optics review march 2009 - ophed | the pitt ophthalmology …€¦ · · 2011-05-30as well as...
TRANSCRIPT
Various materials
Crown glass 1.523Flint glass 1.616Plastic polymerized methylmethacrylate 1.495CR-39 plastic 1.498Index 8 glass 1.805Polycarbonate thin-Lite plastic 1.586
Higher index glass makes the spectacles lighter and thinner
Writing prescriptions
• Sphere first • Cylinder next• Axis third• Plus or minus cylinder• To convert, add sphere to the cylinder for
the new sphere, change the sign of the cylinder for the new one and change the axis by 90 degrees
answer
• Draw a power diagram for the cornea• The amount of the cylinder is 2.00• The spherical equivalent is equal to the
sphere plus one-half the cylinder, or• -3.00=sphere + ½(2.00), or• Sphere = -4.00, so • In plus cylinder form: -4.00+2.00 x 170• In minus cylinder: -2.00-2.00 x 80
answer
• Make a power diagram• Draw lines with the degrees first• Then add the powers• The chose one of the powers as your
sphere and then add cylinder at the correct axis to give yourself the other power
• Then start with the other power
answer
• Draw power diagram of the cornea• Draw power diagram of the lens• 2.00 x 90 will make the cornea spherical
by steepening the horizontal meridian• The other 2.00 put in the glasses would
steepen a flat horizontal crystalline lens surface
• The lens correction is +2.00 x 90
answer
• S.E. is -.5+1/2(-3.00)=-2.00• Claculate the S.E. of the answers:• -1.25• -3.25• -2.00• -2.00• -3.25• But, c. is the answer with the correct axis
answer
• S.E. = one• If the new cylinder is two, then• One=1/2(2.00) + sphere, so• The sphere is ZERO, and the Rx is• Plano + 2.00 x 180
VERGENCES
• U + P = V• Object vergence plus power of the lens is
equal to the image vergence• Frame of reference is at the lens• Light travels from left to right• Diverging light to your eye is negative• Converging light to your eye is positive
answer
• ¼ meter (25 cm) means vergence of -4• -4+6=+2• So, image is ½ meter or 50 cm to the right
of the lens
answer
• U+P=V• -2 -10 = -12• This is 1/12 meter to the right, or 8.33 cm
to the left, it’s negative, virtual, and the rays are diverging to the eye
answer
• U+P=V• -5 +10 = +5• The image is located 20 cm to the right of
the lens• Magnification is the image distance
divided by the object distance• It is 20cm/20cm = one!• Thus, same size!
Answer
• Draw it out first• U+P=V• -1-3=-4, so the image is 25 cm to the left
of the first lens• This new object is 50 cm (25+25) to the
left of the second lens.• So, -2-2=-4, and the image is 25 cm to the
left of the second lens, which is in the plane of the first lens!
answers• Zero• Zero• 0+P=2, (one over distance in meters), P=2• One over 1/10th meter, or 10 diopters• Infinity• 25 cm;1/4th meter, minus 4 diopters• Right surface of lens is 10 diopters, so lens is
minus 6• 33cm is 1/3 meter and thus -3 diopters
answer
• -4+2= ?• -2,so the image is 50 cm to the left of the
lens.• The new object is 50 + 50 to the left of the
second lens, so vergence is -1.00, so• -1+3= ?• +2, so the image is 50 cm to the right of
the second lens, it’s inverted and real.
POWER OF CURVED SURFACES
• Examples include the anterior surface of the cornea and a plano convex intraocular lens
formula
• Power=(index n1 – index n2)/radius of curvature
• Power is in diopters• Radius is in meters
• P=(1.336-1.00)/0.008=42 diopters
Change in refractive index of crystalline lens
• If the refractive index of the crystalline lens increases, does the patient become more myopic or hyperopic?
answer
• More myopic, because the lens has more power.
