Intensity & Inverse-Square Law - Physics 2

1. A point source emits light of total power $P$ isotropically. A flat, perfectly absorbing detector of area $A_d$ is oriented so that its normal makes an angle $\theta = 60°$ with the radial direction from the source, and the detector is at distance $r$ from the source. A student argues: 'The power intercepted by the detector is $P A_d \cos\theta / (4\pi r^2)$.' A second student counters: 'The cosine factor should not appear here because intensity from a point source is the same in all directions, so orientation is irrelevant.' Which of the following correctly evaluates both students' claims?
   Answer: The first student's formula is correct, and the second student's conclusion is wrong despite a correct premise. Intensity $I = P/(4\pi r^2)$ is indeed isotropic, but the power intercepted is $I \ cdot A_d\cos\theta$ because the solid angle subtended by a tilted flat detector is $A_d\cos\theta/r^2$, not $A_d/r^2$. Isotropy of the source does not imply orientation-independence of the detector's projected area. Whenever you see a question mixing source isotropy with detector orientation, you need to carefully separate two independent ideas: how the source radiates, and how the detector collects. An isotropic point source creates intensity $I = P/(4\pi r^2)$ that is the same in every direction at distance $r$ — this is what "isotropic" means. But the power a detector intercepts depends on how much of that intensity it actually captures, which is determined by its projected area facing the source, not its physical area. When a flat detector of area $A_d$ is tilted so its normal makes angle $\theta$ with the radial direction, the area it presents to incoming radiation is $A_d \cos\theta$. The intercepted power is therefore: $P_{\text{det}} = I \cdot A_d\cos\theta = \frac{P\,A_d\cos\theta}{4\pi r^2}$ This confirms that C is correct. The first student's formula is right. The second student's premise — that intensity is isotropic — is actually true, but the conclusion is wrong. Isotropy describes the source, not the detector's geometry. These are completely independent. A is wrong because the solid angle subtended by a tilted flat detector is $A_d\cos\theta/r^2$, not $A_d/r^2$. The projected area shrinks with tilt. B identifies the right formula and correctly calls out the second student's error, but says the first student's reasoning is "incomplete" — in fact, the first student's formula and implicit reasoning are fully correct, making B's framing inaccurate. D invents a nonexistent $\cos^2\theta$ dependence; path length through a thin flat detector is irrelevant to power interception. Study tip: Always ask separately — "What does the source do?" and "What does the detector see?" Isotropy fixes intensity at a given $r$; the detector's projected area determines how much power it captures.
2. An astronomer observes two stars, Star A and Star B, that are known to be identical in absolute luminosity (same total power output $L$). Star A appears 16 times more intense than Star B as measured at Earth. The astronomer also knows that Star B is at a distance of $d_B = 200 \text{ pc}$ from Earth. Assuming no interstellar absorption and that both stars radiate isotropically, what is the distance $d_A$ to Star A, and what is the ratio of the solid angle subtended by Star A to that subtended by Star B as seen from Earth (assuming both stars have the same physical radius $R_s$)?
   Answer: $d_A = 50 \text{ pc}$, and the solid angle ratio $\Omega_A/\Omega_B = 16$, because both intensity and solid angle scale as $1/d^2$, so the fourfold decrease in distance produces a 16-fold increase in both quantities. Whenever you see a question involving the apparent intensity of identical light sources at different distances, your anchor concept is the inverse-square law: intensity falls off as $I \propto L/d^2$. Since both stars have the same absolute luminosity $L$, their intensities at Earth are $I_A = L/(4\pi d_A^2)$ and $I_B = L/(4\pi d_B^2)$. Setting their ratio equal to 16 gives: $\frac{I_A}{I_B} = \frac{d_B^2}{d_A^2} = 16 \implies \frac{d_B}{d_A} = 4 \implies d_A = \frac{200}{4} = 50 \text{ pc}$ Now for solid angle. A star of physical radius $R_s$ at distance $d$ subtends $\Omega \approx \pi R_s^2 / d^2$, so solid angle also scales as $1/d^2$. Since $d_A$ is four times smaller than $d_B$, the solid angle ratio is: $\frac{\Omega_A}{\Omega_B} = \frac{d_B^2}{d_A^2} = 4^2 = 16$ This confirms answer A is correct — both intensity and solid angle follow the same $1/d^2\$ dependence, so a fourfold decrease in distance produces a 16-fold increase in both. **B** is wrong because it claims solid angle scales as$1/d$rather than$1/$d^2$$— this confuses angular *diameter* (which scales as$1/d$) with angular *area*, or solid angle (which scales as$1/$d^2$$). **C** inverts the logic entirely; a star appearing *brighter* must be *closer*, not farther. **D** describes the same correct numerical setup as A but then misapplies the scaling, claiming the solid angle ratio is only 4 — contradicting the correct$1/$d^2$$geometry. Your study tip: distinguish angular diameter ($\propto 1/d$) from solid angle ($\propto 1/$d^2$$$) — exams frequently exploit this confusion. Both intensity and solid angle share the same inverse-square dependence on distance.
