Waves, Sound and Physical Optics: Unit Diagnostic (Physics 2 Unit 14)

What this diagnostic is for

Use this check to find which Unit 14 topics (Waves, Sound, and Physical Optics) to revisit. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real results and give no predicted score. What matters is which questions you miss.

Work without notes, in about 30 minutes, and answer before opening each explanation.
A scientific calculator is allowed. c = 3.00 × 10⁸ m/s; on a string v = √(F_T/μ); filters are ideal; light meets thin films at normal incidence. All data are invented.

Question 1 (multiple choice · 14.1)

A pulse takes 0.20 s to travel the length of a stretched string. The tension is made 9 times larger; the string is unchanged. How long does a pulse now take?

(A) 0.022 s
(B) 0.60 s
(C) 0.067 s
(D) 0.20 s

Answer and explanation

Answer: (C). v = √(F_T/μ), so 9 times the tension gives √9 = 3 times the speed, and one third of the time: 0.20 s ÷ 3 = 0.067 s.

(A) divides by 9. It forgets the square root.
(B) multiplies by 3, as if more tension slowed the pulse.
(D) ignores the tension, which helps set the speed.

If you missed this: read the Topic 14.1 study guide.

Question 2 (multiple choice · 14.2)

A displacement–time graph shows that one point on a string takes 4.0 ms to move from its highest to its lowest position. The wave speed is 50 m/s. What is the wavelength?

(A) 0.20 m
(B) 0.40 m
(C) 2.5 m
(D) 6250 m

Answer and explanation

Answer: (B). Highest to lowest is half a cycle, so T = 8.0 ms and f = 125 Hz. λ = v/f = 50 ÷ 125 = 0.40 m.

(A) uses 4.0 ms as the full period.
(C) divides f by v; (D) divides v by T. Neither has the unit of a length.

If you missed this: read the Topic 14.2 study guide.

Question 3 (multiple choice · 14.3)

A wave passes from string A into string B across a knot. On B its wavelength is 1.5 times its wavelength on A. Which row is correct?

(A) f on B is 1.5 × f on A; v unchanged; a pulse reflected at the knot is upright
(B) f unchanged; v on B is 1.5 × v on A; a pulse reflected at the knot is inverted
(C) f unchanged; v on B is 1.5 × v on A; a pulse transmitted into B is inverted
(D) f unchanged; v on B is 1.5 × v on A; a pulse reflected at the knot is upright

Answer and explanation

Answer: (D). The knot moves both strings at the same rate, so f is unchanged and λ ∝ v. The wave is 1.5 times faster on B. Going into a faster medium, the reflection is upright.

(A) changes f. The source sets f; the medium sets v.
(B) uses the rule for a slower medium.
(C) inverts the transmitted pulse, which is always upright.

If you missed this: read the Topic 14.3 study guide.

Question 4 (multiple choice · 14.4)

Which statement about microwaves and ultraviolet radiation in a vacuum is correct?

(A) Microwaves have the longer wavelength and lower frequency; both travel at the same speed.
(B) Microwaves have the longer wavelength and travel faster.
(C) Ultraviolet has the longer wavelength, because it lies beyond violet.
(D) Ultraviolet needs a medium to travel through, but microwaves do not.

Answer and explanation

Answer: (A). By decreasing wavelength: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. All EM waves travel at c in a vacuum, so from c = fλ the longer wavelength has the lower frequency.

(B) gives EM waves different speeds in a vacuum.
(C) reverses the order. Ultraviolet lies beyond the short-wavelength (violet) end.
(D) No EM wave needs a medium.

If you missed this: read the Topic 14.4 study guide.

Question 5 (multiple choice · 14.5)

An ambulance sounds its siren (rest frequency f₀) while driving along a straight road at 20 m/s. A cyclist follows it at 8 m/s. A pedestrian stands still on the road behind both. What does the cyclist hear?

