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Single-Slit Diffraction

Single-slit diffraction is the spreading of light after it passes through a narrow slit, producing a broad central maximum and weaker secondary maxima. Minima occur at a sin θ = n λ for n = 1, 2, 3, ...

In one line: Single slit uses a; double slit fringe width uses d.

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Understand it

When the slit width is comparable to the wavelength of light, different parts of the same wavefront act as secondary sources and interfere at the screen. The central maximum is brightest and widest because contributions around the straight-through direction mostly reinforce. On either side, minima occur where light from different parts of the slit cancels in pairs.

Example

A narrow beam passing through a very thin slit forms a central bright band with faint bands on both sides instead of only a sharp geometrical shadow.

Formula and rule

Minima: a sin θ = n λ, where a is slit width in metre, θ is angular position, λ is wavelength in metre and n = 1, 2, 3, ....

For small angles, yn = n λ D/a.

Central maximum width on screen = 2 λ D/a.

Use these for Fraunhofer diffraction with a narrow single slit and distant screen or lens arrangement.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

A slit of width a = 0.20 mm is illuminated by light of wavelength 500 nm.

  1. 1The screen is D = 2.0 m away.
  2. 2Central maximum width = 2 λ D/a = 2 × 5.0 × 10−7 × 2.0 / 2.0 × 10−4 = 1.0 × 10−2 m = 1.0 cm.
  3. 3The central bright band is about 1.0 cm wide.

Method you can reuse

Identify slit width a, wavelength λ and screen distance D, convert to metre, use a sin θ = n λ for angular minima or yn = n λ D/a for screen position, then use twice the first-minimum distance for central maximum width.

Sketch it in your answer

Draw a single slit of width a, central axis, screen at distance D, central maximum, first minima on both sides and secondary maxima. Mark angular positions θ and linear distance y from the centre.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Why is the central maximum in single-slit diffraction wider than the other maxima?

Think first, then show answer
The central maximum extends between the first minima on both sides, so its width is twice the distance from the centre to the first minimum. Secondary maxima lie between later minima and are narrower and much less intense.

Common wrong answer

A common wrong answer says diffraction disappears when slit width is reduced. Actually, reducing slit width increases spreading, provided light can still pass through the slit.

Don't mix this up

Students often write the central maximum width as λ D/a. The angular width from the centre to the first minimum is about λ/a, but the full width of the central maximum is 2 λ D/a on a distant screen.

Exam tip

Mention that diffraction is more prominent when slit width is small and comparable to wavelength. Compare it with interference only when asked; do not mix slit separation d with slit width a.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Questions usually ask for the condition of minima, width of central maximum, effect of changing slit width or wavelength, and comparison with YDSE fringes.
  1. 1

    Enough for 1 mark

    Single-slit diffraction is the spreading of light after it passes through a narrow slit, producing a broad central maximum and weaker secondary maxima. Minima occur at a sin θ = n λ for n = 1, 2, 3, ...

  2. 2

    Add for 2 marks

    Example: A narrow beam passing through a very thin slit forms a central bright band with faint bands on both sides instead of only a sharp geometrical shadow.

  3. 3

    Add for 3 marks

    When the slit width is comparable to the wavelength of light, different parts of the same wavefront act as secondary sources and interfere at the screen.

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