Thin Film Interference Notes FE Sem 1 Applied Physics | MU

Soap bubbles and oil films show bright colours because light reflected from the top and bottom surfaces of a thin film interferes. This module derives the path difference for a film of uniform thickness in reflected light, gives the conditions for maxima and minima, and describes the wedge-shaped film qualitatively. It then applies these ideas to two engineering uses: Newton's rings for measuring an unknown wavelength and the refractive index of a liquid, and anti-reflection coatings on lenses and solar cells. It maps to course outcome CO3, and the Newton's rings derivation and numericals are regular exam questions.

Summary

Interference by division of amplitude: When a light wave falls on a thin transparent film, part of it is reflected from the upper surface and part is refracted, reflected from the lower surface and comes out again. These two reflected beams come from the same incident wave, so they are coherent, and they overlap to produce interference. This is called interference by division of amplitude, as opposed to division of wavefront in Young's double slit. Thin film means a thickness comparable to the wavelength of light, so that the path difference is small enough to give visible fringes.

Path difference in a film of uniform thickness (reflected system): Consider a film of thickness t and refractive index \(\mu\) in air, with light incident at angle i and refracted at angle r. The geometrical path difference between the ray reflected from the top and the ray reflected from the bottom works out to an optical path difference of \(2\mu t\cos r\). In addition, reflection at the top surface (air to denser film) causes a phase change of \(\pi\), equivalent to an extra path of \(\lambda/2\) (Stokes' law), while reflection at the bottom (film to air) does not. Hence the effective path difference is \(\Delta = 2\mu t\cos r - \lambda/2\).

Conditions for maxima and minima in reflected light: Constructive interference (bright) occurs when \(\Delta = n\lambda\), giving \(2\mu t\cos r = (2n-1)\lambda/2\), n = 1, 2, 3 and so on. Destructive interference (dark) occurs when \(\Delta = (2n+1)\lambda/2\), giving \(2\mu t\cos r = n\lambda\), n = 0, 1, 2. With white light, different wavelengths satisfy the bright condition at different thicknesses and angles, so the film appears coloured. A very thin film (\(t \ll \lambda\)) appears dark in reflected light because the path difference is only the \(\lambda/2\) from reflection. The conditions are reversed in transmitted light, so the reflected and transmitted patterns are complementary.

Wedge-shaped film (qualitative): If two glass plates touch at one edge and are separated by a thin spacer at the other, the air film between them has thickness increasing uniformly from zero, forming a wedge of small angle \(\theta\). For near-normal incidence the path difference is \(2\mu t\cos(r+\theta) - \lambda/2 \approx 2\mu t - \lambda/2\). Bright fringes occur where \(2\mu t = (2n-1)\lambda/2\) and dark fringes where \(2\mu t = n\lambda\). Since t is constant along lines parallel to the edge of contact, the fringes are equally spaced straight lines parallel to that edge, with a dark fringe at the edge itself (t = 0). The fringe width is \(\beta = \lambda/(2\mu\theta)\). Wedge fringes are used to test the flatness of surfaces and to measure very small thicknesses such as a thin wire or paper.

Newton's rings: A plano-convex lens of large radius of curvature R is placed with its convex surface on a flat glass plate, enclosing an air film whose thickness increases from zero at the point of contact. Monochromatic light falling normally is reflected from the top and bottom of this film and interference gives concentric circular bright and dark rings, with a dark centre in reflected light. The film thickness at radius r is \(t = r^2/2R\). Using the dark ring condition \(2\mu t = n\lambda\), the diameter of the nth dark ring is \(D_n^2 = 4n\lambda R/\mu\); for air \(\mu = 1\). The rings get closer together as n increases because \(D_n\) is proportional to \(\sqrt{n}\).

