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Michelson Interferometer

The Michelson interferometer is an optical instrument that divides a beam of light into two paths, reflects them back, and recombines them to produce interference fringes. It is used for precise measurements of wavelengths, small refractive index changes, and standard lengths.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Basic working principle, types of fringes (circular/equal inclination and localized fringes), and wavelength determination using monochromatic sources.

PGT extension

Detailed path difference calculations, visibility of fringes, phase change on reflection, white light fringes, and determination of wavelength difference for closely spaced spectral lines.

Prerequisites

Wave theory of light and superposition principle
Conditions for sustained interference
Path difference and phase difference relationship

Principle of Operation

A monochromatic light beam is incident on a beam splitter (a partially silvered glass plate) which divides the amplitude into two mutually perpendicular paths.
The two split beams travel towards two plane mirrors, get reflected back, and reunite at the beam splitter to enter the detector or observer's eye.
Interference occurs due to division of amplitude, producing fringes whose shape depends on the alignment and separation of the mirrors.

Types of Fringes

Circular fringes (Haidinger fringes) are formed when the two reflecting mirrors are strictly parallel, corresponding to interference by equal inclination.
Localized fringes are formed when the mirrors are inclined at an angle, creating an air wedge between one mirror and the virtual image of the other.
White light interference produces only a few colored fringes and a central achromatic dark/bright fringe, useful for finding the zero path difference position.

Role of Compensator Plate

In white light interferometry, the beam reflected from the beam splitter passes through the glass plate thrice, whereas the transmitted beam passes through it only once.
A compensator plate of identical thickness and material is placed in the path of the transmitted beam to equalize the optical path in glass for both arms.
Without the compensator plate, white light fringes cannot be observed due to dispersion and unequal optical paths.

Applications

Measurement of the wavelength of monochromatic light with high precision.
Determination of the refractive index of thin transparent films or gases.
Historical significance in proving the non-existence of the luminiferous ether via the Michelson-Morley experiment.

Derivations

Path Difference for Circular Fringes

Conditions: Mirrors M1 and M2 are exactly perpendicular, and the effective air film thickness between M1 and the virtual image M2' is d.

Consider a ray of light entering at an angle of incidence i to the air film of thickness d.
The path difference between the two interfering beams is given by Delta = 2d cos(theta), where theta is the angle of refraction inside the film.
Accounting for a phase change of pi (or lambda/2) due to reflection from the denser medium (the silvered surface of the beam splitter), the effective path difference becomes Delt$a = 2d \cos(\theta) +/- \lambda/2$.

Result: 2d cos(theta) = m\lambda for bright fringes and 2d cos(theta) = (m + 1/2)\lambda for dark fringes

Wavelength Determination Derivation

Conditions: Mirror M1 is translated by a distance delta d to shift N circular fringes past the crosshair.

For the center of the interference pattern (theta = 0, cos(theta) = 1), the condition for a bright fringe of order m is 2d = m\lambda.
When mirror M1 is moved by delta d, the thickness changes to d + delta d, and a new fringe order m - N is observed at the center: 2(d + delta d) = (m - N)\lambda.
Subtracting the two equations yields 2 delta d = N\lambda, which allows calculation of lambda.

