Transcript 13 Optics - Animated Science

AS Physics Unit 2
13 Optics
Mr D Powell
Chapter Map
Mr Powell 2008
Index
13.1 Refraction of light
Specification link-up 3.2.3: Refraction at a plane Surface
1.
Know that “refraction” happens when a light wave
passes into a different medium (AO1a)
2.
Be able to explain that light changes direction
towards the normal when it slows down (enters a
prism) (AO1a)
3.
Be able to explain that light changes direction away
from the normal when it speeds up (exits a prism)
(AO1a)
4.
Be able to apply formulae for refractive index of a
substance is 1n2 = sin1/sin2 (Snells or refraction
law) to simple situations AO2b
n1 sin 1  n2 sin 2
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Refraction of Water – revisited
When the direction of travel of a wave changes as it travels from one medium
to another. This is due to a change in wave speed.
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Verification of Snell's Law
Snell's Law
Why not try it out for a glass prism?
0.7
0.6
0.5
Sin 1
If we alter the incident angle (i) what
happens to r
We find that a strange thing happens,
the waves seem to have a constant
ratio each time.
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
Sin2
This leads us a formula…..
1
2
sin 1
sin 2
sin1 /sin2 = n
10
6.6
0.174
0.115
1.5
20
13.2
0.342
0.228
1.5
30
19.5
0.5
0.334
1.5
40
25.4
0.643
0.429
1.5
50
30.7
0.766
0.511
1.5
60
35.3
0.866
0.578
1.5
70
38.8
0.94
0.627
1.5
n1 sin 1  n2 sin  2
sin 1 n2

 const
sin  2 n1
sin 1
1 n 2
sin  2
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Extra Examples…
Material
Vacuum
Air @ STP
Gases @ 0 °C and 1 atm
Air
Carbon dioxide
Liquids @ 20 °C
Water
Arsenic trisulfide and
sulfur in methylene
iodide
Solids @ room
temperature
Diamond
Strontium titanate
Amber
Water ice
cornea (human)
n
1 (per definition)
1.000277
1.000293
1.00045
1.3330
1.9
2.419
2.41
1.55
1.31
1.3375
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b) nwater = 1.33, nair  1 , a = 40 
nwsinw= nasina
Going out of water (speeding up)
1.33sin40= 1sina
a)
nwater = 1.33, nair  1
nwsinw= nasina
Going into water (slowing)
1.33sinw= 1sin20
1.33sin40= 0.854
Sin-1(0.854) = a
a = 58.7 (3 s.f.)
ii) See textbook but should have normal,
angle inside water from normal is 40,
in air other side 58.7.
sinw= sin20 / 1.33
sinw= 0.257
w = 14.9
w = 14.9 (3 s.f.)
ii) 28.9
iii) 40.6
n1 sin 1  n2 sin 2
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a)
nair  1
ngsing= nasina
Going into glass (slowing)
ng/na= sina/sing
nair  1
ng = sin50/sin30
b) ng/na= sina/sing
ng = 1.532
ng = 1.5 (2 sig fig)
n1 sin 1  n2 sin 2
1.532/1= sin60/sing
sing = sin60/1.532
g = 34.4
g = 34 (2 sf)
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Hint:
1. Draw a scale diagram 5cm
each side.
2. 60 angles.
3. First refracted angle 22
4. Measure 2nd incident 38
5. Calc 2nd refracted 72.6
n1 sin  2

n2 sin 1
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Starter Quick Test – Print!
The diagram shows a rectangular
glass fish tank containing water.
Three light rays, P, Q and R from
the same point on a small object
O at the bottom of the tank are
shown.
(a) (i) Light ray Q is refracted
along the water-air surface. The
angle of incidence of light ray Q
at the water surface is 49.0°.
Calculate the refractive index of
the water. Give your answer to an
appropriate number of significant
figures. (1)
(ii) Draw on the diagram above
the path of light ray P from the
water-air surface. (3)
n1 sin 1  n2 sin 2
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n1 sin 1  n2 sin 2
Starter Quick Test - Answers
The diagram shows a rectangular
glass fish tank containing water.
Three light rays, P, Q and R from
the same point on a small object
O at the bottom of the tank are
shown.
(a) (i) Light ray Q is refracted
along the water-air surface. The
angle of incidence of light ray Q
at the water surface is 49.0°.
Calculate the refractive index of
the water. Give your answer to an
appropriate number of significant
figures. (1)
(ii) Draw on the diagram above
the path of light ray P from the
water-air surface. (3)
a)
nair  1
nwsinw= nasina
nw/na= sina/sinw
nw/1 = sina/sinw
nw = sin90/sin49
nw = 1.325
nw = 1.33 (3 sig fig)
ray P shown in the air to
right of vertical (1)
refracted away from the
normal in the correct
direction (1)
correct partial reflection
shown (1)
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13.2 More about refraction
1.
Mathematically see when a
wavefront slows it must change
angle (AO1a)
2.
Mathematically link this velocity
change to Snells law (previous
lesson) (AO1a)
3.
Be apply formulae for refractive
index of a substance is
1n2
= sin1/sin2 & n c/cs = /s
(c is speed outside of prism, cs is speed
inside the prism) (AO2b)
n1 sin 1  n2 sin  2
n2 sin 1

