Geometrical OpticsGeometrical Optics

Geometrical Optics

• Optics is usually considered as the study of the behavior of visible light (although all electromagnetic radiation has the same behavior, and follows the same rules).

• The propagation of light can be described in two alternative views:

a) As electromagnetic waves

b) As rays of light

• When the objects with which light interacts are larger than its wavelength, the light travels in straight lines called rays, and its wave nature can be ignored.

• This is the realm of geometrical optics.

Geometrical Optics

Light can be described using geometrical optics, as long as the objects with which it interacts, are much larger than the wavelength of the light.

This can be described using geometrical optics

This requires the use of fullwave optics (Maxwell’s equations)

Propagation of Light

Light propagates in straight lines (rays).

This is valid as long as the light does not change the medium through whichit propagates (air, water, glass, plastic),or finds an obstacle (interface).

The velocity of light in air is cc = 3x108 m/s

The velocity of light in other media may be different from c (less than c).

Reflection and Transmission

Most materials reflect light (partially or totally). For example, metals reflect light (almost totally) because an incident oscillating light beam causes the metal’s nearly free electrons to oscillate, setting up another (reflected) electromagnetic wave.

Opaque materials absorb light (by, say, moving electrons into higher atomic orbitals).

Transparent materials transmit light. These are usually insulators whose electrons are bound to atoms, and which would require more energy to move to higher orbitals than in materials which are opaque.

Geometrical Optics

1

1 = angle of incidence

Surface

Normal to surface

Incident ray

Angles are measured with respect to the normal to the surface

Reflection

The Law of Reflection

Light reflected from a surface stays in the plane formed by the incident ray and the surface normal

and

the angle of reflection equals the angle of incidence (measured to the normal)

1 ’1

1 = ’1

Reflection

1 ’1

1 = ’1

Smooth specular shinyRough diffuse dull

Specular and Diffuse Reflection

Two mirrors are placed at right angles as shown.An incident ray of light makes an angle of 30° with the x axis.Find the angle the outgoing ray makes with the x axis.

Refraction

More generally, when light passes from one transparent medium to another, part is reflected and part is transmitted. The reflected ray obeys 1 = ’

1.

1 ’1

2

Medium 1

Medium 2

Refraction

1 ’1

2

Medium 1

Medium 2

More generally, when light passes from one transparent medium to another, part is reflected and part is transmitted. The reflected ray obeys 1 = ’

1.

The transmitted ray obeys

Snell’s Law of Refraction:

It stays in the plane, and the angles are related by

n1sin1 = n2sin2

Here n is the “index of refraction” of a medium.

Refraction

n index of refractionni = c / vi

vi = velocity of light in medium i

Incident ray

1 ’1

2

Medium 1

Medium 2

Reflected ray

Refracted ray

1 = angle of incidence

’1= angle of reflection

2 = angle of refraction

Law of Reflection1 = ’1

Law of Refractionn1 sin1= n2 sin2

Index of Refraction

The speed of light depends on the medium trough which it travels.

The speed of light in a given medium is determined by the medium’s index of refraction n.

8

index of refraction

velocity of light in medium

3 10

cn

vn

v

mc s

Air, n = 1.000293Glass, 1.45 n 1.66Water, n = 1.33

Refraction

The little shaded triangles have the same hypoteneuse: so 1/sin1= 2/sin2, or

v1/sin1=v2/sin2

1=v1T

2=v2T

1

2

1

2

1

2

Define the index of refraction: n=c/v.Then Snell’s law is: n1sin1 = n2sin2

The period T doesn’t change, but the speed of light can be different. in different materials. Then the wavelengths 1 and 2 are unequal. This also gives rise to refraction.

Example: air-water interface

If you shine a light at an incident angle of 40o onto the surface of a pool 2m deep, where does the beam hit thebottom?

air

water

40

2m

Air: n=1.00 Water: n=1.33

(1.00)sin40 = (1.33)sinsin=sin40/1.33 so =28.9o

Then d/2=tan28.9o which givesd=1.1 m.

d

Example: air-water interface

If you shine a light at an incident angle of 40o onto the surface of a pool 2m deep, where does the beam hit thebottom?

air

water

40

2m

Air: n=1.00 Water: n=1.33

(1.00)sin40 = (1.33)sinsin=sin40/1.33 so =28.9o

Then d/2=tan28.9o which givesd=1.1 m.

d

Example: air-water interface

If you shine a light at an incident angle of 40o onto the surface of a pool 2m deep, where does the beam hit thebottom?

air

water

40

2m

Air: n=1.00 Water: n=1.33

(1.00) sin(40) = (1.33) sinSin= sin(40)/1.33 so = 28.9o

Then d/2 = tan(28.9o) which gives d=1.1 m.

d

Turn this around: if you shine a light from the bottom atthis position it will look like it’s coming from further right.

Some commonrefraction effects

Air-water interface

air

water

1

2

Air: n1 = 1.00 Water: n2 = 1.33

When the light travels from air towater (n1 < n2) the ray is bent towards the normal.

When the light travels from waterto air (n2 > n1) the ray is bent away from the normal.

n1 sin1 = n2 sin2 n1/n2 = sin2 / sin1

This is valid for any pair of materials with n1 < n2

c

2 2

11

1

n2

n1

Some light is refracted and some is reflected

Total internal reflection:no light is refracted

Total Internal Reflection

n2sin = n1sin 1

... sin 1 = sin c = n2 / n1

n1 > n2

Total Internal Reflection

• The critical angle is when 2 = / 2,

which gives c = sin-1(n2/n1).

• At angles bigger than this “critical angle”,

the beam is totally reflected.

Find the critical angle for light traveling from glass (n = 1.5) to:a) Air (n = 1.00)b) Water (n = 1.33)

Example: Fiber Optics

An optical fiber consists of a core with index n1 surrounded by a cladding with index n2, with n1 > n2. Light can be confined by total internal reflection, even if the fiber is bent and twisted.

Example: Fiber Optics

Find the minimum angle of incidence for guiding in the fiber,for n1 = 1.7 and n2 = 1.6

Example: Fiber Optics

sin C = n2 / n1

C = sin-1(n2 / n1)

sin-1(1.6/1.7) 70o.

(Need to graze at < 20o)

Find the minimum angle of incidence for guiding in the fiber,for n1 = 1.7 and n2 = 1.6

Reflection and Transmission at Normal Incidence

Geometrical optics can’t tell how much is reflected and howmuch transmitted at an interface. This can be derived fromMaxwell’s equations. These are described in terms of thereflection and transmission coefficients R and T, which are,respectively, the fraction of incident intensity reflected andtransmitted. For the case of normal incidence, one finds:

Notice that when n1=n2 (so that there is not really anyinterface), R = 0 and T = 1.

I RI

TI

R n nn n

T R n nn n

2 1

2 1

21 2

2 12

1 4,( )