ECE 5616Curtis

Ray Optics

• Thin lens• imaging condition

• Graphical Ray Tracing• Real and Virtual Images• Magnification• Scheimpflug condition

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What is a lens?Paraxial Approximation

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Power of a lens Physical Meaning

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Question

Two thin lenses with F1=10mm and F2=10mm are in contact – what is the total focal length ?

ϕ =1/f = 1/10 + 1/10 = 1/5

F = 5mm

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Simple example

F=100mm

t=zo=150 What is t’= z1?

'111ttf

+−

=

1/100 = 1/150 + 1/t’

t’ = 300mm

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General Types of Lens+, - -, -∞, -

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Graphical Ray TracingPositive Lens

1. A ray through the center of the lens is undeviated2. An incident ray parallel to the optic axis goes through the back focal point3. An incident ray through the front focal point emerges parallel to the optic axis 4. Two rays that are parallel in front of the lens intersect at the back focal plane.5. Corollary: two rays that intersect at the front focal plane emerge parallel.

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Example – 2 positive lens

F F F F

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Graphical Ray TracingNegative Lens

1. A ray through the center of the lens is undeviated2. An incident ray parallel to the optic axis appears to emerge from the front focal point3. An incident ray directed towards the back focal point emerges parallel to the optic axis.

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Example

F F F

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Real and Virtual Images and Objects

If you can’t see the light by placing a screen at the planewhere it is in focus, then it is virtual.Equivalently, an image is virtual if you need another lens(e.g. an eye) to make the image real.Equivalently, real (virtual) objects are to the left (right) ofthe surface and real (virtual) images are to the right (left).

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Ray tracing 2 lenses2 Positive lenses F1,F2 intermediate image

f1

f1 f2 f2

Can even use the rays to design system by extending distances toget desired magnification in first lens

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Mirror

PACP

SASC

=

C R

S

P

so

θi θr

si

θi=θr and is bisected by CA

A

SC = so + R, R<0CP = -(si + R)SA ~ soPA ~ si

Ri

i

o

o

sRs

sRs +

−=+

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Mirror Formula

C CF F

1/so + 1/si = -2/R

R > 0 R < 0

-> F = -R/2

R

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Ray Tracing MirrorsReal Images

Mirror System

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Ray Tracing MirrorsVirtual Images

-t’-t

-t

-t’

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Thin Lens Equations

(Focal length of lens is constant)

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Thin Lens Equations

Note: Smith usesz->xt->z

Note: lens in same index both sidesotherwise

y/-y’ = -z/f

-y’

zz’ = -nn’F2

y/-y’ = f/z’

y

M = -z’/fM = f/zt’ = f(1-M)t = f(1/M-1)

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Newton’s Equation ExampleGiven image is 5mm high, F=10mm and image is located 40mm left of focal point of lens (50mm from lens): What is the image position, size and magnification ?

x’ = -f2/xx’ = -102/40 = 2.5mm to the right of right focal point

40 10

M= f/x = 10/-40 = -0.25

h’= Mh = (-0.25)(5mm) = -1.25mm

Ml = M2 = (-0.25)2 = 0.0625

hh’

x’

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Angular Magnification

Remember the radiometric unit L = radiance (or photometric “brightness”) in units of [W/(sr m2)]? We just found that as size of an object goes up, it’s angular extent decreases by the same amount. Brightness is conserved.

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Another ExampleObject is 5mm high, located 2mm to right of front focal plane oflens F=10mm. Where is the image located, what is the image height, magnification (x,l,θ) of system ?

First lens ray trace to see what to expect…

ff

Magnified (positive) virtual image at a negative distance !

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Another ExampleObject is 5mm high, located 2mm to right of front focal plane oflens F=10mm. Where is the image located, what is the image height, magnification (x,l,θ) of system ?

Now lets do the math…

ff

x’ = -f2/xx’ = -102/2 = - 50mm to the right of right focal point

M= f/x = 10/2 = 5

h’= Mh = (5)(5mm) = 25mm

Ml = M2 = (5)2 = 25, if image is 0.1mm thick it is ~2.5mm thick

Mθ = 1/M= 1/5

-50mm

Like a magnifying glass…

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Scheimpflug conditionImaging from a tilted plane

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Scheimpflug conditionImaging from a tilted plane

Angles of planes are therefore given by:

tan θ = M tanθ’

θ θ’

Can be used to determine the tilt of the object or image plane

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ThrowAssume you need to deliver an image a fixed distance, T from the object

Two solutions (t and t’ interchange roles) with the minimum throw being T = 4f and t=t’=2f. 1:1 imaging condition same T as 4F system

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Single Thin Lens Summary

See page 30 in Smith

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Throw ExampleWe need a projection lens with a slide to screen distance of

10meters and a real image on the screen that is 1.6m tall from a 2 cm slide ? What is the focal length of the lens required ?

First, real image so the lens must be positive lens…

Second, M = -160/2 = -80 single lens will have negative mag

F = -TM/(M-1)2 = -(10)(-80)/(-81)2 ~ 12.2cm focal length

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Throw Example 2We need a mirror that produces a real image that is 3x larger than

the real object and the image is 1 meter away from the object. What is the power of the mirror ?

First, for real image both the object and image must lie to the left of the mirror meaning that both the distances are negative !!!! Remember the ray diagrams… Otherwise you get virtual images. So the throw is -1 meter !!!

Second, M = -3 single lens/mirror will have negative mag

Now with these definitions we can writeM =-t’/t and T = -t’ – tThus:

t’ = -1.5m, t =0.5mFor mirror1/t’ + 1/t = Power = -2x curvature of mirror x no =1/FSo

Power = 2.66 m-1

Curvature = -1.33m-1

F = 0.3759 m

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Cartesian Oval

Why do we use spherical lens mostly ?

Fermat’s principle

lon1 + + lin2 = son1 +sin2

Then

lon1 + + lin2 = constant

This is the formula for a Cartesian oval …