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Resolving Power of Diffraction Grating
The resolving power of a diffraction grating is its ability to distinctly separate two closely spaced spectral lines. It is quantitatively defined as the ratio of the wavelength of the light to the smallest resolvable wavelength difference.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Basic formula R = lambda / d(lambda) = nN, meaning of resolving power, and dependence on order and total number of lines.
PGT extension
Detailed derivation of resolving power using Rayleigh's criterion, angular dispersion, free spectral range, and comparison with prism resolving power.
Prerequisites
Fraunhofer diffraction at a single slit and multiple slits (grating equation)
Rayleigh's criterion for optical resolution
Angular dispersion of a diffraction grating
Definition of Resolving Power
Resolving power measures the capacity of an optical instrument to form separate images of close object points or resolve close spectral wavelengths.
It is expressed mathematically as $R = \lambda / dlambda$, where dlambda is the minimum resolvable wavelength difference at wavelength lambda.
A higher resolving power implies the ability to distinguish smaller wavelength differences.
Rayleigh's Criterion for Resolution
Two spectral lines of equal intensity are considered just resolved if the principal maximum of one wavelength coincides with the first minimum of the other wavelength.
At this limit, the resultant intensity curve shows a distinct dip (about 80% of the peak intensity) between the two maxima.
If the separation is smaller than this limit, the combined intensity profile appears as a single broad, unresolved maximum.
Dependence on Grating Parameters
The resolving power is directly proportional to the diffraction order (n) in which the spectrum is observed.
It is directly proportional to the total number of lines (N) on the grating surface, not the grating element ($d = a + b)$.
It is independent of the grating spacing (d) directly, but depends on the total ruled width ($W = N(a+b)) through the relation R = nW(\sin\theta + \sin\theta_i) / \lambda$.
Comparison with Prism Resolving Power
The resolving power of a diffraction grating depends on the order and total number of lines, whereas a prism's resolving power depends on its base length and the material's dispersive power (dt/dlambda).
Grating spectra are rational and normal, with resolving power remaining uniform across orders for equal widths, unlike prisms where dispersion varies non-linearly with wavelength.
Derivations
Derivation of Grating Resolving Power Formula
Conditions: Normal incidence or general oblique incidence, N lines, order n
Write the principal maxima condition for wavelength lambda in order n: (a + b)sin\theta = nlambda.
Consider a slightly different wavelength (lambda + dlambda) forming its nth principal maximum at an angle (\theta + d\theta): (a + b)sin(\theta + d\theta) = n(lambda + dlambda).
According to Rayleigh's criterion, the nth principal maximum of (lambda + dlambda) falls at the first minimum of wavelength lambda, given by (a + b)sin(\theta + d\theta) = (nN + 1)\frac{lambda}{N} ... wait, more precisely, the condition for the first minimum adjacent to the nth principal maximum is N(a + b)sin\theta' = lambda + lambda/N? Let's use the phase difference method: total phase difference between the first and last corresponding points for the first minimum is 2pi. Therefore, N((a+b)sin\theta_1 - (a+b)sin\theta) = lambda.
Equating the angular position of the first minimum of lambda to the nth maximum of lambda + dlambda gives: n(a + b)d\theta = n dlambda and from grating equation nd$\theta = \frac{\lambda}{N(a+b)\cos\theta}$... simplifying yields R = nN.
Result: R = \frac{\lambda}{d\lambda} = nN
Maximum Resolving Power of a Grating
Conditions: Limit of diffraction angle \theta = 90 degrees
Start with the grating equation: (a + b)(sin\theta + sin\thet$a_i) = nlambda$.
For normal incidence (sin\thet$a_i = 0)$, the maximum possible diffraction angle is $\theta = 90^circ$, so sin\theta = 1.
Substitute $n = (a + b) / \lambda into the resolving power expression R = nN = \frac{N(a + b)}{\lambda} \sin\theta$.
Since N(a + b) = W (total width of the grating), the maximum resolving power is $R_{max} = \frac{W}{\lambda}(\sin\theta + \sin\theta_i)$.
Delta\theta = angular separation between two close spectral lines radians
Just resolved condition
Resolving Power via Total Width
$$R = \frac{W(\sin\theta)}{\lambda} \text{ (for normal incidence)}$$
W = ruled width, lambda = wavelength Dimensionless
Normal incidence
Graphs and experiments
Intensity Distribution for Resolved and Unresolved Spectral Lines
Draw: Three subplots showing intensity vs angle or position: (a) two peaks well separated, (b) two peaks just resolved showing a dip to 80% height matching Rayleigh's criterion, (c) a single broad peak representing unresolved lines.
