Chapter 2

Geometrical optics: the underlying principles

Outline

• The law of reflection

• Refraction: Snell’s law

• Total internal reflection

• Optical fibers and optical waveguides

• The geometrical model utilizes three basic principles: • the law of rectilinear or straight-line propagation,

• the law of reflection

• the law of refraction

• The straight-line paths are called light rays

Fig. 2.1

Example 2.1

• To find the height of Washington Monument, a student compared the length of its shadow, 296.3 ft, with the length of the shadow of a 3 ft yardstick, 1.6 ft. What is the height of the Washington Monument?

The Law of Reflection

• A light ray striking a smooth plane surface will be scattered in a perfectly regular fashion

• The law of reflection requires that the angle AO makes with OP, i, is equal to the angle OB makes with OP, r, and both rays as well as the perpendicular lie in the same plane

The law of reflection

• Fig. 2.2

• Reflection is often characterized as specular, spread, or diffuse, according to the surface roughness.

• Fig 2.3

Example 2.2

• A man stands in front of a wall entirely covered by a plane mirror. At what angle of incidence must be view the mirror to see a friend standing 2 m to his left if both he and the friend are 4 m from the mirror?

Reflectivity

• The reflectivity R normal to the surface between two media with indices n1 and n2 is given by

• An important consideration in the design of optical instruments

• Reflection effect can be reduced by using a non-reflecting surface treatment

2

21

2

21

)(

)(

nn

nnR

Example 2.3

• Window glass has an index of refraction of 1.50, and that of air can be taken to be 1.0. What is the reflectivity normal to an air-glass interface?

• This reflectivity explains why windows act like mirrors at night

Example 2.4

• Mirrors for the infrared spectral region are often made with semiconductor materials. The index of refraction of germanium in the infrared region is 4.0, and that of air is 1.0. What is the infrared reflectivity of a germanium surface?

Example 2.5

• A telescope system consists of three lenses, each with an index of 1.5. What fraction of the incident light energy passes through the system? Assume normal incidence on each surface.

• If the light is obliquely incident, the reflectivity falls off rapidly with angle and becomes a small fraction of the normal values.

Refraction; Snell’s Law

• As light strikes any interface between two different media both reflection and refraction occur.

• Fig. 2.5

Refraction; Snell’s Law

constant

sin

sinsin 2211

n

nn

A layered medium with parallel surfaces:

662211 sin...sinsinsin nnnn 00

Example 2.6

• A light ray from a light at the bottom of a swimming pool strikes the surface at an angle 35o with respect to the surface normal. At what angle does it emerge from the water?

Example 2.7

• At what angle of incidence at the water surface will the ray in the air travel off parallel to the water surface?

Optical Reversibility

• If the direction of a ray traveling between two points is reversed, the ray will following the identical path but in the reversed direction.

Figure 2.7

• A ray displaced parallel to itself in passing through a medium with parallel faces

• The parallel displacement, di, is given by

i

i

ii

n

n

td

2

2

1

2

2

sin

sin11sin

Example 2.8

• What is the parallel displacement of a ray incident at 45o from air onto a 1-cm-thick glass plate with an index of 1.523?

Fig. 2.8: A prism

• Prism can be used to change the direction of a light ray

• A penalty: because the index of refraction is dependent on the wavelength, the angle of rotation varies with the wavelength

• The light separates into a spectrum, a process known as dispersion

• So, to change the direction of rays, reflection is better (nondispersive)

• A prism refracts rays twice, it’s an ideal device for dispersing light into its component wavelengths or colors (spectroscopy)

Total Internal Reflection

• Consider a ray arising in a medium of high index and traveling into a medium of lower index

• It’s possible that the ray could strike the surface at an angle so that the ray is refracted parallel to the interface

Figure 2.9: Total internal reflection (TIR)

• The sine of refraction angle is

• In the case of a glass-air interface where ni is 1.5 and nr is 1.0, sinr exceeds 1.0 whenever i exceeds 41.81o

• Critical angle: the angle for which the angle of refraction is 90o

• Rays incident on the interface at an angle greater than this are totally internally reflected, and these rays will remain in the medium of incidence – the higher index medium

• TIR is an important issue in fiber-optic systems

i

r

ir

n

n sinsin

• TIR occurs only when the light is coming from a region of greater refractive index into one of lesser index, never the reverse

• TIR is an advantage in cutting gems like diamond

Example 2.9

• The index of refraction of diamond is 2.42. What is the critical angle for total internal reflection?

Optical Fibers and Optical Waveguides

• One of the most important developments during the past 50 years – the use of light in communication systems

• In such systems, the light is generally directed along cylindrical fibers

• Step-index (SI) fiber: the most common form of optical fibers –consists of a core with index n1 and a cladding surrounding the core with index n2

• n2 < n1, the traveling rays along the fiber striking the surface at an angle greater than the critical angle are totally reflected and will remain in the fiber

Two concerns with SI fiber

1. The oil from the fingers of the person handling the fiber will change the critical angle and allow for light leakage

2. Desirable to have the light traveling as close as possible to the fiber axis and thus to have a critical angle of 80o or more• To have the refractive index of cladding be only slightly less than that of

core (typically 0.02 or so)

• Fractional refractive index change

core

cladcore

n

nn

• Core diameter may be as small as 5 um to as large as 1000 um

• Optical fiber size spec: 50/125, 5/100, 100/140, or 200/230• Core diameter in um / cladding diameter in um

Example 2.10

• What would be the maximum angle for a light ray traveling down and SI fiber system with core index 1.52 and cladding index 1.50? What would be the fractional refractive index change?

Example 2.11

• The critical angle at an air-glass interface was earlier determined to be about 42o. If a coating of oil with n=1.41 is applied to the surface, what is the new critical angle?

One problem

• the injection of light into the rather small aperture of the fiber’s core

• Any ray injected into the fiber at an angle < or = will propagate along the fiber, while those rays at an angle > will escape into the cladding and be lost

Numerical Aperture (NA)

• From Snell’s law:

• NA: a quantity that specifies the acceptance aperture of an optical system (n: refractive index of external medium)

2

core

clad2

core

clad

core

1sin1cos find we

sin

cossin

sinsin

n

n

n

n

nn

cc

c

cr

r

2

clad

2

core

2

core

caldcore 1sinNA nn

n

nnn

Example 2.12

• What is the numerical aperture of the fiber in Example 2.10?

Homework

• Problems: 2, 3, 7, 9, 11, 14, 15

• Due day: 3/22