Result: $\lambda = \frac{2 \delta d}{N}$

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Path Difference with Oblique Incidence	$$\Delta = 2d \cos\theta$$	d = thickness of air film, $\theta = angle of incidence/refraction$ m	Parallel mirror configuration
Condition for Bright Fringes (Center)	$$2d = m\lambda$$	d = mirror displacement, m = integer order, \lambda = wavelength m (lengths), dimensionless (order)	Normal incidence at the center of the pattern
Wavelength Formula from Fringe Shift	$$\lambda = \frac{2 \Delta d}{N}$$	\Delta d = distance moved by mirror, N = number of fringes shifted m	Counting N fringes for mirror displacement \Delta d
Refractive Index of Gas	$$(\mu - 1)t = \frac{N\lambda}{2}$$	\mu = refractive index, t = length of gas tube, N = shifted fringes Dimensionless (mu), m (t and lambda)	Evacuated tube filled with gas of refractive index mu
Wavelength Difference for Doublet	$$\delta\lambda = \frac{\lambda^2}{2d}$$	\lambda = mean wavelength, \delta\lambda = wavelength difference, d = mirror displacement between visibility minima m	Using a source with two closely spaced spectral lines
Fringe Width for Wedge-shaped Air Film	$$\beta = \frac{\lambda}{2\theta_w}$$	\beta = fringe width, \thet$a_w = wedge angle$ m (beta), radians (theta_w)	Localized fringes with inclined mirrors
Phase Change on External Reflection	$$\phi = \pi$$	\phi = phase shift radians	Reflection from optically denser medium
Effective Optical Path Difference with Glass Plate	$$\Delta_{opt} = 2t(\mu - 1)$$	t = thickness of plate, \mu = refractive index m	Normal incidence through transparent sheet

Graphs and experiments

Circular Fringes Pattern

Draw: Concentric bright and dark rings whose spacing decreases outward from the center.

Interpret: Represents loci of constant angle of incidence (Haidinger fringes); rings crowd together at larger angles because cos(theta) varies non-linearly.

Exam use: Tested via conceptual questions on fringe spacing variation and mirror alignment.

Localized Straight Fringes

Draw: Parallel, equidistant straight fringes formed when mirrors are inclined.

Interpret: Represents an air wedge interference pattern localized in the plane of the mirrors.

Exam use: Used to explain measurement of wedge angle and surface flatness testing.

Fringe Visibility Curve

Draw: A periodically modulating amplitude curve showing peaks and nodes as mirror displacement increases for a doublet source.

Interpret: Demonstrates the superposition of two sets of interference patterns going in and out of phase, used to measure hyperfine splitting or doublet separation.

Exam use: Advanced PGT numerical and conceptual graph analysis.

Solved numericals

TGT core

In a Michelson interferometer, when the movable mirror is displaced by 0.294 mm, 1000 fringes cross the field of view. Calculate the wavelength of light used.

Identify given values: Delta $d = 0.294 mm = 0.294 \times 10^-3 m$, N = 1000.
Use the formula lambd$a = (2 * \delta d) / N$.
Substitute the values: lambd$a = (2 * 0.294 \times 10^-3) / 1000$.
Calculate result: lambd$a = 5.88 \times 10^-7 m = 588 nm$.

Answer: 588 nm

PGT extension

A transparent tube of length 10 cm filled with air is placed in one arm of a Michelson interferometer using light of wavelength 589 nm. When the air is evacuated, 70 fringes shift past the crosshair. Calculate the refractive index of air.

Identify given values: t = 10 cm = 0.1 m, lambd$a = 589 \times 10^-9 m$, N = 70.
Use the relation (mu - 1)$t = (N * \lambda) / 2$.
Rearrange for refractive index: m$u = 1 + (N * \lambda) / (2 * t)$.
Substitute the values: m$u = 1 + (70 * 589 \times 10^-9) / (2 * 0.1)$.
Calculate result: m$u = 1 + (4.123 \times 10^-5) / 0.2 = 1 + 2.0615 \times 10^-4 = 1.000206$.

Answer: 1.000206

PGT extension

Calculate the wavelength difference delta lambda for the sodium yellow doublet (mean wavelength 589 nm) if the visibility of the fringes drops to a minimum when the mirror is moved by 0.294 mm.

Identify given values: lambd$a = 589 \times 10^-9 m$, $d = 0.294 mm = 2.94 \times 10^-4 m$.
Use the formula for doublet separation: delta lambd$a = \lambda^2 / (2 * d)$.
Substitute the values: delta lambd$a = (589 \times 10^-9)^2 / (2 * 2.94 \times 10^-4)$.
Calculate result: delta lambd$a = (3.469 \times 10^-13) / (5.88 \times 10^-4) = 5.90 \times 10^-10 m = 0.59 nm$.