Determining wavelength and refractive index with Newton's rings: In air, measuring the diameters of the nth and (n+p)th dark rings with a travelling microscope gives \(\lambda = (D_{n+p}^2 - D_n^2)/(4pR)\); taking the difference removes the error due to dust or imperfect contact at the centre. If a transparent liquid of refractive index \(\mu\) is introduced between lens and plate, every ring shrinks, and \(\mu = (D_{n+p}^2 - D_n^2)_{air}/(D_{n+p}^2 - D_n^2)_{liquid}\). R is found with a spherometer.

Anti-reflecting coating: A glass surface reflects about 4 percent of light, and a multi-element lens loses a lot of light and gets glare. A thin transparent layer, usually magnesium fluoride (\(\mu \approx 1.38\)), is coated on the glass. Its index lies between air and glass, so both reflections (air-film and film-glass) have the same \(\pi\) phase change and cancel the \(\lambda/2\) terms. For destructive interference of the two reflected beams, the film's optical thickness must be a quarter wavelength: \(\mu_f t = \lambda/4\), i.e. \(t = \lambda/4\mu_f\). For complete cancellation the two beams must also have equal amplitude, which needs \(\mu_f = \sqrt{\mu_g}\). The coating is designed for about 550 nm (the eye's peak sensitivity), giving coated lenses a purple tint. Uses: camera and spectacle lenses, binoculars, solar cells and display screens.

Mind map

Newton's rings: deriving the dark ring diameter and finding λ

Important definitions

Formulas

Key points to remember

• Thin film interference is interference by division of amplitude.
• Extra path λ/2 arises from reflection at a denser medium (top surface of a film in air).
• Reflected light: bright when 2μt cos r = (2n−1)λ/2, dark when 2μt cos r = nλ; transmitted light has the opposite conditions.
• An extremely thin film looks black in reflected light.
• Wedge film gives straight, parallel, equally spaced fringes; the edge of contact is dark.
• Newton's rings centre is dark in reflected light and bright in transmitted light.
• Ring diameters are proportional to √n, so rings get closer outward.
• Liquid between lens and plate makes rings contract: D(liquid)² = D(air)²/μ.
• AR coating: quarter-wave optical thickness and μf = √μg; MgF2 with μ = 1.38 is common.
• Using the difference D(n+p)² − D(n)² avoids errors from dust at the point of contact.

Solved examples

Example 1. A soap film of refractive index 1.33 and thickness 5000 Å is illuminated by white light at an angle of 45 degrees. Which wavelength in the visible region is absent in the reflected light?

\(\sin r = \sin45^\circ/1.33 = 0.7071/1.33 = 0.5317\), so \(r = 32.12^\circ\) and \(\cos r = 0.8470\). \(2\mu t\cos r = 2\times1.33\times5000\times0.8470 = 11265\) Å. Dark condition: \(\lambda = 2\mu t\cos r/n\). n = 1: 11265 Å (infrared); n = 2: 5632 Å (visible); n = 3: 3755 Å (ultraviolet). So 5632 Å (about 563 nm, yellow-green) is absent.

Example 2. An oil film (μ = 1.4) of thickness 0.4 μm on water is seen at normal incidence in white light. Find the wavelengths strongly reflected and absent in the visible range (400 to 700 nm). Assume μ(oil) > μ(water).

Reflection at the top (air to oil) gives a λ/2 change, but reflection at the bottom (oil to water, 1.4 to 1.33) does not, so the standard reflected-light conditions apply. \(2\mu t = 2\times1.4\times400 = 1120\) nm. Bright: \(\lambda = 2(2\mu t)/(2n-1)\): n = 1, 2240 nm; n = 2, 747 nm; n = 3, 448 nm. Visible bright: 448 nm (blue); 747 nm lies just beyond the red end. Dark: \(\lambda = 2\mu t/n\): n = 2, 560 nm (visible); n = 3, 373 nm. Absent: 560 nm. The film looks bluish-violet.