• This is what happens with nuclear sclerosis and a previously emmetropicpatient with has worn just some readers will find that he can throw them away for awhile until his cataract gets more clouded as well as having a higher refractive index
Power change of a lens placed in water
• A plano-convex lens made of crown glass has a power of 5 diopters. The index is 1.5
• What is the power of the lens in water of index 1.33?
answer
First calculate the radius5=(1.5-1)/radius, so, radius is 1/10 meter.Now in water, p=(1.33-1)/(1/10)=3.3 diopters
Reflecting power of the cornea
• Reflecting power is= 1/focal length• Focal length=radius/2• This is = 1/one half the radius• The radius of the cornea is .008• So the reflecting power of the cornea is
one over .004 = -250 diopters! (minus because it is a convex surface)
method• Obtain best V.A. using spheres only• Fog the eye to 20/50 by adding plus sphere• Find the blackest and sharpest line on the dial• Add NEGATIVE cylinder with the axis
PERPENDICULAR to the blackest line until all lines appear equal
• Reduce the plus sphere until maximum visual acuity is obtained
• In the fogged state, the patient is “myopic” and the focal lines are in the vitreous
Tilting an intraocular lens• A 20 diopter intraocular lens tilted 10
degrees about the horizontal axis will require about 0.75 diopters of a plus correcting cylinder with axis at 90 degrees
Several points here….
• First draw a power diagram• The curvature is greater vertically• This is astigmatism with the rule, or the
shape of a football with the laces running horizontally
• The flattest K is 42.75• The 12:30 o’clock suture would have to be
cut
Astigmatism in the different planes
• In a myopic patient the far point is in front of the eye; the power of a minus lens in the glasses is greater than a contact at the cornea because the glasses are CLOSER to the far point. The same is true of amounts of cylinder lenses as well.
• If there is residual corneal astigmatism after cataract surgery and the patient is left with some degree of myopia, then there will be more cylinder in the glasses than is measured at the cornea.
• Just the opposite is true in hyperopia, because the far point of the hyperopic eye is behind the eye.
Method of solution• First draw a power diagram of the cornea• Since the contact lens is a hard lens, why does the over-refraction have
cylinder in it?• It is because of the presence of crystalline lens astigmatism• How much is the corneal cylinder? 2 diopters• How is the corneal cylinder corrected?• It is corrected by a “tear” lens of -2.00 x 90 when the contact is in place.• What is the significance of fitting 0.75 steeper than K?”• It creates a 0.75 “tear” lens• Since the power of the contact is -2.00, its effective power is reduced by the
plus tear lens of 0.75, yielding minus 1.25. Add to that the “tear” lens of -2.00 x 90 and that gives you -1.25-2.00 x 90. This is the total correction provided by the contact lens.
• Now when the contact lens IS worn, an over-refraction was done yielding +1.00 + 1.00 x 90. If the contact lens is NOT worn, then, combine this to the total correction that was provided by the contact.
• That gives –0.25-1.00 x 90.
reasoning• A partially contact-lens corrected myope (using a
negative lens) will be more myopic when the contact is removed, in terms of doing an over-refraction.
• If this negative contact was placed on a hyperopic patient, he would be less hyperopic when it is removed.
• So, this problem makes sense for a myope, especially since the K readings are a bit higher than average.
Further reasoning
• So do you add or subtract the contact correction to the over-refraction to get the final answer?
• By adding the negative, you are actually subtracting from the plus.
• In order words, this patient was a very mild myope that was overcorrected.
• Consider the equation: A –B + C= K where A=eye itself, B is the contact and is a negative value, C is the positive over-refraction and K is a constant.
• Rewrite as A + C = K + B• A is also a constant• If B is deleted, then C must be reduced.• Thus, the contact will be less plus.
