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Rayleigh Criterion for Resolution
The Rayleigh criterion is a widely accepted geometric convention that defines the minimum resolvable limit of separation between two point sources of light. It states that two images are just resolved when the central maximum of the diffraction pattern of one source falls exactly over the first minimum of the diffraction pattern of the other source.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of resolving power, Rayleigh's criterion condition for circular apertures and optical instruments like microscopes and telescopes, conceptual dependency on wavelength and aperture diameter.
PGT extension
Mathematical derivations of resolving power of telescopes, microscopes, and diffraction gratings, angular separation limits, and numerical problems involving Rayleigh's criterion.
Prerequisites
Fraunhofer diffraction from a single slit and circular aperture
Formation of central maximum and secondary minima in diffraction patterns
Wave theory of light and superposition principle
Physical Basis of Rayleigh Criterion
Due to diffraction, light from a point object does not converge to a point image, but forms a central bright disk (Airy disk) surrounded by alternating dark and bright rings.
When two point objects are very close, their diffraction patterns overlap, making it difficult to distinguish them as separate entities.
Rayleigh established a quantitative limit: two sources are resolvable if their angular or linear separation allows the intensity dip between their central maxima to drop to about 81% of the peak intensity.
Resolving Power of a Telescope
The resolving power of a telescope is defined as the inverse of the smallest angular separation (dθ) between two distant point objects that can be resolved by the telescope.
Given by Rayleigh criterion for a circular aperture of diameter D, d$\theta = 1.22 \lambda / D$.
Resolving power (RP) = 1 / d$\theta = D / (1.22 \lambda)$, showing that larger objective diameters improve angular resolution.
Resolving Power of a Microscope
The resolving power of a microscope is defined as the inverse of the minimum linear distance (Δx) between two point objects that can be resolved.
Expressed by the Abbe-Rayleigh relation Δ$x = \lambda / (2 n \sin \theta)$, where n sin θ is the numerical aperture (NA).
Immersing the objective lens in a medium of high refractive index n (oil immersion) decreases λ in the medium, thereby increasing the resolving power.
Resolving Power of a Diffraction Grating
According to Rayleigh criterion, two adjacent spectral wavelengths λ and λ + dλ are just resolved by a grating if the principal maximum of order n for wavelength λ+dλ coincides with the first minimum of the same order n for wavelength λ.
The resolving power in this case is given by $R = \lambda / d\lambda = N n$, where N is the total number of ruled lines on the grating and n is the order of the spectrum.
Higher orders and a greater number of illuminated lines yield finer spectral line separation capability.
Derivations
Angular Resolution Limit for a Circular Aperture
Conditions: Fraunhofer diffraction through a circular aperture of diameter D
Write the condition for the first diffraction minimum for a circular aperture: sin $\theta = 1.22 (\lambda / D)$, where θ is the angle of diffraction.
Apply the Rayleigh criterion: the central maximum of the second source coincides with the first minimum of the first source, meaning the angular separation between the two sources equals θ.
Assume the angle θ is very small, so that sin θ ≈ θ (in radians).
Substitute the small angle approximation to obtain the limiting angle of resolution dθ.
Result: d$\theta = 1.22 \lambda / D$
Resolving Power of a Diffraction Grating
Conditions: Normal incidence of light containing two close wavelengths λ and λ + dλ
The position of the nth principal maximum for wavelength λ is given by the grating equation: a sin $\theta = n \lambda$.
The position of the first minimum adjacent to this maximum occurs at a path difference of (λ / N) added to nλ, so a sin(θ + dθ) = n λ + (λ / N), where N is total lines.
Differentiating the grating equation gives a cos θ d$\theta = n d\lambda$.
Equating the angular width of the principal maximum to the separation between the two peaks yields the resolving power formula.
