NivaarExam PrepOfficial exam papers ↗

17-Phys-A7 Optics · Undated paper

Question 10 of 10: Air-wedge thin-film fringes (choice with Q9)

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

Notes on this paper

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).

Note on question choice
Q7/Q8 and Q9/Q10 are each an either/or pair on the printed paper (only one of each counts toward the mark total, and the printed 67-mark total matches the mandatory questions plus the Q8+Q10 pairing). As a complete study resource, all four (7, 8, 9, 10) are solved in full below.
Note on the numbers
The numbers used here are internally consistent (mark totals sum correctly, angle/critical-angle values cross-check, see the Concept boxes below).

Question 10: Air-wedge thin-film fringes (choice with Q9) (6 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.

a) Interference at the contact edge. Ray A reflects off the underside of the inclined (wedge) plate — a glass→air reflection, no phase shift. Ray B reflects off the top surface of the horizontal plate — an air→glass reflection, off a higher-index medium, a $\pi$ phase shift. This is the same asymmetric pairing as Q9(a). At the line of contact the air gap thickness is zero, so the only difference between the two rays is the single $\pi$ shift; they interfere destructively — the observer sees a dark fringe exactly at the contact line.

horizontal plate inclined plate (wedge) ray A: off wedge's underside — glass→air, no shift ray B: off horizontal plate's top — air→glass, π shift contact (t=0) x (distance from contact) → t(x) increases linearly, t/L = 0.5 mm / 2 cm
The two rays that interfere near the wedge's contact edge: ray A off the wedge's underside (glass→air, no shift), ray B off the horizontal plate's top (air→glass, π shift).

b) Distance to the first dark fringe.

Given. Wedge thickness at the far end $t_{\max}=0.5$ mm; plate length $L=2$ cm; $\lambda=500$ nm.

Find. $X$, the distance from the contact line to the first dark fringe.

Approach. With the single $\pi$ shift from (a), dark fringes occur at $2t=m\lambda$ ($m=0,1,2,\dots$; $m=0$ is the contact line itself). The gap thickness grows linearly with distance $x$ from the contact line, at the wedge’s slope $t_{\max}/L$.

  1. Dark-fringe condition, first fringe away from contact ($m=1$). $2t_1=\lambda\ \Rightarrow\ t_1=\lambda/2=250$ nm.
  2. Wedge slope. $\dfrac{t_{\max}}{L}=\dfrac{0.5\ \text{mm}}{2\ \text{cm}}=\dfrac{5\times10^{-4}\ \text{m}}{2\times10^{-2}\ \text{m}}=0.025$ (dimensionless).
  3. Solve for $X$. $X=\dfrac{t_1}{t_{\max}/L}=\dfrac{2.5\times10^{-7}}{0.025}=\boxed{1.0\times10^{-5}\ \text{m}=10\ \mu\text{m}}.$
QuantityValue
Distance to first dark fringe, $X$$10\ \mu$m

c) Rainbow-like colours on a soap film/bubble. A soap film is a thin layer of soapy water (index $n\approx1.33$) bounded by air on both sides. White light reflecting off its front and back surfaces interferes exactly as in Q9(c) — here the front (air→film) reflection carries a $\pi$ shift while the back (film→air) reflection does not, so the bright-fringe condition is again $2nt=(m+\tfrac12)\lambda$. Gravity drains the film unevenly, and evaporation/surface currents constantly reshape it, so the local thickness $t$ varies continuously across the film (and with time) — each thickness satisfies the bright condition for a different visible wavelength, so different regions (and the swirling, shifting bands seen on a real bubble) show different, constantly-changing spectral colours: a thin-film interference effect that mimics a rainbow’s spread of colour without any refraction/dispersion inside water droplets being involved.

Back to the paper →