light and optics - 2/file/...huygen’s principle allows us to deduce law of refraction for a large...
TRANSCRIPT
1
Light and Optics - 2Propagation of light
Electromagnetic waves (light) in vacuum and matterReflection and refraction of lightHuygens’ principlePolarisation of light
Geometric opticsPlane and curved mirrorsThin lenses
InterferenceDouble slits
DiffractionSingle slitDouble slits
Luke Wilson (Luke.wilson@... Room E17)
2
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/
3
Huygen’s principle
vt
4
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
5
Huygen’s principle - refraction
ba vv ≠OQ = vat, AB = vbt
From AOQ:
From AOB:So,
bbaa
a
b
b
a
b
a
a
b
a
b
b
a
b
a
bb
aa
nnnn
cc
nn
AOt
AOt
θθθθ
θθ
θ
θ
sinsinsinsin
sinsin
sin
sin
=
=
==
=
=
=
vv
vv
vv
v
v
(1)
(1) →
6
Total internal reflection
Snell’s Law
It is possible for all light to be reflected back from transparent material!
bbaa sinsin θθ nn =
• nb < na
• For θa < θcrit , partial reflection and transmission
• For θa > θcrit , total internal reflection
na
nb
θcrit
θb = 90°
θa
θb
7
bbaa sinsin θθ nn =
Total internal reflection
For θb = 90° , sinθb = 1, so the ‘critical angle’ θcrit is given by:
a
bcrita sinsin
nn
== θθ
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)
°=== − 1.41658.0sin ,658.052.11sin 1
critθ
8
Refraction at a spherical surface
Sign rules – these apply to all systems we will consider !
Object distanceObject on same side of a surface as the incoming light, the object distance s is positive.
Image distanceImage on the same side of a surface as the outgoing light, the image distance s’ is positive.
Curvature of spherical surfaceWhen the centre of curvature is on the same side as outgoing light, the radius of curvature is positive.
9
Refraction at a spherical surfaceSpherical 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
10
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
( ) ( )
bbaa
ba
nn θθ
θβφφαθ
sinsin
2 1
=
+=+=
Snell’s law:
11
Refraction at a spherical surface
( )
( )
Rnn
sn
sn
nnnnnnnn
abba
abba
b
ab
bbaa
−=+
−=+
+=
=
'
φβα
φαθ
θθSmall angle
From (1)
Rh
sh
sh
=== φβα '
Small angle, ignore δ
12
Refraction at a spherical surface - magnification
Small angles:
snsn
yym
syn
syn
nnsy
sy
b
a
ba
bbaa
ba
'''
'
''
−==
−=
=
−==
θθ
θθ
13
Refraction at a spherical surface - exampleA 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.
cm8.8'0.25.0
'0.1
115.1
'
+=−−
=+
−=+
ss
Rnn
sn
sn abba
2.1110.1
8.85.1
''
−=××
−=
−==snsn
yym
b
a(a) (b)
s=11cm s’
R=2.0cm
14
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
We considered refraction at spherical surfaces and found a relationship between object and image distance in terms of refractive indices and radius of curvature
Lateral magnification given by
a
bcritsin
nn
=θ
Rnn
sn
sn abba −
=+'
snsn
yym
b
a ''−==