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Spherical Thin Lenses
Optics for Residents
Amy Nau O.D., F.A.A.O
Spherical LensesFormulas
Thin lens formula 1/do+1/di=1/f F=1/f2
Linear Mag Image size/Object size= di/do
Prentice’s Rule P=F (diopters) xd (cm)
Thin Spherical Lenses
The shape of the refracting surface determines the type of lens and power
Object space Image space
A1 A2 A1 A2
C1C2 C2 C1
Object space Image space
Lens is thin if the thickness is small enough not to influence the powerAll refraction is thus considered to occur in one plane between the two surfaces
A1 front surface vertex; A2 back surface vertex
Thin Lens Power
F1=nlens-nobject/r and F2=nimage-nlens/r The total dioptric power of the lens (F) is
F1+F2
By substitution, another way to write this is : F=(n’-n)/(1/r1-1/r2) This is called the Lens Makers Formula, and it assumes that the surrounding media is air.
Thin Lens Problem
A plastic biconcave lens (n=1.49) has a surface radii of 40cm and 20 cm. Calculate the surface and total lens powers.
Solve for F1 F1=nlens-nobject/r
Solve for F2 F2=nimage-nlens/r
Solve for total power F=F1+F2
Image Object Relationships: Position Assumptions:
surrounding media is air Imaging occurs through single plane
Formulas The Gaussian Imaging formula is used to calculate image-
object power relationship L’=L+F
F=total power of the lens
Incident vergence (L)=no/l=ns/l=1.0/l (in air)
Emergent vergence (L’)=ni/l’=ns/l’=1.0/l’(in air) Where l=object distance and l’=image distance
Object Image Relationships
The types of images and objects as well as sign convention are the same as for single refracting surfaces
REAL OBJECTSVIRTUAL IMAGES
VIRTUAL OBJECTSREAL IMAGES
negative positive
Problem
Virtual image is formed 25cm from a thin lens. If the object is real and positioned 50 cm from the lens, what is the lens power?
L’=L+F
Image-Object RelationshipsFocal Points The primary and secondary focal points are
calculated the same was as that for single curved surfaces. Assume the surrounding media is air (n=1)
F=-n/f=n’/f’ becomes F=-1.0/f=1.0/f’ The equation shows that for thin lenses f=f’,
but they are opposite signs
Focal Points
Recall that an infinite axial object forms an image at f’
Recall that an object placed at f, forms an image at infinity
Secondary focal point – convex surface
n n’
f’ secondary focal length (+)
F’
Secondary focal point
Rays converge towards secondary focal point
Object at infinity
Secondary focal point –concave surface
n n’
f’ secondary focal length (-)
F’
Secondary focal point
Rays diverge as if they came from secondary focal point
Object at infinity
Primary focal point
The object position that yields the image at infinity
Emergent vergence (L’) is zero
Primary focal point- positive surface
F
C F’ (infinity)
f (primary focal length) f’ secondary focal length
n n’
Primary focal point – negative surface
cF’
f’’ secondary focal length
F
f primary focal length
Infinite image rays
n n’
Problem
What are the primary and secondary focal lengths for a lens (n=1.49) with a power of -12D in air?
F=-n/f F=n’/f’
Image-Object RelationshipsSize and Orientation This is LATERAL MAGNIFICATION and the
ratio is image size/object size - invereted/+erect
These equations are the basis for the lensometer
LM=h’/h=nl’/n’l=nsl’/nsl=l’/l LM=-f/x=-x’/f’ (x= extrafocal dist)
Where x=dist from primary focal point to object and x’=dist from seconary focal point to image
Just as for single refracting surfaces, set up similar triangles!!!!!!!!!!!!!!!!!!!!!!
Lateral Magnificationxx’=ff’ xx’=-f’f’ xx’=-(f)2
h
h’
c
c
c
c
l l’
f f’x x’
Problem LM
A real image moves to 5cm away from a lens when an object is moved from infinity to a position 40cm in front of the lens. What is the power of the lens?
xx’=-(f)2
Distant Object Size When an object is at an infinite distance, the LM cannot be
determined. To solve for the object size use the chief ray
h’F
F’
f f’
h’
tanh’/f=-h’/f’
h’=-f’tan
Who cares?