Answer: 0.59 nm

Assumptions and limitations

Assumption: Light sources are assumed to possess sufficient temporal coherence to produce stable interference over the path difference introduced.
Limitation: Extended sources are required to observe circular fringes, and path difference cannot exceed the coherence length of the source.

Common misconceptions

Common error: The Michelson interferometer relies on division of wavefront.

Correct: It relies on division of amplitude.

A beam splitter divides the amplitude of a single incoming wavefront into two parts, which travel along different arms and then interfere.

Common error: Circular fringes are localized at the surface of the mirror.

Correct: Circular fringes are localized at infinity (or viewed through a telescope focused to infinity).

They are fringes of equal inclination (Haidinger fringes) formed by rays emerging at the same angle.

Quick revision

Michelson interferometer works on division of amplitude to produce circular or localized interference fringes.
Path Difference with Oblique Incidence: \Delta = 2d \cos\theta
Condition for Bright Fringes (Center): 2d = m\lambda

Must remember

The Michelson Interferometer works on the principle of division of amplitude, splitting an incoming light beam into two mutually perpendicular paths using a beam splitter.
Circular fringes (Haidinger fringes) are formed when the mirrors are strictly parallel and correspond to fringes of equal inclination.
Localized fringes are formed when the mirrors are inclined at an angle, creating an air wedge between one mirror and the virtual image of the other.
A compensator plate of identical thickness and material is essential in white light interferometry to equalize the optical path in glass for both arms.
The path difference for oblique incidence is given by $\Delta = 2d \cos\theta$, where $d$ is the thickness of the air film and $\theta$ is the angle of incidence.

Glossary

Beam SplitterA partially silvered glass plate used to divide the amplitude of an incoming light beam into two perpendicular paths.

Haidinger FringesCircular fringes formed by equal inclination when the reflecting mirrors are strictly parallel.

Compensator PlateA plate identical in thickness and material to the beam splitter, placed in the transmitted arm to equalize glass optical paths for white light interferometry.

Wedge AngleThe angle $\theta_w$ between inclined mirrors used to produce localized fringes.

Phase Change on ReflectionA phase shift of $\pi$ radians (or $180^\circ$) experienced by light when reflected from an optically denser medium.

Circular Fringes vs. Localized Fringes

Parameter	Circular Fringes (Haidinger)	Localized Fringes
Mirror Alignment	Strictly parallel	Inclined at an angle
Localization	At infinity	Localized near the mirrors/air wedge
Nature of Fringes	Fringes of equal inclination	Fringes of equal thickness

Practice Quiz

Choose an option to reveal the explanation.

1. What is the fundamental optical phenomenon responsible for fringe formation in a Michelson Interferometer?

Explanation: The Michelson Interferometer divides the amplitude of a single incoming wavefront into two parts using a partial reflector, which then interfere after reflection.

2. Which type of fringes are produced in a Michelson Interferometer when the two reflecting mirrors are strictly parallel to each other?

Explanation: Circular fringes, also known as Haidinger fringes or fringes of equal inclination, are formed when the two mirrors are strictly parallel.

3. What is the primary purpose of the compensator plate in a Michelson Interferometer?

Explanation: In white light interferometry, the reflected beam passes through the beam splitter plate thrice while the transmitted beam passes through it only once. A compensator plate equalizes this glass path.

4. What is the formula for the path difference $\Delta$ in a Michelson Interferometer for parallel mirror configuration at oblique incidence?

Explanation: For parallel mirrors with air film thickness $d$ and angle of incidence/refraction $\theta$, the path difference is given by $\Delta = 2d \cos\theta$.

5. How are localized fringes formed in a Michelson Interferometer?

Explanation: Localized fringes are formed when the mirrors are inclined at an angle, creating an air wedge between one mirror and the virtual image of the other.