Aspects of retinoscopy
• Plano-mirror effect means the light coming out is either divergent, parallel or convergent to a point BEHIND the patient
• When set this way, the light ON THE RETINA always is in “with” motion, regardless of the refractive error
continuing
• The light that the examiner sees is actually coming from, or heading toward, the far point of the eye, even though it seems to be located in the patient’s pupil
• The far point in this -1.75 myope is 1 divided by 1 ¾ = 4/7 meter or 57 cm in front of the eye.
• This is behind the examiner, who is at 50 cm from the patient
• The examiner will see “with” motion
Far point• The far point is the point that is conjugate to the
retina with all accommodation relaxed.• The purpose of retinoscopy is to locate the far
point• A far point between the patient and the examiner
indicates “against” motion• A far point behind the examiner or the patient
indicates “with” motion• There is no motion at neutralization at the
retinoscopist’s eye
On axis
• When the streak is on axis, it is defined and bright and it aligns with the steak outside the pupil. There is no break or skew.
• It is easier to go from “with” to “against”motion by adding minus lenses, rather the other way around.
To neutralize
• Add plus sphere until the first line neutralizes• Add plus cylinder at axis parallel to the second
far line until it is also neutralized• The plus cylinder axis is the position of the
second far line• The amount of astigmatism is the difference
between the two lines• Add -2 diopters to compensate for a working
distance of 50 cm
2 conjugacy relationships
• At neutralization, • The far point of the patient’s eye coincides
with the peephole of the retinoscope• The patient’s pupil and the examiner’s
retina are also conjugate
answer
• The peephole is about 1/20 the size of the blind spot on a tangent screen at the same distance.
diagramNote the width of the intercept at the sleeve height that gives the mostEnhancement. If you can’t enhance, hyperopia is less than one.
Handle is attached 45 degrees2 cylinders of equal power and of opposite sign and at right angles toEach other.
Always check the axis before the power
Adjust the circle of least confusion
• The circle of least confusion is ½ way between the focal lines
• When you add cylinder, you move one focal line and the circle of least confusion therefore also moves
• You must adjust the sphere to compensate for this effect
rationale
• Let A=distance from circle of least confusion to each focal line
• If you add Y, the total distance is 2A+Y and one half that distance is A+Y/2
• So shift Y/2 to keep the circle of least confusion on the retina
• So, when the power of the cylinder is increased, the sphere must be changed by one-half the opposite amount. Ans d
Draw power diagrams
a. +.25 -.50 x 180b. +.50 -1.00 x 180c. -.50 +1.00 x 90d. +.25 -0.75 x 180
Write out the four examples as prescriptions
Look at wavelength characteristics
The retina should straddle the red and green, and this patient is overcorrected and needs less plus lens.
The index of refraction is lower for red light
• You should start in a fogged condition such that the letters on the red side should be blacker to start with
• If the green side is sharper, accommodation is not controlled
• If you are color blind, the test should still work adequately
Astigmatism of oblique incidence
• Tilting a lens increases the power of the lens and it also introduces CYLINDER power axis PARALLEL to the axis of the tilt
• There is an induced sphere and cylinder that are both of the same sign as the lens
• There is a greater amount of induced cylinder as opposed to sphere
What is the effective power through the bifocal?
• For the right eye, the effective power in the bifocal is -4+2=-2
• For the left eye, the effective power in the bifocal is -1+2=+1
• With the tilt, the right eye will be more minus and the left eye more plus, creating more anisometropia and worse vision.
• A is the INcorrect statement and the answer to the question.
tableAge Accommodative Amplitude(+ or – 2 diopters)
8 1412 1316 1220 1124 1028 932 836 740 644 4.548 352 2.556 260 1.564 168 0.5
facts
• At age 40, there are 6 diopters of accommodation
• If normal reading distance is 33 cm, then 3 diopters of accommodation is required
• RULE: To be comfortable, leave one-half the accommodative amplitude in reserve
• So, at age 40 with 33 cm reading distance, one half of the accommodation is left, beyond that the patient will start to get uncomfortable
At what distance can a emmetropicpatient age 48 read comfortably
without glasses?• His or her accommodative amplitude is 3
diopters total, and we we allow the use of 1.5 diopters, leaving 1.5 diopters
• The reciprocal of the diopters ( in meters) gives a reading distance of 1 divided by 1.5=1/(3/2)=2/3meter=67cm.
