Physics 2 Quiz: Intensity And Inverse Square Law

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Intensity And Inverse Square LawQuestion 1 of 15

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?

The second student is correct. Intensity from an isotropic source is uniform in all directions, so the power intercepted depends only on the solid angle subtended by the detector. Since the solid angle is

A_d/r^2

regardless of orientation, the intercepted power is

PA_d/(4\pi r^2)

with no cosine factor, and the first student's formula is wrong.

The first student is correct, but for an incomplete reason. The effective collecting area of the detector is

A_d\cos\theta

, giving intercepted power

PA_d\cos\theta/(4\pi r^2)

. The second student's error is equating the isotropy of the source with orientation-independence of the detector, which are unrelated properties.

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.

Neither student is fully correct. The intercepted power depends on

\cos^2\theta

rather than

\cos\theta

, because both the projected area and the effective path length through the detector change with tilt angle, each contributing one factor of

\cos\theta

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Physics 2 Quiz

Physics 2 Quiz: Intensity And Inverse Square Law

Practice Intensity And Inverse Square Law in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Intensity And Inverse Square Law, 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.

All questions

Question 1

The second student is correct. Intensity from an isotropic source is uniform in all directions, so the power intercepted depends only on the solid angle subtended by the detector. Since the solid angle is $A_d/r^2$ regardless of orientation, the intercepted power is $PA_d/(4\pi r^2)$ with no cosine factor, and the first student's formula is wrong.
The first student is correct, but for an incomplete reason. The effective collecting area of the detector is $A_d\cos\theta$ , giving intercepted power $PA_d\cos\theta/(4\pi r^2)$ . The second student's error is equating the isotropy of the source with orientation-independence of the detector, which are unrelated properties.
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. (correct answer)
Neither student is fully correct. The intercepted power depends on $\cos^2\theta$ rather than $\cos\theta$ , because both the projected area and the effective path length through the detector change with tilt angle, each contributing one factor of $\cos\theta$ .

Explanation: 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.

Question 2

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 $P$ ?

The intensity increases by a factor of $\dfrac{16}{5}$ , because collapsing the sources doubles the combined power at the near distance while eliminating the weaker contribution from the far source.
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. (correct answer)
The intensity decreases by a factor of $\dfrac{5}{8}$ , because placing both sources at the same point causes destructive interference that reduces the net power delivered to the detector.
The intensity increases by a factor of $4$ , because both sources are now at the closest possible distance and the inverse-square law predicts a fourfold gain relative to the configuration when $S_2$ was at distance $2d$ .

Explanation: 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.

Question 3

A spherical surface of radius $R$ surrounds an isotropic point source of electromagnetic radiation with total radiated power $P$ . The medium filling the sphere has an absorption coefficient $\alpha$ (in units of $\text{m}^{-1}$ ), so that intensity falls off as $I(r) = \frac{P}{4\pi r^2}e^{-\alpha r}$ for $r > 0$ .

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 $\alpha r$ ?

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. (correct answer)
The fractional reduction is approximately $1 - e^{-\alpha r} \approx \alpha r - (\alpha r)^2/2$ , indicating that absorption and geometric spreading contribute comparably when $\alpha r \sim 1$ , but absorption is the dominant loss mechanism for $\alpha r \ll 1$ .
The fractional reduction is approximately $(\alpha r)^2/2$ , indicating that absorption contributes a second-order correction and is far less significant than geometric spreading for small $\alpha r$ , so the $1/r^2$ law is essentially exact in this limit.
The fractional reduction is exactly $e^{-\alpha r}$ , which cannot be further simplified without knowing $r$ explicitly; therefore no statement about the relative importance of spreading versus absorption can be made in this limit.

Explanation: 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.

Question 4

A 60 W point source radiates uniformly. At what distance is the intensity 0.15 W/m^2?

5.6 m (correct answer)
11.3 m
20.0 m
31.8 m

Explanation: 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.

