6.2 Diffraction and polarisation — Practise

1. The slit spacing

Answer: $250\,\text{lines mm}^{-1} = 2.50 \times 10^5\,\text{lines m}^{-1}$, so $d = \dfrac{1}{2.50 \times 10^5} = 4.0 \times 10^{-6}\,\text{m}$.

2. The first-order angle

Answer: $\sin\theta = \dfrac{500 \times 10^{-9}}{4.0 \times 10^{-6}} = 0.125$, so $\theta = 7.2^\circ$.

3. The highest order

Answer: $\dfrac{d}{\lambda} = \dfrac{4.0 \times 10^{-6}}{500 \times 10^{-9}} = 8.0$. The order must be a whole number with $n\lambda \le d$, so the highest order is 8.

4. The wavelength

Answer: $\lambda = \dfrac{d\sin\theta}{n} = \dfrac{3.33 \times 10^{-6} \times \sin 20.0^\circ}{2} = 5.70 \times 10^{-7}\,\text{m}$, or $570\,\text{nm}$.

5. Explain why you can hear a person through a doorway from around a corner, but you cannot see them.

• The wavelength of sound is about a metre, which is close to the width of the doorway.
• The wavelength of light is about 500 nm, which is far smaller than the doorway.
• Sound therefore spreads out strongly at the doorway, so it reaches you around the corner.
• Light spreads very little, because the gap is far larger than its wavelength, so it travels almost straight on and you cannot see the person.

6. The fraction of the intensity

Answer: $\dfrac{I}{I_0} = \cos^2 30^\circ = 0.75$, so three quarters of the intensity passes.

7. The transmitted amplitude

Answer: $E = E_0\cos\theta = 25 \times \cos 40^\circ = 19.2\,\text{V}\,\text{m}^{-1}$.

8. Describe what happens to the light when the second of two polarising filters is turned until its axis is at 90 degrees to the first, and explain why.

• No light passes through the second filter when the axes are at 90 degrees. The filters are crossed.
• The first filter leaves the light polarised along one line. The transmission axis of the second filter is at right angles to that line.
• The component of the electric field along the axis of the second filter is then zero, so there is nothing for it to transmit.

9. You are given a diffraction grating labelled 300 lines per millimetre and a laser of unknown wavelength. Describe how to find the wavelength, and state the measurements you need.

• The number of lines per millimetre gives the slit spacing from d = 1 / N.
• Shine the laser through the grating and measure the angle from the straight-through direction to a bright maximum.
• Note the order n of that maximum, counting the straight-through direction as order 0.
• Use $d\\\\sin\\\\theta = n\\\\lambda$, rearranged to give $\\\\lambda = \\\\dfrac{d\\\\sin\\\\theta}{n}$.

10. The fringe spacing

Answer: $w = \dfrac{\lambda D}{s} = \dfrac{600 \times 10^{-9} \times 1.5}{0.25 \times 10^{-3}} = 3.6 \times 10^{-3}\,\text{m}$, or $3.6\,\text{mm}$.

11. The wavelength and its uncertainty

Answer: The wavelength is $\lambda = d\sin\theta = 2.00 \times 10^{-6} \times 0.300 = 6.00 \times 10^{-7}\,\text{m}$. The grating spacing is exact, so only the sine contributes: the percentage uncertainty is 1.0%. The absolute uncertainty is $6.00 \times 10^{-7} \times \dfrac{1.0}{100} = 6.0 \times 10^{-9}\,\text{m}$.

12. The angle to three significant figures

Answer: $18.46^\circ = 18.5^\circ$ to three significant figures.