n1 sin  2
cair
 ns
cs
c 
ns  
c s s
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What happens?
air
glass / water
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Explaining it...
REFRACTIVE INDEX FROM 1 TO 2
c1
1 n2 
c2
Optically less dense
medium (1)
Optically denser
medium (2)
c1 f1 1


1 n2 
c2 f2 2
Waves travel SLOWER
Wavelength REDUCED
Frequency UNCHANGED
NB: If c = f if speed is less as light finds it harder to get through glass the 
must go down as well (colour does not change)
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Problem drawn out...
 If we try and think about the problem by drawing out the wave front
and looking at triangles..
 Then considering the constant velocity of the wave which then changes
to a new constant velocity.
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Similar Triangles...
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Similar Triangles...
ct=YY’
i
ct
 sin i
XY '
XY’
cst=XX’ XY’
r
ct  XY ' sin i
ct
XY ' sin i

cs t XY ' sin r
cs t
 sin r
XY '

Think about how the wave slows when it reaches the medium.

We can express the distance it travels YY’ or XX’ simply by s/v=t
(constant vel)
cs t  XY ' sin r
c sin i

cs sin r

Then similar triangles result in a sin function for the angle
calculations and an expression for a ratio of velocities.

Finally we can express this ratio as what we call a refractive index
or “how much the light slows down”. Higher is slower!
sin i
ns 
sin r
c 
ns  
c s s
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More Maths….
n1 sin 1  n2 sin  2
n2 sin 1

n1 sin  2
So we now have an expression to express the
ratios of angles for a ray that refracts through
from air (n=1) into a block (n) resulting in
something call the refractive index….
Then we can also express this idea in
terms of the ratio of speed in air to speed
in the substance. Or the change in
wavelengths to also give ns.
They are in fact the same things and can
combine to give….
sin 1
1 n2 
sin  2
 cair

 ns
s c s
c
c
sin 1 
sin  2
cs
cair
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Speed of Light More Details...
The speed of light, usually denoted by c, is a physical
constant, so named because it is the speed at which light
and all other electromagnetic radiation travels in vacuum.
Its value is exactly 299,792,458 ms-1. The refractive index
is exactly 1 by definition.
The speed at which light propagates through transparent
materials, such as glass or air, is less than c.
Wavelength
REDUCED
Waves travel
SLOWER
Frequency
UNCHANGED
The ratio between c and the speed v at which light travels
in a material is called the refractive index n of the
material (n = c / v).
For example, for visible light the refractive index of glass is
typically around 1.5, meaning that light in glass travels at
c / 1.5 ≈ 200,000 km/s; the refractive index of air for
visible light is about 1.0003, so the speed of light in air is
very close to c.
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So nw/ng = 1.52 / 1.33 = 1.142
nw x sinw = ng x sing
1.33 x sin55 = 1.52 x sing
Extra help see p190
worked example…
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Whiteboard Knowledge Check?
1.
Why does a wave change direction when it enters a
denser medium?
2.
What formula can express the idea of the slowing in
terms of  or c or v or ?
3.
1) It slows
down
c1 f1 1


