Diffraction & Resolving Power - Physics 2

A grating with 5000 lines/cm has slit width 1.0 μm. At which order is the first missing maximum?
Answer: 2nd The grating spacing d is 1/5000 cm = 2 μm, twice the slit width a = 1 μm. A maximum vanishes when an interference order lands on a diffraction minimum: d sinθ = mλ and a sinθ = nλ give m/n = d/a = 2. The first such coincidence is n = 1, so m = 2, the 2nd order. The 1st order is tempting but it falls before the first diffraction minimum, so it isn't missing.
A 3.0 cm grating has 2000 lines/cm. In second order at 600 nm, the smallest resolvable Δλ is:
Answer: 0.05 nm Total lines on the grating are 3.0 cm times 2000 lines/cm = 6000. Resolving power is mN = 2 times 6000 = 12,000, so Δλ = 600 nm / 12,000 = 0.05 nm. The tempting 0.15 nm comes from using 2000 lines/cm instead of the full 6000 lines in the grating.
A laser of wavelength $\lambda$ passes through a circular aperture of diameter $D$ and illuminates a screen at distance $L$ . An observer notes that the radius of the central Airy disk on the screen is $r$ . The observer then submerges the entire apparatus in a liquid of refractive index $n > 1$ (the same laser operates at the same frequency). Which of the following correctly describes what happens to the radius $r$ of the central Airy disk and the angular resolution $\theta_R$ of the aperture?
Answer: Both $r$ and $\theta_R$ decrease by a factor of $n$ , because the wavelength in the medium is $\lambda/n$ , which reduces both the diffraction angle and the physical size of the Airy disk on the screen. Whenever you see a diffraction question involving a medium with refractive index $n$ , your first move should be to identify how the wavelength changes — because wavelength, not frequency, governs diffraction. When light enters a medium of refractive index $n$ , its frequency stays constant (it's set by the source), but its wavelength shrinks to $\lambda_n = \lambda/n$ . The Rayleigh criterion for angular resolution is $\theta_R = 1.22\,\lambda/D$ , so substituting the new wavelength gives $\theta_R' = 1.22\,\lambda/(nD)$ — a factor of $n$ smaller. Since the screen is still physically at distance $L$ , the Airy disk radius is $r' = \theta_R' \cdot L = 1.22\,\lambda L/(nD)$ , also reduced by $n$ . Both quantities shrink together, confirming B is correct. A is wrong because it invents a fictitious "effective decrease in L." The physical distance to the screen doesn't change; only the wavelength does, so there is no cancellation. C is the most tempting trap — it sounds sophisticated to say frequency is unchanged — but diffraction is a wave-interference phenomenon that depends explicitly on wavelength, not frequency. Plugging frequency into the Rayleigh formula would give you the wrong units entirely. D gets the radius right but incorrectly claims $\theta_R$ is purely geometric. Angular resolution directly involves $\lambda/D$ ; the medium absolutely affects it. Your study tip: always separate what changes (wavelength) from what doesn't (frequency, physical geometry) when a wave crosses into a new medium. Diffraction formulas use $\lambda$ — substitute $\lambda/n$ immediately and carry it through every quantity.
What is the minimum resolving power required to separate the sodium doublet, and what is the lowest diffraction order in which this grating can achieve it?
Answer: The required resolving power is $R \approx 982$ . With $N = 30{,}000$ illuminated slits, the grating achieves $R = mN = 30{,}000$ in first order — far exceeding the requirement — so the doublet is resolved in $m = 1$ . When a question asks about resolving a spectral doublet with a diffraction grating, you need two formulas: the required resolving power and the grating's actual resolving power. The required resolving power is defined as $R = \lambda / \Delta\lambda$ , where $\lambda$ is the average wavelength and $\Delta\lambda$ is the separation between the two lines. Here, $\lambda \approx 589.0$ nm and $\Delta\lambda = 0.6$ nm, giving $R = 589.0/0.6 \approx 982$ . A grating's actual resolving power in order $m$ is $R = mN$ , where $N$ is the total number of illuminated slits. With 600 lines/mm over 5.0 cm = 50 mm, you get $N = 600 \times 50 = 30{,}000$ slits. In first order, $R_1 = 1 \times 30{,}000 = 30{,}000$ , which vastly exceeds 982 — so the doublet is resolved in $m = 1$ . That makes D correct. Choice A makes a critical arithmetic error: it multiplies 600 lines/mm by 5.0 (treating the width as 5.0 mm instead of 50 mm), yielding only 3,000 slits — off by a factor of 10. Choice B correctly finds $N = 30{,}000$ but then invents a free spectral range argument that doesn't apply here. The FSR concern arises when two lines from different orders overlap — not when both lines sit well within the same order. Choice C misunderstands the Rayleigh criterion: you only need one resolving power calculation for the pair of lines, not one per line. Study tip: Always convert grating width to millimeters before computing $N$ — unit errors here are the most common trap, as seen in choice A.