Result: $R = \lambda / d\lambda = N n$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Rayleigh Criterion for Circular Aperture
$$d\theta = 1.22 \frac{\lambda}{D}$$
d\theta = minimum angular separation, lambda = wavelength of light, D = diameter of aperture Radians
Circular aperture, small angle approximation
Resolving Power of Telescope
$$RP_{telescope} = \frac{D}{1.22 \lambda}$$
D = diameter of objective lens, lambda = wavelength Dimensionless (or rad^-1)
Distant point objects
Limit of Resolution for Microscope
$$\delta x = \frac{\lambda}{2 n \sin\theta}$$
Delta x = minimum linear separation, lambda = wavelength, n = refractive index, $\theta = semi-vertical angle of objective$ Meters (m)
Object close to the objective lens
Numerical Aperture (NA)
$$NA = n \sin\theta$$
n = refractive index of medium between object and objective, $\theta = half-angle of the cone of light entering the lens$ Dimensionless
Microscope optics
Resolving Power of Grating
$$R = \frac{\lambda}{dlambda} = N n$$
lambda = wavelength, dlambda = wavelength difference, N = total number of lines, n = order of spectrum Dimensionless
n = order, a = grating element, \theta = angle of diffraction Radians per meter (rad/m)
Normal incidence
Grating Equation
$$a \sin\theta = n \lambda$$
a = grating element (slit width plus opaque space), \theta = diffraction angle, n = order, lambda = wavelength Meters for a, radians for \theta
Normal incidence
Linear Separation for Telescope
$$d = f d\theta = 1.22 \frac{f \lambda}{D}$$
d = linear separation between images, f = focal length, D = aperture diameter Meters (m)
Objects at focal plane of objective
Graphs and experiments
Intensity Distribution Curves for Rayleigh Criterion
Draw: Three curves: (1) Unresolved case where peaks are far apart, (2) Just resolved case where the central peak of one pattern sits on the first minimum of the other (81% intensity dip), (3) Completely unresolved single broad peak.
Interpret: Illustrates the visual limit of separation where the central dip allows distinct identification of two separate point sources.
Exam use: Directly tested in conceptual questions asking for the distinct intensity profile of 'just resolved' point sources.
Airy Disk Diffraction Pattern
Draw: A circular central bright spot surrounded by concentric dark and bright rings with decreasing intensity.
Interpret: Represents the realistic image of a point source formed by a lens with a circular aperture due to aperture diffraction.
Exam use: Basis for understanding why circular apertures use the 1.22 factor instead of 1.00.
Resolving Power of Grating Experiment (Spectrometer Setup)
Draw: A schematic diagram showing a mercury source, collimator, diffraction grating table, and telescope focused on sodium D-lines (D1 and D2).
Interpret: Demonstrates the practical separation of close spectral doublet lines (589 nm and 589.6 nm) using a high line-density grating.
Exam use: Asked in practical viva and PGT theoretical derivations of spectral resolution.
Solved numericals
TGT core
Calculate the minimum angular separation of two point sources that can be resolved by a telescope with an objective lens diameter of 1.0 m, using light of wavelength 550 nm.
Identify given values: D = 1.0 m, lambd$a = 550 nm = 5.5 \times 10^-7 m$.
Use Rayleigh criterion formula for a telescope: d\theta = 1.22 \frac{lambda}{D}.
Calculate the result: d$\theta = 6.71 \times 10^-7 rad$.
Answer: 6.71 \times 10^-7 radians
PGT extension
Light of wavelength 600 nm falls normally on a diffraction grating 2 cm wide with 4000 lines per cm. Find the resolving power of the grating in the second order and the smallest wavelength difference dlambda that can be resolved near 600 nm.
Calculate total number of lines N: Grating width = 2 cm, lines per cm = 4000, so N = 2 \times 4000 = 8000 lines.
Identify the order n = 2.
Calculate resolving power R = N n = 8000 \times 2 = 16000.
Use $R = \lambda / dlambda$ to find dlambda: dlambd$a = \lambda / R = (600 \times 10^-9 m) / 16000$.
Compute final value: dlambd$a = 3.75 \times 10^-11 m = 0.0375 nm$.
Answer: Resolving power = 16000, dlambda = 0.0375 nm
PGT extension
A microscope has an objective with a numerical aperture of 0.12. If light of wavelength 500 nm is used, what is the limit of resolution (minimum distance between two resolvable points)?
Identify the given values: NA = n sin\theta = 0.12, lambd$a = 500 nm = 5 \times 10^-7 m$.
Use the microscope resolution formula: Delta x = \frac{lambda}{2 n sin\theta} = \frac{lambda}{2 cdot NA}.
Microscope linear resolution: Delta $x = \lambda / (2 n \sin\theta)$
Grating resolving power: R = N n
Shorter wavelengths and larger apertures improve optical resolution.
Must remember
According to Rayleigh criterion, two point sources are just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other.
The angular separation for a circular aperture is given by $d\theta = 1.22 \frac{\lambda}{D}$.
Resolving power of a telescope is directly proportional to objective diameter $D$ and inversely proportional to wavelength $\lambda$.
Resolving power of a microscope depends on numerical aperture ($n \sin \theta$) and wavelength $\lambda$ through the relation $\Delta x = \frac{\lambda}{2n \sin \theta}$.
Resolving power of a diffraction grating is given by $R = \frac{\lambda}{d\lambda} = N n$, where $N$ is the total number of lines and $n$ is the order of spectrum.
Magnification does not improve the optical resolution of an instrument; resolution is strictly governed by wave optics parameters.
Glossary
Airy DiskThe central bright diffraction disk formed when light from a point source passes through a circular aperture.