1 n2 
c2 f2 2
n1 sin 1  n2 sin 2
What happens to a water waves velocity as it gets
closer to shore?
3) Slows down
4.
As you move from red to blue end of visible light
what happens to the refractive index and why
(harder)
4) Increases due to
shorter
wavelength
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Starter….
You have used this formula from
previous lessons and should
understand how it works. Can you
demonstrate your knowledge of
what it is used for? (5 marks)
1 = 40.0
n1sin1 = n2sin2
2 = Sin40 * (1/1.33) = 0.48
2 = 28.9
Hint…
Try drawing the rays
n1 = 1.00
Try an example calculation.
n2 = 1.33
Extension: what happens
when refracted ray is at 90 to
normal.
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13.3 Total Internal Reflection
• When the angle of refraction is 90 or /2 then Snells law simplifies from
n1sin1 = n2sin2 n1sin1 = n2 (out of a prism) (AO1a)
• Appreciate that the higher the refractive index or optical density of a
substance the more light will refract for the same entry angle. (AO2b)
• Gain understanding of these key terms and be able to apply to an optical fibres
exam question : Core, Cladding, coherence, Spectral Dispersion, Multipath
Dispersion, Critical Angle (TIR), Mono-mode (AO2b)
• AO3a – Take precise practical readings to support the theories during the
lesson.
http://en.wikipedia.org/wiki/Optical_fiber
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TIR
weak reflected
ray
As the angle of incidence
increases towards the critical
angle ( glass = 420 ) the refracted
ray gets weaker and the reflected
ray gets stronger.
NB: When light travels from an optically denser medium to a less dense medium,
rays are bent away from the normal. The incident substance has a larger refractive
index than the other substance
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Critical Angle?
n prism
sindepends
 prism  nupon
 air
Critical
angle
the
air sin
refractive indices of the media
n prism sin  prism  nair sin 90
n prism 
2
1
C
n prism 
nair
sin  prism
1
sin  prism
HENCE
If medium 2 is air, n2 = 1, and so….
sin C 
1
n prism
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Critical Angle Example?
For Water
1
sin C 
1.33
2
AIR
WATER
1
C
For Crown Glass
C = 48.80
1
sin C 
1.50
C = 41.80
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Diamonds Sparkle the most!
sin C 

Diamond has a refractive index of 2.417.

This means that colours are spread out more.

