Physics 2 Quiz: Diffraction And Resolving Power

What this quiz covers

This quiz focuses on Diffraction And Resolving Power, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 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 $\theta_{\text{radio}}/\theta_{\text{optical}} \approx 1.4 \times 10^{-3}$. (correct answer)
2. The optical telescope is superior by a factor of approximately $10^3$, because the much shorter wavelength overcomes the advantage of the large radio baseline, yielding $\theta_{\text{optical}}/\theta_{\text{radio}} \approx 10^{-3}$.
3. The two instruments have nearly equal angular resolution because the longer wavelength of the radio array is offset by its proportionally larger baseline, and both yield $\theta \approx 10^{-8}$ rad.
4. The radio array is superior by a factor of approximately $10^6$, because the baseline of $3 \times 10^9$ m utterly dominates the wavelength difference, giving $\theta_{\text{radio}} \approx 10^{-13}$ rad.

Question 2

1. 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. (correct answer)
2. The student is incorrect. The resolving power is $R = d/\lambda$, which depends on slit spacing and wavelength. Doubling the number of slits changes the aperture but not the spacing, so $R$ is unchanged.
3. The student is correct only if the grating is used in first order ($m = 1$). For higher orders, the resolving power saturates at $R = N$ regardless of $m$, so doubling $N$ still doubles $R$ only in first order.
4. The student is correct in principle, but the resolving power is $R = mN/2$ for a grating used in even orders and $R = mN$ for odd orders, so the effect of doubling $N$ depends on which order is being used.

Question 3

1. 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, there are 6 observable bright fringes (orders $-2, -1, 0, +1, +2$ are visible, while orders $\pm 3$ are partially suppressed near the envelope edge).
2. 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, there are 8 observable bright fringes (orders $-3$ through $+3$ plus two half-intensity fringes at $\pm 4$, which are suppressed but not fully absent).
3. The condition is $a/d = 4$, meaning the slit width is four times the slit separation, which places the first single-slit minimum at the position of the 4th double-slit maximum. Within the central envelope there are only 3 observable fringes (orders $-1, 0, +1$) because the very wide slits suppress all higher orders rapidly.
4. 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. (correct answer)

Question 4

1. Quadruples
2. Doubles
3. Octuples (correct answer)
4. Unchanged

Question 5

1. 0.10 nm
2. 0.30 nm
3. 0.15 nm
4. 0.05 nm (correct answer)

Question 6

1. 2nd (correct answer)
2. 1st
3. 3rd
4. 4th

Question 7

1. Using blue instead of red
2. Enlarging the objective
3. Using a stronger eyepiece (correct answer)
4. Decreasing the wavelength

Question 8

1. 1.1 m
2. 1.4 m (correct answer)
3. 2.8 m
4. 0.69 m

Question 9

1. 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. (correct answer)
2. Use a longer X-ray wavelength, because larger $\lambda$ increases the Bragg angle $\theta$ for the same crystal spacing, spreading diffraction spots further apart on the detector and making them easier to measure accurately, which improves resolution.
3. Rotate the crystal more slowly during data collection, because slower rotation increases the exposure time per reflection, improving the signal-to-noise ratio of each Bragg peak and therefore the phase information needed for high-resolution maps.
4. Increase the crystal-to-detector distance, because placing the detector farther from the crystal increases the angular spread of the diffraction pattern on the detector, improving the ability to distinguish closely spaced Bragg peaks and hence the resolution.

Question 10

1. The minimum diameter is $D_{\min} \approx 1.7$ m; at half this aperture ($D = 0.85$ m), the minimum resolvable angle doubles to $4.0 \times 10^{-7}$ rad, correctly placing the stars at the new Rayleigh limit and confirming they are unresolved.
2. The minimum diameter is $D_{\min} \approx 3.4$ m; at half this aperture, the minimum resolvable angle increases by a factor of $\sqrt{2}$, because the diffraction-limited intensity at the Rayleigh separation drops by half, and the angular criterion scales with the square root of intensity.
3. 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. (correct answer)
4. The minimum diameter is $D_{\min} \approx 3.4$ m; at half this aperture, the minimum resolvable angle increases by a factor of four, because diffraction peak intensity scales as $D^2$, so halving $D$ quarters the intensity and quadruples the effective angular resolution limit.

