Lecture 3.3
Geometrical Optics
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Reflection
Refraction
Critical angle
Total internal reflection
Wave nature of light
Interference
Diffraction
Polarisation
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 ≈ 50µm
Geometrical Optics
 ray optics
 beam of light
Light can travel through
•empty space,
• air,
•glass,
• water,
Each referred to
•cornea,
as a medium
•eye lens etc.
Light rays will travel in a straight line if they
remain in the same medium
Specular reflection
At the boundary between two media,
the light ray can change direction by
reflection or refraction
Normal
Reflected Ray
Incident Ray
θi θr
Metal surface
Laws of reflection
1. angle of incidence(θi) = angle of reflection(θr)
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
Diffuse 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.
Refraction
At the surface of a 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
Normal
Medium 1 (Air)
Medium 2 (glass)
θ1 θ1
Reflected Ray
θ2
Refracted Ray
Light ray changes direction going from one
medium to another.
Which way does it bend and by how much?
Is θ2<θ1 or is θ2>θ1
Answer
Depends on the speed of light in both media
Refraction
Refraction
Analogy:
Rolling barrel
Smooth concrete
grass
Index of 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 of refractive index 1.52
Calculate the speed of light in glass
c
v=
n
3 ×108 ms −1
−1
8
=
= 1.97 ×10 ms
v
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
n2 > n1
θ1
Normal
n2 < n1
θ1
θ2
θ2
Medium 2
Sinθ1 v1
=
Sinθ 2 v2
Sinθ1 c / n1
=
Sinθ 2 c / n2
where v1 and v2 are the
speeds of light in media
1 and 2 respectively
Sinθ1 n2
=
Sinθ 2 n1
n1Sinθ1 = n2 Sinθ 2
Law of refraction or Snell’s law
(can be derived from Maxwell’s equations)
Refraction
Monochromatic light (one colour or frequency)
Incident
Ray
Medium 1
Medium 2
Normal
n2 > n1
θ1
Incident
Ray
n2 < n1
or
n2 > n1
θ2
Sinθ1 n2
=
Sinθ 2 n1
n1Sinθ1 = n2 Sinθ 2
Normal incidence θ1 = 0
therefore θ2 = 0.
transmitted ray is not deviated
independent of the materials on either
side of the interface.
Refraction
n1Sinθ1 = n2 Sinθ 2
Law of refraction or Snell’s law
Incident and refracted rays and the normal
are all in the same plane
Index of refraction changes for different
wavelength. This is called dispersion.
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
n1Sinθ1 = n2 Sinθ 2
air
n1
water
n2
θ1
1.0003Sinθ1 = 1.33Sin350
35º
1.33Sin35
Sinθ1 =
1.0003
Sinθ1 = 0.76
θ1 = 49.7
0
0
If light enters the water at an angle of 49.70,
what is its refraction angle in the water?
Refraction
Real and apparent depth
Ruler partially
immersed in water
Apparent position
of ruler end
n1 air
water
ruler
n2
End of ruler
d= real depth
d’= apparent depth
 n1 
d =  d
 n2 
n1
1
d’
d
n2
Refraction
Real and apparent depth
Snell’s law:
Geometry:
𝒏𝟏 𝒔𝒔𝒔(𝜷)
=
𝒏𝟐 𝒔𝒔𝒔(𝜶)
𝑳
𝐭𝐭𝐭(𝜶) =
𝒅𝒅
For small angles:
𝑳
𝐭𝐭𝐭(𝜷) =
𝒅
𝐭𝐭𝐭(𝜶) ≈ 𝒔𝒔𝒔(𝜶)
𝒕𝒕𝒕 𝜷
𝒅𝒅
𝒏𝟏 𝒔𝒔𝒔(𝜷)
=
≈
=
𝒅
𝒏𝟐 𝒔𝒔𝒔(𝜶)
𝒕𝒕𝒕 𝜶
n1
n2
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
θ1
θc
θ2
θ2 =900
θc is critical angle
2
as θ1 is increased θ2 increases
Angle of incident for which refracted ray emerges
tangent to the surface is called the critical angle
in this case θ2 = 90o or Sin θ2 =1
Sinθ c n2
=
Sinθ 2 n1
n2
Sinθ c =
n1
Refraction
Total internal reflection
1
n2 < n1
θ1
θc
θ2
2
>θc
Ray undergoes
total internal
reflection
θ2 =900
θc is critical angle
incident ray undergoes total internal
reflection at boundary and cannot
θ1 > θ c pass into the material with the lower
refractive index
when
maximum value of the sine of any angle is one
Sinθ c n2
= = Sinθ c
Sinθ 2 n1
total internal reflection occurs
at interface when n2 <n1
Refraction
Example
Determine the critical angle for water and
diamond with respect to air.
water
n2
−1 1.0003
=
θ c sin
= sin = 490
1.33
n1
−1
n2
−1 1.0003
diamond=
= sin = 24.40
θ c sin
2.42
n1
−1
Diamond Ring
Diamond has large refractive index and
consequently small critical angle
Light enters from any direction
(no TIR on entering)
Large number of facets:
TIR from facets on back surface:
exits from many front facets all of which
receive some light at angles <24.40
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
θ c sin
=
= sin = 410
1.52
n1
−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
=
θ c sin
= sin= 610
1.52
n1
−1
45º
θc > 45º
Total internal reflection
at glass-water interface does not occur
Refraction
Rainbow formation is due to a combination of
refraction and reflection. Incoming white light
(broad spectrum of wavelength) is separated
into its component colours.