• That’s pretty far away!• Quite a “trombone effect”• What power reader would bring him to 33 cm?• 1.5 reader
Apply pressure over the punctae to reduce side effects, especially with
atropine in childrenCyclopentolate !% have a good onset and duration
Dilating someone who is going back to work
Use Paramyd, a mix of 0.25% tropicamide and 1% hydroxyamphetamine. It works well and canalso be reversed with Rev Eyes
answer
• A near point of 20 cm translates to 5 diopters
• If he can accommodate 4 diopters, then he must have 1 diopter of myopia
• The far point of a 1 diopter myope is 1 meter or 100 cm in front of the eye
• Answer is c.
answer
• The accommodation that translates from 20 cm is 5 diopters, which includes the use of a 3 diopter lens
• Without the glasses, the accommodation is 5-3=2 diopters.
• This translates to 50 cm as the near point• Answer c.
Some more practice
• Without correction,if the far point is 20 cm in front of the eye, and the near point is 12.5 cm in front of the eye, what is the distance refractive correction and the amplitude of accommodation?
• -5 diopters and 3 diopters• What is the near point with correction?• 33 cm
Amplitude of accommodation
• If the far point of the eye is 25 cm behind the eye and the near point is 50 cm in front of the eye, what is the amplitude of accommodation?
• 4+2=6
Calculate some points• Where is best focus with 2.50 add?
1/2.50=1/(5/2)=2/5=40 cm• Where is the best focus for the 1.25 add?
1/1.25=1/(5/4)=4/5=80cm• Add one diopter to each calculation:
1/3.50=1/(7/2)=2/7meter=28.5 cm 1/2.25=1/(9/4)=4/9meter=44.4cm
• There is clear vision from 28.5 to 40 and again from 44.4 to 80 and also from 100cm to infinity, leaving two small gaps of blurred vision
• Answer c.
Accommodation in uncorrected hyperope
• What is the range of accommodation of an uncorrected 4 D hyperope with an accommodative amplitude of 6 D?
• 4 D of accommodation are needed just to infinity and the other 2 D would bring his near point to 50 cm
• So, the range is from infinity to 50 cm
Draw a diagram or use a formula
• Logically draw it out • Use the formula: Correction=Shift in
meters toward the eye multiplied by the square of the power of the lens
Calculate with the formula
• Correction=0.003 x (+11) squared=0.003 x 121= +0.36 diopters
• So the lens power is +11 + 0.36= +11.36
Using the formula
• Correction = 0.015 x (+10) squared• .015 x 100=1.5 diopter• Subtract this from 10 gives -8.5 diopter
A contact problem with astigmatism
Always convert the prescription to minus cylinder form first to account for the “tear” lens and thendrop the cylinder from the calculation
Explanation
• Light rays from the anterior chamber angle are incident on the tear cornea interface at angles greater than the critical angle and therefore totally internally reflected
Two prism positions
• There are tow orientations for holding prisms: minimum deviation position and Prentice position.
• This is important because the deviation of a light ray striking a prism varies with the angle of incidence
Prisms are made of plastic or glass
• The two materials are calibrated differently• Plastic prisms are calibrated for the
minimum deviation position• Glass prisms are calibrated for the
Prentice position• Use glass prisms with the back surface
perpendicular to the visual axis• Use plastic prisms with the back surface
parallel to the facial plane
Prentice’s Rule
• The prism effect of any lens is equal to the power of the lens multiplied by the distance from its optical center (in cm)
answer
• Draw a diagram• 5 cm over 10 meters is .5 cm over one
meter so the prism effect is 0ne-half prism diopter
• Prism effect = power x 1.5 cm• .5=power x 1.5• Power=.5/1.5=.33 diopters
answers
• Power is 12 prism diopters• The image will move down toward the
apex of the prism (on the retina it moves up)
• A right hyper (think of how you point the apex of a prism in the direction of a deviation to correct it)
If a base out prism is held in front of an orthophoric patient, what
phoria is induced?
