Interference by Thin Film

Q: What is thin film interference?

Thin film interference is the interference of light reflected from the top and bottom surfaces of a transparent layer whose thickness is comparable to the wavelength. The two reflected waves come from one incident wave by division of amplitude, and their path difference 2 mu t cos r, plus any reflection phase change, decides which colours are reflected strongly.

Q: Why does light suffer a phase change of pi on reflection?

When light in a rarer medium reflects from a denser medium, the reflected wave is inverted, just as a pulse on a string reflects upside down from a fixed end. Inversion is a phase change of pi, equal to half a wavelength of path. Reflection from a rarer medium and transmission produce no phase change.

Q: Why do soap bubbles show colours?

The soap film has different thicknesses at different places. At each point, light reflected from the outer and inner surfaces interferes constructively for some wavelengths and destructively for others, so that point shows the colour of the reinforced wavelengths. As the film drains and thins, the colours move, and a very thin film looks black.

Q: What is the minimum thickness of an anti-reflection coating?

For a coating whose index lies between air and glass, both reflections have a pi shift, so destructive interference needs 2 mu t = lambda/2. The least thickness is lambda/(4 mu), a quarter-wave layer. For magnesium fluoride (mu = 1.38) at 550 nm this is about 100 nm.

Q: Why does a soap film look black just before it bursts?

When the film becomes much thinner than the wavelength, the path difference between the two reflected waves is almost zero. Only the top reflection has a pi phase change, so the two waves are exactly out of phase for every colour and cancel. Almost no light is reflected and the film appears black.

Q: Are the conditions for transmitted light the same as for reflected light?

No, they are opposite. Transmitted rays have no reflection phase change at the surfaces they cross, and energy not reflected is transmitted. A wavelength that is strongly reflected by a film is weakly transmitted, so a film that looks yellow in reflected light looks bluish when you look through it.

Q: Which thin film questions are asked in NEET?

NEET asks for the phase change on reflection, the condition 2 mu t = (m + 1/2) lambda for strong reflection from a film in air, the minimum thickness lambda/(4 mu) of a coating or soap film, and why thin films show colours. Remember to multiply the thickness by the refractive index.

Q: How is thin film interference tested in JEE?

JEE mixes thin films with reflection phase rules: films on substrates with different indices, oblique incidence with cos r, finding all visible wavelengths reflected strongly or weakly, direct and reflected light near a mirror, air wedges and Newton's rings. Counting the pi shifts correctly is the step most students get wrong.

PhysicsWave OpticsFor JEE aspirants

Interference by a thin film explains the colours of soap bubbles and oil on water, and the purple glint of coated camera lenses. Light reflected from the top and bottom surfaces of a film interferes, with optical path difference $2 μ t cos r$ plus any phase change of $π$ on reflection. Thin film interference uses division of amplitude, unlike YDSE, and it appears in JEE Main and NEET as minimum-thickness, coating and "which colour is reflected" questions.

On this page1What a thin film is2Phase change3Path difference4Bright and dark5Colours in white light6Anti-reflection coatings7Wedge film8Newton's rings

Key Formulas - Quick Reference

Reflection from a denser medium: phase change $π$ (extra path $\frac{λ}{2}$ ); from a rarer medium, or on transmission: no change
Direct and reflected light at a height $d$ above a mirror: $Δ x = 2 d cos θ + \frac{λ}{2}$
Optical path difference in a film: $Δ = 2 μ t cos r$ (normal incidence: $2 μ t$ )
★ Must learnOne $π$ shift (film in air, soap bubble, oil on water): bright $2 μ t cos r = (m + \frac{1}{2}) λ$ , dark $2 μ t cos r = mλ$
★ Must learnZero or two $π$ shifts (coating on glass): bright $2 μ t cos r = mλ$ , dark $2 μ t cos r = (m + \frac{1}{2}) λ$
Transmitted light: conditions opposite to reflected light (energy is conserved)
Minimum thickness for anti-reflection coating, or for strong reflection from a film in air: $t = \frac{λ}{4 μ}$
Wedge film: $β = \frac{λ}{2 μ θ}$ ; Newton's rings (reflected, dark): $r_{n} = \frac{nλ R}{μ}$

1. What Is a Thin Film?

A thin film is a transparent layer whose thickness is comparable to the wavelength of light (from a few tens of nanometres to a few micrometres). Examples: a soap bubble, a layer of oil on a wet road, a magnesium fluoride coating on a camera lens, and the air gap between two glass plates.

