Diffraction-grating maxima and wavelength
Key idea: H2 Physics lessons on progressive-wave models, standing waves, interference, diffraction and resolution.
Continue where you stopped
The core idea
Build the idea
Learn the idea
Big question: Why does a diffraction grating produce sharp, widely separated maxima?
Many equally spaced slits reinforce at angles satisfying d sinθ = nλ, where d is grating spacing. Convert lines per metre to d by taking the reciprocal. The largest possible order follows sinθ ≤ 1; a missing calculated order is often a unit or integer error.
Build the principal-maximum condition
A grating contains many equally spaced slits. For light leaving at angle θ to the normal, adjacent rays have path difference d sinθ. They reinforce as a principal maximum when d sinθ = nλ, where n is an integer order.
If line density is N lines per metre, spacing d = 1/N. Convert lines per millimetre before taking the reciprocal. The central maximum is n = 0; orders occur symmetrically on either side for normal incidence.
Check your understanding: A grating has 500 lines mm⁻¹. What is d?
500 lines mm⁻¹ = 5.00×10⁵ m⁻¹, so d = 2.00×10⁻⁶ m.
Interpret orders and sharpness
Because sinθ cannot exceed 1, the largest possible order is the greatest integer no larger than d/λ. Equality at 90° is a limiting direction and may not be observable in a practical setup.
Many slits make principal maxima narrow because small departures from the correct angle accumulate large phase disagreement across the grating. More slits sharpen maxima; the spacing d determines their angles. Detailed spectrometer construction is not needed here.
Check your understanding: Why should nmax be rounded down, not to the nearest integer?
Only whole orders satisfying nλ/d ≤ 1 are physically possible.
Key ideas to keep
- Lines per millimetre must be converted to lines per metre before finding d.
- Order n is an integer and the central maximum is n = 0.
- Greater line density means smaller d and larger angular separation.
See the reasoning
Worked example
Find wavelength and available orders
Question: A grating of 600 lines mm⁻¹ gives first-order maxima at 22.0°. Find wavelength and the greatest possible order.
Step 1: Convert line density
Why: Spacing must be in metres.
Working: N = 6.00×10⁵ m⁻¹, so d = 1/N = 1.667×10⁻⁶ m.
Step 2: Use the measured order
Why: The maxima satisfy d sinθ = nλ.
Working: λ = d sin22.0° = 6.24×10⁻⁷ m.
Step 3: Apply the sine limit
Why: A real angle requires nλ/d ≤ 1.
Working: d/λ = 2.67, so nmax = 2.
Answer: The wavelength is 624 nm and orders 0, ±1 and ±2 are possible.
Check: The calculated wavelength is visible and order 3 would require sinθ > 1.
Another worked model
Question
A 600 lines mm⁻¹ grating receives 500 nm light normally. Find first-order angle and highest possible order, and state how wavelength is determined experimentally.
Check the worked solution
a = 1/(6.00 × 10⁵) = 1.67 × 10⁻⁶ m. First order: sinθ = λ/a = 0.300, so θ = 17.5°. Since nλ ≤ a, nmax = 3. Measure a principal-maximum angle for known a and n, then calculate λ = a sinθ/n; spectrometer structure is not required.
Use a hint if needed
Practise with support
Try this
A first-order maximum for 450 nm light is at 15.0°. Find the grating spacing and line density in lines mm⁻¹.
Hint: Find a from a sinθ = λ, then use N = 1/a and convert metres to millimetres.
Check your answer
a = 450 × 10⁻⁹/sin15.0° = 1.74 × 10⁻⁶ m. N = 5.75 × 10⁵ m⁻¹ = 575 lines mm⁻¹.
Now work without the hint
Practise independently
Your turn
A grating spacing is 2.50 μm. For 625 nm light, find the first two principal-maximum angles and state why no fifth order exists.
Check your answer
For n = 1, sinθ = 0.250, so θ = 14.5°. For n = 2, sinθ = 0.500, so θ = 30.0°. A fifth order would require sinθ = 5λ/a = 1.25, which is impossible.
Avoid these traps
Common mistakes
Common mistake
Lines per metre can be inserted directly as a in a sinθ = nλ.
What is wrong with this reasoning?
Show better thinking
Convert line density N to adjacent-line spacing a = 1/N before applying the grating equation.
Write for the examiner
Exam guidance
Check that sinθ is no greater than one and quote the physically possible integer orders.
Exam-style practice [7 marks]
Light containing wavelengths 450 nm and 650 nm is incident normally on a grating. Explain how their first-order maxima compare and how increasing the number of illuminated slits affects what is observed.
Plan before you answer
- Apply d sinθ = nλ at fixed d and n.
- Compare angles monotonically.
- Separate maximum angle from maximum width.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
For first order, sinθ = λ/d, so 650 nm appears at a larger angle than 450 nm. Both have n = 0 at the central maximum. Illuminating more slits makes the principal maxima narrower because phase mismatch grows across more sources away from the exact condition; it does not change the angles given by d sinθ = nλ. Narrower peaks make close wavelengths easier to distinguish.
Come back in three days
Check what stayed with you
Recall question 1
How is grating spacing related to line density?
Check the answer
d = 1/N when N is in lines per metre.
Recall question 2
State the grating equation.
Check the answer
d sinθ = nλ.
Recall question 3
What sets the maximum possible order?
Check the answer
The condition nλ/d ≤ 1.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 10 states no explicit exclusions. Topic 11(j) does not require knowledge of spectrometer structure or use. Graph interpretations distinguish time traces at one position from spatial profiles at one instant, and all small-angle equations are used only with their stated geometry.
- GCE A-Level H2 PhysicsTopic 11(i) / Topic 11(j) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027