Studyledge

Superposition: Question 9

Syllabus 8.4

Structured AS 8 marks

White light, containing all wavelengths from violet (400 nm400\text{ nm}) to red (700 nm700\text{ nm}), is incident normally on a diffraction grating that has 300300 lines per millimetre. A spectrum is formed in the first order on each side of the central (zero-order) maximum.

(a) Show that the spacing dd between adjacent lines of the grating is 3.33×10−6 m3.33\times10^{-6}\text{ m} (to 3 significant figures). [2]

(b) Calculate the angular separation, in the first order, between the violet end (400 nm400\text{ nm}) and the red end (700 nm700\text{ nm}) of the spectrum. [4]

(c) State and explain the effect on the angular separation calculated in (b) if a grating with more lines per millimetre is used instead. [2]

Show worked solution Hide worked solution

Worked solution

Part (a): Grating spacing

The grating has 300300 lines per millimetre, so the number of lines per metre is: 300×1000=3.00×105 lines per metre300 \times 1000 = 3.00\times10^{5}\text{ lines per metre}

The spacing dd between adjacent lines is the reciprocal of this: d=13.00×105=3.33‾×10−6 m≈3.33×10−6 md = \frac{1}{3.00\times10^{5}} = 3.3\overline{3}\times10^{-6}\text{ m} \approx 3.33\times10^{-6}\text{ m}

Check (independent method): 300300 lines per mm means each line is 1300 mm=0.0033‾ mm\frac{1}{300}\text{ mm}=0.003\overline{3}\text{ mm} from the next; converting to metres, 0.0033‾ mm=3.33×10−6 m0.003\overline{3}\text{ mm}=3.33\times10^{-6}\text{ m} (3 s.f.), the same result.

So d≈3.33×10−6 md \approx \boxed{3.33\times10^{-6}\text{ m}}, as required.

Part (b): Angular separation of the first-order spectrum

Using the diffraction grating equation dsin⁡θ=nλd\sin\theta = n\lambda with n=1n=1 and the unrounded spacing d=3.33‾×10−6 md = 3.3\overline{3}\times10^{-6}\text{ m}:

Violet end (λ=400 nm=4.00×10−7 m\lambda = 400\text{ nm} = 4.00\times10^{-7}\text{ m}): sin⁡θviolet=λd=4.00×10−73.33‾×10−6=0.1200\sin\theta_{\text{violet}} = \frac{\lambda}{d} = \frac{4.00\times10^{-7}}{3.3\overline{3}\times10^{-6}} = 0.1200 θviolet=arcsin⁡(0.1200)=6.89° (3 s.f.)\theta_{\text{violet}} = \arcsin(0.1200) = 6.89° \text{ (3 s.f.)}

Red end (λ=700 nm=7.00×10−7 m\lambda = 700\text{ nm} = 7.00\times10^{-7}\text{ m}): sin⁡θred=λd=7.00×10−73.33‾×10−6=0.2100\sin\theta_{\text{red}} = \frac{\lambda}{d} = \frac{7.00\times10^{-7}}{3.3\overline{3}\times10^{-6}} = 0.2100 θred=arcsin⁡(0.2100)=12.12° (4 s.f.)\theta_{\text{red}} = \arcsin(0.2100) = 12.12° \text{ (4 s.f.)}

Check (recompute independently): using the rounded d=3.33×10−6 md=3.33\times10^{-6}\text{ m}, sin⁡θviolet=4.00×10−7/3.33×10−6=0.1201≈0.1200\sin\theta_{\text{violet}}=4.00\times10^{-7}/3.33\times10^{-6}=0.1201\approx0.1200 and sin⁡θred=7.00×10−7/3.33×10−6=0.2102≈0.2100\sin\theta_{\text{red}}=7.00\times10^{-7}/3.33\times10^{-6}=0.2102\approx0.2100, consistent with the unrounded values; and sin⁡(6.89°)≈0.1200\sin(6.89°)\approx0.1200, sin⁡(12.12°)≈0.2099≈0.2100\sin(12.12°)\approx0.2099\approx0.2100, confirming both angles.

The angular separation between the two colours is: Δθ=θred−θviolet=12.12°−6.89°=5.23°\Delta\theta = \theta_{\text{red}} - \theta_{\text{violet}} = 12.12° - 6.89° = 5.23°

Part (c): Effect of using more lines per millimetre

Using a grating with more lines per millimetre means there are more lines per metre, so the spacing dd between adjacent lines becomes smaller. Since sin⁡θ=nλd\sin\theta = \dfrac{n\lambda}{d}, a smaller dd makes sin⁡θ\sin\theta larger for both the violet and red wavelengths, so each colour is diffracted through a larger angle. Because this increase is greater for the longer (red) wavelength than for the shorter (violet) wavelength, the angular separation between the colours also increases. The spectrum is spread out over a wider range of angles (greater dispersion).

Final answers

  • (a) d≈3.33×10−6 md \approx \boxed{3.33\times10^{-6}\text{ m}}
  • (b) θviolet=6.89°\theta_{\text{violet}} = 6.89°, θred=12.12°\theta_{\text{red}} = 12.12°, angular separation =5.23°= \boxed{5.23°}
  • (c) A grating with more lines per mm has smaller dd, so both angles increase and the angular separation (dispersion) increases