PHYS7021
Burdwan · Electricity, Magnetism & Wave Optics · 2018–2024

8E · Interference — Thin Films, Stokes, Haidinger & Fizeau

Parallel-film interference in reflected and transmitted light, Stokes relations, wedge films, Haidinger and Fizeau fringes, soap-film numeric, all solved.

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A soap bubble shows colours without any pigment. Light reflects off the front surface of the thin film and off the back surface; the two reflections interfere. Film thickness decides which colour survives. This is interference by division of amplitude.

Formula 1. Parallel film, reflected light — bright: $2\mu t\cos r = (2n+1)\dfrac{\lambda}{2}$; dark: $2\mu t\cos r = n\lambda$
Formula 2. Parallel film, transmitted light — bright: $2\mu t\cos r = n\lambda$; dark: $2\mu t\cos r = (2n+1)\dfrac{\lambda}{2}$ (opposite of reflected)

Here $\mu$ = film index, $t$ = film thickness, $r$ = angle of refraction inside the film, $\lambda$ = wavelength. $2\mu t\cos r$ = optical path difference between the two reflections.

Why reflected and transmitted are opposite (Stokes’ treatment)

Step 1 — Two reflections. In reflected light, ray 1 bounces off the top surface (air→film: rarer→denser) and ray 2 bounces off the bottom surface (film→air: denser→rarer, after crossing the film twice).

Step 2 — One flip. By Stokes’ rule, reflection at the rarer→denser face flips the phase by $\pi$ (= $\lambda/2$ path); reflection at the denser→rarer face flips nothing. So the reflected pair differs by one $\lambda/2$.

Step 3 — Stokes relations. Let $r$, $t$ be the amplitude reflection/transmission coefficients going in, and $r’$, $t’$ coming back out. Reversing any ray must give back the original beam, which forces $r = -r’$ (the minus IS the $\pi$ flip) and $tt’ + r^2 = 1$ (energy conserved). These are the Stokes relations.

Step 4 — Effective path. Geometric round trip $2\mu t\cos r$ plus the one flip: $\Delta = 2\mu t\cos r + \lambda/2$. Bright when $\Delta = n\lambda$, dark when $\Delta = (2n+1)\lambda/2$ — giving Formula 1.

Step 5 — Transmitted light. Both transmitted rays cross faces in the same sense with no net flip, so $\Delta = 2\mu t\cos r$ with no added $\lambda/2$ — bright and dark swap (Formula 2). Energy check: bright-in-reflection + bright-in-transmission would exceed the input; the swap keeps total = input ✓.

Unit check: $[2\mu t\cos r] = L$ ✓. Limit: $t \to 0$ gives $\Delta = \lambda/2$ (reflected) = dark — an extremely thin film looks dark in reflection, which is why a breaking soap film goes black before it pops.

Trap: the flip happens at the RARER→DENSER face only. Count flips before writing any condition.

Wedge-shaped film

If the film thickness grows linearly ($t = x\tan\alpha \approx x\alpha$ for small wedge angle $\alpha$), fringes of equal thickness appear — each fringe marks one contour of constant $t$. Dark fringes (reflected): $2\mu t = n\lambda$. Spacing between adjacent dark fringes:

Formula 3. Wedge fringe width — $\beta = \dfrac{\lambda}{2\mu\alpha}$ ($\alpha$ in radians).

Straight, parallel, equally spaced fringes — the air wedge between two glass plates is the standard lab example.

Haidinger fringes (equal inclination)

With a parallel film viewed in reflected light, $2\mu t\cos r = $ constant decides brightness. For fixed $t$ and $\lambda$, each angle $r$ gives one ring: fringes of EQUAL INCLINATION. Seen by the naked eye (or a telescope focused at infinity) as concentric circular rings centred on the normal — these are Haidinger fringes. Each ring = one value of $r$.