Question 11

1. The required resolving power is $R \approx 982$. The number of illuminated slits is $N = 600 \times 5.0 = 3{,}000$, giving $R_1 = 3{,}000$ in first order. This exceeds 982, so the doublet is resolved in $m = 1$.
2. The required resolving power is $R \approx 982$. With $N = 30{,}000$ slits, first-order resolving power is $R_1 = 30{,}000$, which exceeds 982; however, the free spectral range in first order is $\Delta\lambda_{\text{FSR}} = \lambda/m \approx 589$ nm, which is too narrow to include both sodium lines without order overlap, so $m = 2$ is actually required.
3. The required resolving power is $R \approx 1{,}964$, because both lines must each individually satisfy the Rayleigh criterion, effectively doubling the resolving power needed. With $R_1 = 30{,}000$ in first order, $m = 1$ still suffices.
4. 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 $m = 1$. (correct answer)

Question 12

1. The claim is entirely correct: the angular half-width $\theta \approx \lambda/a$ is unchanged by the simultaneous doubling, and the on-axis intensity scales as $a^2$, which quadruples — contradicting the doubling claim for intensity but confirming the angular part.
2. 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. (correct answer)
3. The claim is entirely wrong: doubling both $\lambda$ and $a$ narrows the central maximum by a factor of two because intensity redistributes into higher-order maxima, and on-axis intensity actually decreases.
4. The claim is partially correct: the central maximum angular width is unchanged because $\lambda/a$ is constant, but the on-axis intensity scales as $a^2/\lambda$, which doubles when both are doubled — so both parts of the claim are valid in this specific scenario.

Question 13

1. The first student is correct in conclusion and reasoning. The wavelength relevant to the Abbe limit is the free-space wavelength $\lambda_0$, not the in-medium wavelength, so $\delta = \lambda_0/(2\,n\sin\alpha)$. Increasing $n$ by 1.5 at fixed $\alpha$ decreases $\delta$ by exactly 1.5, improving resolution by that factor.
2. The second student is correct. Because the wavelength in the medium is $\lambda = \lambda_0/n$, the Abbe formula gives $\delta = \lambda_0/(2\,n\sin\alpha)$, but the increased optical path length introduces a phase lag that partially offsets the resolution gain, reducing improvement to a factor of $\sqrt{n} \approx 1.22$.
3. 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×. (correct answer)
4. Neither student is correct. Immersion oil changes both $n$ and $\alpha$ simultaneously via Snell's law at the coverslip interface, so the effective NA is not simply $n\sin\alpha$. The resolution improvement can be greater or less than 1.5 depending on the specific coverslip and objective geometry.

Question 14

1. $\theta_S \approx 0.85\, \theta_R$, 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. (correct answer)
2. $\theta_S = \theta_R$ exactly, because both criteria locate the point at which the overlap of the two Airy disks produces a detectable feature, and the flat-top condition coincides with the first zero of the Airy pattern of each source.
3. $\theta_S \approx 1.22\, \theta_R$, because the Sparrow criterion is more conservative than Rayleigh — requiring a flat top demands more separation between sources to eliminate any dip, effectively doubling the minimum separation relative to Rayleigh.
4. $\theta_S \approx 0.5\, \theta_R$, because the Sparrow criterion only requires the midpoint intensity to equal the peak intensity, which occurs when the sources are separated by exactly half the Rayleigh distance, making it far more permissive than any other criterion.

Question 15

1. $\theta_R$ decreases by a factor of $n$ (angular resolution improves), but $r$ remains unchanged because the physical distance $L$ also effectively decreases by $n$ in the medium, and the two effects cancel.
2. 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. (correct answer)
3. Neither $r$ nor $\theta_R$ changes, because while the wavelength decreases by $n$, the speed of light also decreases by $n$, and the Rayleigh criterion depends on frequency (which is unchanged), not wavelength.
4. $r$ decreases by a factor of $n$ because the reduced wavelength shrinks the diffraction pattern, but $\theta_R$ remains the same because angular resolution is a geometric property of the aperture that does not depend on the medium's refractive index.