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Lecture 16:
Geometrical Optics
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Reflection
Refraction
Critical angle
Total internal reflection
Polarisation of light waves
Geometrical Optics
Optics—Branch of Physics,
concerning the interaction of light with matter
Geometrical Optics- subset of optics concerning
interaction of light with macroscopic material
Dimension larger than a human hair ≈ 50mm
Geometrical Optics
ray optics
Light Ray –beam of light
Light can travel through
•empty space,
•air, glass, water,
•cornea,
•eye lens etc.
each one referred to as a medium
Light rays will travel in a straight line if they
remain in the same medium
Reflection
At the boundary between two media, the light
ray can change direction by
reflection or refraction
Reflection
For a mirror (smooth metal surface)
“all” light will be reflected
Normal
Incident Ray
Reflected Ray
qi qr
Smooth
Metal surface
Reflection
Normal
Reflected Ray
Incident Ray
qi qr
Metal surface
Laws of reflection
1. angle of incidence(qi) = angle of reflection(qr)
2. Angles measured with reference to the
normal to the surface
3. Incident and reflected rays and normal all
lie in the same plane
Smooth surface: reflection at a definite angle
--Specular reflection
Incidence Rays
Reflected Rays
Metal surface
Reflection
Diffuse reflection
Rough Surface
No unique angle of reflection for all rays
Light reflected in all directions
Majority of objects (clothing, plants, people)
are visible because they reflect light in
a diffuse manner.
Normal
Reflected Ray
Incident Ray
qi qr
Refraction
At the surface of transparent media, glass,
water etc both reflection and refraction occur.
Refraction (deflection from a straight path in
passing obliquely from one medium
( such as air) into another (such as glass)
Incident Ray
Medium 1
Medium 2
Normal
q1 q1
Reflected Ray
q2
q2
Refracted Ray
Light ray changes direction going from one
medium to another.
Which way does it bend and by how much?
Is q2<q1 or is q2>q1
Answer
Depends on the speed of light in both media
Refraction
Speed of light in a vacuum: c = 3x108 ms-1
The amount by which a medium reduces the
speed of light is characterised by
Index of refraction (n) of the medium
c
n
v
speed of light in
the material = v
Indices of Refraction
Vacuum
1(by definition)
Air
1.0003
Glass
1.52
Water
1.33
Diamond
2.42
Example
Calculate the speed of light in diamond
v =c/n =(3x108 ms-1)/2.42 = 1.24 x108ms-1
Refraction
Example
How long does it take light to travel
394cm in glass
Calculate the speed of light in glass
c
v
n
3 108 ms 1
8
1
v
 1.97 10 ms
1.52
d
t
v
3.94m
8
t

2

10
s
8
1
1.97 10 ms
Refraction
Monochromatic light (one colour or frequency)
Incident
Ray
Medium 1
Normal
Incident
Ray
Normal
n2 > n1
q1
q1
n2 < n1
q2
q2
Medium 2
Sinq1 v1

Sinq 2 v2
c
n
v
where v1 and v2 are the
speeds of light in media
1 and 2 respectively
Sinq1 c / n1

Sinq 2 c / n2
Sinq1 n2

Sinq 2 n1
n1Sinq1  n2 Sinq2
Refraction
n1Sinq1  n2 Sinq2
Law of refraction or Snell’s law
Incident and refracted rays and the normal
are all in the same plane
product nSinq remains constant as
light crosses a boundary from
one medium to another
Example
A laser beam is directed upwards from below
the surface of a lake at an angle of 35º to the
vertical. Determine the angle at which the light
emerges into the air. n1(air) =1.0003 and
n2 (water) =1.33
Snell’s law
Normal
n1Sinq1  n2 Sinq2
air
n1
water
n2
q1
1.0003Sinq1  1.33Sin350
35º
1.33Sin35
Sinq1 
1.0003
Sinq1  0.76
0
q1  49.70
If light enters the water at an angle of 49.70,
what is its refraction angle in the water?
Refraction
Monochromatic light (one colour or frequency)
Incident
Ray
Medium 1
Medium 2
Normal
n2 > n1
q1
Incident
Ray
n2 < n1
or
n2 > n1
q2
Sinq1 n2

Sinq 2 n1
n1Sinq1  n2 Sinq2
Normal incidence q1 = 0
therefore q2 = 0.
transmitted ray is not deviated
independent of the materials on either
side of the interface.
Refraction
Real and apparent depth
Ruler partially immersed in water
Apparent position
of ruler end
air
water
ruler
End of ruler
Refraction
Setting sun appears flattened (top to bottom)
because light from lower part of the sun
undergoes greater refraction upon passing
through denser air (higher refractive index)
in lower part of the Earth’s atmosphere.
Refraction
Critical Angle
1
n2 < n1
q1
qc
q2
2
>qc
q2 900
qc is critical angle
as q1 is increased q2 increases
Angle of incident for which refracted ray emerges
tangent to the surface is called the critical angle
in this case q2 = 90o or Sin q2 =1
Sinqc n2

