9478 / 2027
Superposition overview

Lesson 7 of 9

Diffraction gratings and wavelength

  • I can use a sin(theta) = n lambda for the principal maxima of a diffraction grating.Syllabus

    Syllabus 9478, 11(i). Use a sin(theta) = n lambda for the principal maxima of a diffraction grating.

  • I can describe a diffraction-grating measurement of light wavelength.Syllabus

    Syllabus 9478, 11(j). Describe a diffraction-grating measurement of light wavelength.

A grating's regularly spaced slits reinforce light in particular directions. Use the spacing between slits, the order and the actual angle from the normal.

A transmission grating has many parallel slits. With monochromatic light normally incident, adjacent slits start in phase. Towards an outgoing angle θ, their path difference is a sin θ, where a is the slit spacing.

Locate principal maxima

When adjacent contributions differ by an integer number of wavelengths, they reinforce to form a principal maximum:

a sin θ = nλ

The central direction has n = 0. Matching positive and negative orders lie on opposite sides of the normal. Use the angle from that normal, not from the grating surface. The spacing a is a length; it is not the quoted number of lines per millimetre.

Estimate the optical scales

A few hundred lines per millimetre means a slit spacing of order micrometres. Visible wavelength is of order half a micrometre, so a/λ is only a few. Expect a few possible orders and first-order angles of order a few tenths of a radian. The exact sine bound, not this estimate, decides the highest integer order.

Worked orders and angles

Convert line density before using the equation

A grating has 600 lines/mm and is illuminated normally by light of wavelength 500 nm.

Line density = 600 000 m-1
a = 1/600 000 m
≈ 1.67 × 10-6 m
sin θ = nλ/a = 0.300n
Principal directions for the supplied grating model; negative orders have opposite angles
Order nsin θθ / °
000
10.30017.46
20.60036.87
30.90064.16

Keep the unrounded spacing in the calculation. Since |sin θ| cannot exceed 1:

|n| ≤ a/λ = 3.33...
Highest integer order = 3

The allowed directions are 0, ±1, ±2 and ±3: seven in total. Order 4 would require sin θ = 1.2 and has no real direction. These are geometrically allowed maxima, not a promise that every spot has equal brightness or can be detected.

The third-order angle is about 64°. Using sin θ ≈ θ there would be inappropriate; use the inverse sine.

Try it step by step

Work it out in steps: Slit spacing d = 1 / (lines per metre) in micrometres, then sin theta = n lambda / d for the second order, then Angle theta in degrees.

  1. Problem 1. A grating has 300 lines per mm. Light of wavelength 600 nm falls on it normally. Find the angle of the second-order maximum.
  2. Problem 2. A grating has 500 lines per mm. Light of wavelength 500 nm falls on it normally. Find the angle of the second-order maximum.
  3. Problem 3. A grating has 400 lines per mm. Light of wavelength 650 nm falls on it normally. Find the angle of the second-order maximum.
Show the answer

Problem 1.

Slit spacing d = 1 / (lines per metre) in micrometres: d = 1 / 300 mm = 3.33 x 10-6 m = 3.33 micrometres

sin theta = n lambda / d for the second order: sin theta = 2 x 600 x 10-9 / (3.33 x 10-6) = 0.360

Angle theta in degrees: theta = inverse sin of 0.360 = 21.1 degrees (3 sf)

Problem 2.

Slit spacing d = 1 / (lines per metre) in micrometres: d = 1 / 500 mm = 2.00 x 10-6 m = 2.00 micrometres

sin theta = n lambda / d for the second order: sin theta = 2 x 500 x 10-9 / (2.00 x 10-6) = 0.500

Angle theta in degrees: theta = inverse sin of 0.500 = 30.0 degrees (3 sf)

Problem 3.