When light meets such a film, part of it reflects at the top surface and part enters, reflects at the bottom surface and comes back out. The two reflected waves come from the same incident wave, so they are coherent. Splitting one wave into two weaker waves in this way is called division of amplitude.

	YDSE	Thin film
Coherent sources made by	Division of wavefront (two slits)	Division of amplitude (two surfaces)
Light source	Must be narrow (a slit)	Can be broad (sky, lamp)
Path difference decided by	Position on the screen, $\frac{y d}{D}$	Film thickness, $2 μ t cos r$
Fringes	Equally spaced straight lines	Fringes of equal thickness (colours, rings, bands)

2. Phase Change on Reflection

★ Must learnWhen light travelling in a rarer medium is reflected from a denser medium, the reflected wave suffers a phase change of

π

, equivalent to an extra path of

\frac{λ}{2}

. There is no phase change on reflection from a rarer medium, and none on transmission.

Figure 1: Reflection from a denser medium inverts the wave (phase change

π

, equivalent to an extra path

\frac{λ}{2}

); reflection from a rarer medium does not. Transmitted waves never change phase.

Event	Phase change	Equivalent extra path
Reflection at a denser medium (air to glass, air to water, film to glass)	$π$	$\frac{λ}{2}$
Reflection at a rarer medium (glass to air, soap film to air)	0	0
Refraction (transmission) into any medium	0	0

This is exactly what a pulse on a string does: at a fixed end (a "denser" boundary) it comes back upside down; at a free end it comes back upright.

2.1 Direct and reflected light near a mirror

The simplest use of this rule is a point P that receives light directly and after reflection from a plane mirror (as in Lloyd's mirror). The reflected wave travels further and also gains $\frac{λ}{2}$ at the mirror.

Figure 2: Direct and reflected light interfere at P. The reflected wave travels further (

2 x

, or

2 d cos θ

measured from the common wavefront PF) and gains

\frac{λ}{2}

from reflection at the denser mirror.

For light at angle $θ$ to the normal and P at height $d$ , the extra path measured from the common wavefront PF is $F M + M P = d sec θ cos 2 θ + d sec θ = 2 d cos θ$ . So

Δ x = 2 d cos θ + \frac{λ}{2}; maximum at P when 2 d cos θ = (n - \frac{1}{2}) λ, n = 1, 2, \dots

Key idea

Reflection from a denser medium adds a phase of

π

(half a wavelength); reflection from a rarer medium and transmission add nothing.

3. Path Difference in a Thin Film

Consider a film of thickness $t$ and refractive index $μ$ , with light incident at angle $i$ and refracted at angle $r$ .

Figure 3: Division of amplitude. Rays 1 and 2 come from the same incident ray (drawn with Snell's law,

i = 4 2^{\circ}

μ = 1.33

, so

r = 30. 2^{\circ}

). After CN they travel equal paths, so

Δ = μ (A B + B C) - A N = 2 μ t cos r

Ray 1 reflects at A. Ray 2 refracts at A, reflects at B and leaves at C, parallel to ray 1.
Draw CN perpendicular to ray 1. After N and C the two rays travel equal distances, so only the paths before CN matter.
Optical path of ray 2 inside the film: $μ (A B + B C) = μ \cdot \frac{2 t}{cos r}$ .
Path of ray 1 in air: $A N = A C sin i = (2 t tan r) sin i = 2 t tan r \cdot μ sin r$ , using Snell's law $sin i = μ sin r$ .
Subtract:
$Δ = \frac{2 μ t}{cos r} - \frac{2 μ t sin ^{2} r}{cos r} = 2 μ t cos r$

★ Must learnOptical path difference between the two reflected rays:

Δ = 2 μ t cos r

. For normal incidence (

r = 0

Δ = 2 μ t

. Add

\frac{λ}{2}

for each reflection that happens at a denser medium.

4. Conditions for Bright and Dark Reflection

Everything depends on how many of the two reflections suffer a $π$ change.