Is the student's reasoning correct, and what is the resolving power $R$ of a diffraction grating expressed in terms of the relevant parameters?
Answer: The student is correct. The resolving power is $R = mN$ , which depends only on the order $m$ and number of illuminated slits $N$ . Doubling $N$ at constant $m$ precisely doubles $R$ , regardless of grating spacing or aperture width. When you encounter questions about diffraction grating resolution, the central formula to anchor on is the resolving power: $R = mN$ , where $m$ is the diffraction order and $N$ is the number of illuminated slits. This formula comes directly from the Rayleigh criterion applied to grating diffraction — the minimum resolvable wavelength difference is $\Delta\lambda = \lambda/mN$ , so $R = \lambda/\Delta\lambda = mN$ . Notice what's in that formula and what isn't: no slit spacing $d$ , no aperture width on its own, just order and slit count. The student's reasoning in choice A is exactly right. If you keep $m$ fixed and double $N$ by widening the beam to illuminate more slits, you double $R$ directly and precisely. The relationship is linear and universal — it holds for any order, any spacing. Choice B is wrong because it invents a false formula $R = d/\lambda$ . Slit spacing $d$ determines where principal maxima appear (via $d\sin\theta = m\lambda$ ), not how well the grating resolves nearby wavelengths. Confusing the grating equation with the resolving power formula is a classic trap. Choice C introduces a fictitious "saturation" effect that doesn't exist in grating physics. The formula $R = mN$ applies at all orders — there is no special behavior at $m = 1$ . Choice D fabricates an order-parity dependence (even vs. odd orders) that has no physical basis whatsoever. Study tip: Memorize $R = mN$ as a standalone fact. On exam questions, watch for distractors that smuggle in $d$ or $\lambda$ — those belong to the grating equation, not the resolving power.
The Sparrow criterion for resolving two point sources states that two sources are just resolved when the combined intensity distribution has a flat top — i.e., the second derivative of total intensity at the midpoint equals zero. For two incoherent point sources of equal intensity imaged through a circular aperture, how does the minimum resolvable angular separation $\theta_S$ (Sparrow) compare to the Rayleigh criterion $\theta_R = 1.22\lambda/D$ ?
Answer: $\theta_S \approx 0.85\, \theta_R$ , because the Sparrow criterion requires less separation than Rayleigh — the dip in intensity between the two peaks need not reach any specific depth, so sources can be closer than the Rayleigh limit and still be declared resolved. When comparing resolution criteria, think about what each one demands from the intensity profile. The Rayleigh criterion requires the intensity maximum of one Airy disk to fall exactly on the first zero of the other, producing a modest ~26% dip between the two peaks. The Sparrow criterion is fundamentally more permissive: it asks only that the combined intensity profile shows a flat top at the midpoint — meaning the second derivative of total intensity equals zero there. No dip is required at all. This means two sources can be brought closer together than the Rayleigh limit before the Sparrow condition is triggered. Numerically, for two incoherent point sources imaged through a circular aperture, the Sparrow limit works out to approximately $\theta_S \approx 0.85\,\theta_R$ . Answer A is correct: the Sparrow criterion is less demanding (more permissive) because it declares sources resolved the moment any central feature — even just a flattening — disappears, rather than requiring a measurable valley. Answer B is wrong because the flat-top condition does not coincide with the first zero of the Airy pattern — that's the Rayleigh criterion's defining feature, not Sparrow's. Answer C inverts the logic entirely: requiring a flat top is less restrictive than requiring a dip, so the Sparrow limit is smaller than Rayleigh, not larger. Answer D confuses "flat top" with "equal midpoint and peak intensities at half the Rayleigh separation" — the actual geometry gives ~0.85, not 0.5, and the reasoning about what triggers the criterion is flawed. A useful mnemonic: Sparrow < Rayleigh < Dawes in terms of minimum resolvable separation — each successive criterion is stricter. When a question names a criterion, ask yourself whether it requires a dip, a zero, or just a flat profile.
