📝 Unit 4: Wave Nature of Light — Unit Test

Assessment OF Learning — Summative
✅ Graded — Counts toward 70% term mark
⏱️ Duration: 75 minutes  |  Total: /60 marks. Use \(c=3.00\times10^8\,\text{m/s}\), \(h=6.63\times10^{-34}\,\text{J·s}\).
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15

📋 Formulae

\(c=f\lambda\) | \(n_1\sin\theta_1=n_2\sin\theta_2\) | \(\sin\theta_c=n_2/n_1\) | Double slit (max): \(d\sin\theta=m\lambda\), \(\Delta y=\lambda L/d\) | Single slit (min): \(a\sin\theta=m\lambda\) | Grating: \(d\sin\theta=m\lambda\) | Thin film: \(2nt = (m+\tfrac12)\lambda\) (constructive, soft-hard) | \(E=hf\) | Malus: \(I=I_0\cos^2\theta\)
Part A: Knowledge & Understanding [15 marks]
1. [2]
A radio wave has wavelength \(3.0\,\text{m}\). Calculate its frequency.
2. [3]
Light travels from air (\(n=1.00\)) into water (\(n=1.33\)) at an angle of incidence of \(40°\). (a) Find the angle of refraction. (b) Compute the speed of light in water. (c) State Snell's law.
3. [2]
Determine the critical angle for light passing from glass (\(n=1.50\)) into air.
4. [3]
In a Young's double-slit experiment, slits separated by \(0.20\,\text{mm}\) produce fringes \(6.6\,\text{mm}\) apart on a screen \(2.5\,\text{m}\) away. Calculate the wavelength of the light.
5. [2]
Choose the best answer. Diffraction is most pronounced when:
6. [3]
Briefly state how each of the following provides evidence for the wave nature of light: (a) diffraction, (b) interference, (c) polarization.
Part B: Thinking & Investigation [15 marks]
7. [5]
A diffraction grating with \(6000\) lines/cm is illuminated with monochromatic light. The 2nd-order maximum is observed at \(43°\). (a) Find the slit spacing \(d\). (b) Calculate the wavelength. (c) How many maxima are observable in total?
8. [5]
An anti-reflective coating on a camera lens has refractive index \(n_{film}=1.38\) on glass (\(n_{glass}=1.52\)). Find the minimum thickness of the coating that minimizes reflection of light with vacuum wavelength \(550\,\text{nm}\). Show reasoning about phase changes.
9. [5]
Two polarizing filters are crossed (axes at \(90°\)). A third polarizer is inserted between them with its axis at \(\theta\) to the first. (a) Derive the transmitted intensity in terms of \(I_0\) and \(\theta\). (b) Find the value of \(\theta\) that maximizes transmission and the maximum fraction transmitted.
Part C: Communication [15 marks]
10. [5]
Using a labelled diagram, explain how Young's double-slit experiment provides quantitative evidence for the wave nature of light. Include path difference and the conditions for constructive/destructive interference.
[ Diagram: source → slits → screen with fringe pattern; mark path difference \(d\sin\theta\) ]
11. [5]
A student says: "Total internal reflection happens whenever light hits a boundary going from a less dense to a more dense medium." Correct this statement, defining the critical angle and giving one technological example (e.g., fibre optics) where TIR is exploited.
12. [5]
Compare and contrast single-slit diffraction and double-slit interference patterns. Address central maximum width, fringe spacing formulas, intensity envelope, and a simple sketch of each pattern.
[ Sketch single-slit envelope vs double-slit fringes ]
Part D: Application [15 marks]
13. [5]
An optical fibre core has \(n=1.50\); the cladding has \(n=1.46\). (a) Compute the critical angle at the core-cladding boundary. (b) Compute the maximum acceptance angle in air for a ray that will undergo TIR. (c) Discuss one social impact of fibre-optic networks.
14. [5]
A soap film (\(n=1.33\)) on a window appears to reflect green light strongly at \(\lambda=540\,\text{nm}\). Find the minimum non-zero thickness consistent with constructive reflection. Identify the phase changes at each surface.
15. [5]
Polarized sunglasses are useful for reducing glare from horizontal surfaces (water, roads). (a) Explain using Brewster's idea why glare is preferentially polarized. (b) Calculate Brewster's angle for water (\(n=1.33\)). (c) Discuss the orientation of the transmission axis of polarized sunglasses and its effect on LCD screen visibility.
📖 Complete Answer Key (Click to reveal)

Q1.

