The resolving power of an optical instrument is its ability to distinguish between two closely spaced point objects as separate entities. It is fundamentally limited by the diffraction of light waves at the aperture of the instrument.
Rayleigh's criterion for resolution, resolving power of microscopes and telescopes, dependence on wavelength and aperture size.
Quantitative derivation of resolving power for rectangular and circular apertures, diffraction grating resolution, Rayleigh limit for spectral lines.
Conditions: Fraunhofer diffraction by a circular aperture of diameter D
Result: R$P = 1 / d\theta = D / (1.22 \lambda)$
Conditions: Light of wavelength λ entering an objective of semi-vertical angle u in a medium of refractive index μ
Result: R$P = 1 / d = (2 \mu \sin(u)) / \lambda = (2 NA) / \lambda$
| Formula / law | Equation | Variables and units | Conditions |
|---|---|---|---|
| Rayleigh's Angular Resolution for Telescope | $$d\theta = \frac{1.22 \lambda}{D}$$ | d\theta = angular separation, \lambda = wavelength of light, D = diameter of objective aperture Radians | Small angle approximation, circular aperture |
| Resolving Power of a Telescope | $$RP_{telescope} = \frac{D}{1.22 \lambda}$$ | D = aperture diameter, \lambda = wavelength Dimensionless (or rad^-1) | Circular aperture |
| Limit of Resolution for Microscope | $$d = \frac{\lambda}{2 \mu \sin(u)} = \frac{\lambda}{2 \text{NA}}$$ | d = minimum distance, \lambda = wavelength, \mu = refractive index, $u = semi-vertical angle$, NA = numerical aperture Metres (m) | Object in medium of refractive index \mu |
| Resolving Power of a Microscope | $$RP_{microscope} = \frac{2 \mu \sin(u)}{\lambda} = \frac{2 \text{NA}}{\lambda}$$ | \mu = refractive index, $u = semi-vertical angle$, \lambda = wavelength Metres^-1 (m^-1) | Normal incidence, circular aperture |
| Resolving Power of Diffraction Grating | $$R = \frac{\lambda}{d\lambda} = N n$$ | \lambda = mean wavelength, d\lambda = smallest wavelength difference, N = total number of lines on grating, n = order of spectrum Dimensionless | Diffraction grating operating at order n |
| Numerical Aperture | $$\text{NA} = \mu \sin(u)$$ | \mu = refractive index of medium between object and lens, $u = semi-vertical angle of cone of light$ Dimensionless | Microscope objective |
| Angular Dispersion of Grating | $$\frac{d\theta}{d\lambda} = \frac{n}{a + b \cos(\theta)}$$ | n = order, (a+b) = grating element, \theta = angle of diffraction Radians per metre | Normal incidence |
| Limit of Resolution for Rectangular Aperture | $$d\theta = \frac{\lambda}{a}$$ | \lambda = wavelength, a = slit width Radians | Single slit aperture |
Draw: Two overlapping single-slit diffraction intensity curves where the central maximum of one coincides with the first minimum of the other, showing a clear dip of about 20% in the center.
Interpret: Represents the Rayleigh criterion limit where two point sources are just resolved by an optical instrument.
Exam use: Directly tested in conceptual questions regarding the intensity profile at the limit of resolution.
Draw: Two diffraction curves whose angular separation is less than λ/D, resulting in a single broad peak with an insufficient central dip.
Interpret: Indicates that the optical instrument cannot distinguish the two sources as separate objects.
Exam use: Contrasted with the resolved state in multiple-choice questions.
Draw: Schematic setup showing two movable mirrors receiving light from a distant star to create interference fringes.
Interpret: Used to determine the angular diameter of distant stars by finding the fringe disappearance separation.
Exam use: Advanced PGT application of resolving power concepts to stellar astronomy.
Answer: 1.49 \times 10^6 \text{ rad}^{-1}
Answer: 2.5 \mu\text{m}
Answer: 5000
Common error: Magnification and resolving power mean the same thing for an optical instrument.
Correct: Magnification enlarges the image size, whereas resolving power allows two close details to be distinguished separately.
An image can be made arbitrarily large (high magnification) without adding any new structural details if the resolving power is insufficient.
Common error: Decreasing the wavelength of light decreases the resolving power of a microscope.
Correct: Shorter wavelengths increase the resolving power.
Resolving power is inversely proportional to wavelength (λ); hence, ultraviolet or electron beams provide higher resolution than visible light.
| Feature | Telescope | Microscope |
|---|---|---|
| Target Objects | Distant point objects (e.g., binary stars) | Close microscopic specimens |
| Governing Parameter | Diameter of objective lens ($D$) | Numerical aperture ($\text{NA} = \mu \sin u$) |
| Expression for RP | $RP = \frac{D}{1.22 \lambda}$ | $RP = \frac{2 \mu \sin u}{\lambda}$ |
| Key Determinant | Aperture size | Refractive index and wavelength |
Choose an option to reveal the explanation.
Explanation: Rayleigh's criterion states that two point sources are just resolved if the principal maximum of the diffraction pattern of one falls exactly over the first minimum of the diffraction pattern of the other.
Explanation: The resolving power of a telescope is given by $RP = \frac{D}{1.22 \lambda}$. It is directly proportional to the diameter $D$ of the objective aperture.
Explanation: The limit of resolution for a microscope is $d = \frac{\lambda}{2 \mu \sin(u)} = \frac{\lambda}{2 \text{NA}}$. It represents the minimum distance between two distinguishable point objects.
