Thin film interference — reflected and transmitted

WHAT is happening

The two key physical facts we will USE:

Optical path length. Light slows inside a medium of refractive index $n$ . A geometric distance $d$ inside the film "costs" an optical distance $nd$ , because what matters for phase is the number of wavelengths, and inside the medium $\lambda_{film} = \lambda/n$ .
Reflection phase flip. When light reflects going from a rarer to a denser medium (low $n$ → high $n$ ), it gains an extra phase of $\pi$ , i.e. an extra half wavelength $\lambda/2$ . Reflecting off a rarer medium (high $n$ → low $n$ ) gives no flip.

Figure — Thin film interference — reflected and transmitted

HOW we derive the condition (from scratch)

Take a film of index $n$ , thickness $t$ , with light incident at angle $\theta_i$ and refracting to angle $\theta_r$ inside.

Step 1 — Geometric extra path. Ray 2 goes down and back up inside the film. The extra path over ray 1 (after accounting for the wavefront geometry) is

$\Delta_{geo} = 2t\cos\theta_r$

Why this step? The ray inside travels a slant length $t/\cos\theta_r$ each way (down and up), giving $2t/\cos\theta_r$ of slant path. But the two emerging rays must be compared along a common wavefront (perpendicular to the rays). Dropping that wavefront from the exit point cuts off a piece of the slant path; the part of the in-film travel that actually counts is its projection onto the film normal, $2t\cos\theta_r$ . Geometrically: $\big(2t/\cos\theta_r\big) - \big(2t\sin\theta_r\cdot\tan\theta_r\big) = 2t\cos\theta_r$ . The cosine (not secant) survives.

Step 2 — Convert to optical path. Inside the film light is slowed, so multiply by $n$ :

$\Delta_{opt} = 2nt\cos\theta_r$

Why? Phase counts wavelengths; inside the film wavelengths are shorter by factor $n$ , so the optical path is $n$ times the geometric path.

Step 3 — Add the reflection phase term. Consider a film (index $n$ ) sitting on a less dense backing, with denser film than the medium above (typical: air–soap–air, or air–oil–water style cases differ — check each interface!).

For air–film–air (e.g. soap bubble): top reflection air→film is rarer→denser ⇒ flip $\lambda/2$ . Bottom reflection film→air is denser→rarer ⇒ no flip. Net extra: one $\lambda/2$ .

So the reflected total path difference is:

$\boxed{\Delta = 2nt\cos\theta_r \pm \tfrac{\lambda}{2}}$

WHY the $\lambda/2$ flip is the whole game

Worked Examples

Worked example 2) Anti-reflection coating (MgF₂ on glass)

Coating $n_c=1.38$ between air ( $1.0$ ) and glass ( $1.52$ ). We WANT reflection to cancel (so more light transmits into a lens).

Here both reflections flip: air→coating (rarer→denser) and coating→glass ( $1.38<1.52$ , rarer→denser). Two flips cancel ⇒ net no extra $\lambda/2$ . So destructive reflection needs $2n_c t = (m+\tfrac12)\lambda$ . Thinnest ( $m=0$ ): $t = \frac{\lambda}{4n_c} = \frac{550}{4(1.38)} = 99.6\text{ nm}$ Why no $\lambda/2$ term? Two flips = full $\lambda$ = no relative shift. Steel-man check: the conditions look swapped vs Example 1 — that's correct, because the interface ordering changed the number of flips.

Worked example 3) Wedge film / oil on water at an angle

Oil $n=1.4$ on water, $\lambda=600$ nm, viewed at $\theta_r$ with $\cos\theta_r=0.8$ , thickness $t=500$ nm. Reflected bright or dark?

Air→oil flips; oil→water ( $1.4>1.33$ ) denser→rarer, no flip ⇒ net one flip. $2nt\cos\theta_r = 2(1.4)(500)(0.8) = 1120\text{ nm} = 1.87\lambda$ Compare to bright condition $(m+\tfrac12)\lambda$ : $m+\tfrac12 = 1.87 \Rightarrow$ close to $m=1$ ( $1.5$ ) and $m=2$ ( $2.5$ )? Nearest integer-plus-half is $1.5$ → not exact; $1.87$ is closest to $m=2\lambda$ (within $0.13$ ), which is the dark condition ( $m\lambda$ ). So it appears nearly dark. Why this step? We match the computed multiple to whichever condition (integer vs half-integer) it's nearest to.