In X-ray crystallography, Bragg diffraction from crystal planes with spacing $d$ follows $2d\sin\theta = m\lambda$ . A crystallographer wants to improve the spatial resolution of the technique (i.e., resolve finer features in the electron density map). Which of the following strategies correctly achieves this and for the correct physical reason?
Answer: Use a shorter X-ray wavelength, because smaller $\lambda$ allows diffraction at larger angles $\theta$ for a given $d$ , populating higher spatial-frequency components in reciprocal space and improving real-space resolution by the crystallographic analogue of Abbe's diffraction limit. Whenever you see a question about X-ray crystallography resolution, think in terms of reciprocal space: real-space resolution improves when you can measure diffraction data at higher spatial frequencies, meaning larger scattering angles and higher-order reflections. From Bragg's law $2d\sin\theta = m\lambda$ , you can solve for the minimum resolvable spacing: $d_{min} = \frac{\lambda}{2\sin\theta_{max}}$ . To resolve finer features (smaller $d_{min}$ ), you need either a smaller $\lambda$ or a larger accessible $\theta_{max}$ . Using shorter-wavelength X-rays directly shrinks $\lambda$ , which means reflections from smaller $d$ -spacings now satisfy Bragg's law within physically accessible angles ( $\theta \leq 90°$ ). This populates higher spatial-frequency terms in the Fourier reconstruction of the electron density, directly analogous to Abbe's diffraction limit in optical microscopy. Answer A is correct for exactly this reason. Answer B is tempting because larger $\lambda$ does increase $\theta$ for a fixed $d$ , but this works against resolution—it actually excludes high-frequency data from being collected at all, since those reflections would require $\sin\theta > 1$ , which is impossible. Spreading low-order spots apart does not improve resolution. Answer C confuses signal-to-noise with resolution. Better SNR improves the accuracy of measured intensities and phases, which helps map quality, but it does not extend the resolution limit. The two are related but distinct concepts. Answer D is similarly flawed: increasing crystal-to-detector distance spreads spots physically, aiding indexing, but it does not extend the maximum angle captured—it actually reduces the angular coverage of the detector, potentially losing high-angle data. Study tip: Always distinguish between factors that extend the resolution limit (access to higher spatial frequencies) versus those that improve data quality at existing resolution. These are a common point of confusion on physics and crystallography exams.
Two stars subtend 0.10 arcsec. What aperture diameter just resolves them at 550 nm?
Answer: 1.4 m Convert 0.10 arcsec to radians: 0.10 / 206265 = 4.85 x $10^{-7}$ rad. The Rayleigh criterion gives theta = 1.22 lambda / D, so D = 1.22 (550 x $10^{-9}$ ) / (4.85 x $10^{-7}$ ) = 1.4 m. The 1.1 m trap comes from omitting the 1.22 factor for a circular aperture.
A single slit of width $a$ is illuminated by monochromatic light of wavelength $\lambda$ . An experimenter claims that doubling the slit width while simultaneously doubling the wavelength will keep the central diffraction maximum the same angular width AND double the intensity at the center. Which of the following correctly evaluates this claim?