Interpret: Visualizes Rayleigh's criterion where the central maximum of one pattern coincides with the first minimum of the adjacent pattern.
Exam use: Frequently asked conceptual diagrams in PGT descriptive questions and TGT multiple-choice visual identification.
Variation of Resolving Power with Order (n)
Draw: A linear straight line graph plotting Resolving Power (R) on the y-axis against diffraction order (n) on the x-axis passing through the origin.
Interpret: Demonstrates that resolving power increases linearly as observation is shifted to higher orders for a fixed grating.
Exam use: Interpreting slope as total number of lines N.
Spectrometer Setup for Grating Resolving Power
Draw: A schematic diagram of an optical spectrometer showing collimator, diffraction grating on a turntable, and telescope.
Interpret: Experimental layout used to measure sodium doublet lines (D1 and D2) to calculate experimental resolving power.
Exam use: Practical viva voce and experimental physics numerical problems in PGT.
Solved numericals
TGT core
A diffraction grating has 500 lines per cm and is 5 cm wide. Find the total number of lines and the resolving power in the second order for sodium light of wavelength 5890 Å.
Calculate total number of lines N: N = lines per cm × total width in cm = 500 × 5 = 2500 lines.
Identify given parameters: order n = 2, wavelength lambda = 5890 Å.
Apply resolving power formula: R = nN = 2 × 2500 = 5000.
Answer: Resolving power = 5000
PGT extension
Calculate the minimum wavelength difference dlambda that can be resolved by a grating having 4000 lines per cm over a ruled width of 5 cm in the 3rd order, for light of average wavelength 6000 Å.
Find total number of lines N: N = (lines/cm) × W = 4000 cm^-1 × 5 cm = 20,000 lines.
Calculate resolving power R = nN = 3 × 20,000 = 60,000.
Use the definition $R = \lambda / dlambda$ to find dlambda: dlambd$a = \lambda / R$.
What is the maximum theoretical resolving power of a grating of width 2.5 cm for light of wavelength 5000 Å at normal incidence?
Identify given values: W = 2.5 cm = 0.025 m, lambd$a = 5000 Å = 5 × 10^-7 m$.
Use the maximum resolving power formula for normal incidence: $R_{max} = \frac{W}{\lambda}(\sin\theta + \sin\theta_i) = \frac{W}{\lambda}(1 + 0) = \frac{W}{\lambda}$.
Assumes monochromatic light sources of equal intensity when applying Rayleigh's criterion.
Assumes uniform spacing (a+b) and parallel rulings across the entire optical aperture of the grating.
Limitation: Overlapping of orders can occur in white light spectra at higher orders, reducing practical resolvable range.
Common misconceptions
Common error: Resolving power depends directly on the grating element ($d = a + b)$.
Correct: Resolving power depends on the total number of lines N and order n, not on individual spacing d.
The formula R = nN shows dependence solely on order and total lines; spacing determines angular dispersion and positions of maxima.
Common error: Rayleigh's criterion states that two lines are resolved when their maxima overlap completely.
Correct: Two lines are just resolved when the maximum of one coincides with the first minimum of the other.
Complete overlap would result in a single unresolved peak, whereas Rayleigh's limit provides an identifiable dip in intensity.
Quick revision
Resolving power formula: $R = nN = \lambda / dlambda$.
Resolving Power Definition: R = \frac{\lambda}{d\lambda}
Resolving Power of Grating: R = nN
Must remember
The resolving power of a diffraction grating is expressed as $R = \frac{\lambda}{d\lambda} = nN$.
Resolving power is directly proportional to the diffraction order $n$ and the total number of lines $N$ on the grating surface.
According to Rayleigh's criterion, two spectral lines are just resolved when the principal maximum of one coincides with the first minimum of the other.
Resolving power is independent of the grating element $d = a + b$, but depends on the total ruled width $W$.
Unlike prism spectra where dispersion varies non-linearly, grating spectra are rational and normal with uniform resolving characteristics across orders for equal widths.
Glossary
Resolving Power ($R$)The capacity of an optical instrument to form separate images of close object points or resolve close spectral wavelengths, given by $R = \frac{\lambda}{d\lambda}$.
