ECE 5616Curtis

Thick Optics (Guassian)

• Thick lens• 2 lens equations•What are EFL, BFL, FFL, PP•What are focal lengths of lenses given EFL, BFL, d•What are power of lenses given distances and mag•Where are the two lenses given T,mag, and powers

• Example calculations

ECE 5616Curtis

Thick Lens

F is 20mm, image is 10mm to left of FFP, where is image and what is magnification ?

xx’ = -F2

x’ = -F2/x = -(400)/-10 = 40mm left of BFP

M=si/so= 60/-30 = -2

Can use thin lens equations to analyze system from PP

ECE 5616Curtis

Remember: Lens Example

bflfflF

R1 = 40mmR2 =20mmn =1.5d = 5mm

ϕ1= (n-1)/R1 = 0.0125ϕ2= 0.025

1/F = ϕ1 + ϕ2 - d ϕ1ϕ2F = 27.8mm compared to 26.67mm for d=0

bfl = F(1-d ϕ1 ) = 26.1mmffl = F(d ϕ2 -1) = -24.3mm

H2= 27.8-26.1 = 1.7mmH1=27.8-24.3=3.5mm

Given

ECE 5616Curtis

Lens Example

bflfflF

F=30mm biconcave lens (symmetric)n =1.5d = 5mm

ϕ= (n-1)/R1

What is R ?1/F = ϕ1 + ϕ1 - d ϕ1ϕ1F = 30mm, ϕ1 = 0.0174 or 0.3825(not physical)ϕ= (n-1)/R = .0174R= 28.73mm

Given

ECE 5616Curtis

2 Lens ExampleWhat is EFL?

Same result from Matrix approach earlier

Ray trace axial ray

ECE 5616Curtis

2 Lens ExampleWhere are PP and FFD and BFD?

Find Back Focal Distance from 2nd lens 2nd PP (B)

Similarly

From these PP locations of system are known

ECE 5616Curtis

2 Lens ExampleWhat are fa and fb?

Given EFL, BFD, and distance between lens, what are the focal lengths of the system?

Substituting fa into

Solve for fa yields

ECE 5616Curtis

ExampleGiven• EFL of 60mm• BFD of 50mm• d is 15mm

fa = 10(40)/(60-50) = 40mm

fb = -(10)(50)/(60-50-15) = 100mm

FFD = -fab(fb-d)/fb = -60(100-15)/100 = -51mm

15mm50mm

60mm

P2

-51mm

P160mm

ECE 5616Curtis

2 Lens ExampleWhat are the powers of the lenses?

Given (s, s’, d, M) find the power of the lenses.

Write equations for rays through systemuA’ – uA = -hAϕA, refractive equation for lens Ah2=h1+duA’, Transfer equation to find height at lens BuB’-uA’=-h2ϕB, - refractive equation for lens BM=u1/u2’uA = - h1/s – paraxial expression tanθ = θuB’ = -h2/s’

ECE 5616Curtis

2 Lens ExampleWhat are the powers of the lenses?

Given (s, s’, d, M) find the power of the lenses.

Start with MM = uA/uB = h1s’/h2sUse h2=h1+d(-h1/s – h1ϕ1) => h2/h1 = 1-d/s-d ϕ1

∴ )1/('Ad

sd

ssM ϕ−−=

Results in

mdssmdms

A)'( −−

=ϕSimilarly

')'(

dssmsd

B+−

=ϕ

ECE 5616Curtis

2 Lens ExampleWhat are the position of the lenses?

Given (ϕ1, ϕ2, T, and M) find the d, s, and s’.

just addition with sign convention

then use the expression for ϕA and substitute in s’

mdsdsTmdms

mdssmdms

A))(()'( −+−−

=−−

=ϕ

Solve for s results in

AmdmTdmsϕ−−

+−=

)1()1(

dsTs −+='

ECE 5616Curtis

2 Lens ExampleWhat are the position of the lenses?

Given (ϕ1, ϕ2, T, and M) find the d, s, and s’.

Using EFL formula and magnification

Use formula for s’ and s results eventually in equation for d

0)1()(2

2 =−

+++−m

ffmffTdTd BABA

This can have 2, 1 or no real solutions

ECE 5616Curtis

Example: Surgeon’s Glasses• Wants M=2x • object distance 25cm => s = -25cm• Want object and virtual image same place T=0, (wants to see

what he is cutting !) T = s + s’ - d• Wants them compact – so set d = 3cm

-> using track length constraint then s’ = -28cm

mdssmdms

A)'( −−

=ϕ

Use formula for power

')'(

dssmsd

B+−

=ϕ

ϕA = 0.187cm-1FA = 5.36cm

ϕB = -0.298cm-1FB =-3.36cm

3cm

ECE 5616Curtis

Using Thick Lens on Optical Systems

• Replace multiple elements with equivalent lens – Principle Planes, EFL, BFL, FFL, etc

ECE 5616Curtis

Equivalent Lens Example

Given 2 thin lenses f1=-30cm, f2=20cm and d is 10cm.