\(f=c/\lambda = 3.0e8/3.0 = 1.0\times10^8\,\text{Hz} = 100\,\text{MHz}\).

Q2.

(a) \(\sin\theta_2 = (1/1.33)\sin 40°=0.483 \Rightarrow \theta_2=28.9°\). (b) \(v = c/n = 2.26\times10^8\,\text{m/s}\). (c) \(n_1\sin\theta_1=n_2\sin\theta_2\).

Q3.

\(\sin\theta_c = 1/1.50 \Rightarrow \theta_c = 41.8°\).

Q4.

\(\lambda = d\Delta y/L = (2e{-}4)(6.6e{-}3)/2.5 = 5.28\times10^{-7}\,\text{m}=528\,\text{nm}\).

Q5.

(c).

Q6.

(a) Bending around obstacles → wave behaviour. (b) Bright/dark fringes from path-difference interference (only waves superpose). (c) Transverse waves can be filtered by axis orientation; particles cannot be polarized.

Q7.

(a) \(d=1/6000\,\text{cm}=1.667\times10^{-6}\,\text{m}\). (b) \(\lambda = d\sin 43°/2 = (1.667e{-}6)(0.682)/2=5.68\times10^{-7}\,\text{m}=568\,\text{nm}\). (c) \(m_{max}=d/\lambda=2.93 \Rightarrow m=0,\pm1,\pm2 \Rightarrow 5\) maxima visible.

Q8.

Both reflections (air→film, film→glass) introduce \(\pi\) phase changes (low→high \(n\) at each), so destructive condition is \(2nt = (m+\tfrac12)\lambda\). Min: \(t_{min} = \lambda/(4n) = 550/(4\cdot1.38)=99.6\,\text{nm}\).

Q9.

(a) After P1: \(I_0/2\). After P2 (at θ): \(I_0/2 \cdot \cos^2\theta\). After P3 (at \(90°-\theta\) to P2): \(I_0/2 \cdot \cos^2\theta\cdot\cos^2(90°-\theta) = I_0/2 \cdot \cos^2\theta\sin^2\theta = I_0/8 \cdot \sin^2(2\theta)\). (b) Max at \(\theta=45°\), maximum fraction = \(1/8\) = 12.5%.

Q10.

Bright fringes where \(d\sin\theta=m\lambda\); dark where \(d\sin\theta=(m+\tfrac12)\lambda\). Fringe spacing \(\Delta y = \lambda L/d\). Equal-spaced bright/dark fringes are predicted only by superposition of waves.

Q11.

TIR requires going from higher \(n\) to lower \(n\) AND \(\theta>\theta_c\). Example: fibre-optic communication.

Q12.

Single slit: central max twice as wide, minima at \(a\sin\theta=m\lambda\), intensity envelope decreases. Double slit: equally spaced bright fringes inside diffraction envelope.

Q13.

(a) \(\sin\theta_c = 1.46/1.50 \Rightarrow \theta_c = 76.7°\). (b) Acceptance: \(n_0\sin\theta_a = \sqrt{n_1^2-n_2^2}=\sqrt{1.50^2-1.46^2}=0.343 \Rightarrow \theta_a=20.1°\). (c) High-bandwidth global telecom; "digital divide" if access is uneven.

Q14.

One π phase change (air→soap), none soap→air → constructive condition: \(2nt=(m+\tfrac12)\lambda\). Min: \(t=\lambda/(4n)=540/(4\cdot1.33)=101\,\text{nm}\).

Q15.

(a) Reflection off horizontal surfaces preferentially polarizes light horizontally (especially near Brewster's angle). (b) \(\tan\theta_B = 1.33 \Rightarrow \theta_B=53.1°\). (c) Glasses block horizontal polarization (vertical axis). LCD screens emit polarized light; rotating head can dim the display.

Evaluation Rubric

LevelDescription%
4Thorough, insightful80–100%
3Considerable (provincial standard)70–79%
2Some effectiveness60–69%
1Limited50–59%
RInsufficient<50%