Slit spacing d = 1 / (lines per metre) in micrometres: d = 1 / 400 mm = 2.50 x 10-6 m = 2.50 micrometres

sin theta = n lambda / d for the second order: sin theta = 2 x 650 x 10-9 / (2.50 x 10-6) = 0.520

Angle theta in degrees: theta = inverse sin of 0.520 = 31.3 degrees (3 sf)

Use path difference for maxima and exact angles for large orders

Adjacent-slit path difference

A perpendicular construction gives path difference a sine thetaTwo adjacent slits are separated vertically by 110 drawing units, representing spacing a. Parallel outgoing rays make the actual first-order angle, about 17.4576 degrees, to the horizontal normal. A perpendicular from the upper slit meets the lower outgoing ray at Q. The lower-slit-to-Q segment is 33 drawing units, exactly 0.3a, so its length is a sine theta. The right-angle mark is at Q. Horizontal and vertical geometry scales are equal. Incident wavefronts reach both slits in phase under normal incidence. This is an adjacent-path construction for a many-slit grating, not a two-slit replacement for the grating.aθNormalQLower slit to Q:a sin θ = 0.3aNormal incidence; θ = 17.46°

Seven geometrically allowed directions

Orders zero and plus or minus one, two and three are allowedFor 600 lines per millimetre and wavelength 500 nanometres, the outgoing directions satisfy sine theta equals 0.3 times order. The exact first three angles are approximately 17.4576, 36.8699 and 64.1581 degrees on both sides of the normal. The rays use equal horizontal and vertical geometry scales and equal drawing lengths solely to show directions, not equal intensities. The central ray is order zero. Fourth order would require a sine greater than one, so it has no real outgoing direction.-364.16°-236.87°-117.46°0 (normal)+117.46°+236.87°+364.16°Equal x/y scale; directions onlyGrating

Measure a first-order screen displacement

The screen geometry gives a tangent ratio before the grating sine calculationThe grating is one metre from a screen perpendicular to the central normal. The first-order centres lie 0.3145 metres above and below the zero-order centre. Horizontal and vertical distances share a scale of 220 drawing units per metre, so the drawn ray angle is arctangent 0.3145, about 17.4584 degrees. This rounded measurement gives approximately 500 nanometres when its angle is inserted into the grating sine equation. The screen displacement divided by distance is a tangent, not an exact sine.D = 1.000 m+0.3145 m-0.3145 m0ScreenGratingNormal incidence; equal distance scales

Use θ = arctan(0.3145/1.000), then λ = a sin θ for first order. Measurements on both sides help locate the centre and reveal alignment problems.

The first view constructs the adjacent-slit path difference. The separate direction fan uses the calculated angles, and the screen view shows a first-order wavelength measurement. The different views have their own stated geometry and scales.

Check your understanding

Normally incident 500 nm light falls on a grating with 600 lines/mm. What is the highest geometrically allowed principal order and the total number of allowed order directions?

  1. Highest order 4; nine directions in total.
  2. Highest order 3; three directions in total.
  3. Highest order 3; seven directions in total.
Show the answer

Highest order 3; seven directions in total.

The sine bound gives n no greater than 3.33, hence integer orders through 3. Counting zero and both sides gives 1 + 2 x 3 = 7 directions. This does not imply equal brightness.

Determine wavelength from a screen measurement

Use a monochromatic source and a grating of known line density. Direct the beam normally onto the grating. Establish the zero-order direction, then measure a chosen order's angle or obtain it from a calibrated perpendicular screen.

For a screen distance D = 1.000 m and a first-order displacement y = 0.3145 m from the zero-order centre:

θ = arctan(y/D)
= arctan(0.3145/1.000) ≈ 17.46°
λ = a sin θ/n
≈ 5.00 × 10-7 m = 500 nm

Here y/D is the tangent, not the exact sine. First find the angle, then use its sine in the grating equation.

Record the order, line density, screen geometry and spot-centre positions with units. Measure matching spots on both sides to help establish the centre and check alignment. Repeated centre readings assess variation, but averaging cannot automatically correct a tilted grating. In a school laser experiment, use the approved equipment and beam-control procedure.