Rayleigh's CriterionCondition for resolution stating that two spectral lines of equal intensity are just resolved when the principal maximum of one coincides with the first minimum of the other.
Grating Element ($a+b$)The sum of the width of a transparent ruling and an opaque space on a diffraction grating.
Total Ruled Width ($W$)The total width of the grating covered by rulings, given by $W = N(a+b)$.
Diffraction Order ($n$)The integer order of the principal maximum in the diffraction pattern.
Diffraction Grating vs. Prism Resolving Power
Parameter
Diffraction Grating
Prism
Governing Formula
$R = nN$
$R = t \frac{d\mu}{d\lambda}$
Primary Dependency
Total number of lines $N$ and order $n$
Base length $t$ and material dispersive power $\frac{d\mu}{d\lambda}$
Dispersion Nature
Rational and normal spectrum
Non-linear dispersion varying with wavelength
Practice Quiz
Choose an option to reveal the explanation.
1. What is the general mathematical expression for the resolving power of an optical instrument resolving two close spectral lines?
Explanation: The resolving power is defined as $R = \frac{\lambda}{d\lambda}$, where $d\lambda$ is the minimum resolvable wavelength difference at wavelength $\lambda$.
2. The resolving power of a diffraction grating is given by which of the following expressions in terms of order $n$ and total lines $N$?
Explanation: For a diffraction grating, the resolving power is $R = nN$, directly proportional to the diffraction order and the total number of lines.
3. According to Rayleigh's criterion, two spectral lines of equal intensity are just resolved when:
Explanation: Rayleigh's criterion states that two lines are just resolved if the principal maximum of one wavelength coincides with the first minimum of the other wavelength.
4. At Rayleigh's limit of resolution, what is the intensity of the dip between the two central maxima relative to the peak intensity?
Explanation: At the Rayleigh limit, the resultant intensity curve shows a distinct dip of about $80\%$ of the peak intensity between the two maxima.
5. The resolving power of a diffraction grating depends directly on:
Explanation: Resolving power $R = nN$ depends directly on the total number of lines $N$ on the grating surface, not on the grating element $(a+b)$.
6. How does the resolving power of a grating change with the diffraction order $n$?
Explanation: The formula $R = nN$ indicates that resolving power is directly proportional to the diffraction order $n$.
7. What is the relation between total ruled width $W$, grating element $(a+b)$, and total number of lines $N$?
Explanation: The total ruled width $W$ is equal to $N(a+b)$, meaning the total number of lines is $N = \frac{W}{a+b}$.
8. For normal incidence, what is the maximum possible resolving power of a diffraction grating of total ruled width $W$?
Explanation: For normal incidence ($\theta_i = 0$) and maximum diffraction angle ($\theta = 90^\circ$, so $\sin\theta = 1$), $R_{max} = \frac{W}{\lambda}$.
9. Which of the following parameters does NOT directly determine the resolving power of a diffraction grating?
Explanation: Resolving power is independent of the grating element $d = (a+b)$ directly; it depends on the total number of lines and order.
10. What is the primary factor determining the resolving power of an optical prism?
Explanation: A prism's resolving power depends on its base length and the material's dispersive power, unlike a grating which depends on $nN$.
11. If the number of lines on a grating is doubled while keeping the width constant, what happens to its resolving power?
Explanation: Since resolving power $R = nN$ is directly proportional to the total number of lines $N$, doubling $N$ doubles the resolving power.
12. What is the angular separation $\Delta\theta$ between two close spectral lines at the Rayleigh criterion limit?
Explanation: The angular separation at the just-resolved limit is given by $\Delta\theta = \theta_2 - \theta_1 = \frac{\lambda}{N(a+b)\cos\theta}$.
13. Why are grating spectra termed rational and normal?
Explanation: Grating spectra are rational and normal because resolving power and dispersion behave predictably and uniformly, unlike prisms where dispersion varies non-linearly.
14. A diffraction grating has 500 lines per cm and a total width of 5 cm. What is the total number of lines $N$?
15. Which misconception about grating resolving power is correct?
Explanation: Resolving power depends on total lines $N$ and order $n$, not directly on the individual spacing $d = a+b$.
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 diffraction grating can be increased by:
Explanation: Since $R = nN$, increasing either the diffraction order $n$ or the total number of lines $N$ increases the resolving power.
2. If a diffraction grating is illuminated normally, the resolving power in the $n$-th order for $N$ lines is expressed as:
Explanation: The fundamental formula for grating resolving power is $R = nN$, valid regardless of normal or oblique incidence as long as $N$ lines participate.
