# Optics by yurtgc548

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```									Sight and wave phenomena

Topic A.4 Diffraction
For this section, we will be
dealing with Diffraction at a single
slit
When a He-Ne beam passes through a single
narrow slit, this was observed on a distant screen
What happened?
 Recall that when waves meet an aperture or
barrier, they diffract.
 The image shows the diffraction of the light
at the slit.
 Take a closer look at the image, what can
you observe?
Did you observe 2 things?
 The centre spot is the brightest, and the
spots get dimmer as they go to the sides
 The centre spot is about twice the length of
Sketch the variation with angle of diffraction
of the relative intensity of light diffracted at a
single slit.
Single Aperture Diffraction Pattern
With a narrower aperture, the overall intensity of the
light is reduced and the pattern on the screen is
wider.
This is how the image and the graph coincide
Diffraction Patterns
 When plane wavefronts pass through a
 This is an example of the phenomenon
called diffraction
 Light waves are no exception to this
 However, when we look at the diffraction pattern
produced by light we observe a fringe pattern,
 that is, on the screen there is a bright central
maximum with "secondary" maxima either side of
it.
 There are also regions where there is no
illumination and these minima separate the
maxima.
 This intensity pattern arises from the fact
that each point on the slit acts, in
accordance with Huygen's principle, as a
source of secondary wavefronts.
 It is the interference between these
secondary wavefronts that produces the
typical diffraction pattern.
Derive the formula θ = λ / b for the position
of the first minimum of the diffraction pattern
produced at a single slit.
Consider the diagram below
 In particular we consider the light from
one edge of the slit to the point P where
this point is just one wavelength further
from the lower edge of the slit than it is
from the upper edge.
 The secondary wavefront from the upper
edge will travel a distance λ/2 further
than a secondary wavefront from a point
at the centre of the slit.
 Hence when these wavefronts arrive at
P they will be out of phase and will
interfere destructively.
 The wavefronts from the next point below
the upper edge will similarly interfere
destructively with the wavefront from the
next point below the centre of the slit.
 In this way we can pair the sources across
the whole width of the slit.
 If the screen is a long way from the slit
then the angles θ1 and θ2 become nearly
equal.
 From the previous figure we see
therefore that there will be a minimum
at P if
 sin θ1 = λ / b
 where b is the width of the slit.
 However, both angles are very small, and
 θ1≈ θ2 ≈ θ, so we can write that
 θ =λ/b

 With this, we have linked the size of the
slit to the angle and wavelength of the
wave.
 What is the relationship between the size
of the image formed on the screen and
the wavelength?
tan θ ≈ y/D
tan θ ≈ sin θ ≈ θ ≈ y/D

b

There is a mistake in the diagram, can you spot it?
If the distance of the screen to the slit is D, the
actual width of the maximum along the screen can
be calculated:
y

θ
D

   tan θ = y/D
   For very small values of θ, we approximate
   tan θ ≈ θ
   Therefore θ ≈ y/D , since θ = λ / b
   We have y = λ D / b
Solve problems involving single-slit
diffraction.
Example
 Light from a laser is used to form a single
slit diffraction pattern. The width of the slit is
0.10 mm and the screen is placed 3.0 m
from the slit. The width of the central
maximum is measured as 2.6 cm. What is
the wavelength of the laser light?