Year 12 PhysicsModule 7: The Nature of LightIQ215 MC + 5 written45 min

Checkpoint 2: Wave Evidence and the Wave-Model Synthesis

Checkpoint 2 assesses L03, L04, L05, L10 in the published syllabus sequence.

Stable ID L03

Interference and Young’s Double Slit

Assessment is drawn only from this lesson’s repaired effective pool.

Stable ID L04

Diffraction and Diffraction Gratings

Assessment is drawn only from this lesson’s repaired effective pool.

Stable ID L05

Polarisation of Light

Assessment is drawn only from this lesson’s repaired effective pool.

Stable ID L10

Evidence for the Wave Model of Light

Assessment is drawn only from this lesson’s repaired effective pool.

Coverage boundary

This checkpoint assesses L03, L04, L05, L10 only. Lesson file IDs remain stable; displayed lesson numbers follow the module’s syllabus sequence.

1. In Young's double slit experiment, a dark fringe appears at a point on the screen where the path difference from the two slits is:

Aa whole number of wavelengths, nλ
Bexactly zero
Ca half-integer number of wavelengths, (n + ½)λ
Dexactly one quarter of a wavelength

2. Two light sources are described as coherent. This means they have:

Athe same amplitude and therefore the same brightness
Bthe same polarisation but different frequencies
Cthe same direction of travel through the apparatus
Dthe same frequency and a constant phase relationship

3. Why can two separate ordinary light bulbs not produce a stable interference pattern on a screen?

Atheir atoms emit randomly, so the phase relationship between them fluctuates and they are incoherent
Blight from separate sources cannot overlap in the same region of space
Cbulbs are not bright enough for interference to occur
Dinterference only happens with light that has passed through glass

4. A student running a double slit experiment with red light switches to blue light without changing any other part of the apparatus. The fringes will:

Aspread further apart, because blue light carries more energy
Bmove closer together, because the spacing is proportional to wavelength
Cstay exactly the same, because only the slits control the spacing
Ddisappear, because blue light cannot form an interference pattern

5. For monochromatic light passing through a single slit of width a, the dark bands (minima) in the pattern occur at angles satisfying:

Aa sinθ = (n + ½)λ
Ba sinθ = nλ
Ca cosθ = nλ
Da = nλ sinθ

6. The grating element d of a diffraction grating is best described as:

Athe width of one individual slit in the grating
Bthe total width of the illuminated part of the grating
Cthe distance between adjacent slits, equal to 1/N where N is lines per metre
Dthe number of lines per metre ruled onto the grating

7. Diffraction effects are most pronounced when the width of the aperture is:

Avery much larger than the wavelength
Bexactly ten times the wavelength
Cso small that no wave can pass through at all
Dcomparable to or smaller than the wavelength

8. A diffraction grating is preferred over a prism for precise spectroscopy mainly because:

Aits maxima are very sharp and its dispersion is linear in sinθ, so wavelengths can be measured precisely
Bit transmits far more light than a prism, making the spectrum brighter
Ca prism cannot separate white light into its component colours at all
Dit produces only a single order, so there is no risk of confusion

9. Polarisation of light is only possible because light is a...

Alongitudinal mechanical wave
Blongitudinal electromagnetic wave
Ctransverse electromagnetic wave
Dtransverse mechanical wave

10. When unpolarised light passes through a single ideal polarising filter, the transmitted intensity is...

Aequal to the incident intensity
Bhalf the incident intensity
Czero
Done-quarter the incident intensity

11. Malus's Law gives the intensity of polarised light after passing through an analyser as...

AI = I₀ cosθ
BI = I₀ sinθ
CI = I₀ sin²θ
DI = I₀ cos²θ

12. Young's double-slit experiment demonstrates light interference because...

Alight particles collide at the screen
Btwo beams of light add their intensities
Ccoherent waves from two slits superpose to create constructive and destructive interference
Dlight diffracts around the slits independently

13. In a double-slit experiment, the fringe spacing Δy is given by...

AΔy = d/(λL)
BΔy = λL/d
CΔy = λd/L
DΔy = Ld/λ

14. Huygens' principle states that every point on a wavefront acts as...

Aa mirror
Ba lens
Ca particle source
Da source of secondary spherical wavelets

15. Diffraction is most pronounced when the wavelength of light is...

Acomparable to or larger than the aperture size
Bmuch shorter than the aperture
Cequal to the refractive index
Dmuch longer than the wavelength of red light

SA1. Define coherent sources and state the path-difference conditions for constructive and destructive interference. (4 marks)

SA2. Light of wavelength 500 nm passes through slits 0.25 mm apart onto a screen 2.0 m away. Calculate the fringe spacing and state the approximation used. (4 marks)

SA3. A grating has 600 lines per millimetre. Calculate its slit spacing and the first-order angle for 500 nm light at normal incidence. (4 marks)

SA4. Unpolarised light of intensity 120 W m$^{-2}$ passes through a polariser and an analyser at 60°. Calculate the final intensity and state why the first step is separate from Malus’s law. (4 marks)

SA5. Compare one successful prediction and one limitation of a classical wave model of light. (4 marks)

SA1 Model Answer (4 marks)

Coherent sources have the same frequency and a constant phase difference. Constructive interference occurs for $\Delta r=n\lambda$; destructive interference occurs for $\Delta r=(n+\tfrac12)\lambda$, where $n=0,1,2,\ldots$.

SA2 Model Answer (4 marks)

$\Delta x=\lambda L/d=(500\times10^{-9})(2.0)/(0.25\times10^{-3})=4.0\times10^{-3}$ m = 4.0 mm. This uses the far-screen/small-angle approximation with coherent, effectively monochromatic illumination.

SA3 Model Answer (4 marks)

$d=1/(600\times10^3)=1.67\times10^{-6}$ m. With $d\sin\theta=n\lambda$ and $n=1$, $\sin\theta=0.300$, so $\theta=17.5°$.

SA4 Model Answer (4 marks)

The first ideal polariser transmits half: $I_1=60$ W m$^{-2}$. Then $I=I_1\cos^260°=15$ W m$^{-2}$. The half-intensity rule applies to unpolarised incident light; Malus’s law applies to already plane-polarised light.

SA5 Model Answer (4 marks)

Wave superposition predicts interference and diffraction, and transverse oscillations account for polarisation. A classical continuous-energy wave model cannot account for the photoelectric threshold and the dependence of maximum electron kinetic energy on frequency; a quantum photon model is required for those interactions.

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