TIR occurs many times inside the diamond before
emerging.
1
n1
Can you work out the
critical angle?
24.44
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Index
Answers
Material
Refractive
index n1
Refractive
index n2
2
refracted
1 incident
Water (tank)
1.33
1
90
49
Your Prism
1.39
1 (air)
90
46
Crown Glass
1.5
1
90
42
Diamond
2.417
1
90
24
Un cladded fibre
optic
1.62 (core)
1
90
38
Cladded fibre optic
1.62 (core)
1.52 (clad)
90
70
n prism sin  prism  nair sin  air
sin C 
1
n prism
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Basics of the Fibre…
The typical refractive value for the
cladding of an optical fibre is 1.52.
The core value is typically 1.62.
The larger the index of refraction,
the slower light travels in that
medium. The idea of the core is to
slow the light to make it produce
the TIR effect at lower angles of
incidence.
Light can enter in a range of values which
can lead to signals travelling a longer path
length to the end of the fibre (multipath
dispersion). This can cause merging of
the signal from pulse to pulse.
Use of white light can also cause
problems (spectral dispersion) for the
same reasons. However, in this case the
light turns into a spectrum at the other
end.
n= 1.62
n= 1.52
A solution to the dispersion is to make
the fibre very thing 5 μm in diameter to
reduce the effect. (Monomode)
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Cladding and multipath dispersion (More Detail)
The fibres are coated with a glass of slightly lower refractive index. This is known
as cladding. The cladding increases the critical angle within the core fibre and
also prevents adjacent fibres from touching each other. At every point of contact
light would escape into another fibre. The fewer the reflections the less energy
loss, and the shorter the time of transfer of information down the fibre since the
light travels a shorter distance.
Initially it would seem that the addition of the cladding would allow light to
escape into the surroundings. This is indeed the case but the cladding has
another purpose. It means that only the light that makes a small angle with the
axis of the fibre is transmitted over large distances. The difference in the time of
travel between the individual light rays is therefore smaller and so the spread of
information (known as multipath dispersion) is also reduced.
n= 1.62
n= 1.52
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Applications of Fibre Optics…
A coherent bundle of optical fibres in which
the relative spatial coordinates of each fibre
are the same at the two ends of the bundle.
Such a bundle are used for the transmission
of images. Endoscope
A non-coherent fibre bundle, as you would
expect, does not have this precise matrix
alignment since they need only transmit
light for illumination purposes. They are
cheaper to produce. Christmas Tree.
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More Uses of Optical Fibres….
1. Illuminating models or road signs using only one bulb
2. Endoscopy - seeing down inside a patient’s body
3. Communications – sending information along a light beam. Useful for
telephone, television, radio, computer networks, stereo links, control in
aircraft
4. Security fencing – very difficult to bypass
5. Fibre optic lamp
Advantages of fibre optics over copper wire
1.
2.
3.
Cheap – glass is made from silica, the basic constituent of sand
Light in weight – useful in aircraft
Light beam can carry a huge amount of information
Such fibres can be made to carry information such as TV channels or telephone
conversations. Other applications of fibre optics include its use in medicine to
see inside the human body and in road signs where one light bulb and a set of
fibres is used to illuminate different parts of the sign thus saving electrical
energy. A further recent application is in security fences. The metal strands of
the fence contain a piece of fibre optic material down which a beam of light
passes. If the strand is cut the light beam is interrupted and an alarm sounds. It
is thought that this type of system is impossible to bypass.
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Answers
Feature
Critical Angle
(TIR)
Core
Cladding
Endoscopes
– coherence
Purpose / Related Physics
This is the angle at which then a light ray incident on a boundary becomes great
enough to it will reflect off the internal surface of the material. “Total Internal
Reflection”
Narrow flexible fibre of glass designed to make light TIR at the inside surface. It has
a high refractive index to make sure the maximum amount of light is transmitted.
The cladding has a lower refractive index that the core. It is there to ensure that if
adjacent fibres touch there is no transmission of light. Also as a change of index to
increase the angle at which TIR takes place and keep the light in at greater angles.
The cables in the endoscope specifically must form a pattern or “image” of the
patient. Hence we need to make sure that the same pattern is show at the top and
bottom of the bundle
Multipath
Dispersion
Use of white light can also cause problems (spectral dispersion) for the same
reasons. However, in this case the light turns into a spectrum at the other end.
Spectral
Dispersion
Light can enter in a range of values which can lead to signals travelling a longer path
length to the end of the fibre (multipath dispersion). This can cause merging of the
signal from pulse to pulse.
Mono mode fibres are made very thin to 5 μm in diameter to reduce the effect of
spectral and multipath dispersion (broadening).
Mono-mode
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13.4 Double Slit Interference
What is the danger to our eyes from laser light (AO1a)
How evenly spaced bright and dark fringes form on a screen from interference?
(AO2b)
What factors could be (i) increased, or (ii) decreased, to increase the fringe
spacing? (AO3b)
How the process of interference works on a mathematical basis using similar
triangles. (A01a) A*-B
Newton/ Young/ Huygens & Einstein HSW Wave video – home extension.
http://www.youtube.com/watch?v=AMBcgVlamoU&feature=related
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Water Waves Interference Rules….
To produce a steady interference
pattern….
1. The waves must be of the same
shape (e.g. sinusoidal ) and must
meet at a point.
2. The waves are coherent, i.e. the
waves from each source maintain a
constant phase difference.
3. The waves must have the same
wavelength and roughly the same
amplitude.
NB: Two sources are said to be coherent if
waves from the sources have a constant phase
difference between them.
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Using Lasers - Dangers
Laser safety is safe design, use and implementation of
lasers to minimise the risk of accidents, especially those
involving eye injuries.
Since even relatively small amounts of laser light can lead
to permanent eye injuries, the sale and usage of lasers is
typically subject to government regulations.
The coherence or low divergence angle of laser light and
the focusing mechanism of the eye means that laser light
can be concentrated into an extremely small spot on the
retina.
An increase of only 10 °C can destroy retinal photoreceptor
cells. If the laser is sufficiently powerful, permanent
damage can occur within a fraction of a second, literally
faster than the blink of an eye.
Sufficiently powerful in the visible to near infrared laser
radiation (400-1400 nm) will penetrate the eyeball and may
cause heating of the retina.
http://en.wikipedia.org
/wiki/Laser_safety
Class 2
A Class 2 laser is
relatively safe
because the blink
reflex will limit the
exposure to no more
than 0.25 seconds.
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Young's Double Slit Interference - Introduction
In 1801, an English physicist named Thomas Young
performed an experiment that strongly inferred the wavelike nature of light.
Because he believed that light was composed of waves,
Young reasoned that some type of interaction would occur
when two light waves met.
Young passed sunlight through a red colour filter then single
slit in a screen to produce coherent light. Then another
screen that has a double slit.
The results of interference between the diffracted light
beams can be visualised as light intensity distributions on
the dark film.
His hypothesis was that if light were wave-like in nature, then
it should behave in a manner similar to ripples or waves on a
pond of water. What do you think you would see?
Answer: They should either cancel or add at different points
depending on the distance from the slit.
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3D View….
They should either cancel or add at different points
depending on the distance from the slit. Producing “fringes”
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Wavefront View…
Single source passes through
single then double slits (a few
 apart).
The waves diffract then
interference occurs where
the beams cross.
1st bright
fringe
central
fringe
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1st bright
fringe
Index
ICT Investigation…
We can see the following from
this…
1.
Light passes through a
double slit
2.
It forms an interference
pattern of maxima and
minima
3.
The pattern spreads out to
make smaller fringes if the
slit spacing is made closer.
http://www.walterfendt.de/ph14e/doubleslit.htm
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Young's Double Slit Interference - Results
Young observed that when the slits were large, spaced far apart and close to
the screen, then two overlapping patches of light formed on the screen.
However, when he reduced the size of the slits and brought them closer
together, the light passing through the slits and onto the screen produced
distinct bands of colour separated by dark regions.
Young coined the term interference fringes to describe the bands and realised
that these coloured bands could only be produced if light were acting like a
wave.
Pattern is wider as slit spacing gets larger…
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Young's Double Slits
Young actually worked out a formula for the theory of how far the fringes are separated
(w). It related the distance from S1 or S2 to screen (D) the wavelength of the light ()
and s (slit spacing – distance between centres of slits S1 or S2)
w
D
s
W
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Test yourself….
Use the example data to test out the formula..
 = wavelength
D = slit to screen distance
S = slit separation
W = fringe width
s
= 590nm
= 1m
= 0.4 mm
= 1.48mm
w 