Numerical Aperture (NA)The product $n \sin \theta$, characterizing the acceptance angle and light-gathering power of a microscope objective.
Resolving Power (RP)The ability of an optical instrument to distinguish closely spaced point objects or adjacent spectral lines as separate entities.
Grating ElementThe sum of the width of a ruling and the width of an opaque space ($a$), representing the spatial period of a diffraction grating.
Comparison of Optical Instruments' Resolving Powers
Instrument
Governing Formula
Key Determining Factors
Telescope
$RP = \frac{D}{1.22 \lambda}$
Diameter of objective ($D$), Wavelength ($\lambda$)
1. What is the physical condition for two point objects to be 'just resolved' according to Rayleigh's criterion?
Explanation: Rayleigh's criterion states that two sources are just resolved when the central maximum of the diffraction pattern of one source falls directly over the first minimum of the diffraction pattern of the other source.
2. What is the value of the intensity drop at the dip between two just resolved central diffraction maxima according to Rayleigh?
Explanation: At the Rayleigh limit, the intensity at the midpoint between the two overlapping central maxima dips to about 81% of the peak intensity, meaning a drop of approximately 19%.
3. The angular resolution limit $d\theta$ of a telescope with circular aperture of diameter $D$ for light of wavelength $\lambda$ is given by:
Explanation: For a circular aperture, diffraction calculations yield the minimum angular separation as $d\theta = 1.22 \frac{\lambda}{D}$.
4. If the diameter of the objective lens of a telescope is doubled, its resolving power will:
Explanation: Resolving power of a telescope is given by $RP = \frac{D}{1.22 \lambda}$, which is directly proportional to the aperture diameter $D$. Thus, doubling $D$ doubles the resolving power.
5. The resolving power of a microscope depends on the numerical aperture (NA) and wavelength ($\lambda$) as:
Explanation: The limit of resolution for a microscope is $\Delta x = \frac{\lambda}{2 n \sin \theta}$. Since resolving power is the inverse of $\Delta x$, $RP = \frac{2 n \sin \theta}{\lambda}$, meaning it is proportional to the numerical aperture ($n \sin \theta$) and inversely proportional to wavelength ($\lambda$).
6. Why is oil immersion used in high-power microscope objectives?
Explanation: Immersing the objective in oil increases the refractive index $n$, which increases the numerical aperture ($n \sin \theta$) and decreases the effective wavelength of light, enhancing resolving power.
7. The resolving power of a diffraction grating is expressed as:
Explanation: The resolving power of a grating is given by $R = \frac{\lambda}{d\lambda} = N n$, where $N$ is the total number of lines and $n$ is the order of the spectrum.
8. If the total number of ruled lines on a diffraction grating is doubled, its resolving power in a given spectral order will:
Explanation: Since resolving power $R = N n$ is directly proportional to the total number of lines $N$, doubling $N$ doubles the resolving power.
9. Which of the following optical changes would improve the resolution of a telescope viewing distant stars?
Explanation: Blue light has a shorter wavelength ($\lambda$) than red light. Since $RP = \frac{D}{1.22 \lambda}$, decreasing $\lambda$ increases the resolving power.
10. What is the numerical aperture (NA) if the refractive index of the immersion medium is $1.5$ and the semi-vertical angle of the cone of light is $30^\circ$?
11. Which misconception about resolution is fundamentally incorrect?
Explanation: Magnification only enlarges the image scale; it does not add finer optical detail or improve resolution, which is governed strictly by aperture size and wavelength.
12. According to Rayleigh's criterion for a diffraction grating, two wavelengths $\lambda$ and $\lambda + d\lambda$ are just resolved if:
Explanation: Rayleigh's condition for spectral lines states that two adjacent wavelengths are just resolved when the principal maximum of one wavelength coincides with the first minimum of the other in the same spectral order.
13. What is the unit of angular resolving power of a telescope ($1 / d\theta$)?
Explanation: Angular separation $d\theta$ is in radians, so its inverse (resolving power) has units of inverse radians ($\text{rad}^{-1}$) or can be treated as dimensionless.
14. In the context of microscopes, what does the variable $\theta$ represent in the term $n \sin \theta$?
Explanation: In microscope optics, $\theta$ is the half-angle of the maximum cone of light that can enter the objective lens from the object point.
15. If a telescope objective has a diameter $D$ and focal length $f$, the linear separation $d$ between two point objects at the focal plane is given by:
Explanation: Since linear separation $d = f \cdot d\theta$ and $d\theta = 1.22 \frac{\lambda}{D}$, we get $d = 1.22 \frac{f \lambda}{D}$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The resolving power of a telescope can be increased by:
Explanation: Since $RP_{telescope} = \frac{D}{1.22 \lambda}$, increasing the objective diameter $D$ increases the resolving power.
