Download - Refraction ofLight
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Higher Physics Unit 3
3.2 Refraction of Light
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Refraction
What is Refraction?
Refraction is when the speed and direction of light changes, aslight moves from one material to another.
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a
m
a = angle of incidence in air (larger)m = angle of incidence in material
When light passes from air into a material the ratio is constant.m
a
sin
sin
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The absolute refractive index, n, of a medium is given by:
where 1 is the angle in a vacuum (air is used as an approximation) and
2 is the angle in the medium.
Material Refractive Index n
glass 1.5perspex 1.47
water 1.33
diamond 2.4
2
1
sin
sinn =
m
a
sin
sinn =
*** NOT ON DATA SHEET ***
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Example
A ray of light shines into a block of perspex. Calculate angle x
nsinsin
m
a =
1.47
xsin
20sin=
1.47
20sinxsin =
= 13.5x
200
x
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P&N Tutorial Booklets
Q3.22 3.26
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Outcome 3
Refractive Index of a Perspex Block
normala
pB
A
C
Place the block on white paper and tracearound its outline.
Draw in the normal at the midpoint B.
With incident angle a = 100, measure
the angle p, the refracted angle inthe perspex.
Repeat for other values of incident angle.
Use an appropriate format to determine the refractive index of the perspex block.
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Refractive Index
The refractive index can also be found using:
m
a
v
vn =
m
a
n =
speed of light in air (3x108 ms-1)
speed of light in material (ms-1)
wavelength of light in air (m, nm)
wavelength of light in material (m, nm)
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Example 1
Show that the refractive index of glass 1.50.
-18a ms103v =
-18m ms102v =
?n =
m
a
v
vn =
8
8
102
103
=
1.5n =
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Example 2
The refractive index of water is 1.33.Calculate the speed of light in the water.
-18a ms103v =
?vm =
1.33n =
m
a
vvn=
m
8
v
1031.33
=
1.33103v
8
m
=
18m ms102.26v
=
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Frequency
Light of wavelength 600nm in air is shone through glass ofrefractive index 1.5.
Calculate: a) speed of light in the glassb) wavelength of light in the glass
c) frequency of light in the air
d) frequency of light in the glass
2 x 108 ms-1
400nm
5 x 1014 Hz
5 x 1014 Hz
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Conclusion
The frequency of a wave is determined by its source and does notchange in different media.
fair = fmaterial
Velocity and wavelength change.FREQUENCY DOES NOT CHANGE
Summary
When light is refracted:
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Snells Law
Refractive index depends on the frequency of the incident light.
Refraction occurs because a wave travels at different speeds indifferent media. As the light changes speed, it changes direction.
The refractive index is equal to the ratio of the speeds, giving:
but as frequency is constant this cancels to:
2
1
2
1
2
1
f
f
v
v
sin
sinn ===
2
1
2
1
2
1
v
v
sin
sinn ===
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P&N Tutorial Booklets
Q3.27 3.30
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Critical Angle
When a ray of light is shone into a semi-circular glass block, mostof the light is refracted however this is a partial reflection.
normal
small
incidentangle partial reflection
refracted light
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If the angle within the medium m is increased, a point is reached wherethe angle in air a is 90.
The angle in the medium that causes this is the CRITICAL ANGLE.
normal
largeincidentangle
partial reflection
refracted light
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On increasing the angle within the medium m beyond the criticalangle, all the light is reflected.
This is known as TOTAL INTERNAL REFLECTION.
normal
largerincidentangle
reflected light
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Refraction or Total Internal Reflection
Comparing the size of the incident angle (i) with the critical angle (c)determines whether refraction or total internal reflection occurs.
ci
reflectionpartialandrefraction
reflectioninternaltotal
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normalap
B
C
Critical Angle of a Perspex Block
Make measurements of various incident angles p and thecorresponding refracted angle a to determine the criticalangle c for the perspex block.
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Critical Angle Formula
When the angle in the mediumis equal to the critical angle,the angle in air is 900
So applying Snells Law:
normal
c
900
m
a
sin
sinn =
csin
90sinn =
csin
1n =
But because sin 90 = 1
or
n
1sin c =
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Example
The refractive index of glass is 1.5. Calculate the critical angle.
n
1sin c =
1.51sin c =
.....666.0sin c =
0c 8.41 =
( )0.666.....sin 1c
=
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Example 2
A swimming pool is illuminated by a lamp built into the bottom of thepool.
Three rays of light from the same point in the lamp are incident on thewater-air boundary with angles of incidence of 30, 40 and 50 asshown.
The refractive index of the water in the pool is 1.33.
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(a) Draw a diagram to show clearly what happens to each ray at theboundary.
Indicate on your diagram the sizes of appropriate angles.
All necessary calculations must be shown.
Critical Angle
n
1sin c =
1.33
1sin c =
( )0.752sin 1c
=
= 48.8c
ci reflectioninternaltotal
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(b) An observer stands at the side of the pool and looks into thewater.
Explain, with the aid of a diagram, why the image of the lampappears to be at a shallower depth than the bottom of the pool.
To an observer the light appears to originate from where thebroken rays cross.
This produces a virtual image of the lamp at the apparentlyshallower depth.
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Refraction & Total Internal Reflection in Nature
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P&N Tutorial Booklets
Q3.31 3.36