6. What kind of fringe pattern is observed in a Michelson Interferometer when white light is used as the source?

Explanation: White light has a continuous wavelength spectrum, resulting in a few colored fringes and a distinct central zero path difference fringe.

7. If a mirror is displaced by $\Delta d$ and $N$ fringes are shifted, what is the formula to calculate the wavelength $\lambda$?

Explanation: The relationship between mirror displacement $\Delta d$ and fringe shift $N$ is given by $2 \Delta d = N \lambda$, yielding $\lambda = \frac{2 \Delta d}{N}$.

8. What is the phase change experienced by a light beam when it is reflected from an optically denser medium (external reflection)?

Explanation: Reflection from an optically denser medium introduces a phase shift of $\pi$ radians (equivalent to a path difference of $\frac{\lambda}{2}$).

9. Which historical experiment utilized the Michelson Interferometer to prove the non-existence of the luminiferous ether?

Explanation: The Michelson-Morley experiment used a Michelson interferometer to detect the relative motion of matter through the stationary luminiferous ether, yielding a null result.

10. What is the expression for the fringe width $\beta$ for wedge-shaped air film formed by inclined mirrors in a Michelson Interferometer?

Explanation: The fringe width for localized fringes created by a wedge angle $\theta_w$ is given by $\beta = \frac{\lambda}{2\theta_w}$.

11. How can the refractive index $\mu$ of an evacuated gas tube of length $t$ be determined using a Michelson Interferometer when gas is introduced?

Explanation: Introducing gas of refractive index $\mu$ over path length $t$ changes the optical path by $2(\mu - 1)t$, which equals $N\lambda$, leading to $(\mu - 1)t = \frac{N\lambda}{2}$.

12. What is the formula for the wavelength difference $\delta\lambda$ for a spectral doublet using mirror displacement $d$ between visibility minima?

Explanation: When using a source with two closely spaced spectral lines, visibility minima occur when $2d = \frac{\lambda^2}{\delta\lambda}$, giving $\delta\lambda = \frac{\lambda^2}{2d}$.

13. Where are circular fringes (Haidinger fringes) localized in a Michelson Interferometer setup?

Explanation: Circular fringes are fringes of equal inclination and are localized at infinity, requiring a telescope focused to infinity for observation.

14. What is the effective optical path difference introduced by placing a transparent glass plate of thickness $t$ and refractive index $\mu$ normally in one arm?

Explanation: Normal incidence through a transparent sheet of thickness $t$ introduces a round-trip optical path difference of $\Delta_{opt} = 2t(\mu - 1)$ compared to an equal thickness of air.

15. Which misconception about the Michelson Interferometer is incorrect regarding fringe localization?

Explanation: Circular fringes are NOT localized on the mirror surface; they are localized at infinity and formed by rays emerging at the same angle.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. In a Michelson Interferometer, when the movable mirror is displaced by $0.2945 \text{ mm}$, a shift of $1000$ fringes is observed. What is the wavelength of the monochromatic light used?

Explanation: Using $\lambda = \frac{2 \Delta d}{N}$, $\lambda = \frac{2 \times 0.2945 \times 10^{-3} \text{ m}}{1000} = 5.890 \times 10^{-7} \text{ m}$ (or $5890 \text{ \AA}$), which is the sodium D line.

2. What causes the necessity of a compensator plate in a Michelson Interferometer when working with white light?

Explanation: The reflected beam passes through the glass beam splitter three times, whereas the transmitted beam passes through it only once. A compensator plate equalizes this glass thickness.

3. For a Michelson Interferometer adjusted for circular fringes, the order of interference at the center of the pattern is maximum when:

Explanation: The path difference is $\Delta = 2d \cos\theta$. At the center of the pattern, $\theta = 0$, giving $\cos\theta = 1$, which maximizes the path difference and interference order $2d = m\lambda$.

4. A gas tube of length $10 \text{ cm}$ is placed in one arm of a Michelson Interferometer. As the tube is slowly evacuated, $150$ fringes shift. If $\lambda = 5000 \text{ \AA}$, what is the refractive index of the gas?