Answer: The claim is partially correct: the angular half-width remains the same since $\lambda/a$ is unchanged, but the central-maximum intensity scales as $(a/\lambda)^2$ , which is also unchanged — so intensity does not double. When analyzing single-slit diffraction, you need to track two separate quantities independently: the angular width of the central maximum and the on-axis intensity. Conflating their scaling behavior is the core trap in this question. The angular half-width of the central maximum is given by $\theta \approx \lambda/a$ . When both $\lambda$ and $a$ are doubled, the ratio $\lambda/a$ remains unchanged — so the angular width is indeed preserved. That part of the experimenter's claim holds up. Now for intensity. The on-axis intensity for single-slit diffraction scales as $I_0 \propto a^2/\lambda^2$ , not simply $a^2$ . When both $a$ and $\lambda$ are doubled: the numerator scales by $(2a)^2 = 4a^2$ and the denominator by $(2\lambda)^2 = 4\lambda^2$ , leaving the ratio — and therefore the intensity — completely unchanged. The experimenter's intensity claim fails entirely. This makes B the correct answer: the angular width is preserved, but the intensity does not double. Choice A is wrong because it claims intensity scales as $a^2$ alone, ignoring the $\lambda^2$ dependence in the denominator — a partial formula that leads to an incorrect quadrupling prediction. Choice C is wrong on both counts: the central maximum width doesn't narrow, and intensity doesn't decrease; the $\lambda/a$ ratio governs both behaviors here. Choice D invents a scaling of $a^2/\lambda$ , which has no physical basis in the standard single-slit intensity formula. As a strategy: whenever a diffraction question changes multiple parameters simultaneously, always substitute into the full formula rather than reasoning about one variable at a time — that's where most errors occur.
In a single-slit pattern, screen distance doubles, slit width halves, and wavelength doubles. Central maximum width:
Answer: Octuples Central maximum width scales as wavelength times screen distance divided by slit width: W = 2λL/a. Doubling wavelength and screen distance each double the width, multiplying by 4; halving slit width doubles it again, so the width becomes 8 times larger. The tempting mistake is picking quadruples after doubling wavelength and screen distance but forgetting the slit-width factor.
Two stars separated by an angle $\theta = 2.0 \times 10^{-7}$ rad are observed through a telescope with a circular aperture of diameter $D$ . The telescope operates at $\lambda = 550$ nm. What is the minimum aperture diameter required to resolve the stars by the Rayleigh criterion, and if the aperture is instead set to half this minimum value, by what factor does the minimum resolvable angle increase?
Answer: The minimum diameter is $D_{\min} \approx 3.4$ m; at half this aperture ( $D = 1.7$ m), the Rayleigh limit doubles to $4.0 \times 10^{-7}$ rad, so the stars are no longer resolved and the minimum resolvable angle is twice as large as the stellar separation. When a question asks about resolving two objects through a circular aperture, your first instinct should be the Rayleigh criterion: $\theta_{min} = 1.22 \frac{\lambda}{D}$ . This sets the smallest angular separation a telescope can resolve — and it scales inversely with aperture diameter. To find the minimum diameter needed to resolve stars separated by $\theta = 2.0 \times 10^{-7}$ rad at $\lambda = 550$ nm, set $\theta_{min} = \theta$ and solve: $D_{min} = \frac{1.22\lambda}{\theta} = \frac{1.22 \times 550 \times 10^{-9}}{2.0 \times 10^{-7}} \approx 3.4 \text{ m}$ This confirms C starts correctly. Now, if you halve the aperture to $D = 1.7$ m, the new Rayleigh limit becomes: $\theta_{min}' = \frac{1.22\lambda}{D/2} = 2 \times \frac{1.22\lambda}{D} = 2 \times (2.0 \times 10^{-7}) = 4.0 \times 10^{-7} \text{ rad}$ The minimum resolvable angle doubles, and since the stellar separation is now below this limit, the stars are unresolved. Answer C is correct. A gets the doubling of the angle right but miscalculates $D_{min}$ as 1.7 m — that's actually half the correct minimum, so it contradicts itself internally. B invents a square-root relationship between intensity and the angular criterion. The Rayleigh formula has no such dependence — this is a fabricated distractor. D incorrectly claims that halving $D$ quadruples the angular resolution limit by conflating intensity scaling ( $\propto D^2$ ) with angular resolution. These are separate concepts — the Rayleigh criterion depends linearly on $1/D$ , not $1/D^2$ . Study tip: Memorize $\theta_{min} = 1.22\lambda/D$ and internalize that it's a linear inverse relationship — halving $D$ always doubles $\theta_{min}$ , nothing more complex.