D s
W
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Laser Interference Patterns….
Red (Quoted 656nm)
1.
Take a double slit (0.15mm thick) and spacing
0.5mm apart.
2.
Shine a red and then green laser through it
onto a screen placed 6.75m away
3.
On the paper sketch out the regular blocks of
colour that you get.
4.
Use the formulae to work out the wavelength
of the green and red lights.
ws

D
D = 6.75m
S = 0.5 x 10-3m
w = 9 x 10-3m
 = 666nm
Green (Quoted 532nm)
D = 6.75m
S = 0.5 x 10-3m
w = 7.2 x 10-3m
 = 533nm
 = wavelength
D = distance slit to screen
S = slit separation
w = fringe width
s
w = 9mm
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Factors affecting fringe separation
Another way to express this idea is that
Fringe separation (x = w)….
w

D
s
w
S

w is increased if distance to the
screen D is increased.

w is decreased if slit separation s is
increased.

w is increased as wavelength of light
λ is increased
D
s
= wavelength
= slit to screen distance
= slit separation
= fringe width
S
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Another Example….
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Complex Theory

Consider the diagram above where at P the fringe is observed. Light emitted from
S1 arrives later than S2 as it travels further. The difference in the travel is called the
path difference.

We find that for a wave to add constructively the path difference must be a whole
wavelength i.e. m = 1 when S1P-S2P = m (m = 1,2,3)

Hence: light emitted at same time from S1, S2 will arrive in phase at P if
reinforcement occurs

We find that for a wave to cancel the path difference must be a ½ wavelength i.e.
m = 1 when S1P-S2P = (m+0.5)

Hence: light emitted at same time from S1, S2 will arrive out of phase at P if
cancellation occurs.
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Path difference
Phase difference
For constructive interference at the screen:
( ie a bright fringe )
Wave fronts from S1 and S2 must arrive at the screen in phase
with a path difference of a whole number of wavelengths
For destructive interference at the screen:
( ie a dark fringe )
Wave fronts from S1 and S2 must arrive out of phase
with a path difference of half a wavelength
Complex Theory
S1Q m
sin  

S1S 2
s
OP w
tan  

OM D
w m

D
s
1.
2.
3.
4.
5.
6.
7.
Point P has been selected so that QP = S2P
This means that path difference is S1P – S2P or S1P – S2P = S1Q
Consider the triangles S1S2Q and MOP.
W = OP
M is midpoint between slits and O the midpoint of brightest fringe
The two triangles are similar in terms of angles PMO and QS2S1
We also know that sin = tan  when <5
So we can say that for triangle OMP and S1S2Q that they equal the same ratios (see
eqs)
8. This works for each fringe where P is the m’th bright fringe (m = 0, 1,2,3)
9. w is fringe separation, s spacing, D distance to screen perpendicular.
10. For this to be true s << D.
Mr Powell 2008
Index
Summary Q’s
 = wavelength
D = distance slit to screen
S = slit separation
w = fringe width
mD
w
s
b)
a)
• If the screen is closer from the
formulae….
• D is reduced so the fringes
would get closer
• Dw
• Interference fringes would
change.
• The pattern would change
and not be the same regular
fringes as the interference
cannot take place. (how
exactly it will look is covered
later)
Mr Powell 2008
Index
Summary Q’s
Mr Powell 2008
Index
Summary Q’s
Mr Powell 2008
Index
Summary Q’s
Mr Powell 2008
Index
13.5 More about Interference
Specification link-up 3.2.3: Interference

What are coherent sources?

