DSE Physics | Double-Slit Interference and Grating Diffraction: a tutor guide with fringe spacing and angles

2026-11-05

Students can recite the formulas but cannot say which one belongs to a double slit and which to a grating, and they mislabel the path difference. This lesson separates the two questions on one diagram: a double slit asks for spacing, a grating asks for an angle, and both start from the same path difference.

Objectives — Students can: (1) state the two conditions for coherent sources and decide constructive or destructive interference from a path difference; (2) use fringe spacing equals wavelength times screen distance over slit separation to find the spacing and the position of the nth bright fringe; (3) use d sin theta equals n lambda for a grating and decide the highest visible order.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Wavelength	Convert 589 nm to metres	5.89 x 10-7 m
Coherence	What must two sources share to be coherent?	Same frequency and a fixed phase difference
Trigonometry	If sin theta is 0.589, what is theta?	About 36 degrees

02 Lesson flow (45 minutes)

Time	Stage	Tutor does	Student does
0-6 min	One diagram, two questions	Mark a double slit and a grating on one diagram and ask which measures spacing	Write spacing and angle on the diagram
6-16 min	Path difference and coherence	Use two loudspeakers, draw the path difference, show when it adds	Draw one constructive and one destructive case
16-26 min	Double-slit spacing	Model the spacing formula and insist every unit becomes metres	Repeat the same data themselves
26-36 min	Grating equation and orders	Model d sin theta equals n lambda and test each order against one	Work out order 3 and order 4
36-42 min	White light and single slit	Compare the first-order angle for red and violet, then the central maximum width	Explain why red bends more
42-45 min	Exit check	Collect and mark three questions	Work alone

03 Teaching script (read it out)

Scene 1: One diagram, two questions

Students blend three formulas because they never separated what is being measured. Draw one diagram: on the left a double slit, where the screen measures the distance between fringes; on the right a grating, where the screen measures the angle of a bright fringe from the centre. The same path difference serves both, one in the small-angle approximation and one at larger angles.
Have students write spacing and angle on the diagram, then ask which one the question wants before touching any formula.

Students do: label slit separation, screen distance, spacing and angle with their units.

Scene 2: Double slit: metres first, divide second

Nine out of ten errors are unit errors: nanometres left as they are, millimetres never converted, or the slit separation halved. Fix three steps: write wavelength, screen distance and slit separation in metres, substitute, then convert the answer back to millimetres with its unit.
Model it: wavelength 589 nm, slit separation 0.05 mm, screen 2 m, so the spacing is 0.0236 m or 23.6 mm. The fifth bright fringe sits 117.8 mm from the centre because position is order times spacing.

Students do: write equals blank metres next to all three quantities before substituting.

Scene 3: Grating: test each order until sine exceeds one

Students stop after the first order, or report order four as if it existed. Build a table: order one, two, three, four, with sine theta equals order times wavelength over spacing on each row; the first row whose sine exceeds one does not exist, so the highest order is the row before it.
Model it with 500 lines per millimetre, so spacing is 2.0e-06 m and wavelength is 589 nm: order two gives sine 0.589, angle 36.1 degrees; order three gives sine 0.884, angle 62.1 degrees; order four gives 1.178, above one, so only three orders exist.

Students do: draw a four-row table and write does not exist in the row that fails.

04 Question types and worked steps

Type	Example	Steps	Answer
Fringe spacing	Wavelength 589 nm, slit 0.05 mm, screen 2 m	Convert to metres, divide, convert to millimetres	23.6 mm
Position of the nth fringe	Same data; find the fifth bright fringe	Position is order times spacing	117.8 mm
Grating angle	500 lines per mm; find the second order angle	Spacing is one over 500 mm, converted to metres	36.1 degrees
Highest order	Same grating; find the highest visible order	Test each order until sine exceeds one	Third order; fourth does not exist

Sanity check (teach this): Sanity check: sine theta can never exceed one, so such an answer means the order does not exist rather than a calculator fault; spacing grows with screen distance and shrinks with slit separation, so an answer that shrinks when the screen moves back has the quantities swapped; longer wavelength bends more, so red at 20.5 degrees exceeds violet at 11.5 degrees; the central maximum of a single slit is about twice wavelength times distance over slit width, here 23.6 mm, the same order as the double-slit spacing; and when a question asks why grating fringes are sharp, the answer is the grating, not the double slit.

05 Independent practice (3 questions)

Question	Answer	What to watch
Spacing	Wavelength 589 nm, slit 0.050 mm, screen 2.0 m; find the spacing	Check that all three quantities are in metres
Grating angles	500 lines per mm; find the second and third order angles	Check that spacing starts as one over 500 mm
Highest order	Same grating; find the highest order and justify it	Check that sine is tested against one every row

06 Common mistakes and what to say

Mistake	Student writes	Say this
Nanometres and millimetres unconverted	Spacing = 589 x 2.0 divided by 0.050	“Are all three numbers in the same unit? Write them in metres first.”
Slit separation used as screen distance	Spacing = wavelength times slit over screen	“Slit separation is the gap between slits. How far the screen is has its own letter.”
Grating treated as a double slit	Using the spacing formula for a grating	“Does the question want an angle or a distance? Where is the screen?”
Reporting order four	sine = 1.178, calculator used anyway	“Can a sine exceed one? Does that order exist at all?”

07 Homework (paper, section and question numbers)

Redo this chapter (Paper 1 Q10 style) — Two grating questions with a full order, sine, angle table
Secondary physics index — Other chapters and their paper question types
All prep guides — The same ten-section structure in other subjects

08 Exit check (last 5 minutes)

Question	Answer	Decision
Coherence	Same frequency and fixed phase difference	Colour only means repeat act one
Double-slit work	Convert to metres before dividing	Unconverted units mean repeat act two
Highest order	Test each order until sine exceeds one	Jumping to order four means repeat act three

09 Tutor FAQ

Can this be taught from formulas alone without a demonstration?

No. Use two loudspeakers or a ripple tank once, or students cannot judge constructive against destructive.

Does the double-slit formula work for gratings?

It is close at small angles, but grating angles are large, so use d sin theta equals n lambda.

Must students memorise the speed of light?

Not for this topic, since wavelength is given, but they must convert nanometres with 10 to the minus nine.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.