Sinq 2 n1
n2
Sinqc 
n1
Refraction
Total internal reflection
1
n2 < n1
q1
qc
q2
>qc
Ray undergoes
total internal
reflection
q2 900
qc is critical angle
2
incident ray undergoes total internal
reflection at boundary and cannot
q1 > qc pass into the material with the lower
refractive index
when
maximum value of the sine of any angle is 1
n2
Sinqc 
n1
Sinqc  1
total internal reflection occurs at interface only
when n2 < n1
Refraction
Example
Determine the critical angle for both water and
diamond with respect to air.
n2
Sinqc 
n1
water
diamond
n2
1 1.0003
qc  sin
 sin
 490
n1
1.33
1
n2
1 1.0003
qc  sin
 sin
 24.40
n1
2.42
1
Example
What happens to light ray at the glass-air
interface in prism as shown.
Refractive index of glass =1.52
Refractive index of air =1.0003
45º
Critical angle given by
n2
1 1.0003
qc  sin
 sin
 410
n1
1.52
1
Glass prism
(right angled
isosceles triangle)
Total internal reflection
at glass air interface if
incident angle is >410
What happens the beam if the prism is immersed
in water? Refractive index of water =1.33
n2
1 1.33
qc  sin
 sin
 610
n1
1.52
1
45º
qc > 45º
Total internal reflection
at glass-water interface does not occur
Refraction
Total internal reflection
diameter of core 8mm
Applications
Optical fibre
(end on)
Refractive index of core greater than
refractive index of clading
Light coupled into core will travel extremely
long distances along fibre, undergoing
total internal reflection at core-cladding
interface and exit only at the other end.
Fibre optic cables used for telecommunications
and for diagnostic tools in medicine
Example
Light in air is incident on a glass block at an
angle of 350 The sides of the glass block are
parallel. At what angle does the light emerge
into the air from the lower surface of the glass
block? 350
q2 air
glass block has
parallel sides,
glass
therefore
q3 = q2
air
q3
q4
Let n1 = refractive index of air
& n2 = refractive index of glass
Using Snell’s
0
n1Sin35  n2 Sinq2
Law
n2 Sinq3  n1Sinq4
n2 Sinq2  n1Sinq4
n1Sin35  n1Sinq 4
0
Since q3 = q2
Sinq 4  Sin35
0
q 4  350
Light: Electromagnetic wave
Visible spectrum
Infrared
Wavelength
Electromagnetic wave
Transverse wave
Electromagnetic wave
Ultra violet
v f
V: velocity
f: frequency
: wavelength
Polarised Light
Schematic representation
Polariser
Light beam
Light waves
vertically polarised
Light source
polarised light
viewed along
direction of
propagation
Unpolarised light
viewed along
direction of
propagation
Unpolarised
light
Polaroid
filter
Polarised
light
Polarised Light
Schematic representation
Vertical
polariser
Horizontal
polariser
Unpolarised
Incident beam
Vertically polarised
light wave
Unpolarised
light
Polarising
filter
Polarised
light
Polarised Light
Light can become polarised by
•scattering
•Reflection •refraction
Unpolarised incident
light
Polarised reflected
light
?
?
Polarised incident
light
Polarised incident
light
Polarised reflected
light
No reflected
light
Polarised Light
Applications
3D movies
2 cameras, a short distance apart,
photograph original scene
2 slightly different images projected on screen
Each image linearly polarised in
mutually perpendicular direction
3D glasses have perpendicular polarisation axis
Each eye sees a different image associated with
different viewing angle from each camera
Brain perceives the compound image
as having depth or three dimensions.
Polarisation of light : application
Application to dentistry
Early detection of caries
Visual, mechanical probing, x rays???
Demineralised enamel viewed directly with
unpolarised light
No information
Demineralised enamel is polarisation sensitive
Polarised light incident on the dental tissue
shading may be seen, indicating the early
stages of caries at the tooth’s surface
Example
The wavelength of red light from a HeNe laser is 633 nm
but is 474 nm in the aqueous humor inside an eyeball.
Calculate the index of refraction of the aqueous humor and
the speed and frequency of the light in the substance.
c  f 0
c f 0
n 
v
f
0  633n
0 633nm

 1.34
Refractive index n 
 474nm
Speed in aqueous humor
c 3x108 ms 1
v 
 2.25 x108 ms 1
n
1.34
Frequency of the light in aqueous humor
2.25 x108 ms 1
14
f  

4.75
x
10
Hz
9

474 x10 m
v
Frequency of the light in air
3.00 x108 ms 1
14
f0 


4.75
x
10
Hz
9
0
633x10 m
c