Why are slits used, rather than two separate light sources, in Young’s
experiment?

What are the roles of diffraction, and interference, when producing Young’s
fringes?
Mr Powell 2008
Index
Exam Question….
Singe slit
Double slit
single slit then double slits to the right (1)
single slit and double slits labelled (2)
Mr Powell 2008
Index
Focus on Coherence…
• Double slits and light are a coherent
source as they emit light waves of a
constant with a constant phase
difference. (same as water) so you get a
regular pattern.
• Two light bulbs don’t produce the same
effect as the emission of the light is
random and the phase difference
changes depending on the times when
the waves interfere.
Mr Powell 2008
Index
White Light Fringes
For the double Slit experiment the fringe spacing depends on the
wavelength used (see formulae) w  .
•
Using white light, fringes appear from all the various
wavelengths present and do not overlap exactly, hence
coloured fringes. (laser produce more collimated dots)
•
Inner fringes are tinged with blue on the inside and red on the
outside (red diffracted more as longer wavelength).
•
Where all the colours overlap it produces white light.
λ red » λ blue
D
w
s
Mr Powell 2008
Index
Exam Question - Printed
Mr Powell 2008
Index
Exam Question
Mr Powell 2008
Index
13.6 Diffraction
Specification link-up 3.2.3: Diffraction

Why is diffraction of light important in the design of optical instruments?

How does the single slit diffraction pattern compare with the pattern of
Young’s fringes?

What is the effect of the single slit pattern on the brightness of Young’s
fringes?
Mr Powell 2008
Index
Laser Light
The most common and inexpensive gas laser, the
helium-neon laser is usually constructed to
operate in the red at 632.8 nm.
One of the excited levels of helium at 20.61 eV is
very close to a level in neon at 20.66 eV, so close in
fact that upon collision of a helium and a neon
atom, the energy can be transferred from the
helium to the neon atom.
Helium-neon lasers can still be dangerous! An
unfocused HeNe laser is dangerous to stare at
directly and will blind you. You must be careful to
avoid refracted beams whilst conducting
experiments.
Mr Powell 2008
Index
Double Slit Interference
w
D
s
 = wavelength
D = distance slit to screen
S = slit separation
w = fringe width
a = slit thickness
Mr Powell 2008
Index
Double
Single
Link
Mr Powell 2008
Index
Double & Single Slit Diffraction Compared
The best way to look at this phenomenon is to try some readings. Two lasers were
projected through similar thickness of slit at same distance. Here are the
findings…
Double
Single
5 fringes
w
1 fringe
w
 = wavelength
D = distance slit to screen
S = slit separation
w = fringe width
a = slit thickness
D
s

a
2D
 =
D = 3.1m
s = 0.5 x 10-3m
5w = 20.65 x 10-3m
a = 0.15mm
 =
D = 3.1m
w = 20.65 x 10-3m / 5
a = 0.15 x 10-3m
Task: use my findings to check the
wavelength of the light (as it did)
Mr Powell 2008
Index
Double & Single Slit Interference (Higher quality results)
Same double-slit assembly (0.7mm between slits); in top image, one slit is closed.
Note that the single-slit diffraction pattern — the faint spots on either side of the
main band — is also seen in the double-slit image, but at twice the intensity and
with the addition of many smaller interference fringes.
Mr Powell 2008
Index
Summary of Patterns

The double slit pattern is superimposed on the much broader single slit
diffraction pattern.

The bright central maximum is crossed by the double slit interference
pattern, but the intensity still falls to zero where minima are predicted from
single slit diffraction.