A coherent light source (laser) of wavelength $\lambda$ illuminates two identical narrow slits separated by $d$ (Young's double-slit geometry). Due to the finite width $a$ of each slit ( $a < d$ ), the intensity envelope is modulated by single-slit diffraction. The experimenter observes that the $m = 4$ double-slit maximum is entirely missing from the pattern. Which of the following conditions on $a$ and $d$ must hold, and what does this imply about the effective number of observable bright fringes within the central diffraction maximum?
Answer: The condition is $d/a = 4$ , meaning the 4th-order double-slit maximum coincides with the first single-slit minimum. Within the central diffraction envelope (between the first single-slit minima), there are 7 observable bright fringes (orders $-3, -2, -1, 0, +1, +2, +3$ ), since orders $\pm 4$ are missing. When double-slit and single-slit effects combine, you need to track two separate conditions simultaneously: where double-slit maxima occur ( $d\sin\theta = m\lambda$ ) and where single-slit minima occur ( $a\sin\theta = p\lambda$ ). A "missing order" happens when a double-slit maximum lands exactly on a single-slit minimum, making the envelope zero out that fringe entirely. If the $m = 4$ double-slit maximum is missing, it must coincide with the first ( $p = 1$ ) single-slit minimum. Setting $d\sin\theta = 4\lambda$ equal to $a\sin\theta = \lambda$ gives $d/a = 4$ . This is the governing condition. The central diffraction envelope spans from the $-1$ to $+1$ single-slit minima, meaning it covers all double-slit orders satisfying $|m| < d/a = 4$ . That yields orders $m = -3, -2, -1, 0, +1, +2, +3$ — exactly 7 visible fringes — with $m = \pm4$ zeroed out. This confirms D is correct. A is wrong because it counts only 6 fringes and incorrectly claims orders $\pm3$ are "partially suppressed." They sit well inside the central envelope and are fully visible. B is wrong because it invents "half-intensity fringes at $\pm4$ " — those orders are completely extinguished by the single-slit minimum, not merely dimmed. C has the ratio inverted ( $a/d = 4$ would mean the slits are wider than their separation, a physically nonsensical setup for standard double-slit geometry) and produces an entirely wrong fringe count. A handy rule: when $d/a = N$ , the $\pm N$ orders vanish, and you get $2N - 1$ observable fringes within the central maximum. Memorize this pattern — it appears frequently on diffraction problems.
Which instrument has the finer angular resolution, and by approximately what factor does it outperform the other? Use the Rayleigh criterion $\theta \approx 1.22 \lambda / D$ .
Answer: The radio array is superior by a factor of approximately $700$ (roughly $10^3$ ), because although the radio wavelength is $4.2 \times 10^5$ times longer than optical, the baseline is $3 \times 10^8$ times larger, making the ratio $\theta_{\text{radio}}/\theta_{\text{optical}} \approx 1.4 \times 10^{-3}$ . Whenever you compare angular resolution across instruments, remember that the Rayleigh criterion $\theta \approx 1.22\lambda/D$ means finer resolution (smaller $\theta$ ) comes from a smaller wavelength-to-diameter ratio. The instrument that wins is whichever minimizes $\lambda/D$ . Let's compute both. For the radio array: $\lambda_{\text{radio}} = 0.21 \text{ m}$ , $D_{\text{radio}} = 3 \times 10^6 \text{ m}$ , so $\theta_{\text{radio}} \approx 1.22 \times (0.21)/(3\times10^6) \approx 8.5 \times 10^{-8} \text{ rad}$ . For the optical telescope: $\lambda_{\text{opt}} = 5\times10^{-7} \text{ m}$ , $D_{\text{opt}} = 10 \text{ m}$ , so $\theta_{\text{opt}} \approx 1.22 \times (5\times10^{-7})/10 \approx 6.1\times10^{-8} \text{ rad}$ . The ratio is $\theta_{\text{radio}}/\theta_{\text{optical}} \approx 1.4\times10^{-3}$ ... wait — that means the radio array has the smaller angle and wins by a factor of roughly $700 \approx 10^3$ . Answer A captures this correctly: even though the radio wavelength is $4.2\times10^5$ times longer, the baseline is $3\times10^8$ times larger, so the net ratio strongly favors the radio array. B is wrong because it claims the optical telescope wins — it inverts the correct conclusion. The optical wavelength is shorter, yes, but the 10 m mirror simply cannot match a 3000 km baseline. C is tempting but incorrect: the resolutions are close in order of magnitude, but they are not equal, and neither equals $10^{-8}$ rad exactly — the radio array is meaningfully better. D uses a wildly wrong baseline ( $3\times10^9$ m instead of $3\times10^6$ m) and invents an implausible $\theta \approx 10^{-13}$ rad. Your strategy: always plug in SI units carefully and compute both $\lambda/D$ values before declaring a winner — intuition about "big wavelength" or "big aperture" alone will mislead you.