Question 5

A detector of area A at distance d collects power P. At 2d, with area 2A, collected power is

$P$
$P/4$
$P/2$ (correct answer)
$8P$

Explanation: 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.

Question 6

A point source gives intensity I at distance r. Power doubles and distance triples. New intensity?

$2I/9$ (correct answer)
$2I/3$
$I/9$
$9I/2$

Explanation: 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.

Question 7

A point source moves 20% farther from an observer. The observed intensity decrease is closest to

17%
31% (correct answer)
40%
44%

Explanation: 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.

Question 8

At distance d from a point source, correct exposure time is 1/250 s. At 2d, the correct exposure time is closest to

$1/1000$ s
$1/250$ s
$1/125$ s
$1/60$ s (correct answer)

Explanation: 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.

Question 9

A student investigates how the intensity from a point source varies with distance by taking measurements at $r = 1, 2, 3, 4 \text{ m}$ . Due to instrument noise, each measurement has an independent random uncertainty of $\pm 5\%$ in intensity. The student plans to determine the exponent $n$ in $I \propto r^{-n}$ by plotting $\ln I$ vs. $\ln r$ and finding the slope.

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?

The uncertainty in $n$ is dominated by the $\pm 5\%$ intensity error, which translates directly to a $\pm 5\%$ uncertainty in $n$ via the logarithmic derivative, so collecting more data points at the same four distances would halve the uncertainty in $n$ every time the number of points doubles.
The uncertainty in $n$ is negligible because the $\pm 5\%$ intensity error becomes negligible after the logarithm is taken — since $\ln(1.05I) - \ln(I) = \ln(1.05) \approx 0.049$ , which is very small compared to the typical values of $\ln I$ encountered in the experiment.
The uncertainty in $n$ is completely determined by the number of data points and is independent of the range of distances chosen, because the least-squares fit extracts the slope from all data equally regardless of their spacing in $\ln r$ .
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. (correct answer)

Explanation: 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.

Question 10

The intensity level (in decibels) of a sound is defined as $\beta = 10\log_{10}(I/I_0)$ , where $I_0 = 10^{-12} \text{ W/m}^2$ . A small loudspeaker can be modeled as a point source radiating isotropically with total acoustic power $P$ .

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?

$\beta_2 = 80/9 \approx 8.9 \text{ dB}$ , obtained by dividing the original decibel level by $3^2 = 9$ , because intensity level scales as the square of the inverse-square reduction.
$\beta_2 = 80 - 30 = 50 \text{ dB}$ , obtained by noting that tripling the distance reduces intensity by a factor of 9, and incorrectly associating a factor of 9 in intensity with a 30 dB drop (which actually corresponds to a factor of 1000).
$\beta_2 \approx 70.5 \text{ dB}$ , 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 $\beta_2 = 80 - 9.5 = 70.5 \text{ dB}$ . (correct answer)
$\beta_2 \approx 70.5 \text{ dB}$ , obtained by applying $\Delta\beta = 20\log_{10}(3) \approx 9.5 \text{ dB}$ , because the decibel scale for sound inherently references pressure amplitude rather than intensity, so the distance factor must enter as $20\log_{10}(3)$ rather than $10\log_{10}(9)$ .

Explanation: 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.

Question 11

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?

The point-source intensity increases by a factor of 4 when the distance halves, which equals the factor by which the focused-beam intensity increases, since both situations involve a fourfold increase in intensity.
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. (correct answer)
The point-source intensity increases by a factor of 4 when the distance halves, and the focused-beam intensity also increases by a factor of 16; however, the comparison is invalid because a laser beam does not obey the inverse-square law, so the factor-of-4 point-source result cannot be meaningfully related to beam focusing.
The point-source intensity increases by a factor of 4 when the distance halves, while the focused-beam intensity increases by a factor of 4 as well, because reducing beam radius by 4 is equivalent in energy terms to halving the distance to an isotropic source, both representing the same geometric compression of wavefronts.

Explanation: 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

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.