The brightness of each bright fringe due to the double slit pattern will be
“modulated” by the intensity envelope of the single slit pattern.
NB: The double slit fringes
are still in the same place
Single slit
pattern
Double slit pattern
Mr Powell 2008
Index
Single Slit and Double Slit?
1. A diffraction pattern formed by a real double
slit. The width of each slit is much bigger than
the wavelength of the light. This is a real photo.
2. This idealised pattern is not likely to occur in
real life. To get it, you would need each slit to
be comparable in size to the wavelength of the
light, but that's not usually possible. This is not
a real photo.
3. A real photo of a single-slit diffraction pattern
caused by a slit whose width is the same as the
widths of the slits used to make the top pattern.
Mr Powell 2008
Index
Single Slit
This is an attempt to more clearly visualise the nature of single slit diffraction. The
phenomenon of diffraction involves the spreading out of waves past openings
which are on the order of the wavelength of the wave.
Mr Powell 2008
Index
Slit Width
One of the characteristics of single
slit diffraction is that a narrower slit
will give a wider diffraction pattern
as illustrated in the images.
Mr Powell 2008
Index
Or can view as a water wave (single slit diffraction)
a=λ
a=2λ
Semi circular
wave fronts
First minima & maxima
become visible
a=4λ
Diffraction is the spreading of wavefronts around corners and obstacles.
If the slit gets narrower diffraction increases.
If the wavelength increases diffraction increases.
Mr Powell 2008
Index
Single Slit Diffraction...
Mr Powell 2008
Index
What happens…..
w
Mr Powell 2008

a
2D
Index
What happens when  changes…
w

a
2D
Longer  so
the w is
larger
Shorter  so w is
smaller
Mr Powell 2008
Index
What happens when slit width changes…
w

a
2D
Larger a so
the w is
smaller
smaller a so the
w is larger
Mr Powell 2008
Index
13.7 The Diffraction Grating
Specification link-up 3.2.3: Diffraction

Why does a diffraction grating diffract monochromatic light in certain
directions only?

If a coarser grating is used, what is the effect on the number of diffracted
beams produced and on the speed of each diffracted beam?

How can we determine the grating spacing for any given grating, if it is not
known?

Derive the Generic Diffraction Formulae
n  d sin 
Mr Powell 2008
Index
Light through a diffraction grating...
If we pass light rays through a diffraction grating. Which is made up of very small
slits in a regular pattern we find that there is a pattern of dots formed on a screen.
This is because the wave fronts interfere with each other destructively and
constructively.
Simply put a wave front which emerges at point P is reinforced by another wave
front acting in phase at point Q and then R. The result is a wave front along Y &Z
Mr Powell 2008
Index
Light through a diffraction grating...
If we consider the triangle as shown in the zoomed in
diagram we can see that an angle  can be worked
out from knowing QP and QY and then similar
triangles rule used to find out the angle of the rays as
they appear on a screen.
We can use the idea of constructive interference and
take the idea that from trigonometry.
However, we also know that the length QY should be
the length of a whole wavelength and also that QP is
the slit spacing…..
QY
 sin 
QP

d
 sin 
Mr Powell 2008
Index
Light through a diffraction grating...
We can also now think about the idea that this
happens not only once but several times or orders.
Each time we have a whole wavelength we get the
construction of a new wave front or “n” wave front.
Our formulae can be changed to;

 sin 
d
n  d sin 
d = slit spacing
 = angle from normal
n = order of diffraction
 = wavelength of the light
NB: max number of orders is given by n = d/ where sin(90) = 1
Mr Powell 2008
Index
Practical
Setup a laser light source and use the
formula below to check the wavelength of
the light. Your quoted value is 632.8nm for
red (632.8 x 10-9m).
Your teacher will give you a slit which has
300 lines per mm or a spacing d = 3.33 x
10-6m. Use this one at first to check the
calculation. Then try and take some other
readings for 100, 600 and repeat with the
green laser beam.
n  d sin 
  d sin 1
d = slit spacing
 = angle from normal
n = order of diffraction
 = wavelength of the light
opp dist scattered
tan  

adj
dist screen
Mr Powell 2008
Index
Results.....
n
Distance
(m)
Distance
Diffracted
(m)
1
0.5
0.099
2
0.5
3
0.5
Hyp
opp /hyp =
sin(theta)
lambda d/lambda
(nm) (max orders)
d
Lambda (m)
0.509706778 0.194229318
3.33333E-06
6.47431E-07
647
5
0.208
0.541538549 0.384090847
3.33333E-06
6.40151E-07
640
5
0.348
0.60918306
3.33333E-06
6.3473E-07
5
Ave
635
641
Quoted!
632.8
Within...
98.7
0.571256857
1.30%
Mr Powell 2008
Index
June 2010 Q6 Practice…

d = slit spacing
 = angle from normal
n = order of diffraction
 = wavelength of the light
d