3. A laser beam with circular cross-section and total power $P$ is focused by a lens so that its beam radius decreases from $R_0$ to $R_0/4$ at the focal point. Meanwhile, a student separately considers an isotropic point source at distance $r$ that produces the same intensity as the original (unfocused) beam. If the distance to the point source is halved to $r/2$, by what factor does the point-source intensity change, and how does this compare to the factor by which the focused-beam intensity changes relative to the unfocused beam?
   Answer: The point-source intensity increases by a factor of 4 when the distance halves, but the focused-beam intensity increases by a factor of 16, since beam area scales as $R^2$ and the radius decreased by a factor of 4, so the two scenarios differ by a factor of 4. Whenever you see a question mixing beam focusing with point-source intensity, your job is to apply the correct intensity formula to each scenario separately — they follow different geometric rules. For an isotropic point source, intensity spreads over a sphere: $I = \frac{P}{4\pi r^2}$. Halving the distance gives $I \propto \frac{1}{r^2}$, so $I$ increases by $\left(\frac{r}{r/2}\right)^2 = 4$. That part is consistent across all answer choices. For the focused laser beam, intensity is power divided by the beam's cross-sectional area: $I = \frac{P}{\pi R^2}$. When the radius shrinks from $R_0$ to $R_0/4$, the area shrinks by a factor of $4^2 = 16$. Since power is unchanged, intensity increases by a factor of 16. This confirms answer B — the point-source factor is 4, the focused-beam factor is 16, and the two scenarios differ by a factor of 4. A is wrong because it claims both factors equal 4, incorrectly treating beam focusing as if the radius reduction produces the same result as halving a distance. The radius shrank by 4, so area — and therefore intensity — changes by $4^2 = 16$, not 4. C correctly calculates both factors (4 and 16) but then wrongly concludes the comparison is "invalid." You absolutely can compare intensity ratios from different physical setups — the comparison is meaningful and instructive. D conflates two geometrically distinct processes, incorrectly claiming a fourfold radius reduction is "equivalent in energy terms" to halving source distance. They give different intensity factors precisely because they obey different area-scaling rules. Study tip: Always ask yourself: is intensity spreading over a sphere ($\propto r^{-2}$) or concentrating into a circular cross-section ($\propto R^{-2}$, but factor comes from area = $\pi R^2$)? Both use squaring, but the physical setup determines what gets squared.
4. At distance d from a point source, correct exposure time is 1/250 s. At 2d, the correct exposure time is closest to
   Answer: $1/60$ s Intensity from a point source drops as the inverse square of distance, so at 2d the light is only 1/4 as bright. To get the same exposure, you need 4 times the time: 4 x 1/250 = 1/62.5 s, closest to 1/60 s. The tempting 1/1000 s is wrong because it shortens the exposure, but a dimmer source requires a longer one.
5. Two identical, isotropic point sources $S_1$ and $S_2$ each radiate power $P$ incoherently (no fixed phase relationship). They are separated by distance $d$. A detector is placed on the line connecting them, at distance $d$ from $S_1$ (and therefore distance $2d$ from $S_2$). By what factor does the total intensity at the detector change if both sources are moved to the same location as $S_1$ (i.e., both sources now occupy the position of $S_1$), while each source still radiates power $P$?