Fizeau fringes (equal thickness)

With a film of varying thickness viewed at near-normal incidence, $2\mu t = $ constant decides brightness: fringes of EQUAL THICKNESS. Each fringe traces a contour of constant $t$. Newton rings are the circular example; wedge fringes are the straight example. These are Fizeau fringes.

Solved PYQs

Q1

Explain the formation of interference fringes due to a thin film when observed in reflected light.

20222023×2

Recall

Two reflections (top: flip; bottom: no flip) + round trip $2\mu t\cos r$ give $\Delta = 2\mu t\cos r + \lambda/2$.

Steps

  1. Draw the film: ray hits the top face — part reflects (ray 1, rarer→denser: $\pi$ flip), part enters.
  2. The entered ray crosses the film, reflects at the bottom face (denser→rarer: no flip), and exits as ray 2.
  3. Optical path between them: $2\mu t\cos r$ for the down-and-up trip at refraction angle $r$.
  4. Add the one Stokes flip: effective $\Delta = 2\mu t\cos r + \lambda/2$.
  5. Bright: $\Delta = n\lambda \Rightarrow 2\mu t\cos r = (2n+1)\lambda/2$; dark: $2\mu t\cos r = n\lambda$ — Formula 1.
  6. Note the division-of-amplitude point: one parent ray split by partial reflection, so the pair is coherent.

Answer

$\boxed{\text{Reflected: bright }2\mu t\cos r = (2n+1)\lambda/2,\ \text{dark }2\mu t\cos r = n\lambda}$ — one $\lambda/2$ flip at the top face.

Exam tip

Draw the two rays with the flip marked $\pi$ on ray 1 only — the diagram carries the flip mark.
Q2

Discuss Stokes' treatment of phase change on reflection.

20222023×2

Recall

Reversibility forces $r = -r'$ (the $\pi$ flip) and $tt' + r^2 = 1$ (energy).

Steps

  1. Set up: amplitude $a$ hits a surface; reflected $ar$, transmitted $at$; reversed coefficients $r'$, $t'$.
  2. Reverse the two outgoing rays and demand the original beam $a$ reappears with nothing going the other way.
  3. Collecting amplitudes gives $r^2 + tt' = 1$ and $r + r' = 0$ (for a lossless surface).
  4. Read the physics: $r = -r'$ means reflection one way flips phase by $\pi$ relative to the other way — the Stokes $\lambda/2$ rule.
  5. $tt' + r^2 = 1$ is energy conservation at the surface (reflected + transmitted fractions = 1).
  6. Apply: in Newton rings / thin films the rarer→denser reflection carries the flip, the denser→rarer one does not — hence the dark centre.

Answer

$\boxed{r = -r',\qquad tt' + r^2 = 1}$ — the minus sign is the $\pi$ phase flip on rarer→denser reflection.

Exam tip

Write both relations boxed, then state in words which reflection flips — relation plus application = full marks.
Q3

What are Haidinger fringes? Explain the formation of these fringes.

2023

Recall

Equal-inclination fringes: fixed $t$, varying $r$; rings centred on the normal, seen at infinity.

Steps

  1. Take a parallel film ($t$ fixed) in reflected light: brightness follows $2\mu t\cos r + \lambda/2$.
  2. For fixed $t$ and $\lambda$, each angle of refraction $r$ gives a fixed $\Delta$ — so each direction carries its own bright/dark state.
  3. Rays with the same $r$ form a cone about the normal; on a screen at infinity (telescope / relaxed eye) each cone is one circular ring.
  4. Hence concentric rings centred on the normal — fringes of equal inclination = Haidinger fringes.
  5. Contrast with Fizeau: Haidinger needs a parallel film + varied angle; Fizeau needs a wedge film + fixed (normal) angle.

Answer

$\boxed{\text{Haidinger = equal-inclination rings from a parallel film; each ring is one angle }r.}$

Exam tip

One line separates the pair: "Haidinger varies the angle, Fizeau varies the thickness."
Q4

Explain the formation of Fizeau fringes.