The angular subtense of an object as it falls on the retina is used in vision to describe the size of a distant object. A 20/60 letter on an eye chart subtends a 15
minute arc at 20 feet (1 degree= 60 minutes).
Prismatic Effect of Thin Lenses The amount of bending after refraction is a
function of the distance from the axis Closer to the axis bends less This is considered the prismatic effect of a lens The change in direction is the deviation Unlike a prism which has ONE deviation for any
incident ray, a lens has an infinite number of deviations
Prismatic Effect of Thin Lenses
F’
f’
Prismatic Effect of Thin Lenses
THERE IS NO PRISMATIC EFFECT AT THE OC OF A LENSHowever, if the visual axis is not aligjed with the OC there is prism
Prismatic Effect of Thin Lenses
The image always shiftsTowards the APEX.
Positive lenses displaceImages opposite the directionOf decentration
Prismatic Effect of Thin Lenses
Lens decentered downward = base up. In general, negative lenses displace imagesIn the SAME direction as decentration.
Prismatic Effects
The magnitude of prismatic deviation in prism diopters is expressed in terms of displacement (cm) and the secondary focal length (m)
=ycm/f’m = ycmFD
Prism Problem
Patient with PD of 60mm is given a frame with a 70mm PD. If the pt’s Rx is -3.00D OU, what is the direction and amount of induced prism?
Effective Power
Vergence (power) required for a lens at a new position to have the same effect on the incident rays.
Becomes important if the distance between a lens and an image plane (retina) is changed.
MOVING A LENS AWAY FROM THE IMAGE PLANE INCREASES POSITIVE POWER (DECREASES NEG. POWER)
Effective Power
F’
p1 screenp2
d
f’
f’xd
F’d
Fx=1/f’x=1/f’-dFx=F/1-dF
This is the situation whenYou move a lens closer (i.e. glasses going to contactLenses)
Effective Powerp1p2
F’
d d
f’
f’x
When the lens is movedAWAY, then less powerIs required to put the imageIn the same place
CLS to glasses
OR need more bifocal power
If lens is moved to the rightd is positive, if lens is Moved right to left, d is neg
Which soldier is taller?
Is it moving and shimmering?
Buy a poster!
Look at this illusion for a while and it willappear to be shimmering and moving.
Also: Follow the outermost groove and watch itchange from a groove to a hump as you go around the wheel.
Problem Sets
When rays from the sun pass through a convex lens, it makes a bright point image 0.7m behind the lens on the ground.
What is the focal length of the lens? What is the nature of the image? What is the magnification? What is the power of the lens?
Problem Sets
A light bulb is placed 300 cm from a convex lens with a focal length of 50 cm.
Where is the image located? What is the nature of the image? What is the magnification of the image?
Problem Sets
A light bulb is placed 300 cm from a convex lens with a focal length of 500 cm.
Where is the image? What is the nature of the image? What is the magnification of the image?
Problem Sets
A slide is placed 50mm from a projector lens and no image is formed. Why?
An object of height 7cm is placed 25 cm in front of a thin converging lens with a focal length of 35 cm. What is the height, location and nature of the image?
Problem Sets
Now the object is moved to 90 cm. What is the new image distance, height, and nature of the image?
Problem Sets
A patient comes in having trouble with their new glasses which you prescribed. Their acuity is 20/20 in each eye and you determine that the refraction (-5.50 OD and -2.50+1.00X090 OS) is accurate. However, the patient reports that they glasses make them uncomfortable. You determine that the lenses have not been centered properly. They are decentered 2mm out on the OD and 2mm up on the OS. (how would you do that??). What is the induced prismatic effect?
Problem Sets
A patient has a 10D exophoria at near. The patient needs a +2.00D reading add. What amount of prism would you prescribe? How much decentration needs to be added? In what direction is the base of the prism?
Problem Sets
A patient with ARMD is having trouble reading. You determine that she needs a +4.50 add on top of her distance Rx of +1.00 OU to see the newspaper. How far away does she need to hold the paper to see it clearly? What will be the magnification of an object 10 cm tall held 5 cm from the lens?
Problem Sets
An object 0.08 m high is placed 0.2m from a (+) lens. If the distance of the image from the lens is .40m, what is the height of the image?
Diverging lenses form what kind of images?
Problem Sets
An object is placed 0.2m from a lens with a focal length of 1.0m. How far from the lens will the image be formed?