   Answer: The intensity increases by a factor of $\dfrac{8}{5}$, because the two incoherent sources at the same point produce intensity $2P/(4\pi d^2)$, which is larger than the original combined intensity by that ratio. When dealing with incoherent sources, intensity simply adds — there's no interference term. Your job is to compute the total intensity at the detector before and after, then find the ratio. Original configuration: $S_1$ is at distance $d$, contributing $I_1 = \frac{P}{4\pi d^2}$. $S_2$ is at distance $2d$, contributing $I_2 = \frac{P}{4\pi (2d)^2} = \frac{P}{16\pi d^2}$. The total intensity is: $I_{\text{before}} = \frac{P}{4\pi d^2} + \frac{P}{16\pi d^2} = \frac{4P + P}{16\pi d^2} = \frac{5P}{16\pi d^2}$ New configuration: Both sources sit at distance $d$, so: $I_{\text{after}} = \frac{P}{4\pi d^2} + \frac{P}{4\pi d^2} = \frac{2P}{4\pi d^2} = \frac{P}{2\pi d^2} = \frac{8P}{16\pi d^2}$ The ratio is $\frac{I_{\text{after}}}{I_{\text{before}}} = \frac{8P}{16\pi d^2} \cdot \frac{16\pi d^2}{5P} = \frac{8}{5}$, confirming B is correct. A is wrong because it claims the combined power "doubles at the near distance while eliminating the far source" — a vague miscount that gives the wrong numerator (16 instead of 8) by confusing power with intensity scaling. C is wrong because incoherent sources cannot destructively interfere; phase randomness means only intensities add, never subtract. D is wrong because it ignores $S_1$'s original contribution at distance $d$ — only $S_2$ moved closer, so a factor of 4 for $S_2$ alone doesn't apply to the total. Your strategy: always compute intensities separately using $I = P/(4\pi r^2)$ and add them. Save interference thinking for coherent sources only.
6. A detector of area A at distance d collects power P. At 2d, with area 2A, collected power is
   Answer: $P/2$ Intensity drops as inverse-square: at twice the distance it is one-fourth as strong. Doubling the detector area gives 2 times one-fourth = one-half the original collected power. The tempting error is to forget the area change and choose P/4, but the area increase brings it up to P/2.
7. In the limit $\alpha r \ll 1$ (weak absorption), which of the following expressions best approximates the fractional reduction in intensity at radius $r$ compared to the lossless inverse-square prediction, and what does this reveal about the relative importance of geometric spreading versus absorption at small $\alpha r$?
   Answer: The fractional reduction is approximately $\alpha r$, indicating that at small $\alpha r$ the absorption loss is linear in $\alpha r$ and is negligible compared to the $1/r^2$ geometric spreading, which dominates at all distances in this regime. Whenever you encounter intensity expressions with exponential attenuation factors, your first instinct should be to ask: how does this compare to the baseline (lossless) case, and what happens in limiting regimes? Here, the lossless inverse-square prediction gives $I_0(r) = \frac{P}{4\pi r^2}$, while the actual intensity is $I(r) = \frac{P}{4\pi r^2}e^{-\alpha r}$. The fractional reduction is defined as how much intensity is lost relative to the lossless case: $\frac{I_0 - I}{I_0} = \frac{I_0(1 - e^{-\alpha r})}{I_0} = 1 - e^{-\alpha r}$ Now apply the Taylor expansion $e^{-x} \approx 1 - x$ for $x \ll 1$, giving $1 - e^{-\alpha r} \approx \alpha r$. This is answer A — the fractional reduction is linear in $\alpha r$, confirming absorption is a small perturbation. Crucially, this doesn't mean absorption dominates; it means absorption contributes a small, first-order correction while the $1/r^2$ geometric spreading still governs the overall intensity falloff. Answer B is wrong on two counts: it correctly identifies the Taylor expansion but then makes the false claim that absorption dominates for small $\alpha r$ — the opposite is true. Answer C incorrectly uses a second-order approximation; $1 - e^{-\alpha r} \approx \alpha r$, not $(\alpha r)^2/2$, to leading order. Answer D is a conceptual error — Taylor expansions exist precisely to extract useful limiting behavior, so claiming "no statement can be made" ignores a fundamental mathematical tool. As a study tip: on physics-2 problems involving exponentials in limiting regimes, always reach for $e^{-x} \approx 1 - x + x^2/2 - \cdots$ and keep only the lowest non-trivial order. The leading term tells you everything about which physics dominates.