1/F = 1/f1+1/f2 – d/f1f2 => EFL =30cm

BFL = 30(-30-10)/-30 = 40cm

FFD = -fab(fb-d)/fb = -(30)(20-10)/20 = -15cm

O1H1 = -15+30 = 15cm, or fd/f2 = 300(10)/20) = 15cmO2H2 = 40-30 = 10cm, or -fd/f1 = -300(10)/-30) = 10cm

ECE 5616Curtis

Specify a Telephoto LensGiven: For object at infinity we need a 220mm long (front lens to image) system, with 100mm back working distance, and a EFL of 300 mm and a speed of f/4 with aperture stop at front lens. FOV is 0.1 rad witihout vignetting

Need to find powers of lens since location is mostly given and then need to determine size of first lens.

So d=120mm and B=100mm

Front aperture =300/4 = 75mmBack aperture = 100/4 = 25mm + 2(.1*120)= 49mm

ECE 5616Curtis

Telephoto Lens• Object at infinity then image is

~ at focal plane. Image size will be ~βF for single lens imager. So want F to be large for higher magnification.

• Now look at 2 lens solution separated by F1>d>F1+F2(F2<0) The principle plane will be in front of the first lenses so F can be larger for smaller optical system.

• Can change Feff by changing the spacing d between the lenses. s’ = f(1-m)

ECE 5616Curtis

2 MirrorsGiven: A Gregorian telescope objective with concave primary mirror of radius 200mm, and concave secondary mirror with a radius of 50mm. The separation is 130mm. Find the effective focal length and locate the image.

Remember F=R/2, so Fa = 100mm and Fb= 25.Mirror spacing is negative but so is index so dn=130mm is it is positive in eq.

= 100(25)/(100+25-130) = -500mm

= -500(100-130)/100 = 150mm from second mirror

ECE 5616Curtis

Relay System

Given: image to object distance of 200mm, mag=+0.5, object to lens A distance of 50mm, and a object to second lens distance of 150mm,

What are the power of the element of the image plane.

Image plane = 200-150=50mm

ECE 5616Curtis

Relay System

Given: image to object distance of 200mm, mag= -0.5, object to lens A distance of 50mm, and a object to second lens distance of 150mm,

What are the power of the element of the image plane.

Image plane = 200-150=50mm

ϕA = 0.01 => fa = 100mm

ϕB= 0.025 => fb=40mm

ECE 5616Curtis

Optical InvariantAt image/object plane (special case)

Paraxial Snell’s Law

Triangles

Substitute into M

ECE 5616Curtis

Optical InvariantAt image/object plane (special case)

Invariant – this expression has the same value everywhere in the optical system.

At an object or image plane the invariant is equal to the index times the object/image height times the half convergence/divergence angle of the axial beam

ECE 5616Curtis

Optical invariantaka Lagrange or Helmholtz invariant

is conserved everywhere

At a general surface anywhere in the optical system the invariant is expressed as

The 3D version for throughput is that the product of the object/image area times the solid angle of collection is invariant

Write the paraxial refraction equations for the marginal ray (PMR)and chief or pupil ray (PPR):

With a bit of algebra:

ECE 5616Curtis

Basic Definitions

ECE 5616Curtis

Examples• Given object and image slopes and object height

can find height of image.– uo = .333, ui=-0.04755, h=20mm– M=h’/h => h’= (20)(0.333)/(-0.04755)= -14.0187mm

• Image height for lens with object at infinity– At lens slope of u is zero, up (marginal ray)is half FOV

INV = hn(0) - h’n’u = -y1nupAt image plane, INV =h’n’u’ –(0)n’u

h’ = -upy1/u’ for n=n’F= -y1/u’ soh’ = upF or F tanup for non paraxial case

ECE 5616Curtis

Optical invariantaka Lagrange or Helmholtz invariant

Using the invariant, at the object (or image) of limited field diameter L: y = 0, = edge of field, u = maximum ray angle

Thus we have found the information capacity of the opticalsystem, aka the space-bandwidth product:

Rayleigh Resolution(NA = 0.6λ/Δr)

ECE 5616Curtis

Question

If an object that is 1 cm2 with 1 sr of solid angle is images to 2cm2 area,

What is the solid angle of this image ?