3. Two spectral lines with wavelengths $\lambda = 5000 \text{ \AA}$ and $\lambda + d\lambda$ are just resolved in the 2nd order by a grating with 4000 total lines. What is $d\lambda$?
Explanation: $R = nN = 2 \times 4000 = 8000$. Then $d\lambda = \frac{\lambda}{R} = \frac{5000}{8000} = 0.625 \text{ \AA}$.
4. Which of the following statements regarding Rayleigh's criterion is correct?
Explanation: Rayleigh's criterion for spectral lines states that two lines are just resolved when the principal maximum of one coincides with the first minimum of the other.
5. The total width of a diffraction grating is $W$ and the grating element is $d = a+b$. The maximum resolving power for normal incidence is:
Explanation: For normal incidence, maximum resolving power occurs at $\theta = 90^\circ$, yielding $R_{max} = \frac{W(\sin 90^\circ)}{\lambda} = \frac{W}{\lambda}$.
6. How does the resolving power of a diffraction grating compare with that of a prism of equivalent dimensions?
Explanation: Grating resolving power is given by $nN$ (rational and uniform), whereas a prism's resolving power depends on base length $t$ and $\frac{d\mu}{d\lambda}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Calculate the resolving power of a grating having 6000 lines per cm across a ruled width of 5 cm in the 3rd order.
Explanation: Total lines $N = 6000 \text{ lines/cm} \times 5 \text{ cm} = 30000$. Order $n = 3$. Resolving power $R = nN = 3 \times 30000 = 90000 = 9 \times 10^4$, wait: $3 \times 30000 = 90000$, which is $9 \times 10^4$ (Option A). Let's check: $3 \times 30000 = 90000$. So correct option is A.
2. If the wavelength interval $d\lambda$ that can be resolved by a grating in the 2nd order is $0.1 \text{ nm}$ at $\lambda = 500\text{ nm}$, what is the resolving power $R$?
3. What is the minimum number of lines required on a grating to resolve sodium doublet lines ($\lambda_1 = 5890 \text{ \AA}$ and $\lambda_2 = 5896 \text{ \AA}$) in the first order?
Explanation: Average wavelength $\lambda = 5893 \text{ \AA}$, $d\lambda = 6 \text{ \AA}$. $R = \frac{\lambda}{d\lambda} = \frac{5893}{6} \approx 982$. Since $R = nN$ and $n=1$, $N = 982$ lines.
4. In a diffraction grating experiment, if the angle of incidence is $\theta_i$, the general formula for maximum resolving power is:
Explanation: The maximum resolving power incorporating total width $W$ and angles of incidence and diffraction is $R_{max} = \frac{W}{\lambda}(\sin\theta + \sin\theta_i)$.
5. Which of the following optical components produces a spectrum where the angular dispersion is independent of the order $n$ in the same manner as a grating?
Explanation: Grating spectra have angular dispersion $\frac{d\theta}{d\lambda} = \frac{n}{(a+b)\cos\theta}$, scaling directly with order $n$.
6. If the grating element $(a+b)$ is halved while keeping the total ruled width $W$ constant, the resolving power $R$:
Explanation: Since $W = N(a+b)$, if $(a+b)$ is halved at constant $W$, the number of lines $N$ doubles. Because $R = nN$, the resolving power doubles.
7. At the Rayleigh criterion limit for two spectral lines, the angular separation $\Delta\theta$ is inversely proportional to:
Explanation: From $\Delta\theta = \frac{\lambda}{N(a+b)\cos\theta}$, the angular separation is inversely proportional to $N(a+b)$, which is the total ruled width $W$.
8. A grating has 1000 lines per inch and is 3 inches wide. What is its resolving power in the 2nd order?
Explanation: Total lines $N = 1000 \text{ lines/inch} \times 3 \text{ inches} = 3000$. Resolving power $R = nN = 2 \times 3000 = 6000$.
9. Which statement best contrasts prism dispersion with grating dispersion?
Explanation: Prism dispersion depends on refractive index change with wavelength ($\frac{d\mu}{d\lambda}$) which is non-linear, whereas grating dispersion is rational and normal.
10. If the order of diffraction $n$ is increased from 1 to 3 for a given diffraction grating, the resolving power:
Explanation: Since $R = nN$ and $N$ is constant, increasing order $n$ from 1 to 3 increases the resolving power by a factor of 3.