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Diffraction gratings questions
Swap two slits for thousands and the fuzzy fringes sharpen into thin, brilliant lines at exactly predictable angles. One short equation locates every line, and the geometry of a sine caps how many lines can exist.
18 original questions · 58 marks · the diffraction gratings notes · Waves
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State the diffraction grating equation and define each term in it.
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d sinθ = nλ (1), where d is the grating spacing, the distance between adjacent slits, θ the angle of the maximum, n the order, and λ the wavelength (1).A diffraction grating has 300 lines per millimetre. Calculate the grating spacing d in metres.
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d = 1/(300 × 103 lines per metre) (1)
d = 3.33 × 10−6 m (1)Explain why a diffraction grating produces sharper, brighter maxima than a double slit.
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A grating has many slits, so far more beams interfere (1). Constructive interference occurs only at precise angles where all beams are in phase, giving narrow, intense maxima; away from these angles the many beams cancel (1).State what is meant by the zero-order maximum produced by a diffraction grating, and give the path difference between light from adjacent slits at this maximum.
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The maximum in the straight-through direction, along the normal to the grating (n = 0) (1). The path difference between light from adjacent slits there is zero (1).Explain why, for a given grating and wavelength, only a limited number of orders of maxima can be produced.
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The grating equation gives sinθ = nλ/d, and sinθ cannot exceed 1 (1). Orders therefore exist only while nλ/d ≤ 1, so no order beyond the whole-number part of d/λ has an angle at which it can form (1).Light of wavelength 600 nm is shone normally at a grating with 500 lines per millimetre. Calculate the angle of the first-order maximum.
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d = 1/(500 × 103) (1)
d = 2.00 × 10−6 m (1)
sinθ = nλ/d = (1 × 600 × 10−9)/(2.00 × 10−6) (1)
θ = 17.5° (1)Light of wavelength 600 nm is incident normally on a grating with 500 lines per millimetre. Calculate the angle of the second-order maximum.
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d = 1/(500 × 103) = 2.00 × 10−6 m (1)
sinθ = 2λ/d = (2 × 600 × 10−9)/(2.00 × 10−6) (1)
θ = 36.9° (1)A grating with 600 lines per millimetre gives a first-order maximum at 21°. Calculate the wavelength of the light.
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d = 1/(600 × 103) = 1.67 × 10−6 m (1)
λ = d sinθ = 1.67 × 10−6 × sin 21° (1)
λ = 597 nm (1)Light of wavelength 500 nm gives a first-order maximum at 15° with a certain grating. Calculate the grating spacing and hence the number of lines per millimetre.
Blue light of wavelength 486 nm from a hydrogen discharge lamp is incident normally on a grating with 420 lines per millimetre. Show that the grating spacing is about 2.4 × 10−6 m. Go on to calculate the angle of the third-order maximum.
A laser of wavelength 633 nm shines normally through a grating with 250 lines per millimetre. The maxima form spots on a screen 1.5 m beyond the grating and parallel to it. Calculate the distance on the screen between the two first-order spots.
To reduce the uncertainty in an angle measurement, a student measures the angle between the two first-order maxima on either side of the centre and obtains 40.0°. The grating has 600 lines per millimetre. Calculate the wavelength of the light.
A grating has a spacing of 2.0 × 10−6 m and is used with 600 nm light. Calculate the highest order of maximum that can be seen, and the angle at which it appears.
White light (400 nm to 700 nm) passes normally through a grating of spacing 2.0 × 10−6 m. Calculate the angular width of the first-order spectrum (the angle between the 400 nm and 700 nm maxima).
Give two practical uses of diffraction gratings.
White light with wavelengths from 400 nm to 700 nm passes normally through a grating with 400 lines per millimetre. Deduce whether the second-order and third-order spectra overlap.
A teacher wants to show a class at least three orders of maxima on each side of the centre, using light of wavelength 589 nm from a sodium lamp. Two gratings are available: grating X with 300 lines per millimetre and grating Y with 800 lines per millimetre. Deduce which grating the teacher should use.
White light with wavelengths from 400 nm to 700 nm is incident normally on a diffraction grating with 500 lines per millimetre. Describe and explain the appearance of the pattern produced, up to and including the second order, and show that the second-order spectrum is complete.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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