Explanation: Using $(\mu - 1)t = \frac{N\lambda}{2}$, we have $(\mu - 1)(0.1) = \frac{150 \times 5000 \times 10^{-10}}{2} = 3.75 \times 10^{-5}$, so $\mu - 1 = 3.75 \times 10^{-4}$ and $\mu = 1.000375$.

5. When the mirrors of a Michelson Interferometer are slightly inclined, the resulting fringes are:

Explanation: Inclining the mirrors creates an air wedge, producing localized fringes of equal thickness (straight or curved depending on the wedge configuration).

6. In a Michelson Interferometer experiment using a sodium lamp doublet ($\lambda_1$ and $\lambda_2$), distinct fringe visibility minima occur when the path difference corresponds to:

Explanation: Visibility minima occur when the dark fringes of one wavelength coincide with the bright fringes of the other, giving mirror displacement $d = \frac{\lambda^2}{2\delta\lambda}$, hence path difference $2d = \frac{\lambda^2}{\delta\lambda}$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What happens to the circular fringe pattern in a Michelson Interferometer when the movable mirror is translated parallel to itself towards or away from the beam splitter?

Explanation: Moving the parallel mirror changes $d$, causing fringes to collapse into or expand out from the center as orders change.

2. Why is white light interference in a Michelson Interferometer restricted to only a few fringes around the central fringe?

Explanation: Since fringe spacing depends on $\lambda$, colorful overlapping occurs quickly away from zero path difference where all wavelengths constructively interfere together.

3. Calculate the wedge angle $\theta_w$ if localized fringes of width $\beta = 0.5 \text{ mm}$ are observed using light of wavelength $\lambda = 500 \text{ nm}$.

Explanation: Using $\beta = \frac{\lambda}{2\theta_w}$, we get $\theta_w = \frac{\lambda}{2\beta} = \frac{500 \times 10^{-9}}{2 \times 0.5 \times 10^{-3}} = 5 \times 10^{-4} \text{ radians}$.

4. Which of the following statements is true regarding the division of amplitude in a Michelson Interferometer?

Explanation: The partially silvered beam splitter divides the amplitude of each incident light wave into a reflected wave and a transmitted wave.

5. In the Michelson-Morley experiment, what was the expected fringe shift that experimental results conclusively disproved?

Explanation: The experiment was designed to detect a measurable fringe shift proportional to $(v/c)^2$ caused by the ether drift, but observed no such shift.

6. If the refractive index of a thin transparent film of thickness $t$ placed in one arm is $\mu$, what is the condition for maximum transmission change at normal incidence?

Explanation: The optical path difference introduced by the transparent film is $2t(\mu - 1)$, and interference maxima occur when this equals an integral multiple of $\lambda$ ($m\lambda$).

7. What is the primary reason why Haidinger fringes are circular in shape?

Explanation: Rays striking the parallel mirrors at a specific angle of incidence $\theta$ form a cone, resulting in circular symmetry of equal inclination.

8. If $500$ fringes pass a reference point when a micrometer screw moves the mirror by $0.15 \text{ mm}$, what is the wavelength of light?

Explanation: $\lambda = \frac{2 \Delta d}{N} = \frac{2 \times 0.15 \times 10^{-3}}{500} = \frac{3 \times 10^{-4}}{500} = 6 \times 10^{-7} \text{ m} = 6000 \text{ \AA}$.

9. What distinguishes the central white-light fringe from other fringes in a Michelson Interferometer?

Explanation: At zero path difference, all wavelengths interfere constructively (or destructively) in the exact same manner, forming a central achromatic white-light fringe.

10. Which UP TGT/PGT concept best covers the adjustment of a Michelson Interferometer to observe localized fringes?

Explanation: Localized fringes are produced by tilting one of the mirrors to create an air wedge between the mirror and the virtual image of the other.