Which change does NOT improve the resolution of a diffraction-limited telescope?
Answer: Using a stronger eyepiece Resolution in a diffraction-limited telescope is set by 1.22 lambda / D. Shorter wavelength (blue) or larger objective diameter D shrinks that limit, so those improve it. A stronger eyepiece only magnifies the image; it cannot change lambda or D, so it doesn't improve resolution. The tempting wrong choice is enlarging the objective, which does improve resolution by increasing D.
A microscope objective with numerical aperture $\text{NA} = n\sin\alpha$ (where $n$ is the refractive index of the medium and $\alpha$ is the half-angle of the acceptance cone) is used to image two fluorescent point sources. According to the Abbe diffraction limit, the minimum resolvable separation is $\delta = \lambda/(2\,\text{NA})$ . A student proposes that using a higher refractive-index immersion oil (increasing $n$ from 1.0 to 1.5 with the same objective lens geometry, so $\alpha$ is unchanged) will improve resolution by a factor of 1.5. A second student argues that the improvement is less than 1.5 because increasing $n$ also increases the optical path length, which effectively increases the wavelength seen by the objective. Which student is correct?
Answer: The first student's numerical conclusion is correct, but his physical reasoning is wrong. The Abbe formula uses the in-medium wavelength $\lambda = \lambda_0/n$ , giving $\delta = \lambda_0/(2\,n\sin\alpha) = \lambda_0/(2\,\text{NA})$ . The factor of $n$ appears through NA, not by holding wavelength fixed at $\lambda_0$ . Resolution improves by exactly 1.5×. When tackling resolution questions in wave optics, always trace where each variable comes from physically before plugging into a formula. The Abbe diffraction limit is derived by considering interference of waves inside the medium between the object and objective. The relevant wavelength is the in-medium wavelength $\lambda = \lambda_0/n$ , because that governs the spatial period of the interference pattern. Substituting into the Abbe formula gives $\delta = \lambda/(2\sin\alpha) = (\lambda_0/n)/(2\sin\alpha) = \lambda_0/(2\,n\sin\alpha) = \lambda_0/(2\,\text{NA})$ . So the factor of $n$ enters through NA — not by treating $\lambda_0$ as the relevant wavelength. Increasing $n$ from 1.0 to 1.5 at fixed $\alpha$ scales NA by 1.5, halving $\delta$ by that exact factor. The first student reaches the right numerical answer (1.5× improvement), but for the wrong reason — he claims $\lambda_0$ is the relevant wavelength, which misrepresents the physics. This makes C correct. A is wrong because it endorses the first student's flawed reasoning: the free-space wavelength is not what appears in the Abbe derivation; the in-medium wavelength is. The conclusion happens to be numerically correct, but the justification is not. B is wrong on two counts: the second student's claim that increased optical path length "partially offsets" the gain has no physical basis in the Abbe framework, and the invented $\sqrt{n}$ factor appears nowhere in the theory. D is wrong because standard immersion objectives are designed so that $\alpha$ remains effectively fixed when switching immersion media; Snell's law is already accounted for in the optical design. Study tip: On optics questions, always ask "which wavelength governs this phenomenon — in-medium or free-space?" The answer depends on where the interference occurs.

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