Case	Example	$π$ shifts	Bright (strong) reflection	Dark (weak) reflection
$μ_{1} > μ_{2} > μ_{3}$	Light from dense glass ( $1.7$ ) through a film ( $1.5$ ) into air	0	$2 μ t = mλ$	$2 μ t = (m + \frac{1}{2}) λ$
$μ_{1} < μ_{2} < μ_{3}$	$Mg F_{2}$ coating on glass	2 (cancel)	$2 μ t = mλ$	$2 μ t = (m + \frac{1}{2}) λ$
$μ_{2}$ greater than both, or less than both	Soap film in air, oil on water, air gap between glass plates	1	$2 μ t = (m + \frac{1}{2}) λ$	$2 μ t = mλ$

Here $m = 0, 1, 2, \dots$ and for oblique incidence $2 μ t$ becomes $2 μ t cos r$ .

Figure 4: Count the

π

tags. One tag (soap film in air) swaps the bright and dark conditions; two tags (

Mg F_{2}

on glass,

1 < 1.38 < 1.5

) cancel, so only

2 μ t

matters.

Exam Trick

Count the tags, then choose the formula. Mark $π$ at every reflection from a denser medium. Even number of tags (0 or 2): bright when $2 μ t = mλ$ . Odd number (1): bright when $2 μ t = (m + \frac{1}{2}) λ$ . Oil ( $μ = 1.5$ ) on water ( $1.33$ ): one tag. Oil ( $1.2$ ) on glass ( $1.5$ ): two tags.

4.1 Transmitted light

Transmitted rays suffer no phase change at the surfaces they cross, and energy is shared between reflection and transmission. So a wavelength strongly reflected is weakly transmitted, and vice versa: the conditions for transmitted light are the opposite of those for reflected light. A soap film that looks yellow in reflection looks bluish in transmission.

Reflected light

Conditions depend on the number of $π$ shifts. For a film in air (one shift): bright when $2 μ t = (m + \frac{1}{2}) λ$ , dark when $2 μ t = mλ$ . Fringes have high contrast.

Transmitted light

No phase change at either crossing, so the conditions are swapped: for a film in air, bright when $2 μ t = mλ$ . Fringes are faint, because most of the light passes straight through.

Figure 5: Thin film recipe: path difference

2 μ t cos r

, count the

π

shifts, pick the matching bright and dark conditions, and swap them for transmitted light.

Key idea

Count the

π

tags: one tag swaps the bright and dark conditions, two tags cancel. Transmitted light always shows the opposite pattern.

Quick Recall: tap to check

Soap film in air, normal incidence: condition for strong reflection?

2 μ t = (m + \frac{1}{2}) λ

(one

π

shift).

How many

π

shifts for

Mg F_{2}

(

1.38

) on glass (

1.5

Two (both reflections are from denser media), so they cancel.

Least

Mg F_{2}

thickness that kills reflection at

550 nm

t = \frac{λ}{4 μ} = \frac{550}{4 \times 1.38} = 99.6 nm

Why does a soap film look black just before it bursts?

t ≪ λ

, so only the

π

shift remains and the two reflected waves cancel.

5. Thin Films in White Light: Colours

For a given thickness, $2 μ t$ satisfies the bright condition for only a few wavelengths. Those colours are strongly reflected and the film takes on their colour. Where the thickness changes (a draining soap film, a spreading oil patch), the colour changes too, giving bands of colour called fringes of equal thickness.

Figure 6: Reflectance of the

320 nm

water film of Solved Example 1, computed exactly. Only

m = 1

(

567 nm

) is strongly reflected in the visible, so the film looks yellow-green.

Figure 7: A vertical soap film drains, so thickness increases downwards. Each band is one thickness reflecting its own colour (computed from the exact film reflectance). The top turns black just before it bursts.

Very thin film ( $t ≪ λ$ ): the path difference is almost zero, but one reflection has a $π$ shift, so the two reflected waves cancel for every colour. The film looks black; a soap bubble goes black just before it bursts.
Thick film: the bright condition is met by many wavelengths spread across the spectrum, so the colours mix to white. Also, ordinary light stays coherent only over a few micrometres, so very thick layers (a window pane) show no colours.
The colour seen also changes with the viewing angle, because $2 μ t cos r$ depends on $r$ .

6. Anti-Reflection Coatings

A bare glass surface reflects about $4%$ of the light. In a camera lens with many surfaces this wastes light and causes glare. A thin coating of magnesium fluoride ( $μ = 1.38$ , between air and glass) is chosen so that the two reflected waves cancel.

Both reflections are at denser media (air to $Mg F_{2}$ , $Mg F_{2}$ to glass), so the two $π$ shifts cancel.
For destructive interference: $2 μ t = (m + \frac{1}{2}) λ$ .
Thinnest coating ( $m = 0$ ):
$t_{min} = \frac{λ}{4 μ}$
This is a quarter-wave coating: its optical thickness $μ t$ is $\frac{λ}{4}$ .