Example 3. In a Newton's rings experiment the diameters of the 5th and 15th dark rings are 0.336 cm and 0.590 cm. If the radius of curvature of the lens is 100 cm, find the wavelength of light.

\(\lambda = (D_{15}^2 - D_5^2)/(4pR)\) with p = 10. \(D_{15}^2 = 0.3481\) cm², \(D_5^2 = 0.1129\) cm². Difference = 0.2352 cm². \(\lambda = 0.2352/(4\times10\times100) = 5.88\times10^{-5}\) cm = 5880 Å.

Example 4. The diameter of the 10th dark ring in Newton's rings changes from 1.40 cm to 1.27 cm when a liquid is introduced between lens and plate. Find the refractive index of the liquid.

For the same ring \(D_{air}^2 = 4n\lambda R\) and \(D_{liq}^2 = 4n\lambda R/\mu\). So \(\mu = D_{air}^2/D_{liq}^2 = 1.96/1.6129 = 1.215\).

Example 5. Find the radius of the 10th dark ring in Newton's rings with sodium light of 5890 Å and a lens of radius of curvature 1 m (air film).

\(r_n = \sqrt{n\lambda R} = \sqrt{10\times5890\times10^{-10}\times1} = \sqrt{5.89\times10^{-6}} = 2.43\times10^{-3}\) m = 2.43 mm.

Example 6. Calculate the minimum thickness of a magnesium fluoride (μ = 1.38) anti-reflection coating on glass (μ = 1.52) for light of 5500 Å. What would be the ideal coating index?

\(t = \lambda/4\mu_f = 5500/(4\times1.38) = 996\) Å, about 99.6 nm. Ideal index \(\mu_f = \sqrt{1.52} = 1.233\); MgF2 is used as the nearest durable material.

Example 7. A film of index 1.5 surrounded by air and viewed at normal incidence with light of 600 nm. Find the minimum thickness that appears bright and the minimum non-zero thickness that appears dark in reflected light.

Bright: \(2\mu t = \lambda/2\) (n = 1), so \(t = \lambda/4\mu = 600/6 = 100\) nm. Dark: \(2\mu t = \lambda\) (n = 1), so \(t = \lambda/2\mu = 600/3 = 200\) nm.

Example 8. In a Newton's rings arrangement the diameter of a particular dark ring is 3.0 mm in air and 2.5 mm with a liquid. Find μ of the liquid.

\(\mu = (3.0/2.5)^2 = 1.2^2 = 1.44\).

Important questions and answers

Q1. Derive the conditions for maxima and minima for a thin film of uniform thickness in reflected light. (6 marks)

Let a ray AB be incident at angle i on a film of thickness t and index μ in air. It is partly reflected along BC and partly refracted along BD at angle r; it reflects at D and emerges at E along EF parallel to BC. Draw EG perpendicular to BC. Path difference \(= \mu(BD + DE) - BG\). From geometry \(BD = DE = t/\cos r\), so \(\mu(BD + DE) = 2\mu t/\cos r\). \(BE = 2t\tan r\) and \(BG = BE\sin i = 2t\tan r\sin i = 2t\tan r\,\mu\sin r\) (Snell's law). So path difference \(= 2\mu t/\cos r - 2\mu t\sin^2 r/\cos r = 2\mu t\cos r\). Reflection at B (air to film) adds a phase change π, i.e. λ/2, so \(\Delta = 2\mu t\cos r - \lambda/2\). Maxima: \(\Delta = n\lambda \Rightarrow 2\mu t\cos r = (2n-1)\lambda/2\). Minima: \(\Delta = (2n+1)\lambda/2 \Rightarrow 2\mu t\cos r = n\lambda\). Draw the ray diagram with both reflected rays.

