Wave Nature of Light
Refraction
Interference
Young’s double slit experiment
Diffraction
Single slit diffraction
Diffraction grating
24. Wave Nature of Light
24.1 Waves Versus Particles; Huygens’
Principle and Diffraction
Huygens’ principle:
Every point on a wave
front acts as a point
source; the wavefront
as it develops is
tangent to their
envelope
24.1 Waves Versus Particles; Huygens’
Principle and Diffraction
Huygens’ principle is consistent with
diffraction:
24.3 Interference – Young’s Double-Slit
Experiment
If light is a wave,
there should be
an interference
pattern.
Interference
Two coherent sources producing waves of the same
frequency and amplitude produce an interference pattern.
Sum of 2 waves in phase
Sum of 2 waves in phase
Constructive Interference: waves add up constructively
Interference
Sum of 2 waves out phase
Destructive Interference: waves add up destructively
24.3 Interference – Young’s Double-Slit Experiment
Between the maxima and the minima, the interference varies
smoothly.
24.3 Interference – Young’s Double-Slit
Experiment
The interference occurs because each point on
the screen is not the same distance from both
slits. Depending on the path length difference,
the wave can interfere constructively (bright
spot) or destructively (dark spot).
24.3 Interference – Young’s Double-Slit
Experiment
We can use geometry to find the conditions for
constructive and destructive interference:
Example:
Monochromatic light falling on two slits 0.042 mm apart
produces the fifth-order fringe at a 7.8° angle. What is
the wavelength of the light used?
For constructive interference, the path
difference is a multiple of the wavelength:
d sin  = m, m = 0, 1, 2, 3, … .
For the fifth order, we have
(4.2 x10–5 m) sin 7.8° = (5) ,
= 1.1 x10–6 m = 1.1 µm.
Example:.
3. (I) The third-order fringe of 650 nm light is observed at an angle of 15°
when the light falls on two narrow slits. How far apart are the slits?
For constructive interference, the path difference is a multiple of the
wavelength:
d sin  = m, m = 0, 1, 2, 3, … .
For the third order, we have
d sin 15° = (3)(650 x10–9 m),
d = 7.5 x10–6 m = 7.5 µm.
4.
(II) A parallel beam of light from a He-Ne laser, with a
wavelength 656 nm, falls on two very narrow slits 0.060 mm apart.
How far apart are the fringes in the center of the pattern on a
screen 3.6 m away?
4.
For constructive interference, the path difference is a multiple of the
wavelength:
d sin   m , m  0,1, 2, 3,K .
y  L tan .
sin   tan  , which gives
 m   mL 
We find the location on the screen from
For small angles, we have
yL

 d 
For adjacent fringes,
y 

L  m
so we have
m 1,
;
d
3.6 m 656  109 m 1

d

0.060  10
3
  3.9  10
m
2
m  3.9 cm.
.
24.3 Interference – Young’s Double-Slit
Experiment
Since the position of the maxima (except the
central one) depends on wavelength, the firstand higher-order fringes contain a spectrum of
colors.
24.4 The Visible Spectrum and Dispersion
Wavelengths of visible light: 400 nm to 750 nm
Shorter wavelengths are ultraviolet; longer are
infrared
24.4 The Visible Spectrum and Dispersion
The index of refraction of a material varies
somewhat with the wavelength of the light.
24.4 The Visible Spectrum and Dispersion
This variation in refractive index is why a prism
will split visible light into a rainbow of colors.
24.4 The Visible Spectrum and Dispersion
Actual rainbows are created by dispersion in tiny
drops of water.
Single slit diffraction
destructive interference
D sin  = m
m = 1,2,3, ...
24.5 Diffraction by a Single Slit or Disk
The resulting pattern of light and dark stripes is
called a diffraction pattern.
This pattern arises because different points along
a slit create wavelets that interfere with each
other just as a double slit would.
24.5 Diffraction by a Single Slit or Disk
The minima of the single-slit diffraction pattern
occur when
(24-3b)
Single slit diffraction
Example: Monochromatic light falls on a slit that is 3.00 X 10-3 mm wide.
If the angle between the first dark fringes on either side of the central
maximum is 37.0°, what is the wavelength of the light used?
The angle from the central maximum to the first minimum is 18.5°.
We find the wavelength from
D sin  = m;
(3.00 x 10–6 m) sin (18.5°) = (1) ,
 = 9.52 x10–7 m = 952 nm.
20.
(II) A single slit 1.0 mm wide is illuminated by 450-nm light.
What is the width of the central maximum (in cm) in the diffraction
pattern on a screen 5.0 m away?
We find the angle to the first minimum from
 

9
m  1 450  10 m
sin 1 min 

 0.00045.
3
D
1.0  10 m
We find the distance on the screen from
y  L tan .
For small angles, we have
sin   tan  , which gives



y  L sin   5.0 m 0.00045  0.00225 m.
Thus the width of the central maximum is
2y  0.0045 m  0.45cm.
24.6 Diffraction Grating
A diffraction grating consists of a large number
of equally spaced narrow slits or lines.
The more lines or slits
there are, the narrower
the peaks.
24.6 Diffraction Grating
The maxima of the diffraction pattern are
defined by
(24-4)
Example:
At what angle will 650-nm light produce a second order maximum
when falling on a grating whose slits are 1.15 X 10-3 cm apart?
We find the angle for the second order from
d sin  = m;
(1.15 x10–5 m) sin  = (2)(650 x 10–9 m),
sin  = 0.113, so  =
6.49°.
Example:
A 3500-line/cm grating produces a third-order fringe at a 22.0° angle.
What wavelength of light is being used?
We find the wavelength from
d sin  = m;
[1/(3500 lines/cm)](10–2 m/cm) sin 22.0° = 3,
= 3.57 x10–7 m = 357 nm.
EXTRA
SLIDES
24.7 The Spectrometer and Spectroscopy
A spectrometer makes accurate measurements
of wavelengths using a diffraction grating or
prism.
24.7 The Spectrometer and Spectroscopy
The wavelength can be determined to high
accuracy by measuring the angle at which the
light is diffracted.
Atoms and molecules can be identified
when they are in a thin gas through their
characteristic emission lines.
24.8 Interference by Thin Films
Another way path lengths can differ, and
waves interfere, is if the travel through
different media.
If there is a very thin film of material – a few
wavelengths thick – light will reflect from both
the bottom and the top of the layer, causing
interference.
This can be seen in soap bubbles and oil
slicks, for example.
24.8 Interference by Thin Films
The wavelength of the
light will be different
in the oil and the air,
and the reflections at
points A and B may or
may not involve
reflection.
24.8 Interference by Thin Films
A similar effect takes place when a shallowly
curved piece of glass is placed on a flat one.
When viewed from above, concentric circles
appear that are called Newton’s rings.