• An exophoria
answer
• 5 diopters each way, symmetrical so orient at 45 degrees
• 25+25=50 and square root about 7
facts
• When the target is displaced downward in the lensometer, this means base down prism in the glasses
• Since it’s just in the left lens and assuming it corrects the patient's phoria,it means the patient has a left hyper
• Answer e.
Answer c.
• Small pupils increase depth of field and reduce symptoms
• Uncorrected hyperopes need to use accommodation just to see distance
• No gender or racial differences
Flat top bifocals
• The optical center of the bifocal segment is at or near the top of the segment
• Think of the bifocal as the lower portion of a plus lens cut at or near the center and pasted onto the distance correction
• As you look down into the bifocal there is little or no image jump
Round top bifocals
• The optical center of the segment is far below the top of the segment
• They are brought to a feather edge on the surface of the lens
• They decrease image displacement on plus lenses
Image jump and Displacement
• Both effects are undesirable• Image displacement is usually more of a
problem and should be minimized• Image jump is usually more tolerated
Round top bifocals
Round tops increase jump and displacement on minus lenses
They are preferred on plus lenses
answer
• Based on Prentice’s rule, the amount of jump will depend on the power of the segment and on the distance from the top of the segment to the optical center of the segment.
• Answer a.
answer
• Flat top segments minimize image jump because the optical center is near the top
• Flat tops also reduce image displacement since the base down effect of the distance portion is reduced by the base up effect of the segment
• Answer c.
facts• A series of curves with progressively
changing contours• The transition zone between distance and
near generates some lateral distortion• Answer c.
reasoning
• Since a contact lens can not be flatter than the flattest meridian, then 42 becomes the baseline.
• We will determine the power of the contact to correct this whether the prescription is plus or minus.
• Consider +6.00 + 2.00 x 90• Consider -6.00 + 2.00 x 90
continuing
• If you put on a contact lens to match the flattest curve, 42, the tears will fill in the space creating a positive tear lens and the astigmatism will be automatically corrected
• You must write the prescription in a form such that you can drop the cylinder that has its effect along the steeper vertical meridian.
continuing• The hyperopic prescription can be written two
ways• +6.00+2.00 x 90• +8.00-2.00 x 180• Therefore, you must write it in minus form and
drop the cylinder because the effect of the power of that cylinder is greatest at 90
• The same is true for the myope• So, always write the prescription in minus form
and drop the cylinder to do these kids of contact lens problems
solution• First convert the prescription to minus cylinder form and
drop the cylinder• -8+2x90 becomes -6-2x180 which becomes -6.00• Convert the -6.00 in glasses to contact lens by adjusting
for vertex• 1/6 meter=16.67cm• Add vertex distance of 1 cm giving 17.67cm and take the
reciprocal of that, i.e. 1/17.67=-5.66 contact power• We must compensate for the positive tear lens of 1
diopter, so add more minus giving -6.66 for the contact lens power.
answer
• Steeply fit lenses tend to ride low and move poorly with blinking, so
• Make the lens flatter• Decrease its diamter• Make it thinner and thus lighter• A prism ballast would make it heavier• Answer d.
answer
• There is corneal astigmatism but the refraction shows no cylinder
• Therefore the crystalline lens must have astigmatism that neutralizes the corneal astigmatism
• A hard lens would therefore uncover the astigmatism, so a soft lens would be the lens of choice
• Answer c.
facts
• Contacts can cause decreased oxygen to the epithelium with increased lactic acid and carbon dioxide under the lens causigcorneal edema
• Decreased corneal sensitivity can result• The glucose to the cornea is delivered
from the aqueous humor• Answer c.
facts
• The younger the child, the more likely that full correction will prevent of treat amblyopia
• Contacts lenses can help overcome large differences
• Anisometropic amblyopia is easily missed due to the lack of a cosmetic strabismus
• Answer a.