Q2. Why does a thin film appear coloured in white light? Why does an excessively thin film appear dark? (3 marks)

Path difference \(2\mu t\cos r - \lambda/2\) depends on λ, t and r. For a given point (fixed t and r) only certain wavelengths satisfy the bright condition and others the dark condition, so the reflected light lacks some colours and is enriched in others; as t or viewing angle changes, the colour changes. For an excessively thin film, \(2\mu t\cos r \approx 0\), so the path difference is λ/2 for all wavelengths, which is the condition for destructive interference; every colour cancels and the film looks dark.

Q3. Explain the formation of fringes in a wedge-shaped film qualitatively and state the conditions for bright and dark fringes. (4 marks)

Two plane glass plates touching at one edge and separated by a thin wire at the other enclose a wedge-shaped air film of small angle θ. Light falling normally is reflected from the top and bottom of the air film; the reflected rays interfere. Path difference \(2\mu t\cos(r+\theta) - \lambda/2\), which for normal incidence and small θ is \(2\mu t - \lambda/2\). Bright: \(2\mu t = (2n-1)\lambda/2\); dark: \(2\mu t = n\lambda\). Thickness is constant along lines parallel to the edge, so the fringes are straight, parallel to the edge and equally spaced, with fringe width \(\beta = \lambda/2\mu\theta\). At the edge t = 0, so the edge is dark. Uses: testing optical flatness, measuring the diameter of a thin wire.

Q4. Describe Newton's rings experiment and show that the diameters of dark rings are proportional to the square root of natural numbers. (6 marks)

Set-up: a monochromatic source S, a convex lens to make the beam parallel, a glass plate at 45 degrees reflecting light vertically down onto a plano-convex lens L of radius R resting on a flat plate P, and a travelling microscope above. The air film between L and P is thin at the centre and thickens outward. Reflected rays from the top and bottom of the film interfere, giving concentric rings (locus of constant thickness) with a dark centre. Theory: in triangle geometry \(r^2 = (2R - t)t \approx 2Rt\), so \(t = r^2/2R\). Dark ring: \(2\mu t\cos r = n\lambda\), with normal incidence cos r = 1: \(2\mu r_n^2/2R = n\lambda\), \(r_n^2 = n\lambda R/\mu\), \(D_n^2 = 4n\lambda R/\mu\). Hence \(D_n \propto \sqrt{n}\), so the rings are not equally spaced and crowd together away from the centre. The central spot is dark because at t = 0 the path difference is λ/2.

Q5. How is Newton's rings experiment used to determine the wavelength of monochromatic light? (5 marks)

Focus the travelling microscope on the rings and, starting from say the 20th dark ring on the left, note the readings of successive dark rings through the centre to the 20th ring on the right. Diameter of each ring = difference of left and right readings. For the nth and (n+p)th rings: \(D_n^2 = 4n\lambda R\) and \(D_{n+p}^2 = 4(n+p)\lambda R\) (air film). Subtracting: \(\lambda = (D_{n+p}^2 - D_n^2)/(4pR)\). R is measured with a spherometer. A graph of \(D_n^2\) against n is a straight line of slope \(4\lambda R\), which gives an averaged λ. Using differences removes errors due to dust or imperfect contact at the centre.

Q6. Explain how the refractive index of a liquid is determined using Newton's rings. (4 marks)

First the diameters of the nth and (n+p)th dark rings are measured with an air film: \(D_{n+p}^2 - D_n^2 = 4p\lambda R\). The lens and plate are then placed in a dish and a few drops of the liquid are put between them without disturbing the arrangement, and the same rings are measured again. With a liquid film of index μ: \(D_{n+p}'^2 - D_n'^2 = 4p\lambda R/\mu\). Dividing, \(\mu = (D_{n+p}^2 - D_n^2)/(D_{n+p}'^2 - D_n'^2)\). The rings contract in the liquid since μ > 1.