(A) Lower than f₀, and lower than the pedestrian hears
(B) Exactly f₀, because both move in the same direction
(C) Lower than f₀, but not as low as the pedestrian hears
(D) Higher than f₀, because the cyclist moves toward the ambulance

Answer and explanation

Answer: (C). The cyclist’s separation from the ambulance grows at 12 m/s, so f < f₀. The pedestrian’s grows at 20 m/s, a greater relative speed, so the pedestrian’s shift is greater.

(A) reverses the size rule.
(B) Only equal velocities keep the separation constant.
(D) looks at the cyclist alone; the separation still grows.

If you missed this: read the Topic 14.5 study guide.

Question 6 (multiple choice · 14.6)

Two tuning forks sound together. The loudness peaks once every 0.20 s. One fork is marked 512 Hz. What could the other fork’s frequency be?

(A) 511.8 Hz or 512.2 Hz
(B) 517 Hz only
(C) 509.5 Hz or 514.5 Hz
(D) 507 Hz or 517 Hz

Answer and explanation

Answer: (D). f_beat = 1/0.20 s = 5 Hz, and f_beat = |f₁ − f₂|, so the other fork is 5 Hz above or below 512 Hz.

(A) uses 0.20 as the beat frequency. It is a time: take its reciprocal.
(B) assumes the second fork is higher; the difference has no sign.
(C) halves the beat frequency; each loud moment is one beat.

If you missed this: read the Topic 14.6 study guide.

Question 7 (multiple choice · 14.6)

A pipe 0.85 m long is open at both ends (sound speed 340 m/s). One end is then closed. Which frequency makes a standing wave in the closed pipe but not the open one?

(A) 400 Hz
(B) 300 Hz
(C) 200 Hz
(D) 600 Hz

Answer and explanation

Answer: (B). Open at both ends: fₙ = nv/(2L) = 200, 400, 600 … Hz. Closed at one end: odd n only in fₙ = nv/(4L), so 100, 300, 500 … Hz. Only 300 Hz is in the second list alone.

(A) is an even multiple of 100 Hz: impossible when closed.
(C) is the open fundamental. Closing an end halves it to 100 Hz.
(D) is the open pipe’s third harmonic; the closed pipe’s is 300 Hz.

If you missed this: read the Topic 14.6 study guide.

Question 8 (multiple choice · 14.7)

Light of wavelength 600 nm passes through a single slit onto a screen 2.0 m away. The first dark fringes on either side of the centre are 8.0 mm apart. What is the slit width?

(A) 0.30 mm
(B) 0.15 mm
(C) 0.45 mm
(D) 0.60 mm

Answer and explanation

Answer: (A). Each first dark fringe is 4.0 mm from the centre. a = λL/y = (6.00 × 10⁻⁷ m)(2.0 m) ÷ (4.0 × 10⁻³ m) = 3.0 × 10⁻⁴ m. (θ ≈ 0.1°, so the small-angle form holds.)

(B) uses 8.0 mm as the distance from the centre.
(C) uses (m + ½)λ, the double-slit dark condition.
(D) puts 4.0 mm into 2λL/a, the full width of the central band.

If you missed this: read the Topic 14.7 study guide.

Question 9 (multiple choice · 14.8)

Each slit in a double-slit experiment is made narrower. Their separation, the light and the screen are unchanged. Which describes the new pattern?

(A) Larger fringe spacing; the same number inside the central band
(B) Smaller fringe spacing; fewer inside the central band
(C) Same fringe spacing; more bright fringes inside the central band
(D) Same fringe spacing; fewer bright fringes inside the central band

Answer and explanation

Answer: (C). The fringe spacing λL/d depends on the separation d, which is unchanged. The envelope’s central band, 2λL/a, widens as the slit width a shrinks, so more fringes fit inside it.

(A) and (B) let the slit width change the spacing; only d sets it.
(D) reverses the envelope rule: a narrower slit spreads light more.

If you missed this: read the Topic 14.8 study guide.