8. A listener at distance $r$ from the speaker measures a sound level of $\beta_1 = 80 \text{ dB}$. A second listener is at distance $3r$. Which of the following correctly gives the sound level $\beta_2$ heard by the second listener AND identifies the key algebraic step used to obtain it?
   Answer: $\beta_2 \approx 70.5 \text{ dB}$, obtained by noting that tripling the distance reduces intensity by a factor of 9 via the inverse-square law, then applying $\Delta\beta = 10\log_{10}(9) \approx 9.5 \text{ dB}$ to give $\beta_2 = 80 - 9.5 = 70.5 \text{ dB}$. Whenever you see a question combining the inverse-square law with decibels, resist the temptation to manipulate the dB number directly — decibels are logarithmic, so distance changes must be handled through intensity ratios first. Here's the right approach (choice C): For a point source, intensity follows $I \propto 1/r^2$. Tripling the distance multiplies the denominator by $3^2 = 9$, so the new intensity is $I_2 = I_1/9$. The change in sound level is then: $\Delta\beta = 10\log_{10}\!\left(\frac{I_2}{I_1}\right) = 10\log_{10}\!\left(\frac{1}{9}\right) = -10\log_{10}(9) \approx -9.5 \text{ dB}$ Therefore $\beta_2 = 80 - 9.5 = 70.5 \text{ dB}$. The key step is converting the intensity ratio into a log difference before subtracting. Choice A commits a fundamental error: you cannot divide a decibel value by 9 (or any factor) directly. Decibels are already on a log scale, so arithmetic on the dB number itself is only valid for addition and subtraction of log-derived changes — never division. Choice B gets the intensity ratio right (factor of 9) but then wrongly equates that to a 30 dB drop. A 30 dB drop corresponds to an intensity factor of $10^3 = 1000$, not 9. Choice D reaches the correct numerical answer but with flawed reasoning. While $10\log_{10}(9) = 20\log_{10}(3)$ algebraically, the justification — that dB "inherently" uses pressure amplitude — is wrong here. The problem defines $\beta$ explicitly in terms of intensity, so the $10\log_{10}$ form is the direct and correct route. Study tip: Always convert distance changes to intensity ratios first, then apply $\Delta\beta = 10\log_{10}(\text{ratio})$. Never scale a dB number multiplicatively.
9. Two radio transmitters, Transmitter 1 with power $P_1$ and Transmitter 2 with power $P_2 = 9P_1$, are located at the same point and radiate at the same frequency with a fixed phase difference of $\phi = \pi$ (coherent and perfectly out of phase). The net radiated field is the superposition of the two individual fields. A receiver is located at distance $r$. Which of the following correctly gives the intensity at the receiver?
   Answer: $I = (\sqrt{P_2} - \sqrt{P_1})^2/(4\pi r^2) = 4P_1/(4\pi r^2)$, because coherent sources superpose by field amplitude; the net amplitude is proportional to $\sqrt{P_2} - \sqrt{P_1}$, and intensity is proportional to the square of the net amplitude spreading over area $4\pi r^2$. When two coherent sources overlap, you must superpose their electric field amplitudes — not their powers — before computing intensity. Since intensity scales as $I \propto E^2$, and radiated power scales as $P \propto E^2$, the field amplitude of each transmitter is proportional to $\sqrt{P}$. With a phase difference of $\phi = \pi$, the fields subtract, giving a net amplitude proportional to $\sqrt{P_2} - \sqrt{P_1} = \sqrt{9P_1} - \sqrt{P_1} = 2\sqrt{P_1}$. The net intensity is then proportional to the square of this amplitude spread over the spherical area $4\pi r^2$: $I = \frac{(\sqrt{P_2} - \sqrt{P_1})^2}{4\pi r^2} = \frac{(2\sqrt{P_1})^2}{4\pi r^2} = \frac{4P_1}{4\pi r^2}$ This confirms D is correct. A is wrong because subtracting powers ($P_2 - P_1$) applies to incoherent sources — it ignores the wave nature of coherent fields. B would be correct only if both amplitudes were equal; here $\sqrt{P_2} \neq \sqrt{P_1}$, so cancellation is incomplete and net power does radiate. C invokes energy conservation incorrectly — summing powers applies to incoherent sources; coherent superposition can reduce (or enhance) the net radiated energy depending on phase, so the total at the receiver is not simply $P_1 + P_2$. Study tip: Whenever you see "coherent sources," your instinct should be to work in amplitudes (proportional to $\sqrt{P}$), combine them with the phase, then square to get intensity. Mixing coherent and incoherent rules is one of the most common traps on wave-interference problems.