Q7. What is an anti-reflection coating? Derive the conditions on its thickness and refractive index. (5 marks)

An anti-reflection coating is a thin transparent film deposited on lenses or solar cells to reduce reflection and increase transmission. Let the film of index \(\mu_f\) and thickness t lie on glass of index \(\mu_g\), with \(1 < \mu_f < \mu_g\). Both reflections (air-film and film-glass) are from denser media, so each has a phase change π; these cancel and the path difference is simply \(2\mu_f t\) at normal incidence. For destructive interference \(2\mu_f t = \lambda/2\) (minimum thickness), so \(t = \lambda/4\mu_f\): the coating is a quarter wave thick optically. Amplitude condition: for complete cancellation the two reflected beams must have equal amplitude, i.e. \((\mu_f - 1)/(\mu_f + 1) = (\mu_g - \mu_f)/(\mu_g + \mu_f)\), which gives \(\mu_f = \sqrt{\mu_g}\). For glass of 1.5, \(\mu_f\) is about 1.22; MgF2 (1.38) is used in practice, designed at 550 nm. The energy not reflected is transmitted, so lenses become brighter and glare-free.

Q8. Why is the centre of Newton's rings dark in reflected light? (3 marks)

At the point of contact the thickness of the air film is t = 0, so the geometric path difference \(2\mu t\) is zero. The ray reflected from the glass plate (air to glass, denser) undergoes a phase change of π, while the ray reflected from the lower surface of the lens (glass to air) does not. The net path difference is λ/2, the condition for destructive interference, so the centre is dark. In transmitted light the centre is bright.

Q9. Why are Newton's rings circular and why do they get closer away from the centre? (3 marks)

Each ring is a locus of points of equal film thickness. Since the curved lens surface touches the plate at one point and the film thickness depends only on the distance r from that point (\(t = r^2/2R\)), points of equal thickness lie on circles. Since \(D_n \propto \sqrt{n}\), the difference \(D_{n+1} - D_n\) decreases as n increases, so the rings get closer together outward.

Q10. Compare thin film interference in a uniform film and in a wedge-shaped film. (4 marks)

Uniform film: thickness constant; colour or brightness depends on the angle of incidence (fringes of equal inclination); with white light the whole film shows a uniform tint at a given angle; path difference \(2\mu t\cos r - \lambda/2\). Wedge film: thickness varies linearly from zero; fringes are of equal thickness, straight, parallel to the edge and equally spaced; edge of contact is dark; path difference \(2\mu t\cos(r+\theta) - \lambda/2\). Both are interference by division of amplitude in reflected light.

Exam tips

• In the uniform film derivation, show the ray diagram, the geometry for BD, DE and BG, and do not forget the λ/2 from Stokes' law.
• Remember: reflected light dark is 2μt cos r = nλ, bright is (2n−1)λ/2. Students often swap them.
• In Newton's rings numericals convert all lengths to the same unit (cm or m) before substituting.
• Use p = (n+p) − n correctly; for rings 5 and 15, p = 10.
• When a liquid is introduced the rings shrink; if your μ comes out less than 1, you have inverted the ratio.
• For AR coating state both conditions: t = λ/4μf and μf = √μg.
• Wedge film is qualitative in the syllabus; conditions plus a neat sketch of straight fringes are enough.

FAQs

Why must the film be thin for interference to be seen?

Light from ordinary sources has a short coherence length. If the film is thick, the path difference exceeds the coherence length and the two reflected beams are no longer coherent, and with white light the colours overlap and wash out.

Why is the λ/2 term not used in the anti-reflection coating condition?

Both reflections occur at denser media (air to coating and coating to glass), so both beams get a π phase change; they cancel and only the path 2μf t remains.

What happens to Newton's rings if white light is used?

Only a few coloured rings are seen near the centre, because each wavelength forms rings of different diameters; further out the colours overlap and give uniform illumination.

Why is a plano-convex lens of large radius of curvature used?

A large R makes the air film thicken slowly, so the rings are large and widely spaced and easy to measure with a travelling microscope.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Physics) and were prepared with AI help and checked.