Use an inverted Galilean telescope
• Reduce the magnification difference by overcorrecting the contact and adding extra minus in the glasses
• Keep the image size difference less than 5%
• This telescope minifies because the plus lens is closer to the eye.
• Answer c.
distortion
• It is due to a differential magnification from the optical axis to the periphery of the lens
• For a plus lens, it is a “pincushion” (there is greater power near the edge of the lens that pulls the “lines in”)
• The doorway looks “bowed in” for an aphake• For a negative lens, it is a “barrel”• It is corrected by grinding the lens aspherical.
answer
• With spherical aberration and retinoscopy, especially in young people, the center part of the reflex moves in the opposite direction from the peripheral part.
• Pay attention to the center and ignore the peripheral part
• As a result neutralization won’t occur uniformly
What causes night myopia?
• With a dilated pupil in low light, the peripheral portion of the lens, which has greater power, will contribute more to the vision
• A -5 myope during the day may be -5.50 at night
How is our eye miraculously made to reduce spherical aberration?
• Our “natural” vision is amazing because:
• The cornea has less power in the periphery compared to the center
• The crystalline lens has a higher refractive index in the nucleus compared to the cortex
• The surface of the lens may be aspheric
What is chromatic aberration?
Longer wavelengths are bent less.
From red to blue, there is 1.25 diopters.
Ring scotoma• The high prismatic power at the edge of the lens
deflects light rays so strongly that light entering from the midperipheral field miss the pupil entirely!
• As the eye rotates the ring scotoma shifts in the opposite direction, causing sudden appearance and disappearance of objects (the jack-in-the-box phenomenon)
• It’s most noticeable between 3 and 12 feet• Thank heavens for contacts or implants.
Primary and secondary focal points
• Parallel light entering an optical system from the left comes to focus at the secondary focal point
• Parallel light entering an optical system from right to left comes to focus at the primary focal point
• The fraction of the entire optical system can be considered to occur at just two planes, the principal planes
• The intersection points of these planes with the optical axis are called the principal points
Back focal length
• In dealing with ophthalmic lenses, use the back focal length
• It is the distance between the posterior surface of the lens and the secondary focal point.
• This is what the lensometer measures
Back vertex power is greater than true vertex power
Because second principal point is INSIDE the surface of the lens
Separation of secondary and primary principal points
• In a complex optical system the primary and secondary principal are separated
• In the human eye, the separation is 0.3mm
Displacement of principal planes
• The principal planes are displaced anteriorly with PLUS meniscus lenses
• The principal planes are displaced posteriorly with MINUS meniscus lenses
• Therefore the posterior vertex power of a plus meniscus lens is greater the true equivalent power because a shorter focal length is being measured
• See the diagrams that follow
Nodal points
• The nodal points are a conjugate pair of axial points with the following property:
• Any light ray striking the primary focal point will leave the secondary nodal point at the same angular inclination to the axis
jokejoke
Nurse: Doctor, there is a man in the Nurse: Doctor, there is a man in the waiting room with a glass eye named waiting room with a glass eye named Brown. Doctor: What does he call his Brown. Doctor: What does he call his other eye?other eye?
Poor visionPoor vision
SobelSobel goes into the optometrist's office. goes into the optometrist's office. He opens the door and says to the He opens the door and says to the receptionist, "I think I need my eyes receptionist, "I think I need my eyes checked." She says, "You're not kidding. checked." She says, "You're not kidding. This is the Ladies Room." This is the Ladies Room."