10. A point source gives intensity I at distance r. Power doubles and distance triples. New intensity?
    Answer: $2I/9$ Intensity depends on power divided by distance squared. Tripling the distance divides intensity by 9, while doubling the power multiplies it by 2. So the new intensity is 2 times I/9, or 2I/9. The tempting error is forgetting the doubled power and choosing I/9, which only accounts for the distance change.
11. A detector of area $A_d$ is placed a distance $D \gg R$ from the aperture, centered on the aperture axis. In steady state, what is the intensity at the detector, treating the aperture as a new isotropic point source of the power that exits through it?
    Answer: $I = \dfrac{P}{4\pi D^2}$, because in steady state all power $P$ must exit through the aperture, and if the aperture is treated as an isotropic point source, it radiates power $P$ uniformly over a sphere of radius $D$. When radiation is trapped inside a perfectly reflecting enclosure with a small opening, the key insight is energy conservation in steady state. Since the shell absorbs nothing, every joule per second emitted by the source must eventually escape through the only exit: the aperture. This makes the aperture the effective source of the full power $P$. Treating the aperture as an isotropic point source radiating total power $P$, the intensity at distance $D$ follows the standard inverse-square law: $I = \frac{P}{4\pi D^2}$ This is answer B, and the reasoning is airtight — steady state + perfect reflection + no absorption = all power exits the aperture. A is tempting but wrong. It calculates only the power that would pass through the aperture on the first pass from the source, as if the rest were lost. It ignores the reflections entirely. The shell keeps bouncing light back until it escapes — the aperture eventually transmits all of $P$, not just the fraction $A_0/(4\pi R^2)$. C makes a similar mistake with extra steps: it computes a fraction of a fraction, essentially treating each reflection as an absorption event. In a perfectly reflecting shell, no power is absorbed, so this underestimates the total exiting power. D misapplies the inverse-square law by using $R + D$ as the effective distance. The aperture, not the original source, is the relevant radiating point for the detector. Since $D \gg R$, the geometry starts at the aperture, not the shell's center. Study tip: When you see "perfectly reflecting" plus "steady state," immediately invoke conservation of energy — all source power must exit through any opening, regardless of its size.
12. A 60 W point source radiates uniformly. At what distance is the intensity 0.15 W/$m^2$?
    Answer: 5.6 m For a point source, intensity equals power divided by sphere area: I = P / (4 pi $r^2$). Set 0.15 = 60 / (4 pi $r^2$), so $r^2$ = 60 / (0.6 pi) = 100 / pi, giving r about 5.6 m. The tempting 11.3 m comes from using circle area pi $r^2$ instead of full sphere area 4 pi $r^2$.
13. Which of the following statements about the uncertainty in the fitted exponent $n$ is most physically accurate, and what does it imply about the experimental strategy?