Note: Colour separation due to dispersion
(refractive index is different for different
wavelengths).
Refraction
Total internal reflection
diameter of core 8µm
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
θ2 air
glass block has
parallel sides,
glass
therefore
θ3 = θ2
air
θ3
θ4
Let n1 = refractive index of air
& n2 = refractive index of glass
Using Snell’s Law
n1Sin35 = n2 Sinθ 2 n2 Sinθ3 = n1Sinθ 4
0
n2 Sinθ 2 = n2 Sinθ3
θ 2 = θ3
n1Sin35 = n1Sinθ 4
0
θ 4 = 35
0
Example
Light in air is incident on a glass block at an angle of θ1 and
is refracted at an angle θ2.The sides of the block are parallel
and a distance T apart. What is the displacement between
the entry and exit rays in term of T, θ1 and θ2?
θ1−θ2
θ1
air
T glass
l
air
d
=
d l sin(θ1 − θ 2 )
T
cos θ 2 =
l
d
θ1
θ2
d
d
sin(θ1 − θ 2 ) =
l
T
l=
cos θ 2
sin(θ1 − θ 2 )
T
T
=
d
θ
θ
−
θ
θ
(sin
cos
cos
sin
)
cos θ 2
1
2
1
2
cos θ 2
T (sin θ1 − cos θ1 tan θ 2 )
T = 5 cm, n =1.52, θ1 =
n1
θ 2 = sin ( Sin350 )
n2
350
n1
Sinθ 2 = Sinθ1
n2
−1
θ 2 = 220
=
d 5cm(sin 350 − cos 350 tan 220 )
d = 1.2cm
Example
A HeNe laser has a wavelength of 633 nm in air
(assume n=1) and 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
n= =
v
f λ0
fλ
λ0 = 633n
λ0 633nm
n =
= 1.34
Refractive index =
λ 474nm
Speed in aqueous humor
c 3 x108 ms −1
v= =
= 2.25 x108 ms −1
1.34
n
Frequency of the light inaqueous humor
v 2.25 x108 ms −1
14
f= =
4.75
x
10
Hz
=
−9
474 x10 m
λ
Frequency of the light in air
c 3.00 x108 ms −1
14
=
f0 =
=
4.75
x
10
Hz
−9
λ0
633 x10 m
Light: Electromagnetic wave
Visible spectrum
Infrared
Wavelength
Electromagnetic wave
Transverse wave
Electromagnetic wave
Ultra violet
v= fλ
V: velocity
f: frequency
L: wavelength
Electromagnetic Waves
Geometrical Optics
 Light represented by rays
 Traveling in straight lines
Not strictly correct
Light has a wave nature
Diffraction
Light waves deviate from straight path and
“spread out” as they pass by obstacle or through
an opening.
Width > λ
Width ≤ λ
Waves
All waves subject to diffraction
e.g. light, sound, water etc.
Electromagnetic Waves
Wave nature of light
Destructive
interference
First proof---Thomas
Young 1801
Beams obtained by passing
sunlight through two closely
spaced narrow slits
Superimposed 2 light beams
and saw constructive and
destructive interference
Interference pattern
(bright & dark regions
Slit widths ≈ λ
r1
laser
r2
r2 = r1+ nλ constructive interference (bright)
r2 = r1+ (n+½)λ destructive interference (dark)
where n is an integer
Electromagnetic Waves
Diffraction and Resolution
Microscopes, telescopes, cameras, eyes
Circular apertures (diameter d)
s1
s1
θ
s2
θ
s2
For circular aperture
1.22λ
θ =
Minimum angle of resolution min
d
Rayleigh criterion
Electromagnetic Waves
Two point like sources viewed through
A circular aperture of size d
For circular aperture
1.22λ
θ =
Minimum angle of resolution min
d
Rayleigh criterion
Electromagnetic Waves
Diffraction
Radio Reception in mountainous area
FM (88-104MHz)
λ=3m
AM (525-1610KHz)
λ = 200 m
Longer wavelength waves
diffracted around and between mountains
-better reception
X-ray diffraction
X-rays
λ 0.1nm
Atomic spacing in crystalline solids
X-ray diffraction used to investigate internal
structure of important biological molecules
- example, proteins and DNA
Polarised Light
Schematic representation
Polariser
Light beam
Light waves
vertically polarised
Light source
polarised light
Unpolarised light
viewed along
viewed along
direction of
direction of
propagation
propagation
Polarisation – orientation of transverse wave
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