L6- More complex refractive optical systems

Question 1

What does it mean when an object is located at infinity (in terms of its power and vergence)?

Answer 1

means that object vergence is = 0D

this is the vergence just before refraction at the first surface

Question 2

What does it mean when an object is located at infinity (in terms of its power and vergence)?

Answer 2

means that object vergence is = 0D

this is the vergence just before refraction at the first surface

Question 3

How does light go through a thick lens when object is located at infinity ?

Answer 3

• parallel rays of light from object at infinity

- beam will converge to form an image at o’ at second focal point

Question 4

What are the 3 stages?

Answer 4

1. Refraction at first spherical surface
2. Displacememnt divergence due to the thickness of lens
3. refraction by second spherical surface

Question 5

Why is L2 not the same as L1’?

Answer 5

• since beam had to travel a certain distance through the lens
• theres a change of vergence happening through displacement
• can only use fundamental paraxial equation with refractions so not useful as toolbox can only deal with refraction and here there is displacement with 2 refractions.

Question 6

What is the solution to attruibuting the power to a single plane then?

Answer 6

• find a power and location of an imaginary thin lens that behave in the same way as the thick lens does.
• To find this plane(imaginary thin lens), we extend the rays as shown on the plane where they intersect which is identified as dashed line (H’) AS IT IS imaginary.
• if the refraction has to happen on one plane
• as it contains all the [power (all the change of vergence happens at this single plane)

Question 7

What needs to happen in order for this imaginary thin lens (plane) (H’) to be equivalent to the thick lens?

Answer 7

-the image it produces must be identical to that of the thick lens.
-therefore require same object rays also require same image rays
-

Question 8

What is the imaginary plane?

Answer 8

H’

• It is equvialne to a thin lens placed at the location with an appropriate power
• belongs to image space

Question 9

What is an image space?

Answer 9

is the region of space to the right of the second principle plain H’.

Question 10

What is the case now when there is an on axis object placed at the first focal point?

Answer 10

• Must give an image formed at infinity
• Therefore rays that diverge from this object point after going through optical system must be parallell to each other
• otherwise the vergence leaving the thick lens L2’ must equal 0
• Have 2 refractions and one displacement through the lens

Question 11

How do we attribute the power of entire lens (Thick lens) to a single plane in this case 2?

Answer 11

• By finding out where object and image rays intersect
• can draw an imaginary plane H if we have the change of vergence for the entire system happens again at this plane it become similar to a thin lens.
• This imaginary plane located at H behave equivalent to a thin lens, located at h with the appropriate power.

Question 12

What is fe?

Answer 12

the focal length equivalent of thin lens

Question 13

What is object space?

Answer 13

region of space to the left of the first principle plane H

Question 14

H and H’ not in the same place, so how can a thin lens be at the same time in a different place?

Answer 14

-change of vergence is taking place at different planes
-depending on whether the object or image rays are parallel
-

Question 15

What is the purpose of these cases?

Answer 15

to find an imaginary thin lens that replicates the behaviour of the thick lens
- SO the location depends on the equivalent thin lens (in case 2)

Question 16

What about these cases in real life?

Answer 16

• cannot replace this complex system with a thin lens
• can’t be in 2 places at the same time
• but on paper you can.

Question 17

What can be in a simple thin lens?

Answer 17

ray can be either in object or image space - since there is no thickness which separates the 2 regions

Question 18

What is the case for the equivalent thin lens?

Answer 18

–There is a region that is neither part of the object or image space called null space

Question 19

What is a equivalent thin lens? And why is this useful?

Answer 19

• any optical system that can be simplified to an equivalent thin lens.
• Now can use the toolbox for a single refracting surface which allows us to examine an optical system to find location, size and type fo image.

Question 20

What is fE and fE’?

Answer 20

first and second equivalent focal length

-represnt distances from a plane that contains all the power of the optical system

Question 21

What is fV and fV’?

Answer 21

front and back focal lengths
distances from the vertex or surface of optical system to focal points
-these focal lengths does not represent the focusing power of the optical system as a whole

Question 22

What are the true focal lengths of an optical system here?

Answer 22

• are the equivalent focal lengths fE and fE’

Question 23

what is the power of the optical system?

Answer 23

-equivalent power FE

Question 24

What are the vertex focal lengths relevant ?

Answer 24

because they are the length we can actually measure respect to a real surface within an optical system
-the principle planes are imaginary surfaces.

Question 25

What is an achromatic doublet lens made up of?

Answer 25

two different types of glass segmented or stuck together with different refractive indices and 3 optical surfaces
-special lens better than a standard lens.

Question 26

What can an image forming system do?

Answer 26

• represented in theory by an equivalent thin lens
• thin lens behaves in the same way as the more complex optical system
• power equal to the equivalent focal powered is located at principle planes H and H’.

Question 27

What do we consider the rays in image space?

Answer 27

the same thin lens is located at H’

Question 28

What do we consider the rays in object space?

Answer 28

thin lens is located at H.

Question 29

What are the cardinal point?

Answer 29

3 pairs of points that are fully characteristic of an optical system.

• A and A’
• P and P’
• N and N’

Question 30

What is A and A’?

Answer 30

A- allows to measure object space parameters and it is un-dashed from the first optical system
A’- allows to measure image space parameters from the last surface

Question 31

What happens to image rays that are parallel to each other and the optical axis?

Answer 31

result in object rays going through the first focal point

Question 32

What happens to object rays that are parallel to each other and the optical axis?

Answer 32

result in image rays going through the second focal point.

Question 33

What do these focal points in an achromatic doublet lens do?

Answer 33

mark special points on the optical axis such that if an object is placed at one side or other side of first focal point, the type and location of images change abruptly.
-these distances and focal length are linked to the power of optical system

Question 34

What are P and P’?

Answer 34

principle points

• defined with respect to the principal planes H and H’
• They are the first and second principle point where optical axis intersects the first and second principle planes H and H’.

Question 35

What are N and N’?

Answer 35

• Consider undeviated ray - does not change direction
• a ray going through the nodal points is the undeviated ray since it makes the same angle with the optical axis before and after going through the optical system.

Question 36

What happens if object space and image space are in the same optical medium i.e refractive index?

Answer 36

nodal points coincide with the corresponding principle points
-when achromatic lens in air , P and N and P’ and N’ are the same point

Question 37

What happens if object and image space are in different refractive indices?

Answer 37

• The nodal points are no longer at the same location as the principle points
• Have shifted towards along the axis towards the medium with a higher refractive index