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Young's Double-Slit Experiment and Fringe Width

Young's double-slit experiment demonstrates interference of light using two coherent sources, producing alternate bright and dark fringes on a screen. The fringe width is beta = lambda D/d.

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Student-friendly explanation

Two narrow slits act as coherent sources when they are derived from the same source. At a point on the screen, waves from the two slits superpose. If the path difference is n lambda, a bright fringe is formed. If the path difference is (n + 1/2) lambda, a dark fringe is formed. Fringe width is the distance between two consecutive bright fringes or two consecutive dark fringes.

How to write this in exams

  1. 1

    Start with the exact idea

    Young's double-slit experiment demonstrates interference of light using two coherent sources, producing alternate bright and dark fringes on a screen. The fringe width is beta = lambda D/d.

  2. 2

    Then show how to use it

    Identify lambda, D and d, convert all to SI units, use beta = lambda D/d, calculate beta, then interpret whether the change makes fringes closer or farther apart. For fringe position questions, apply path difference conditions for bright and dark fringes.

  3. 3

    Add one concrete example

    If red light is replaced by blue light in the same setup, the wavelength decreases, so the fringe width decreases and fringes come closer.

  4. 4

    Avoid this incomplete answer

    A common wrong answer is that the central fringe is dark because the two paths are different. At the central point, path difference is zero, so constructive interference gives a bright fringe.

Definition

Young's double-slit experiment demonstrates interference of light using two coherent sources, producing alternate bright and dark fringes on a screen. The fringe width is beta = lambda D/d.

Example

If red light is replaced by blue light in the same setup, the wavelength decreases, so the fringe width decreases and fringes come closer.

Rule to remember

Fringe width: beta = lambda D/d. beta, lambda, D and d are in metre in SI units. Bright fringe: path difference = n lambda. Dark fringe: path difference = (n + 1/2) lambda. The formula is valid when D is much larger than d and the slits are coherent and narrow.

Memory hook

Wider screen distance widens fringes; wider slit separation squeezes fringes.

Examples and method

Worked example

In YDSE, lambda = 600 nm = 6.0 x 10^-7 m, D = 1.5 m and d = 0.30 mm = 3.0 x 10^-4 m. Fringe width beta = lambda D/d = (6.0 x 10^-7 m x 1.5 m)/(3.0 x 10^-4 m) = 3.0 x 10^-3 m = 3.0 mm. The second bright fringe from the centre is at y2 = 2 beta = 6.0 mm, while the first dark fringe is at beta/2 = 1.5 mm from the central bright fringe.

Method to apply

Identify lambda, D and d, convert all to SI units, use beta = lambda D/d, calculate beta, then interpret whether the change makes fringes closer or farther apart. For fringe position questions, apply path difference conditions for bright and dark fringes.

Diagram support

Draw source, two slits S1 and S2, screen at distance D, a point P at distance y from the central line, slit separation d, path difference and central bright fringe O.

How CBSE asks it

It is asked through fringe-width numericals, identifying bright or dark positions, explaining coherent sources, and predicting changes when lambda, D or d is changed.

Avoid common mistakes

Common confusion

Students often use slit width instead of slit separation in beta = lambda D/d. In this formula, d is the distance between the two slits, not the width of each slit.

Common wrong answer

A common wrong answer is that the central fringe is dark because the two paths are different. At the central point, path difference is zero, so constructive interference gives a bright fringe.

Exam tip

Always convert nanometres to metres and centimetres or millimetres to metres before substitution. State whether the point is bright or dark by checking path difference.

Quick check

What happens to the fringe width in YDSE if the screen is moved farther away from the slits?

The fringe width increases because beta = lambda D/d. When D increases while wavelength and slit separation remain constant, the spacing between consecutive bright or dark fringes becomes larger.

Answer writing and exam use

1-mark answer

Young's double-slit experiment demonstrates interference of light using two coherent sources, producing alternate bright and dark fringes on a screen. The fringe width is beta = lambda D/d.

2-mark answer

Young's double-slit experiment demonstrates interference of light using two coherent sources, producing alternate bright and dark fringes on a screen. The fringe width is beta = lambda D/d. Fringe width: beta = lambda D/d. beta, lambda, D and d are in metre in SI units. Bright fringe: path difference = n lambda. Dark fringe: path difference = (n + 1/2) lambda. The formula is valid when D is much larger than d and the slits are coherent and narrow. If red light is replaced by blue light in the same setup, the wavelength decreases, so the fringe width decreases and fringes come closer.

3-mark answer

Two narrow slits act as coherent sources when they are derived from the same source. At a point on the screen, waves from the two slits superpose. If the path difference is n lambda, a bright fringe is formed. If the path difference is (n + 1/2) lambda, a dark fringe is formed. Fringe width is the distance between two consecutive bright fringes or two consecutive dark fringes. Fringe width: beta = lambda D/d. beta, lambda, D and d are in metre in SI units. Bright fringe: path difference = n lambda. Dark fringe: path difference = (n + 1/2) lambda. The formula is valid when D is much larger than d and the slits are coherent and narrow. In YDSE, lambda = 600 nm = 6.0 x 10^-7 m, D = 1.5 m and d = 0.30 mm = 3.0 x 10^-4 m. Fringe width beta = lambda D/d = (6.0 x 10^-7 m x 1.5 m)/(3.0 x 10^-4 m) = 3.0 x 10^-3 m = 3.0 mm. The second bright fringe from the centre is at y2 = 2 beta = 6.0 mm, while the first dark fringe is at beta/2 = 1.5 mm from the central bright fringe. It is asked through fringe-width numericals, identifying bright or dark positions, explaining coherent sources, and predicting changes when lambda, D or d is changed. A common wrong answer is that the central fringe is dark because the two paths are different. At the central point, path difference is zero, so constructive interference gives a bright fringe.
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