CAIE A-Level Physics AS 8.4 The Diffraction Grating Questions

Practise using d sin θ = nλ to calculate grating order, wavelength and angle, and interpreting measured maxima in a diffraction experiment.

Exam points

calculate wavelength, order or angle from diffraction-grating maxima using d sin θ = nλ
describe a grating experiment and use measured orders, angles and grating spacing to determine light wavelength

Covered syllabus points

8.4.1—D sin θ = nλ 8.4.2—The use of a diffraction grating to determine the wavelength of light

[Maximum number: 6]

(a)

A diffraction grating is used to determine the wavelength of light.

[ 3 ]

(i)

By reference to interference, explain

[ 3 ]

1. waves (from each element/slit) overlap/meet/superpose

with a phase difference/path difference of zero

2. phase difference is $360^{\circ}$ / path difference of $\lambda$

(i)

the zero order maximum,

(ii)

the first order maximum.

[ 3 ]

(b)

A diffraction grating is used with different wavelengths of light. The angle $\theta$ of the second order maximum is measured for each wavelength. The variation with wavelength $\lambda$ of $\sin \theta$ is shown in Fig. 5.1.

Fig. 5.1

[ 3 ]

(i)

Use the gradient determined in (i) to calculate the slit separation d of the diffraction grating.

[ 2 ]

$d \sin \theta=n \lambda$

d=n / gradient
$=2 / 8.0 \times 10^{5}=2.5 \times 10^{-6} \mathrm{~m}$

(ii)

On Fig. 5.1, sketch a line to show the results that would be obtained for the first order maxima.

[ 1 ]

straight line drawn with lower gradient (about $\frac{1}{2}$ ) and all points lower

[Maximum number: 1]

Which property of a light wave can be determined using a diffraction grating?

amplitude

intensity

speed

wavelength

[Maximum number: 4]

(a)

A beam of light of wavelength $4.3 \times 10^{-7} \mathrm{~m}$ is incident normally on a diffraction grating in air, as shown in Fig. 5.3.

Fig. 5.3 (not to scale)

The third-order diffraction maximum of the light is at an angle of $68^{\circ}$ to the direction of the incident light beam.

[ 4 ]

(i)

Calculate the line spacing d of the diffraction grating.

d =m [2

[ 2 ]

$n \lambda=d \sin \theta$ $\begin{aligned} d =\left(3 \times 4.3 \times 10^{-7}\right) / \sin 68^{\circ} =1.4 \times 10^{-6} \mathrm{~m} \end{aligned}$

(ii)

Determine a different wavelength of visible light that will also produce a diffraction maximum at an angle of $68^{\circ}$ .
wavelength = m

[ 2 ]

$1.4 \times 10^{-6} \times \sin 68^{\circ}=2 \times \lambda$ or
$3 \times 4.3 \times 10^{-7}=2 \times \lambda$ $\lambda=6.5 \times 10^{-7} \mathrm{~m}$

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Covered syllabus points

Question 1

Question (a)

Question (i)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

Question 2

Question 3

Question (a)

Question (i)

Question (ii)