Edexcel A-Level Physics AS 2.3.51 Diffraction Grating Equation Questions

Practise Edexcel IAL Physics 2.3.51 by applying nλ=d sinθ, converting line density to grating spacing and using screen geometry to calculate diffraction angles and wavelength.

Exam points

calculate wavelength or grating spacing with nλ = d sin θ and first-order maxima
convert grating line density into slit spacing before substituting in the equation
use screen distances to find diffraction angle and reject Young double-slit substitutions

[Maximum number: 4]

A teacher has a laser pointer. The laser pointer emits a narrow beam of plane polarised light.

(a)

The teacher shines light from the laser pointer onto a diffraction grating as shown. This produces a diffraction pattern on a screen.

Figure for Question (a) — Edexcel A-Level Physics AS

[ 4 ]

(i)

The diffraction grating equation is $n\lambda=d\sin\theta$ . State the quantity represented by d.

[ 1 ]

Distance between (adjacent) slits

Or grating spacing

Or line spacing

Or slit spacing

Or $\frac{1}{\text { lines per } m}$

(ii)

The distance from the central maximum to the first-order maximum is $1.61\,\mathrm{m}$ and the grating-to-screen distance is $2.74\,\mathrm{m}$ . The wavelength is $532\,\mathrm{nm}$ . Calculate the number of lines per millimetre on the diffraction grating.

[ 3 ]

$\tan\theta=1.61/2.74$ , so $\theta=30.4^\circ$ . Using $n\lambda=d\sin\theta$ gives $d=1.05\times10^{-6}\,\mathrm{m}$ . Therefore the grating has $1/d=9.52\times10^5\,\mathrm{m^{-1}}=952$ lines per millimetre (about 950).

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Edexcel A-Level Physics AS 2.3.51 Diffraction Grating Equation Questions question 1

Question (a)

Question (i)

Question (ii)