2. Two spectral lines of wavelengths $5890\text{ \AA}$ and $5896\text{ \AA}$ are to be resolved by a diffraction grating. What minimum number of lines must be illuminated in the first order spectrum?
Explanation: Here $\lambda = 5890\text{ \AA}$, $d\lambda = 6\text{ \AA}$, and $n = 1$. Using $R = \frac{\lambda}{d\lambda} = N n$, we get $N = \frac{5890}{6 \times 1} \approx 981.6$, so at least $982$ lines must be illuminated.
3. What is the limit of resolution of a microscope using light of wavelength $500\text{ nm}$ with a numerical aperture of $0.5$?
Explanation: Using $\Delta x = \frac{\lambda}{2 n \sin \theta} = \frac{500\text{ nm}}{2 \times 0.5} = \frac{500}{1.0} = 500\text{ nm}$
4. If the aperture of a telescope is $2\text{ m}$ and the wavelength of light is $5000\text{ \AA}$, the angular limit of resolution is approximately:
5. A diffraction grating has $500$ lines per centimeter and is used in the second order. What is the resolving power in this order for light around $5000\text{ \AA}$?
Explanation: Resolving power is given by $R = N n$, where $N$ is the total number of lines on the entire grating. Thus, resolving power depends on the total width of the grating determining $N$.
6. In microscopy, immersing the objective lens in cedar wood oil instead of air results in:
Explanation: Cedar wood oil has a high refractive index $n$ ($~1.5$), which increases the numerical aperture ($n \sin \theta$) and decreases the linear resolution limit $\Delta x$, thereby increasing resolving power.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. According to Rayleigh's criterion, two point sources are considered resolvable if:
Explanation: Rayleigh criterion sets the minimum angular separation for resolvability at $d\theta = 1.22 \frac{\lambda}{D}$. If separation is greater than or equal to this limit, they are resolved.
2. Calculate the resolving power of a diffraction grating having $4000$ lines/cm across a total width of $5\text{ cm}$ when used in the 2nd order.
Explanation: Total lines $N = 4000\text{ lines/cm} \times 5\text{ cm} = 20,000$. Order $n = 2$. Resolving power $R = N n = 20,000 \times 2 = 40,000$.
3. If the wavelength of light used in a microscope is halved, the limit of resolution $\Delta x$:
Explanation: Since $\Delta x = \frac{\lambda}{2 n \sin \theta}$, $\Delta x$ is directly proportional to $\lambda$. Halving $\lambda$ halves the limit of resolution (improving resolution).
4. Which instrument's resolving power is independent of the wavelength of light used?
Explanation: All optical instruments governed by wave optics and diffraction (telescopes, microscopes, diffraction gratings) have resolving powers that depend directly or inversely on wavelength $\lambda$.
5. What happens to the angular separation limit $d\theta$ of a telescope if it is taken from air into water ($n = 4/3$)?
Explanation: Inside water, the wavelength becomes $\lambda' = \lambda / n$. Since $d\theta' = 1.22 \lambda' / D$, the angular separation limit decreases by the factor of refractive index $n$ ($4/3$).
6. The minimum distance between two point objects that a microscope can resolve is $250\text{ nm}$. If the numerical aperture is doubled, the minimum resolvable distance becomes:
Explanation: $\Delta x = \frac{\lambda}{2 NA}$. Since $\Delta x$ is inversely proportional to NA, doubling NA halves the limit of resolution to $250 / 2 = 125\text{ nm}$.
7. For a diffraction grating, what is the maximum possible theoretical resolving power for a given wavelength $\lambda$ and grating width $W$? ($a$ is grating element)
Explanation: Since $N = W/a$ and $a \sin \theta = n \lambda$, substituting $N$ and $n$ into $R = N n$ gives $R = \frac{W}{\alpha} \frac{a \sin \theta}{\lambda} = \frac{W \sin \theta}{\lambda}$, with maximum $\sin \theta \le 1$ giving $W / \lambda$ limits.
8. Why are large diameter objective lenses preferred in astronomical telescopes?
Explanation: Larger aperture diameter $D$ increases light gathering power and improves angular resolving power ($RP \propto D$).
9. If two stars cannot be resolved by a telescope, what adjustment can a stargazer make to resolve them?
Explanation: To resolve closer objects, decreasing the wavelength (using bluer light) improves resolution according to $d\theta = 1.22 \frac{\lambda}{D}$. Magnification does not help.
10. The Rayleigh criterion is fundamentally a consequence of which wave phenomenon?
Explanation: The finite size of the central Airy disk and the overlapping of ring patterns are direct consequences of the diffraction of light passing through apertures.