Light and Optics - 2
Propagation of light
Electromagnetic waves (light) in vacuum and matter
Reflection and refraction of light
Huygens’ principle
Polarisation of light
Geometric optics
Plane and curved mirrors
Thin lenses
Interference
Double slits
Diffraction
Single slit
Double slits
Luke Wilson (Luke.wilson@... Room E17)
1
Huygen’s principle
Every point of a wave front is a source
of secondary wavelets.
These spread out in all directions with a
speed equal to the speed of propagation
of the wave.
We can draw a new wave front, some
time later, by constructing a surface
tangent to the secondary wavelets…
Christiaan Huygens, FRS
(1629 –1695)
http://www.phys.uu.nl/~huygens/
2
Huygen’s principle
vt
3
Huygen’s principle - reflection
Wavelets striking the reflecting surface
change direction
Consider wavelet with origin at surface
(A)
AQ = OP = vt
OQA and APO congruent – right angles,
AO common
Therefore θa = θr
Note, angle between wave front and surface is the
same as angle between ray and normal to surface
4
Huygen’s principle - refraction
va ≠ vb
OQ = vat,
AB = vbt
From AOQ:
sin θ a =
From AOB:
So,
va t
AO
vt
sin θb = b
AO
sin θ a va
=
sin θb vb
(1)
nb c vb va
=
=
na c va vb
(1) →
sin θ a nb
=
sin θb na
na sin θ a = nb sin θb
5
Total internal reflection
Snell’s Law
na sin θ a = nb sin θ b
It is possible for all light to be reflected back from transparent material!
• nb < na
θb
• For θa < θcrit , partial
reflection and transmission
• For θa > θcrit , total internal
reflection
nb
θb = 90°
na
θa
θcrit
6
Total internal reflection
na sin θ a = nb sin θ b
For θb = 90° , sinθb = 1, so the ‘critical angle’ θcrit is given by:
sin θ a = sin θ crit
nb
=
na
Total internal reflection occurs if the angle of incidence is
larger than or equal to the critical angle
e.g. from glass (n=1.52) to air (n=1)
sin θ crit
1
=
= 0.658,
1.52
sin −1 0.658 = 41.1°
7
Refraction at a spherical surface
Sign rules – these apply to all systems we will consider !
Object distance
Object on same side of a surface as the incoming light, the
object distance s is positive.
Image distance
Image on the same side of a surface as the outgoing light, the
image distance s’ is positive.
Curvature of spherical surface
When the centre of curvature is on the same side as outgoing
light, the radius of curvature is positive.
8
Refraction at a spherical surface
Spherical interface between 2 materials with different
refractive index
Refraction angles θa and θb measured from surface normal
(nb > na here)
R, the radius of curvature is positive (centre of curvature on
outgoing side)
Object and image distances are both positive
9
Refraction at a spherical surface
For small α, all rays from P intersect at point P’
P’ is the real image of P
Let’s show this:
Exterior angle of triangle = sum of opposite interior angles
θ a = α + φ (1)
Snell’s law:
na sin θ a = nb sin θb
φ = β + θb (2 )
10
Refraction at a spherical surface
Small angle naθ a = nbθb
na
θb = (α + φ )
nb
From (1)
naα + nb β = (nb − na )φ
na nb nb − na
+ =
R
s
s'
Small angle, ignore δ
α=
h
s
β=
h
s'
φ=
h
R
11
Refraction at a spherical surface - magnification
Small angles:
y
θa =
s
naθ a = nbθb
− y'
θb =
s'
na y
n y'
=− b
s
s'
y'
na s '
m= =−
y
nb s
12
Refraction at a spherical surface - example
A small LED is embedded in a plastic rod (n=1.5) and emits light with a small
angular spread along the axis of the rod. The LED is 11cm away from one end of
the rod, which is formed into a hemispherical surface with R = 2.0 cm. Find (a)
the image distance on the axis of the rod and (b) the lateral magnification.
(a)
na nb nb − na
+ =
s
s'
R
1.5 1.0 − 0.5
+
=
11 s ' − 2.0
s ' = +8.8 cm
(b)
y'
na s '
m= =−
y
nb s
1.5 × 8.8
= −1.2
=−
1.0 × 11
R=2.0cm
s=11cm
s’
13
Refraction summary
Huygen’s principle allows us to deduce law of refraction
For a large enough angle of incidence, when light goes from
material with higher refractive index to lower refractive index it is
possible for total internal reflection to occur
The critical angle for this is given by sin θ crit
nb
=
na
We considered refraction at spherical surfaces and found a
relationship between object and image distance in terms of
refractive indices and radius of curvature n
a + nb = nb − na
s
s'
R
na s '
y'
Lateral magnification given by m = = −
y
nb s
14