Figure 8: A

Mg F_{2}

layer of thickness

\frac{λ}{4 μ} = 99.6 nm

cuts glass reflection from

4%

to about

1.4%

550 nm

(exact calculation). Red and violet are reflected a little more, so coated lenses look purple.

JEE Advanced

Perfect cancellation needs the two reflected waves to have equal amplitude as well as opposite phase. This happens when $μ_{co a t in g} = μ_{g l a ss}$ (about $1.22$ for $μ = 1.5$ ). $Mg F_{2}$ is close, so reflection falls to about $1.4%$ , not zero. The coating is designed for $550 nm$ (green, where the eye is most sensitive), so a little red and violet is still reflected: coated lenses look purple.

7. Wedge-Shaped Film

Two flat glass plates touching along one edge and separated slightly at the other enclose a thin air wedge of small angle $θ$ . At distance $x$ from the edge the air thickness is $t = x θ$ .

Figure 9: An air wedge. Thickness grows linearly from the contact edge, so equal-thickness fringes are straight, parallel to the edge and equally spaced,

β = \frac{λ}{2 θ}

; the edge is dark (one

π

shift).

The air film has one $π$ shift (reflection at the lower glass plate), so in reflected light (near normal incidence):

dark: 2 μ x_{n} θ = nλ, β = x_{n + 1} - x_{n} = \frac{λ}{2 μ θ}

The fringes are straight, parallel to the edge and equally spaced, and the line of contact ( $t = 0$ ) is dark. One fringe corresponds to a thickness change of $\frac{λ}{2 μ}$ , which is why wedge fringes are used to test the flatness of surfaces and to measure very thin wires.

Exam Trick

One fringe = one step of $\frac{λ}{2 μ}$ in thickness. A wire of diameter $D_{w}$ at the open end of an air wedge produces $N = \frac{2 D _{w}}{λ}$ dark fringes, so $D_{w} = \frac{N λ}{2}$ . Example: $20$ fringes in $600 nm$ light mean $D_{w} = 6 μ m$ . The same count works for Newton's rings: the $n$ th dark ring sits where the air gap is $\frac{nλ}{2}$ .

8. Newton's Rings

Figure 10: Newton's rings. The air film under the lens has constant thickness on circles, so the fringes are rings; their radii grow as

n

, so rings crowd outwards. The centre is dark in reflected light.

A plano-convex lens of large radius $R$ resting on a flat plate traps an air film of thickness $t \approx \frac{r ^{2}}{2 R}$ at distance $r$ from the contact point. Thickness is constant on circles, so the fringes are rings.

JEE Advanced

With one $π$ shift, dark rings in reflected light satisfy $2 μ t = nλ$ , so

r_{n}^{2} = \frac{nλ R}{μ}, r_{n} = \frac{nλ R}{μ}

The centre ( $t = 0$ ) is dark, ring radii grow as $n$ (rings crowd outwards), and filling the gap with a liquid of index $μ$ shrinks every radius by $μ$ . Measuring $r_{n + k}^{2} - r_{n}^{2} = \frac{k λ R}{μ}$ gives $λ$ or $R$ .

Seen from below (in transmitted light) the pattern is the complement of the reflected one: the centre is bright and every ring that is dark in reflection is bright in transmission. The transmitted rings are faint, because only a few per cent of the light is reflected at each glass surface.

Figure 11: Newton's rings computed from

t = \frac{r ^{2}}{2 R}

(dark centre and 8 dark rings in reflection). Transmitted light shows the complementary pattern: where reflection is dark, transmission is bright, because

I_{R} + I_{T} = I

(no absorption). The transmitted rings are faint because only about

4%

of the light is reflected at each glass surface.

Key idea

Equal-thickness fringes follow lines of constant

t

: straight lines for a wedge, circles for Newton's rings, with

r_{n} = nλ R

for the dark rings.

Quick Recall: tap to check

Fringe width of an air wedge of angle

θ

β = \frac{λ}{2 θ}

; in a liquid of index

μ

\frac{λ}{2 μ θ}

Radius of the

n

th dark ring in reflected light?

r_{n} = nλ R

(air film).

A liquid of index

μ

fills the gap. What happens to the rings?

