Light and Optics - 2/file/...Huygen’s principle allows us to deduce law of refraction For a large...

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 ''−==