Explanation: The resolving power of a diffraction grating is expressed as $R = \frac{\lambda}{d\lambda} = N n$, where $N$ is the total number of lines and $n$ is the spectral order.
Explanation: Oil immersion objectives use a liquid of high refractive index $\mu$ between the specimen and lens, which increases the numerical aperture ($\text{NA} = \mu \sin u$) and thus improves the resolving power.
Explanation: Since resolving power is inversely proportional to wavelength ($RP \propto \frac{1}{\lambda}$), halving the wavelength doubles the resolving power.
Explanation: At the exact limit of resolution, the dip between the two central maxima in the combined diffraction intensity curve drops by about 20% relative to the peak intensity.
Explanation: Magnification enlarges the visual size, but without sufficient resolving power (which reveals fine details), an enlarged image remains blurry and resolves no new structural details.
Explanation: Rayleigh's angular resolution for a circular aperture telescope is given by $d\theta = \frac{1.22 \lambda}{D}$ in radians.
Explanation: Numerical aperture is defined as $\text{NA} = \mu \sin(u)$. Since refractive index $\mu$ and sine of an angle are dimensionless, NA is dimensionless.
Explanation: Resolving power required is $R = \frac{\lambda}{d\lambda} = \frac{589.0 \text{ nm}}{0.6 \text{ nm}} = 981.6$.
Explanation: Since $d\theta = \frac{1.22 \lambda}{D}$, the angular limit of resolution is inversely proportional to objective diameter $D$. Doubling $D$ halves $d\theta$ (improving resolution).
Explanation: The criterion for resolution by a grating equates its resolving power $R = Nn$ to the ratio $\frac{\lambda}{d\lambda}$.
Explanation: Resolving power is inversely proportional to wavelength ($\lambda$). Shorter wavelengths, such as ultraviolet rays or electron beams, yield superior resolution.
Explanation: For a rectangular slit aperture (single slit diffraction), the first minimum occurs at angle $d\theta = \frac{\lambda}{a}$.
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
Explanation: The resolving power of a telescope is defined as the reciprocal of its angular separation limit, yielding $RP = \frac{D}{1.22 \lambda}$.
Explanation: Oil immersion increases the refractive index $\mu$, which directly increases numerical aperture ($\text{NA} = \mu \sin u$) and enhances the microscope's resolving power.
Explanation: Grating resolving power is $R = Nn$. Assuming a standard grating width or total lines $N = 5000 \text{ lines/cm} \times 2 \text{ cm} = 10000$ (or total lines consideration where $N = 5000$ if width is 1 cm), with $n=2$, $R = 5000 \times 2 = 10000$. Wait, if total lines on the utilized width is considered, let's verify standard values: $R = Nn = 5000 \times 2 = 10000$. Let's check options: $1.0 \times 10^4$ is option B. Wait, if $N$ is lines per cm and width is assumed 5cm or similar, let's use standard formula $R = Nn$. If $N=5000$ total lines and $n=2$, $R=10000$.
Explanation: Two point objects can be distinctly resolved by a microscope if their linear separation is greater than or equal to the minimum limit of resolution $d = \frac{\lambda}{2 \text{NA}}$.
Explanation: From the formula $R = Nn$, the resolving power of a diffraction grating depends directly on the total number of lines $N$ and the spectral order $n$.
Explanation: Using $d\theta = \frac{1.22 \lambda}{D} = \frac{1.22 \times 500 \times 10^{-9}}{2} = \frac{610 \times 10^{-9}}{2} = 3.05 \times 10^{-7} \text{ radians}$.
Attempt these questions under a ten-minute limit using a separate timer.
Explanation: Rayleigh's angular separation is $d\theta = \frac{1.22 \lambda}{D}$, which depends on wavelength $\lambda$ and aperture diameter $D$, but is independent of focal length.
Explanation: $\text{NA} = \mu \sin(u) = 1.5 \times \sin(30^\circ) = 1.5 \times 0.5 = 0.75$.
Explanation: $RP = \frac{D}{1.22 \lambda} = \frac{1.22}{1.22 \times 500 \times 10^{-9}} = \frac{1}{500 \times 10^{-9}} = 2.0 \times 10^6 \text{ rad}^{-1}$.
Explanation: Total lines $N = 4000 \text{ lines/cm} \times 5 \text{ cm} = 20000$. Resolving power $R = Nn = 20000 \times 2 = 4.0 \times 10^4$.
Explanation: A compound microscope uses numerical aperture and short-wavelength illumination to resolve microscopic, sub-micron biological and material structures.
Explanation: Since $d = \frac{\lambda}{2 \mu \sin(u)}$, an increase in refractive index $\mu$ decreases the limit of resolution $d$, meaning smaller distances can be resolved.
Explanation: Longer wavelengths increase $d\theta$ ($d\theta = \frac{1.22 \lambda}{D}$), making the angular resolution limit larger, so the stars will become unresolved.
Explanation: $R = \frac{\lambda}{d\lambda} \implies d\lambda = \frac{\lambda}{R} = \frac{5000 \text{ \AA}}{50000} = 0.1 \text{ \AA}$.
Explanation: At Rayleigh's limit of resolution, the combined intensity dip at the center falls to approximately 80% of the maximum peak intensity (i.e., 20% lower).
Explanation: De Broglie wavelength of electrons ($\lambda = \frac{h}{p}$) is thousands of times smaller than visible light wavelengths, leading to exceptionally high resolving power.