Question 12

A point source of light is located at the center of a hollow, perfectly reflecting spherical shell of radius $R$ . A small circular aperture of area $A_0 \ll 4\pi R^2$ is cut in the shell, allowing light to escape. The source radiates total power $P$ isotropically. Because the shell is perfectly reflecting, all power that does not exit through the aperture is reflected back and eventually exits through the aperture (assume steady state with no absorption).

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?

$I = \dfrac{P A_0}{4\pi D^2 \cdot 4\pi R^2}$ , because the aperture re-radiates only the fraction $A_0/(4\pi R^2)$ of $P$ isotropically, and that power spreads over a sphere of radius $D$ .
$I = \dfrac{P}{4\pi D^2}$ , 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 $D$ . (correct answer)
$I = \dfrac{P}{4\pi R^2} \cdot \dfrac{A_0}{4\pi D^2}$ , because the intensity at the shell surface is $P/(4\pi R^2)$ , the aperture intercepts a fraction $A_0/(4\pi R^2)$ of that, and this intercepted power then spreads isotropically to the detector.
$I = \dfrac{P}{4\pi(R+D)^2}$ , because the effective source is at the center of the shell, so the total distance from source to detector is $R + D$ and the inverse-square law must use this combined distance.

Explanation: 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.

Question 13

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 $R_s$ )?

$d_A = 50 \text{ pc}$ , 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. (correct answer)
$d_A = 50 \text{ pc}$ , and the solid angle ratio $\Omega_A/\Omega_B = 4$ , because solid angle scales as $1/d$ rather than $1/d^2$ , so the fourfold decrease in distance produces only a fourfold increase in solid angle, not a 16-fold increase.
$d_A = 800 \text{ pc}$ , and the solid angle ratio $\Omega_A/\Omega_B = 1/16$ , because if Star A is 16 times brighter it must be 16 times farther away to produce equivalent radiation per unit area at Earth, and the greater distance makes its solid angle smaller.
$d_A = 50 \text{ pc}$ , and the solid angle ratio $\Omega_A/\Omega_B = 4$ , because while the distance ratio $d_B/d_A = 4$ , solid angle is proportional to the physical area of the star divided by $d^2$ , and one must account for the factor of 4 in distance linearly before squaring, yielding a net factor of 4 rather than 16.

Explanation: 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.

Question 14

A point source of sound emits waves isotropically in a medium with negligible absorption. A detector at distance $r_1 = 2 \text{ m}$ from the source registers intensity $I_1$ . A second detector is placed at distance $r_2 = 6 \text{ m}$ .

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?

The claim is correct because intensity decreases linearly with distance for a point source, so the midpoint distance yields the average intensity.
The claim is incorrect; the intensity at $r_3$ is greater than the arithmetic mean of $I_1$ and $I_2$ because intensity follows an inverse-square law, which is concave downward over this distance range, making the midpoint intensity exceed the linear average.
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. (correct answer)
The claim is correct only if the medium is non-dispersive; in a dispersive medium the arithmetic mean relationship holds automatically due to phase cancellation effects at intermediate distances.

Explanation: 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

\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.

Question 15

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?

$I = (P_2 - P_1)/(4\pi r^2) = 8P_1/(4\pi r^2)$ , because coherent out-of-phase sources combine by subtracting their powers directly, and this net power spreads isotropically via the inverse-square law.
The intensity is zero at all distances, because two coherent sources with a $\pi$ phase difference always produce complete destructive interference, and no net power reaches the receiver regardless of the individual power levels.
$I = (P_1 + P_2)/(4\pi r^2) = 10P_1/(4\pi r^2)$ , because energy must be conserved, so the total power at the receiver is always the sum of both transmitter powers regardless of phase relationship.
$I = (\sqrt{P_2} - \sqrt{P_1})^2/(4\pi r^2) = 4P_1/(4\pi r^2)$ , 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 $4\pi r^2$ . (correct answer)

Explanation: 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.