10 min readConcept sheet1-mark MCQ and assertion-reason, 3-mark explanation of the pattern with the intensity graph, sometimes a 5-mark question with interference; also a NEET favourite

Diffraction of light at a single slit

Light bends round the edges of a narrow slit. The pattern has a bright central maximum, with dark minima where a sin θ = nλ.

DiffractionBending of light round the edges of an obstacle or a narrow opening into the geometrical shadow

Minimaa sin θ ≈ aθ = nλ, n = ±1, ±2, ... (a = slit width)

Secondary maximaθ ≈ (n + ½)λ/a, getting much fainter as n increases

When it is noticeableWhen the slit width a is comparable to the wavelength λ

What diffraction is

Look closely at the shadow of a sharp edge in light from a small source. The edge of the shadow is not perfectly sharp; faint bright and dark bands appear. Diffraction is the bending of waves round the edges of an obstacle or a narrow opening, so that light enters the region of geometrical shadow. All waves diffract. Sound bends round a door easily, because its wavelength is close to a metre. Light has a tiny wavelength (about 0.5 μm), so its diffraction is seen only with very narrow slits or very small obstacles. Diffraction is the superposition of secondary wavelets from different parts of the same wavefront.

Explaining the single slit pattern

A plane wavefront of monochromatic light falls normally on a slit of width a. The pattern is seen on a distant screen.

1
Centre (θ = 0): every point of the slit is the same distance from the centre of the screen. All wavelets arrive in phase, so the centre is the brightest: the central maximum.
2
At an angle θ, the path difference between wavelets from the two edges of the slit is a sin θ ≈ aθ.
3★
First minimum (aθ = λ): divide the slit into two equal halves. Each point in the upper half has a partner in the lower half with path difference λ/2. The pairs cancel, so the point is dark: θ = λ/a.
4
Other minima: for aθ = 2λ, 3λ, ..., divide the slit into 4, 6, ... parts that cancel in pairs. So minima lie at θ = nλ/a.
5
Secondary maximum (aθ = 3λ/2): divide the slit into three equal parts. Two parts cancel each other, and only one third is left. So the point is bright, but much fainter than the centre.
6
In general, weak maxima lie at θ ≈ (n + ½)λ/a, and each is fainter than the one before.

★ key step

How to draw the intensity graph for a single slit

Draw a horizontal axis for θ (or distance on the screen) and a vertical axis for intensity I.
Draw a tall peak at θ = 0; label it central maximum.
Mark zero intensity on both sides at θ = ±λ/a, ±2λ/a, ±3λ/a.
Between the zeros, draw small humps; label them secondary maxima. Make each smaller than the one before.
Show the central peak twice as wide as each secondary maximum.

Worked example: slit width from the first minimum

Problem

Light of wavelength 500 nm falls normally on a slit. The first minimum of the diffraction pattern is at 30° from the centre. Find the width of the slit.

Working

For the first minimum: a sin θ = λ (n = 1).
a = λ/sin θ = 500 nm ÷ sin 30°.
a = 500 nm ÷ 0.5 = 1000 nm = 1 × 10⁻⁶ m.

Answer

a = 1 μm (1 × 10⁻⁶ m)

Worked example: angles of a minimum and a secondary maximum

Problem

A slit of width 0.3 mm is lit by light of wavelength 600 nm. Find the angular positions of the first minimum and the first secondary maximum.

Working

Given: a = 3 × 10⁻⁴ m, λ = 6 × 10⁻⁷ m.
First minimum: θ₁ = λ/a = (6 × 10⁻⁷ m) ÷ (3 × 10⁻⁴ m) = 2 × 10⁻³ rad.
First secondary maximum: θ ≈ (1 + ½)λ/a = 1.5 × 2 × 10⁻³ rad = 3 × 10⁻³ rad.

Answer

First minimum at 2 × 10⁻³ rad; first secondary maximum at about 3 × 10⁻³ rad

In real life

You can hear a friend talking round a corner but cannot see them. Sound (wavelength near a metre) diffracts easily, while light (about half a micrometre) bends only round very small edges. Look at a tube light through a very narrow gap between two fingers and you will see faint dark lines.

Exam angle

2 marks: “What is diffraction of light? Why is it not easily noticed in daily life?” Bending of light round edges into the geometrical shadow. The wavelength of light (about 5 × 10⁻⁷ m) is very small compared with everyday openings and obstacles.

3 marks: “Using Huygens' wavelets, explain why the centre of a single slit pattern is bright and why a minimum forms at θ = λ/a. Sketch the intensity distribution.” At θ = 0 all wavelets are in phase, so bright. At aθ = λ, split the slit into two halves; partner points differ by λ/2 and cancel, so dark. Graph: tall central peak, zeros at ±λ/a, ±2λ/a, fainter side humps.

1 marks: “How does the angle of the first minimum change if the slit is made narrower?” θ = λ/a increases, so the minimum moves outward and the pattern spreads.

Remember it

Single slit: halves cancel at λ; thirds leave a little light at 3λ/2. Minima at θ = nλ/a.