Every radius shrinks by

μ

Why do Newton's rings crowd together further out?

r_{n} \propto n

, so the gap between neighbouring rings falls as

n

grows.

Figure 12: Revision map of thin films: phase changes,

Δ = 2 μ t cos r

, the two sets of conditions, coatings, wedges and Newton's rings.

9. Solved Examples

Solved Example 1

White light, uniform over

430

690 nm

, falls perpendicularly on a water film (

μ = 1.33

, thickness

320 nm

) suspended in air. At what wavelength is the reflected light brightest?

Solution:

Film in air: one $π$ shift (top surface only), so bright reflection needs $2 μ t = (m + \frac{1}{2}) λ$ .

$λ = \frac{2 μ t}{m + 1/2} = \frac{2 \times 1.33 \times 320}{m + 1/2} = \frac{851 nm}{m + 1/2}$ .

$m = 0$ : $1702 nm$ (infrared). $m = 1$ : $567 nm$ (yellow-green). $m = 2$ : $340 nm$ (ultraviolet).

Answer: $λ = 567 nm$ (Figure 6).

Solved Example 2

Light falls normally on a plane mirror. Find the minimum height

x

of a point P above the mirror at which a maximum is obtained.

Solution:

The reflected wave travels an extra $2 x$ and gains $\frac{λ}{2}$ on reflection: $Δ x = 2 x + \frac{λ}{2}$ .

For a maximum, $Δ x = nλ$ ; the smallest non-zero case is $Δ x = λ$ : $2 x + \frac{λ}{2} = λ \Rightarrow x = \frac{λ}{4}$ .

Answer: $x_{min} = \frac{λ}{4}$ (Figure 2, left).

Solved Example 3

Parallel light falls on a plane mirror at angle

θ

to the normal. Point P is at height

d

above the mirror. Find

θ

for which the first maximum is obtained at P.

Solution:

From Figure 2 (right), the reflected ray travels an extra $F M + M P = d sec θ cos 2 θ + d sec θ = d sec θ (1 + cos 2 θ) = 2 d cos θ$ , and gains $\frac{λ}{2}$ at the mirror.

First maximum: $2 d cos θ + \frac{λ}{2} = λ \Rightarrow cos θ = \frac{λ}{4 d}$ .

Answer: $θ = cos^{- 1} (\frac{λ}{4 d})$ .

Solved Example 4

A glass lens (

μ = 1.50

) is coated with magnesium fluoride (

μ = 1.38

) to reduce reflection. What is the least coating thickness that eliminates reflection at

λ = 550 nm

for nearly normal light?

Solution:

Air ( $1.00$ ) $<$ $Mg F_{2}$ ( $1.38$ ) $<$ glass ( $1.50$ ): both reflections are from denser media, so the two $π$ shifts cancel.

Destructive interference: $2 μ t = (m + \frac{1}{2}) λ$ . Least thickness, $m = 0$ :

$t = \frac{λ}{4 μ} = \frac{550 nm}{4 \times 1.38} = 99.6 nm$ .

Answer: about $100 nm$ ( $99.6 nm$ ; Figure 8).

Solved Example 5

White light may be taken as

4000 \overset{˚}{A}

7500 \overset{˚}{A}

. An oil film (

μ = 1.40

) of thickness

1 0^{- 6} m

is in air. Find the visible wavelengths for which reflection along the normal is (i) weak, (ii) strong.

Solution:

Film in air: one $π$ shift. $2 μ t = 2 \times 1.40 \times 1 0^{- 6} = 2.8 \times 1 0^{- 6} m = 28000 \overset{˚}{A}$ .

(i) Weak reflection: $2 μ t = nλ \Rightarrow λ = \frac{28000}{n} \overset{˚}{A}$ . Visible for $n = 4, 5, 6, 7$ : $7000$ , $5600$ , $4667$ , $4000 \overset{˚}{A}$ .

(ii) Strong reflection: $2 μ t = (n + \frac{1}{2}) λ \Rightarrow λ = \frac{56000}{2 n + 1} \overset{˚}{A}$ . Visible for $n = 4, 5, 6$ : $6222$ , $5091$ , $4308 \overset{˚}{A}$ .

Answer: weak: $7000, 5600, 4667, 4000 \overset{˚}{A}$ ; strong: $6222, 5091, 4308 \overset{˚}{A}$ . (In transmitted light the roles swap.)

Solved Example 6

Find the least thickness of a soap film (

μ = 1.33

) in air that strongly reflects light of wavelength

600 nm

at normal incidence.

