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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 theta = n lambda for n = 1, 2, 3, ...

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Student-friendly explanation

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.

How to write this in exams

  1. 1

    Start with the exact idea

    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 theta = n lambda for n = 1, 2, 3, ...

  2. 2

    Then show how to use it

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

  3. 3

    Add one concrete 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.

  4. 4

    Avoid this incomplete 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.

Definition

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 theta = n lambda for n = 1, 2, 3, ...

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.

Rule to remember

Minima: a sin theta = n lambda, where a is slit width in metre, theta is angular position, lambda is wavelength in metre and n = 1, 2, 3, .... For small angles, y_n = n lambda D/a. Central maximum width on screen = 2 lambda D/a. Use these for Fraunhofer diffraction with a narrow single slit and distant screen or lens arrangement.

Memory hook

Single slit uses a; double slit fringe width uses d.

Examples and method

Worked example

A slit of width a = 0.20 mm is illuminated by light of wavelength 500 nm. The screen is D = 2.0 m away. Central maximum width = 2 lambda D/a = 2 x 5.0 x 10^-7 x 2.0 / 2.0 x 10^-4 = 1.0 x 10^-2 m = 1.0 cm. The central bright band is about 1.0 cm wide.

Method to apply

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

Diagram support

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 theta and linear distance y from the centre.

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.

Avoid common mistakes

Common confusion

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

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.

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.

Quick check

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

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.

Answer writing and exam use

1-mark answer

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 theta = n lambda for n = 1, 2, 3, ...

2-mark answer

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 theta = n lambda for n = 1, 2, 3, ... Minima: a sin theta = n lambda, where a is slit width in metre, theta is angular position, lambda is wavelength in metre and n = 1, 2, 3, .... For small angles, y_n = n lambda D/a. Central maximum width on screen = 2 lambda D/a. Use these for Fraunhofer diffraction with a narrow single slit and distant screen or lens arrangement. 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-mark answer

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. Minima: a sin theta = n lambda, where a is slit width in metre, theta is angular position, lambda is wavelength in metre and n = 1, 2, 3, .... For small angles, y_n = n lambda D/a. Central maximum width on screen = 2 lambda D/a. Use these for Fraunhofer diffraction with a narrow single slit and distant screen or lens arrangement. A slit of width a = 0.20 mm is illuminated by light of wavelength 500 nm. The screen is D = 2.0 m away. Central maximum width = 2 lambda D/a = 2 x 5.0 x 10^-7 x 2.0 / 2.0 x 10^-4 = 1.0 x 10^-2 m = 1.0 cm. The central bright band is about 1.0 cm wide. Questions usually ask for the condition of minima, width of central maximum, effect of changing slit width or wavelength, and comparison with YDSE fringes. 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.
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