Cambridge Syllabus Notes

The Diffraction Grating

What is a Diffraction Grating? 🔬

A diffraction grating is a surface with many equally spaced slits or grooves. When light hits the grating, each slit acts like a tiny source of waves that interfere with one another, producing a pattern of bright and dark spots. Think of it like a row of tiny doors in a fence – light passes through each door and creates a beautiful pattern on the other side.

The Key Equation 📐

The positions of the bright spots (called maxima) are given by the simple formula:

$$d \sin \theta = n \lambda$$

d – distance between adjacent slits (the grating spacing)
θ – angle at which a bright spot appears, measured from the normal (straight‑on) direction
n – order of the maximum (1, 2, 3, …)
λ – wavelength of the light (in metres)

How to Use the Formula 🧪

Measure or look up the grating spacing d (often given in lines per millimetre).
Choose the order n you want to examine (usually the first order, n = 1).
Measure the angle θ of the bright spot using a protractor or a screen.
Rearrange the equation to solve for the wavelength: $$\lambda = \frac{d \sin \theta}{n}$$
Plug in the numbers and calculate λ.

Example Problem 🚀

Suppose a grating has 600 lines per millimetre (so d = 1/600 mm = 1.67 µm) and the first‑order maximum (n = 1) appears at an angle of 30°. What is the wavelength of the light?

Parameter	Value
d (spacing)	1.67 µm
θ (angle)	30°
n (order)	1
λ (wavelength)	$$\lambda = \frac{1.67\,\mu\text{m} \times \sin 30^\circ}{1} = 0.835\,\mu\text{m}$$

Exam Tip Box 📌

Remember:

Always check the units – d should be in metres if you want λ in metres.
Use the sine of the angle; if you only have the tangent, convert using sin θ = tan θ / √(1 + tan²θ).
For higher orders (n > 1), the angle increases – be careful not to exceed the physical limits of the grating.
When the problem asks for λ, rearrange the formula to isolate it.
Check your answer against typical visible wavelengths (400–700 nm) to see if it’s reasonable.

recall and use d sin θ = nλ