Questions on diffraction gratings

Question 1

State the diffraction grating equation and define each term in it.

Accepted Answer

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).

Question 2

A diffraction grating has 300 lines per millimetre. Calculate the grating spacing d in metres.

Accepted Answer

d = 1/(300 × 10³ lines per metre) (1) d = 3.33 × 10⁻⁶ m (1)

Question 3

Explain why a diffraction grating produces sharper, brighter maxima than a double slit.

Accepted Answer

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).

Question 4

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.

Accepted Answer

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).

Question 5

Explain why, for a given grating and wavelength, only a limited number of orders of maxima can be produced.

Accepted Answer

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).

Question 6

Light of wavelength 600 nm is shone normally at a grating with 500 lines per millimetre. Calculate the angle of the first-order maximum.

Accepted Answer

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)

Question 7

Light of wavelength 600 nm is incident normally on a grating with 500 lines per millimetre. Calculate the angle of the second-order maximum.

Accepted Answer

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)

Question 8

A grating with 600 lines per millimetre gives a first-order maximum at 21°. Calculate the wavelength of the light.

Accepted Answer

d = 1/(600 × 10³) = 1.67 × 10⁻⁶ m (1) λ = d sinθ = 1.67 × 10⁻⁶ × sin 21° (1) λ = 597 nm (1)

Question 9

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.

Accepted Answer

d = λ/sinθ = (500 × 10⁻⁹)/sin 15° (1) d = 1.93 × 10⁻⁶ m (1) Lines per mm = 1/d per metre ÷ 1000 = 518 lines per mm (1)

Question 10

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.

Accepted Answer

d = 1/(420 × 10³ lines per metre) (1) = 2.38 × 10⁻⁶ m ≈ 2.4 × 10⁻⁶ m (1). sinθ = 3λ/d = (3 × 486 × 10⁻⁹)/(2.38 × 10⁻⁶) = 0.612 (1), so θ = 37.8° (allow 37.4° to 37.8°) (1).

Question 11

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.

Accepted Answer

d = 1/(250 × 10³) = 4.00 × 10⁻⁶ m (1). sinθ = λ/d = (633 × 10⁻⁹)/(4.00 × 10⁻⁶), so θ = 9.1° (1). Distance of each spot from the centre x = 1.5 × tanθ = 0.24 m (1). Separation of the two first-order spots = 2x = 0.48 m (1).

Question 12

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.

Accepted Answer

First-order angle from the centre θ = 40.0°/2 = 20.0° (1). λ = d sinθ = 1.67 × 10⁻⁶ × sin 20.0° (1) = 5.70 × 10⁻⁷ m = 570 nm (1).

Question 13

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.

Accepted Answer

The maximum order is where sinθ = 1, so n_max = d/λ (1)
= (2.0 × 10⁻⁶)/(600 × 10⁻⁹) = 3.33 (1)
The highest order is therefore n = 3 (1)
At that order sinθ = 3 × 600 × 10⁻⁹/(2.0 × 10⁻⁶), giving θ = 64.2° (1)

Question 14

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).

Accepted Answer

First order: sinθ = λ/d (1) For 400 nm, θ = 11.5° (1) For 700 nm, θ = 20.5° (1) Angular width = 9.0°, with red diffracted more than violet (1)

Question 15

Give two practical uses of diffraction gratings.

Accepted Answer

1. Measuring the wavelength of light or analysing spectra in a spectrometer (1). 2. Identifying elements from their line spectra, for example in the light from stars (1).

Question 16

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.

Accepted Answer

d = 1/(400 × 10³) = 2.5 × 10⁻⁶ m (1). Red end of the second order: sinθ = (2 × 700 × 10⁻⁹)/d = 0.56, so θ = 34.1° (1). Violet end of the third order: sinθ = (3 × 400 × 10⁻⁹)/d = 0.48, so θ = 28.7° (1). The third order begins at a smaller angle than the angle at which the second order ends, so the two spectra overlap (1).

Question 17

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.

Accepted Answer

Highest order = whole-number part of d/λ (1). Grating X: d = 3.33 × 10⁻⁶ m, d/λ = 5.66, so 5 orders are visible each side (1). Grating Y: d = 1.25 × 10⁻⁶ m, d/λ = 2.12, so only 2 orders are visible (1). Only X gives at least three orders, so the teacher should use grating X (1).

Question 18

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.

Accepted Answer

The central (zero-order) maximum is white, because at n = 0 the path difference is zero for every wavelength, so all colours arrive in phase there (1). Each non-zero order is drawn out into a continuous spectrum with violet nearest the centre and red furthest out (1), because sinθ = nλ/d, so the angle of a maximum increases with wavelength (1). d = 1/(500 × 10³) = 2.0 × 10⁻⁶ m, and the most demanding maximum is red in the second order: sinθ = (2 × 700 × 10⁻⁹)/(2.0 × 10⁻⁶) = 0.70, which is below 1, so the second-order spectrum runs the whole way to the red end (1). The first order spans 11.5° to 20.5° and the second 23.6° to 44.4°, so with this grating the two do not overlap (1). The second-order spectrum is the wider of the two and lies further from the centre, since doubling n doubles sinθ for every wavelength (1).