Young's slits experiment demonstrates the wave-like nature of light. Light passes through two parallel slits and spreads out, producing a diffraction pattern on a screen beyond the slits with bright and dark interference fringes. Bright fringes occur when the path difference between waves from the two slits is an integer multiple of the wavelength, producing constructive interference. Dark fringes occur when the path difference is an odd integer multiple of half the wavelength, producing destructive interference. The wavelength of light can be calculated from the known slit separation, fringe spacing on the screen, and distance between the slits and screen.

1. 16 Young’s slits experiment
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2. 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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3. 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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4. 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 = λ
Therefore:
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. 4

5. 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 / D
It may be easier to remember the formula as:
λ.D = a.x
The 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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6. Worked example
Laser 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:
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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.

7. 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?
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