2023

Recall

Equal-thickness fringes: fixed (normal) incidence, varying $t$; each fringe = one contour of constant $t$.

Steps

  1. Take a film of varying thickness (wedge, or Newton-rings air film) at near-normal incidence: brightness follows $2\mu t + \lambda/2$ (reflected).
  2. For fixed $\lambda$ and normal incidence, each thickness $t$ gives a fixed $\Delta$ — so each contour of constant $t$ is one fringe.
  3. Examples: Newton rings (circular contours of the air film), wedge fringes (straight parallel lines, $\beta = \lambda/2\mu\alpha$).
  4. These are fringes of equal thickness = Fizeau fringes — localized at the film surface (seen by focusing on the film).

Answer

$\boxed{\text{Fizeau = equal-thickness fringes; each fringe traces constant }t\text{ (Newton rings, wedge lines).}}$

Exam tip

Name both examples (Newton rings + wedge) — examiners expect the pair.
Q5

A soap film of refractive index $1.333$ and thickness $5\times 10^{-7}$ m is observed at an angle of incidence of $40°$. Find the wavelength of light in the visible spectrum which will be strongly reflected.

2019

Recall

Reflected bright: $2\mu t\cos r = (2n+1)\lambda/2$. Get $r$ from Snell: $\sin r = \sin i/\mu$.

Steps

  1. Given: $\mu = 1.333$, $t = 5\times 10^{-7}$ m, $i = 40°$.
  2. Snell: $\sin r = \sin 40°/1.333 = 0.6428/1.333 = 0.4823$, so $r = 28.86°$ and $\cos r = 0.8761$.
  3. Path: $2\mu t\cos r = 2\times 1.333\times 5\times 10^{-7}\times 0.8761 = 1.1677\times 10^{-6}$ m.
  4. Bright condition: $\lambda = 2(1.1677\times 10^{-6})/(2n+1) = 2.3354\times 10^{-6}/(2n+1)$ m.
  5. Try $n$: $n=0 \to 2335$ nm (IR); $n=1 \to 778$ nm (deep red edge); $n=2 \to 467$ nm (blue, visible ✓); $n=3 \to 334$ nm (UV).
  6. Visible answer: $\lambda \approx 467$ nm — blue light is strongly reflected.

Answer

$\boxed{\lambda \approx 467\ \text{nm (blue)}}$ — from $2\mu t\cos r = (2n+1)\lambda/2$ with $n = 2$.

Exam tip

Always test $n = 0,1,2,3$ and keep only 400–700 nm — the paper asks for the visible one, not all of them.
Q6

Explain the phenomenon of interference in thin films.

2024

Recall

Division of amplitude: front + back reflections interfere; colour/thickness rules from Formula 1.

Steps

  1. State the setup: a thin transparent film (soap, oil) — light reflects partially at the front face and at the back face.
  2. Coherence: both reflections come from the same parent ray (division of amplitude), so they interfere.
  3. Path difference: $2\mu t\cos r$ plus one Stokes $\lambda/2$ flip at the front (rarer→denser) face.
  4. Conditions: reflected bright $2\mu t\cos r = (2n+1)\lambda/2$, dark $2\mu t\cos r = n\lambda$ — Formula 1.
  5. Colours: white light contains all $\lambda$; at each $t$ only the $\lambda$ satisfying bright is reinforced — hence soap-bubble colours; transmitted colours are complementary.
  6. Needs: film thin ($t \sim \lambda$) else the two reflections separate beyond the coherence length and no fringes form.

Answer

$\boxed{\text{Front + back reflections (one with }\lambda/2\text{ flip) interfere; }2\mu t\cos r\text{ picks the surviving colour.}}$

Exam tip

Mention division of amplitude by name and the coherence-length limit — both are easy marks most students miss.

Reference. Brij Lal, Subrahmanyam & Avadhanulu — Optics Ch-15 (thin films, Stokes, Haidinger/Fizeau); Ajoy Ghatak — Optics Ch-14.

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