Common Mistakes

Recall Feynman: explain to a 12-year-old

Imagine light is two runners that split at a soap bubble. One bounces off the front; the other dives in, bounces off the back, and comes out. The diver ran a little extra distance. Also, bouncing off a "harder wall" (denser stuff) gives the runner a tiny hiccup — half a step out of rhythm. When both runners come back in step, the color shines bright; when they're out of step, that color disappears. That's why bubbles show rainbows: different colors get in-step at different bubble thicknesses. And a bubble so thin it's about to pop turns black — because the hiccup alone puts the runners exactly out of step.

Flashcards

What is the geometric extra path for the lower ray in a film of thickness

t

at internal angle

\theta_r

2t\cos\theta_r

Why does the geometric path equal

2t\cos\theta_r

and not

2t/\cos\theta_r

Because rays are compared along a common wavefront; only the projection of the in-film travel onto the film normal contributes, giving

2t\cos\theta_r

Why does optical path use

n

as in

2nt\cos\theta_r

Phase counts wavelengths and inside the film

\lambda_{film}=\lambda/n

, so optical path =

n\times

geometric path.

When does a reflection add a

\pi

(half-wave) phase shift?

When light reflects going from a rarer (low

n

) to a denser (high

n

) medium.

For a soap film in air (one net flip), the reflected BRIGHT condition is?

2nt\cos\theta_r=(m+\tfrac12)\lambda

For a soap film in air, the reflected DARK condition is?

2nt\cos\theta_r=m\lambda

Why is a very thin soap film black in reflection?

Geometric path →0, only the

\lambda/2

flip remains, giving destructive interference for all colors.

Reflected and transmitted patterns are related how?

They are complementary (negatives) — where reflection is bright, transmission is dark (energy conservation).

Thinnest anti-reflection coating thickness for wavelength

\lambda

t=\lambda/(4n_c)

(quarter-wave, since two flips cancel and we want destructive reflection).

Which angle goes in the formula, incident or refracted?

The refracted angle

\theta_r

inside the film.

\lambda

in the formula the film wavelength or vacuum wavelength?

Vacuum/air wavelength; the medium is already accounted for by

n

Connections

Interference of light — superposition principle behind it all
Young's double slit — same coherence/path-difference logic, different geometry
Newton's rings — thin air film of varying thickness (wedge → circular fringes)
Refractive index and Snell's law — give $\theta_r$ and optical path
Phase change on reflection — the π-flip rule
Anti-reflection coatings — engineering application
Conservation of energy — why reflected/transmitted are complementary

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek soap bubble par light girti hai. Light do hisson mein bant jaati hai: ek part upar wali surface se reflect hota hai (Ray 1), aur doosra part andar jaata hai, neeche wali surface se reflect hokar wapas aata hai (Ray 2). Ray 2 ne thoda extra distance travel kiya — yeh extra optical path hai $2nt\cos\theta_r$ . Yahan $n$ isliye lagta hai kyunki film ke andar light slow ho jaati hai aur wavelength chhoti ho jaati hai. Aur dhyan rakho: angle $\theta_r$ (refracted angle) use karna hai, incident wala nahi.

Geometry samajhne ke liye: film ke andar slant path $2t/\cos\theta_r$ hota hai, par dono rays ko ek common wavefront (rays ke perpendicular) par compare karna padta hai. Wavefront drop karne par jo part bachta hai wo film ke normal par projection hai, yani $2t\cos\theta_r$ . Isliye cosine aata hai, secant nahi. Yeh ek subtle but important point hai.

Sabse important cheez hai phase flip. Jab light rarer (kam $n$ ) se denser (zyada $n$ ) medium par reflect hoti hai, to ek extra $\lambda/2$ ka jhatka lagta hai (phase $\pi$ se ulat). Isliye condition likhne se pehle dono surfaces par flips count karo. Soap film (air–soap–air) mein sirf top par flip hota hai — ek odd flip — to bright aur dark conditions swap ho jaati hain: bright tab jab $2nt\cos\theta_r=(m+\tfrac12)\lambda$ . Yahi reason hai ki bahut patli bubble (jo phootne wali hai) reflection mein kaali dikhti hai.

Exam tip: pehle flips gino (odd ya even), phir formula $2nt\cos\theta_r$ likho, phir $\lambda$ ko hamesha air/vacuum wavelength rakho. Reflected vs transmitted pattern ek doosre ke opposite hote hain (energy conservation). Anti-reflection coating mein dono surfaces flip karti hain (even), isliye thinnest coating $t=\lambda/4n$ hoti hai jo reflection cancel karti hai.

Thin film interference — reflected and transmitted

WHAT is happening

HOW we derive the condition (from scratch)

WHY the $\lambda/2$ flip is the whole game

Worked Examples

Common Mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Optics

Connections