    Answer: The uncertainty in $n$ depends on both the intensity measurement error and the span of distances used; extending measurements to larger $r$ (increasing the lever arm in $\ln r$) reduces the uncertainty in the slope more effectively than simply repeating measurements at the existing four distances, because the slope uncertainty scales inversely with the range of the independent variable. Whenever you see a question about fitting a power law from experimental data, think carefully about what controls the precision of the fitted slope — it's not just about measurement error, but also about the geometry of your data in the linearized space. In a $\ln I$ vs. $\ln r$ plot, the slope gives you $-n$. From linear regression theory, the uncertainty in the slope scales as $\sigma_n \propto \frac{\sigma_{\ln I}}{\sqrt{\sum(\ln r_i - \overline{\ln r})^2}}$. The denominator is essentially the spread of your $\ln r$ values — the "lever arm." A $\pm 5\%$ intensity error translates to $\sigma_{\ln I} \approx 0.05$ (since $\delta(\ln I) \approx \delta I / I$), which stays fixed regardless of distance range. So reducing $\sigma_n$ requires *maximizing the spread in $\ln r$*, not just adding more points at the same locations. Extending to larger $r$ dramatically increases that denominator, making D the correct and most physically complete answer. Choice A contains a grain of truth — more points do help — but averaging repeated measurements at the same four distances improves $\sigma_n$ only slowly (as $1/\sqrt{N}$), and misses the far more powerful strategy of extending the range. Choice B commits a subtle error: a small absolute value of $\ln(1.05) \approx 0.049$ is not "negligible" — its size relative to the slope uncertainty, not to $\ln I$ itself, is what matters. Choice C is wrong because it ignores the spacing of $\ln r$ values entirely; the range of the independent variable is critical to slope precision. Your takeaway: in any log-log fitting experiment, extending the measurement range is usually more powerful than repeating existing measurements — always consider the lever arm of your independent variable.
14. A student claims that if a third detector is placed at distance $r_3 = 4 \text{ m}$, the intensity at $r_3$ will be exactly the arithmetic mean of $I_1$ and the intensity at $r_2$. Which of the following best evaluates this claim?
    Answer: The claim is incorrect; the intensity at $r_3$ is less than the arithmetic mean of $I_1$ and $I_2$ because intensity follows an inverse-square law, which is convex (concave upward) with respect to distance, so by Jensen's inequality the function value at the midpoint falls below the linear average of the endpoints. Whenever you see a point source radiating isotropically, anchor your thinking to the inverse-square law: intensity falls off as $I \propto \frac{1}{r^2}$. The question is really asking you to think about the shape of this function — is it linear, concave up, or concave down? Let's assign $I_1 = \frac{P}{4\pi(2)^2} = \frac{P}{16\pi}$. Then $I_2 = \frac{P}{4\pi(6)^2} = \frac{P}{144\pi}$, and $I_3 = \frac{P}{4\pi(4)^2} = \frac{P}{64\pi}$. The arithmetic mean of $I_1$ and $I_2$ is $\frac{1}{2}\left(\frac{1}{16} + \frac{1}{144}\right)\frac{P}{4\pi} = \frac{P}{4\pi} \cdot \frac{160}{2 \cdot 2304} \approx \frac{P}{57.6\pi}$. Since $\frac{1}{64} < \frac{1}{57.6}$, the actual intensity at $r_3$ is less than the arithmetic mean. This confirms C: because $f(r) = 1/r^2$ is convex (concave upward), Jensen's inequality guarantees the function value at the midpoint lies below the chord connecting the endpoints. A is wrong because intensity does not decrease linearly with distance — it follows an inverse-square relationship, which is fundamentally nonlinear. B gets the concavity backwards: $1/r^2$ curves upward (convex), not downward, so midpoint intensity falls below, not above, the linear average. D invents a fictional relationship between dispersion and arithmetic means — dispersion affects speed and frequency, not the geometric spreading described by the inverse-square law. Your takeaway: when comparing a function's midpoint value to the average of its endpoints, always ask whether the function is convex or concave. For any convex function, the midpoint value undershoots the linear average — a pattern that shows up across waves, optics, and thermodynamics.
15. A point source moves 20% farther from an observer. The observed intensity decrease is closest to
    Answer: 31% Intensity follows the inverse-square law. Moving 20% farther means distance is 1.2 times as great, so intensity becomes 1/($1.2^2$) = 1/1.44 of its original value. That leaves about 69%, so the decrease is about 31%. The tempting 44% comes from treating the 1.44 distance-squared factor as the lost intensity; the loss is 1 - 1/1.44.

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