263A #1

Question 1

Geometric Optics

Answer 1

Approximation whereby light travels along a curve in space, also known as a ray. Assumes that light travels in a line in space.

Question 2

A ray

Answer 2

Travels perpendicular to a wavefront.

Question 3

A beam

Answer 3

is a infinite collection of rays.

Question 4

Apperture

Answer 4

A hole/gap/opening. E.g. pupil.

Question 5

Sign convention

Answer 5

Anticlockwise rotation from reference = positive angle.
Clockwise rotation from reference = negative angle.
Sign of the angle does not depend up on the direction of the ray.

Question 6

Convention in doing problems light ..

Answer 6

travels from left to right. (If looking at a fish. Fish will be to the left as the light comes towards your eyes which is on the right).

Question 7

Normal to the surface is

Answer 7

90 degrees to the plane boundary.

Question 8

Useful rules of thumb

Answer 8

1. The refracted ray always crosses the normal.

2. If n’ > n then i’ i.

Question 9

Critical angle

Answer 9

is the angle of incidence given by the equation
i = sin ^ -1 (n’/n)
At the critical angle the refractive angle is 90 degrees. The ray travels along the boundary.
Only occurs when n’ < n.

Question 10

At the interface between two media of different refractive indices

Answer 10

some of the incident light is reflected.
As the angle of incidence increases the fraction of energy in the reflected beam increases.
When total internal reflection occurs this fraction is 1.

Question 11

Applications of total internal reflection - optical fibres

Answer 11

Glass fibres with diameter as small as 10um. Used for transmitting information (telephone, computer signals and data) rapidly across short and long distances.
If the ray enters too steeply such as the angle of incidence on the inside wall is less than critical angle, light will leak out.

Question 12

Reflection

Answer 12

Say that refractive index changes from positive to negative when it reflects because i’ is clockwise = negative.

Question 13

Rectilinear nature of rays

Answer 13

Relating to moving in straight lines.

Question 14

Beams of light

Answer 14

Light energy given off by a light source. Has typically passed through some sort of optical system to create a diverging cone of energy.
With the exception of the helium neon laser, the other beams are a assembly of beams.
Beam = Cone of light from a point source.

Question 15

Effect of source distance on beam divergence.

Answer 15

Diverging beam is is said to have a negative vergence.
Vergence can be measured the units are Diopter (D)

The greater the distance between the source/focus of the beam and the plane of measurement the less the magnitude of the divergence.
Vergence (D) = 1/distance (m)
Distance is negative (to the left of the plane) –> diverging beam. Negative Vergence.
Converging = positive vergence.

Question 16

1/infinity = 0D.

Answer 16

Collimated beam - not converging and not diverging.
Collimated light is important in optometry because it is how your eye, microscope and telescopes work.
When correcting vision you are trying to make the light that goes into your eye collimated.

An emmetropic eye should be able to focus a collimated beam to the back of the retina.

Question 17

Ideal image formation

Answer 17

Optical system images a point O to point O’.

All the rays in the object beam are concurrent/consistent in the image.

Question 18

One common and important problem is to locate the image plane.

Answer 18

If we can locate O’ then we have located the image plane as everything on the object plane will be imaged onto the image plane if the optical system is ideal.

Question 19

Real Image

Answer 19

The image is formed :

• To the right of the lens system.
• Opposite side of the optical system to the object.
• A projection screen can be placed at the image plane and the image can be seen on the screen.
• Rays are converging after leaving the optical system.

Question 20

Virtual Image

Answer 20

The image is formed :

• To the left of the lens system.
• On the same side of the optical system as the object.
• An projection screen cannot be placed at the image plane.
• Rays are diverging after leaving the system.

Question 21

Paraxial optics

Answer 21

approximation of light where rays travelling close to the optical axis. When this happens the Snell’s law is simplified and leads to simple equations where you can predict where the image is, given the object of vice versa.

Question 22

The optical system is made up of a number of lenses

Answer 22

and so the ray is refracted at each of the lens surface in turn.

Question 23

If we trace real, infinite or exact rays through a surface

Answer 23

get different image points depending on height of ray entering the system. Rays from point O are not concurrent at any point are the refraction thus no clear image of O is formed.

Question 24

Paraxial Ray

Answer 24

Is a ray that travels very close to the optical axis and remains close to the axis.

Question 25

Paraxial approximation

Answer 25

For all paraxial rays, the angles are very small and the ray heights at the surface are very small.
When the angles are small the sine of the angle is very close to the magnitude of the angle when measured in radians.

Question 26

Paraxial refraction equation

Answer 26

n’u’ - nu = -hF

F = C (n’-n).
F is calculated by multiplying the curvature of the lens by the change in refractive index across the curved lens surface.

Question 27

Opening equation

Answer 27

In applying the paraxial refraction equation we prefer to use the known position of the object rather than calculating u.
Problem - the object position doesn’t directly appear in the equation.

Question 28

Closing equation

Answer 28

We also need to find the position of the image (where the ray crosses the axis at distance l’) after refraction.

Question 29

Ray tracing step by step

Answer 29

1. Find the power of the surface (F)
2. Choose a particular ray (any ray providing h = -ul.
3. Refract at a surface (find u’ using paraxial refraction equation).
4. Closing equation. Find l’ using h=-u’l’