition for constructive interference at a given thickness, $2nt=(m+\tfrac12)\lambda$ (accounting for the single phase shift at the higher-index air–oil interface, with no compensating shift at the lower-index oil–water interface), picks out a different visible wavelength as bright for each different local thickness $t$. Because natural oil films have thickness that varies smoothly and irregularly across the puddle, each region reflects strongly at a different colour, producing the swirling multicoloured, rainbow-like pattern — a thin-film interference effect, not a refraction/dispersion effect like a true rainbow, but visually similar because both spread white light into its spectral components.