The CzechThe Czech
A Czech goes to the optician who A Czech goes to the optician who shows him a card with the letters shows him a card with the letters
C Z W X N Q S T A C Z C Z W X N Q S T A C Z
"Can you read this?" the optician "Can you read this?" the optician asks. asks.
"Read it?" the Czech replies, "I "Read it?" the Czech replies, "I even know the guy." even know the guy."
• What's the best thing about having Alzheimer's Disease? You can hide your own Easter eggs. You are always meeting new people. You never have to watch reruns on television.
definition
• Magnification requires a reference distance and for simple magnifiers, the reference distance is defined as 25 cm
• The diopter equivalent is found by translating ¼ meter to………4 diopters
• The “X” magnification of a simple magnifier or low vision aid is therefore the power of the lens divided by 4.
Angular magnification
• In other words a 16 diopter lens held at its focal length from an object that is 25 cm from the eye would cause the object to appear 4 times as large.
• The ANGULAR magnification of the system is 4
Use vergence formula
• U+P=V• -4cm converts to -25 diopters• -25+20=V………V=-5 diopters…..-20cm• The image is on the same side as the
printed page.• Magnification is U/V=-25/-5=5
magnification
• Magnification is – Power eyepiece/power of objective
• M=-20/5=-4• This is 4X magnification and the image is
inverted because the sign is negative.• This is the indirect ophthalmoscope• IT is a combination of two plus lenses
facts
• The surgical loupe is a Galilean telescope with an “add” to set it for a certain distance
• The field size is small• Typically the power is 2X-3X• Asthenopia is produced with any kind of
refractive error• Phorias are easily induced if the PD is off• Working distance must be correct to avoid
using excessive accommodation
facts
• The power of a curved mirror is -2/r• For a concave mirror, r is minus• For a convex mirror, r is plus• The image is behind the mirror as far as
the object is before the mirror
answer
• Power is -2/r• Remember the radius is in meters• -2/(1/5 meter)=-10• It doesn’t matter how far the mirror is from
the driver
Focal length is 17mm
• Set up similar triangles to do visual angle problems
• You can calculate sizes of retinal lesions and scotomas
Characteristics of the optic nerve and blind spot
• The horizontal diamter of the optic nerve is 1.5 mm
• The blind spot is a vertical oval, 15.5 degrees temporal to fixation, 1.5 below the horizontal, 5.5 degrees wide and 7.5 degrees high.
• When looking into the eye the optic nerve is superior to the fovea
Size of the blind spot
• What is the actual size of the blind spot on a tangent screen at a distance of one meter?
Let’s calculate
• We know it is 5.5 degrees wide• Let’s convert that to radians• Actual width/1 meter=angle in radians• Since one radian is 57.3 degrees, then• Angle in radians x 57.3 = 5.5 degrees• Width/1000mm x 57.3 = 5.5• Width=5500/57.3=100mm approx• Convert to inches…100/25=4 in approx
How large does the optic nerve appear with the direct
ophthalmoscope?
• The eye is used as a simple magnifier• Magnification is power/4=60/4=15X• If the optic nerve is about 1.5mm wide,
then 15 x 1.5= about 23 mm• This is about an inch
Catopric images from the major eye surfaces
Three convex and one concave, all minified, three erect and virtual, one invertedand real.
Inverted aerial image
• The magnification is the ratio of the power of the eye to the power of the viewing lens
• 15 D……4X• 20 D…….3X• 30 D…….2X• The axial magnification is the square of
the linear magnification, i.e. 16X, 9X, 4X• Thus, retinal findings can look falsely
elevated with a 15 D lens
What does this mean?
• In healthy young patients, they have to be careful driving home, especially if they have latent hyperopia with decreased accommodative amplitude.
• Some patients may wish to have alternative transportation
• If cataracts are present, the situation is accentuated