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16 Young’s slits experiment

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Introduction Light shows interference.To produce two rays, light is shone through a pair of parallel slits.Where the light falls on a screen beyond the slits, light and dark interference ‘fringes’ are seen.

Single slit acts as a narrow source of light, shining on the double slit.Alternatively a laser can be shone directly on the double slit.

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As light passes through each slit, it spreads out into the space beyond. This is diffraction.

The fringe separation can be measured using a travelling microscope. Increasing the slit – screen distance makes the fringes wider but dimmer.

Explaining the interference fringes

S1

S2

C

B

A

A is a point on the screen directly opposite the mid-point between the two slits S1 and S2.If the waves leave the two slits in phase with one another, they will arrive at point A in phase.

They will interfere constructively and a bright fringe will be seen. Path difference = 0.

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B is the centre of the first dark fringe. Waves from S1 have a shorter distance to travel than light from S2.The two waves arrive out of phase and interfere destructively.Path difference = S1B – S2B = λ / 2

C is the centre of the next bright fringe. The two waves arrive in phase, but one has travelled further than the other.Path difference = λTherefore:A bright fringe is seen where two waves arrive in phase; path difference = nλ

A dark fringe is seen where they arrive out of phase; path difference = (n + ½)λ

In-between positions have in-between path differences which give rise to intermediate brightnesses.

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Measuring the wavelength of light

The Young’s slits experiment provides a method for determining λ, which is related to the screen distance D, slit separation a and fringe width x by:

λ = a.x / DIt may be easier to remember the formula as:

λ.D = a.xThe largest quantity (D) multiplied by the smallest (λ) equals the other two multiplied together.Note that for white light, this can only give an average value of λ since many wavelengths are present.Laser light is monochromatic (a single wavelength) so the fringes are cleaner and a more accurate value of λ can be found.

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Worked exampleLaser light of wavelength 648 nm falls on a pair of slits separated by 1.5 mm. what will be the separation of the interference fringes seen on a screen 4.5 m away from the slits?Solution:Step 1 write down what you know and what you want to know:

λ = 648 nm, a = 1.5 mm, D = 4.5 m, x = ?Step 2 rearrange the equation, substitute values and solve:

x =

λ.D a

=648 x 10-9 m x 4.5 m 1.5 x 10-3 m

= 1.9 x 10-3 m

So the fringe width seen on the screen will be 1.9 mm.

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Questions 1. Give the symbol and approximate size for each of

the following in the Young’s slits experiment: slit-screen distance; slit separation; fringe separation; wavelength of light.

2. What can you say about the path difference between two waves which show destructive interference?

3. If the slit separation a is doubled, how will the fringe width x be changed?

4. White light is directed onto a pair of slits separated by 1.0 mm. interference fringes are observed on a screen at a distance of 1.8 m. five fringes have a width of 5.0 mm. estimate the wavelength of the light. Why is your answer an estimate?