Solution:

One $π$ shift, so strong reflection: $2 μ t = (m + \frac{1}{2}) λ$ ; least thickness for $m = 0$ :

$t = \frac{λ}{4 μ} = \frac{600}{4 \times 1.33} = 112.8 nm$ .

Answer: about $113 nm$ .

Solved Example 7

A film of thickness

0.5 μ m

and

μ = 1.5

in air is viewed so that the angle of refraction in the film is

3 0^{\circ}

. Which visible wavelength (

400

700 nm

) is strongly reflected?

Solution:

$2 μ t cos r = 2 \times 1.5 \times 500 \times cos 3 0^{\circ} = 1299 nm$ . One $π$ shift, so $λ = \frac{1299}{m + 1/2}$ .

$m = 1$ : $866 nm$ (infrared); $m = 2$ : $520 nm$ ; $m = 3$ : $371 nm$ (ultraviolet).

Answer: $520 nm$ (green). At normal incidence the answer would change, because $cos r$ changes: the colour of a film depends on the viewing angle.

Solved Example 8

A soap film looks black just before it bursts because
(A) it absorbs all light
(B) its thickness is much less than

λ

and the two reflected waves differ in phase by

π

Solution:

As $t \to 0$ the path difference $2 μ t \to 0$ , but the top reflection (air to film) has a $π$ shift and the bottom one (film to air) does not. The two reflected waves are exactly out of phase for every wavelength and cancel.

Answer: (B).

Solved Example 9

An air wedge is formed between two glass plates and lit normally with light of

600 nm

. The fringe width is

0.3 mm

. Find the wedge angle. What is the fringe width if the wedge is filled with water (

μ = \frac{4}{3}

Solution:

$β = \frac{λ}{2 μ θ} \Rightarrow θ = \frac{λ}{2 β} = \frac{6 \times 1 0 ^{- 7}}{2 \times 0.3 \times 1 0 ^{- 3}} = 1 0^{- 3} rad$ .

With water, $β^{'} = \frac{β}{μ} = \frac{0.3}{4/3} = 0.225 mm$ .

Answer: $θ = 1 0^{- 3} rad$ (about $0.05 7^{\circ}$ ); $β^{'} = 0.225 mm$ .

Solved Example 10

A thin oil film (

μ = 1.40

) floats on water (

μ = 1.33

) and is lit normally with white light (

400

700 nm

). The film is

500 nm

thick. Which wavelengths are strongly reflected?

Solution:

Count the tags: air to oil is rarer to denser ( $π$ shift); oil to water is denser to rarer (no shift). One $π$ shift, so strong reflection needs $2 μ t = (m + \frac{1}{2}) λ$ .

$2 μ t = 2 \times 1.40 \times 500 = 1400 nm$ , so $λ = \frac{1400}{m + 0.5}$ : $m = 1$ gives $933 nm$ (infrared), $m = 2$ gives $560 nm$ , $m = 3$ gives $400 nm$ (the violet limit).

Answer: $560 nm$ (yellow-green), with $400 nm$ just at the violet end; the patch looks yellow-green. If the oil had a lower index than water (say $1.25$ ) there would be two $π$ shifts and the conditions would swap.

Solved Example 11

Newton's rings are viewed in transmitted light. The centre of the pattern is
(A) dark
(B) bright
(C) coloured even with monochromatic light
(D) absent

Solution:

At the centre $t = 0$ . In reflection the only path difference is the $\frac{λ}{2}$ from the $π$ shift, so the centre is dark. Transmitted light suffers no phase change at the crossings, and energy not reflected must be transmitted (Figure 11).

Answer: (B). The centre is bright in transmitted light; the whole pattern is complementary to the reflected one, with fainter rings.