 sin 
d
n  d sin 
1.
Angle will decrease (get smaller) if  decreases. sin  
2.
Path difference must be smaller as  is less so angle must
also be less for constant d sin  
Mr Powell 2008
(2 marks)
Index
Use of Emission Spectrums?
• How do we know which elements stars are made up from?
• How do we know the age of stars?
• Scientists can begin to answer these questions once they have an
understanding of line spectra.
Mr Powell 2008
Index
What are we talking about?
• When we talk about “Line Spectra” for an atom we simply
mean that an atom can absorb or emit radiation at certain
frequencies. If we look at the frequencies emitted or
absorbed we can see a spectrum with omissions.
656nm
Mr Powell 2008
Index
How are they made?
 There are three types that you might see....
Mr Powell 2008
Index
Other Elements Emission Spectra....
Mr Powell 2008
Index
Spectral Gases
Image we look towards a star and then pass its light through a cold gas then
then a prism to make an absorption spectrum what elements would be present?
Star X
Mr Powell 2008
Index
Spectral Gases & Shift
Here are three spectral diagrams from a star. The first one is a reference diagram.
The second two are shifted across to the red end of the spectrum. Hence, they are
moving away from us. We can tell how fast they are moving by the shift. The middle
band is actually shifted by a distance of 100Å (angstrom) or 100 x 10-10 m. (10nm).
This translates to a speed of 24,000 km/hour or 15,000 mph. The bottom band is
shifted by 760Å or (76nm) which translates to a speed of 136,000km/hour or 84,000
mph.
Mr Powell 2008
Index
Spectral Gases Conclusion
So these pictures show us that the lower stars are moving away from us
Also the further they are away the faster they travel.
Mr Powell 2008
Index
June 2010 Q6 Practice…
(2 marks)
Mr Powell 2008
Index
June 2010 Q6 Practice…
d = slit spacing
 = angle from normal
n = order of diffraction
 = wavelength of the light

Distance between grating spacing….
If the grating has 300 lines per mm….
 sin 
d
n  d sin 
1/ 300 per mm or 300,000 per m. 1/300,000 = 3.33 x 10-6m
The 1st order of diffraction from P is 550 -> 500 so 50nm
Hence. …
(2 marks)
sin =  n/d
 =sin-1(550 x 10-9m x 1 )/ 3.33 x 10-6m
= 9.5
Mr Powell 2008
Index
Laser Diffraction!
• Can you work out the spacing of
lines on a CD if light is 632.8
nanometres. (0.5 to 0.9mW Power)
Model = 1 x 10-6m
1.6 microns ,dvd is .74 microns
Experimental 5.4 microns
L = 229.5cm
Opp =
55.5cm
57.1cm
58.3cm
(three repeats 2nd order)
n
d
sin 
opp
 tan 
adj
Mr Powell 2008
Index
Extra Maths
(Maths Students Only)
Trig Series
Trigonometric functions can be expanded in power series, which facilitates
approximations of the functions in extreme cases. The angle x must be in
radians.
Mr Powell 2008
Index
Small Angle Approximation...
One of the most important applications of trigonometric
series is for situations involving very small angles. For such
angles, the trigonometric functions can be approximated
by the first term in their series. This gives the useful small
angle approximations:
Examples of the use of the small angle approximation are in
the calculation of the period of a simple pendulum, and the
calculation of the intensity minima in single slit diffraction.
This approximation is used in most of the common
expressions of geometrical optics which are built on the
concept of surface power for lenses.
This approximation sin x = x reaches a 1% error at about 14
degrees
Mr Powell 2008
Index
Maths puzzling....
Construct an excel spreadsheet to
explore and prove the concept
that for the situation where 
<10 and measured in radians;
  sin 
Degrees
Theta Rad
Sin Theta
Difference
0
0.00
0.00
0.00
1
0.02
0.02
0.00
2
0.03
0.03
0.00
3
0.05
0.05
0.00
4
0.07
0.07
0.00
5
0.09
0.09
0.00
6
0.10
0.10
0.00
7
0.12
0.12
0.00
8
0.14
0.14
0.00
9
0.16
0.16
0.00
10
0.17
0.17
0.00
20
0.35
0.34
-0.01
30
0.52
0.50
-0.02
40
0.70
0.64
-0.06
50
0.87
0.77
-0.11
60
1.05
0.87
-0.18
70
1.22
0.94
-0.28
80
1.40
0.98
-0.41
Mr Powell 2008
Index
Results….
Here are some possible ideas of what could happen….
When spacing is large
When spacing is small
Mr Powell 2008
Index