Question 10 (multiple choice · 14.9)

A glass window (n = 1.52) is covered by a layer of water (n = 1.33) 250 nm thick. White light falls straight onto it. Which visible wavelength is reflected most strongly?

(A) 443 nm
(B) 665 nm
(C) 500 nm
(D) 376 nm

Answer and explanation

Answer: (B). Air → water and water → glass are both reflections into a higher index: two 180° phase changes, which cancel. Bright needs 2nt = mλ. 2nt = 2 × 1.33 × 250 nm = 665 nm, so m = 1 gives 665 nm (m = 2 gives 333 nm, not visible).

(A) counts one phase change. That condition gives the weakest reflection.
(C) uses 2t = λ, the wavelength in air.
(D) divides 2t by n instead of multiplying.

If you missed this: read the Topic 14.9 study guide.

Question 11 (short answer · 14.3)

One ideal polarizing filter is turned slowly through 360° in each of two light beams, P and Q, in front of a light meter. P reads a steady 12 W/m² at every angle. Q falls smoothly from 30 W/m² to 0 and back, twice per turn.

(a) Which beam was already polarized? Explain. (b) State the intensity of each beam before the filter. (c) The meter’s sensor has an area of 2.0 × 10⁻⁴ m². Find the greatest power it receives from beam Q. (d) The student says: “Any wave that gives a steady reading like P’s must be longitudinal.” Evaluate this claim.

Worked answer

(a) Q. Its reading depends on the angle: all of it passes when the axis lines up with its electric field, none at 90°. P is unpolarized: every angle passes half.
(b) P: 12 W/m² is half, so 24 W/m². Q: the maximum passes all of it, so 30 W/m².
(c) P = IA = (30 W/m²)(2.0 × 10⁻⁴ m²) = 6.0 × 10⁻³ W.
(d) Wrong. Unpolarized light is transverse yet reads steadily. Only a change with angle is evidence, and it shows a transverse wave: longitudinal waves cannot be polarized.

If you missed this: read the Topic 14.3 study guide.

Question 12 (short answer · 14.8)

A laser beam falls straight onto a grating with 400 lines/mm. The first-order spots are at 15.1° on each side of the central spot.

(a) Calculate the wavelength. (b) Find the angle of the second-order spots and the highest order produced. (c) White light replaces the laser. Describe the central maximum and the first order.

Worked answer

(a) d = (1/400) mm = 2.50 × 10⁻⁶ m. λ = d sin θ = (2.50 × 10⁻⁶ m)(sin 15.1°) = 6.51 × 10⁻⁷ m (651 nm).
(b) sin θ₂ = 2λ/d = 0.521, so θ₂ = 31.4°. Orders need m < d/λ = 3.84, so the highest is m = 3 (51.4°). These angles are too large for the small-angle form.
(c) The centre is white (zero path difference for every λ). The first order is a spectrum, red farthest out, because sin θ = λ/d grows with λ.

If you missed this: read the Topic 14.8 study guide.

Your next step

Count a guess or a part-answer as wrong.

Topic	Question(s)	If you missed it, read
14.1 Properties of wave pulses and waves	1	Study guide 14.1
14.2 Periodic waves	2	Study guide 14.2
14.3 Boundary behavior and polarization	3, 11	Study guide 14.3
14.4 Electromagnetic waves	4	Study guide 14.4
14.5 The Doppler effect	5	Study guide 14.5
14.6 Interference and standing waves	6, 7	Study guide 14.6
14.7 Diffraction	8	Study guide 14.7
14.8 Double slits and gratings	9, 12	Study guide 14.8
14.9 Thin-film interference	10	Study guide 14.9

How to use your result

Start with the topic where you missed most. Read its study guide, then do its practice questions.
Look at why you missed it. A slip with units or a factor of 2 needs a habit, not a re-read.
Missed nothing? Try the Unit 14 mixed review.
Retake this check a week later.