Practice Questions

Find the least thickness of an $Mg F_{2}$ coating ( $μ = 1.38$ ) on glass that is non-reflecting for $500 nm$ .Answer: $t = \frac{λ}{4 μ} = 90.6 nm$ .
A soap film ( $μ = 1.33$ ) of thickness $500 nm$ in air is lit normally with white light. Which visible wavelength is strongly reflected?Answer: $λ = \frac{1330}{m + 1/2}$ : $532 nm$ (green, $m = 2$ ).
An oil film ( $μ = 1.5$ ) floats on water ( $μ = 1.33$ ). Find its least thickness for strong reflection of $600 nm$ .Answer: One $π$ shift (top only): $t = \frac{λ}{4 μ} = 100 nm$ .
A film of $μ = 1.25$ on glass ( $μ = 1.5$ ) should give minimum reflection at $500 nm$ . Find its least thickness.Answer: Two $π$ shifts cancel: $2 μ t = \frac{λ}{2}$ , so $t = 100 nm$ .
Twenty dark fringes of an air wedge span $1 cm$ in light of $500 nm$ . Find the wedge angle.Answer: $β = 0.5 mm$ , $θ = \frac{λ}{2 β} = 5 \times 1 0^{- 4} rad$ .
In Newton's rings with $R = 1 m$ and $λ = 600 nm$ (air film), find the radius of the 5th dark ring.Answer: $r_{5} = 5 λ R = 1.73 mm$ .

Common Mistakes to Avoid

Watch out

Writing $2 t$ instead of $2 μ t$ : the path inside the film must be multiplied by $μ$ (or use $λ / μ$ , but never both).
Adding $\frac{λ}{2}$ automatically. Count the reflections from denser media: one shift for a film in air, two (cancelling) for a coating between air and glass.
Getting oil on water wrong: oil ( $1.5$ ) on water ( $1.33$ ) has one shift; oil ( $1.2$ ) on water ( $1.33$ ) has two.
Using the reflected-light condition for transmitted light. They are opposite.
Dropping $cos r$ at oblique incidence: the path difference is $2 μ t cos r$ with $r$ the angle inside the film, not $i$ .
Taking the anti-reflection thickness as $\frac{λ}{2 μ}$ . The least thickness is $\frac{λ}{4 μ}$ (quarter wave).
Expecting a coating to remove reflection at every wavelength; it works fully only near its design wavelength.
Saying the centre of Newton's rings (or the edge of an air wedge) is bright in reflected light. With one $π$ shift and zero thickness, it is dark.

Frequently Asked Questions

What is thin film interference?

Thin film interference is the interference of light reflected from the top and bottom surfaces of a transparent layer whose thickness is comparable to the wavelength. The two reflected waves come from one incident wave by division of amplitude, and their path difference $2 μ t cos r$ , plus any reflection phase change, decides which colours are reflected strongly.

Why does light suffer a phase change of pi on reflection?

When light in a rarer medium reflects from a denser medium, the reflected wave is inverted, just as a pulse on a string reflects upside down from a fixed end. Inversion is a phase change of $π$ , equal to half a wavelength of path. Reflection from a rarer medium and transmission produce no phase change.

Why do soap bubbles show colours?

The soap film has different thicknesses at different places. At each point, light reflected from the outer and inner surfaces interferes constructively for some wavelengths and destructively for others, so that point shows the colour of the reinforced wavelengths. As the film drains and thins, the colours move, and a very thin film looks black.

What is the minimum thickness of an anti-reflection coating?

For a coating whose index lies between air and glass, both reflections have a $π$ shift, so destructive interference needs $2 μ t = λ /2$ . The least thickness is $λ / (4 μ)$ , a quarter-wave layer. For magnesium fluoride ( $μ = 1.38$ ) at 550 nm this is about 100 nm.

Why does a soap film look black just before it bursts?

When the film becomes much thinner than the wavelength, the path difference between the two reflected waves is almost zero. Only the top reflection has a $π$ phase change, so the two waves are exactly out of phase for every colour and cancel. Almost no light is reflected and the film appears black.

Are the conditions for transmitted light the same as for reflected light?

No, they are opposite. Transmitted rays have no reflection phase change at the surfaces they cross, and energy not reflected is transmitted. A wavelength that is strongly reflected by a film is weakly transmitted, so a film that looks yellow in reflected light looks bluish when you look through it.

Which thin film questions are asked in NEET?

NEET asks for the phase change on reflection, the condition $2 μ t = (m + 1/2) λ$ for strong reflection from a film in air, the minimum thickness $λ / (4 μ)$ of a coating or soap film, and why thin films show colours. Remember to multiply the thickness by the refractive index.

How is thin film interference tested in JEE?

JEE mixes thin films with reflection phase rules: films on substrates with different indices, oblique incidence with cos r, finding all visible wavelengths reflected strongly or weakly, direct and reflected light near a mirror, air wedges and Newton's rings. Counting the $π$ shifts correctly is the step most students get wrong.

Previous year questions on Interference by Thin Film

5 questions from past papers, each with a step-by-step solution.

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