Spherical Thin Lenses Optics for Residents Amy Nau O.D., F.A.A.O

Learning Objectives Concept of Vergence How does light travel through thin lenses? What happens if there is more than one surface? What happens when the media changes?

Suggested Reading Guyton pages Loshin p 38, 69, Loshin chapter 9- ray tracing (optional)

Spherical Lenses Formulas Diopter = 1/distance (see Ch1,Loshin) Power U+D=V or L=F+L’ Entering light + power of lens = Vergence leaving lens Thin lens formula 1/d o +1/d i =1/ f F=1/f 2

Thin Lens Power F 1 =n lens -n object /r and F 2 =n image -n lens /r The total dioptric power of the lens (F) is F 1 +F 2 By substitution, another way to write this is : F=(n’-n)/(1/r 1 -1/r 2 ) This is called the Lens Makers Formula, and it assumes that the surrounding media is air.

Thin Spherical Lenses The shape of the refracting surface determines the type of lens and power Object spaceImage space C1 C2 C1 Object space Image space Lens is thin if the thickness is small enough not to influence the power All refraction is thus considered to occur in one plane between the two surfaces

Vergence Converging Diverging Collimated

Diopter Reciprocal of distance (meters) to where the rays intersect on the optic axis -1m -1D

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 or U=D+V (same thing)  Vergence entering + power of lens = vergence leaving  Incident vergence (L)=n o /l=n s /l=1.0/l (in air)  Emergent vergence (L’)=n i /l’=n s /l’=1.0/l’(in air) Where l=object distance and l’=image distance

-33cm=l+50cm= l’ 1/-.33= -3D +5D 2D = 1/.50m N1=air=1N’=air=1 Note that the power will change if the index of thesurrounding media changes

Curve of the surface and power There is a relationship between power and curvature  Important for understanding the cornea!  F=n’-n/r

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. F=n’-n/r Solve for F1  F 1 =n lens -n object /r Solve for F2  F 2 =n image -n lens /r Solve for total power  F=F 1+ F 2

Object Image Relationships The types of images and objects as well as sign convention are the same as for single refracting surfaces REAL OBJECTS VIRTUAL IMAGES VIRTUAL OBJECTS REAL 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

Focal Points An infinite axial object forms an image at f’ An object placed at f, forms an image at infinity

Image-Object Relationships Focal 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

Primary focal point The object position that yields the image at infinity Emergent vergence (L’) is zero

Primary focal point- positive surface F CF’ (infinity) f (primary focal length) f’ secondary focal length nn’

Primary focal point – negative surface c F’ f’’ secondary focal length F f primary focal length Infinite image rays n n’

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

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’

Refraction through parallel sided elements n1 n2 n3 Incident angle Emergent angle Internal angles If n1=n3 then incident angle= emergent angle Opposite internal angles are equal

Lateral displacement The perpendicular distance between an incident and emerging ray after traveling through parallel sided elements d Glass plate n=1.50 air This can affect the apparent Position of objects……..

Why do my feet look weird in the swimming pool…… When an object in one medium is viewed from another medium, the apparent position of the object differs from the actual position because of lateral displacement The refracted angle is different than the incident angle or vice versa depending on the media n/l=n’/l’

Apparent position Looking from air into water object seems closer N’ (air) N (water) This occurs because of the index of refraction!

Looking from water to air object seems further away N’ (water) N (air)

Remember Some general rules: –When light travels from low index to hi index interface, refracted ray bends towards the normal and refracted angle is smaller than incident angle –When light travels from hi to low, ray bends away from normal and the refracted angle is greater than incident angle n n’ n air water cornea air

Apparent position problems Use the formula n/l=n’/l’ Where n= index where real object is located n’= index from which it is viewed l= actual distance of object from interface l’= apparent distance of object from interface (image)

problem A pebble located at the bottom of a fish tank appears to be 22.5cm from the surface. What is the depth of the water? n/l=n’/l’ so 1.33/l=1.0/22.5 and l=(1.33)(22.5)= 30 cm

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’ p1screenp2 d f’ f’x d F’d Fx=1/f’x=1/f’-d Fx=F/1-dF This is the situation when You move a lens closer (i.e. glasses going to contact Lenses)

Effective Power p1 p2 F’ d d f’ f’x When the lens is moved AWAY, then less power Is required to put the image In the same place CLS to glasses OR need more bifocal power If lens is moved to the right d is positive, if lens is Moved right to left, d is neg

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 X090 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 add on top of her distance Rx of 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?