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Young’s Double-Slit Experiment (SL IB Physics)

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Young’s Double-Slit Experiment

  • Interference patterns depend on:
    • The coherent light incident on the slits
    • The distance between the slits and the screen where the interference pattern is observed
    • The separation between the slits
  • In particular, it can be seen that:
    • The longer the wavelength λ, the further apart the bright fringes on the screen become
    • The further the screen is placed from the slits, the further apart the bright fringes on the screen become
    • The greater the separation between the slits, the closer the bright fringes on the screen become

4-4-7-interference-patterns-depend-on-wavelength-of-incident-light_sl-physics-rn

Differences between the interference pattern of red and blue light. Red light has a longer wavelength than blue light, so the fringes are more spaced out

  • Two slits are separated by a length d, and a screen is placed a distance D away

4-4-7-double-slit-equation_sl-physics-rn

Light rays of wavelength λ incident on two slits a distance S apart interfere and form bright fringes on a screen placed a distance D away from the slits

  • When a light ray of wavelength λ is incident on the two slits, the observations are:

1. A central bright fringe forms

    • This is where the two waves have travelled the same distance to the screen and their path difference is zero 

2. Bright fringes are formed on either side of the central fringe

    • This is where the path difference between the waves is equal to exactly one wavelength 
    • This is constructive interference 

3. Dark fringes are formed between the bright fringes

    • This is where the path difference between the waves is some number of wavelengths plus half a wavelength 
    • This is destructive interference 
  • Further bright fringes will be located at each position on the screen where the path difference is exactly equal to
    • For example: 2λ, 3λ, 4λ...
    • This is constructive interference
  • Dark fringes are located in between the bright ones, where the path difference is exactly equal to open parentheses n space plus space 1 half close parentheses space lambda
    • For example, lambda over 2 comma space space fraction numerator 3 lambda over denominator 2 end fraction comma space space fraction numerator 5 lambda over denominator 2 end fraction...
    • This is destructive interference

Double-Slit Equation

  • The separation s of successive fringes is given by:

s space equals space fraction numerator lambda D over denominator d end fraction

  • Where:
    • s = separation between successive fringes on the screen (m)
    • λ = wavelength of the waves incident on the slits (m)
    • D = distance between the screen and the slits (m)
    • d = separation between the slits (m)

3-3-7-double-slit

  • Note that s is the separation between two successive bright fringes or two successive dark ones
  • The above equation shows that the separation between the fringes will increase if:
    • The wavelength of the incident light increases
    • The distance between the screen and the slits increases
    • The separation between the slits decreases

4-4-7-interference-patterns-depend-on-slit-separation_sl-physics-rn

Dependence of the interference pattern on the separation between the slits. The further apart the slits, the closer together the bright fringes

Worked example

In a double-slit experiment, two slits are placed a distance of 0.40 mm apart and a screen is located 0.50 m away from the slits.

Coherent electromagnetic waves incident on the slits produce an interference pattern on the screen. The separation between dark fringes is 0.50 cm.

Determine the wavelength and state the type of electromagnetic waves used in the experiment.

Answer:

Step 1: List the known quantities 

  • d = 0.40 mm = 4.0 x 10–4 m
  • D = 0.50 m
  • s = 0.50 cm = 5.0 x 10–3 m

  • Note that you must convert all lengths into metres (m)

Step 2: Write down the double-slit equation

Step 3: Rearrange the above equation to calculate the wavelength λ

Step 4: Substitute the numbers into the above equation 

λ = 4.0 × 10–6 m = 4.0 μm

  • This corresponds to the infrared area of the electromagnetic spectrum

Exam Tip

Remember that the separation between dark fringes is exactly the same as the separation between bright fringes. Whether a question gives you or asks about the separation between dark fringes instead of bright ones it makes no difference.

Some tricky questions might give you the separation between a bright and a dark fringe. This is equal to half the value of s!

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Ashika

Author: Ashika

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.