VISION

XIV.Links to Previous Posts

2006-09-24T13:45:00.000+05:30

I.Refraction

II.Far Point

III.Far Point - Contd.

IV.Theoretical Approach

V.Practical Aspects

VI.Practical Aspects - Contd.

VII.Optics of Retinoscopy

VIII.Optics - Contd.

IX.Practice of Retinoscopy

X.Effect of Cycloplegia

XI.Cycloplegia - Contd.

XII.Practical Tips

XIII.Retinoscopy Reflex

XIII. Retinoscopy Reflex

2006-07-22T22:29:00.000+05:30

Clues we get from features of retinoscopy reflex -----

1. Intensity
In high refractive errors we get a faint reflex and in low refractive errors we get a brighter reflex.

2. Speed
In high refractive error we get a slow movement, and in low refractive error a rapid movement of the reflex.
As the neutral point is reached the movement of the reflex is fast.

3. Size
In high refractive error we get a narrow reflex.
Reflex will fill the pupil when the neutral point is reached.

In very high refractive errors we may not get a reflex
or may get a faint reflex with negligible movement. We
will not get a good reflex with low power lenses in these cases.
So try with high plus or minus lenses.
I

XII. Practical Tips

2006-06-20T19:06:00.000+05:30

To simplify the calculation step –

Keep the “Correction Factor lens" in the trial frame
before starting Retinoscopy.
After reaching the point of reversal you remove this lens.
The power of the remaining lens is the measure of refraction.
No need of correction for the Observer factor and cycloplegic factor.
So you can avoid the confusion of calculation.
(Correction Factor depends on the distance you are from the subject and the cycloplegic used)

To refine your result –

If your assessment is correct you will get the following results--
After reaching the neutral point (end point) in Retinoscopy
(with plane mirror)
you move slightly towards the subject, then you will get
a ‘with movement’
(because the Far Point is now behind the Observer).
If you move slightly away from the subject you will get
an ‘against movement’
(because now the Far Point is between the observer and the subject).

XI. Cycloplegia – Contd.

2006-06-09T17:58:00.000+05:30

Retinoscopy is the Objective method of assessing the refraction of the eye. To measure the Static Refraction the accommodation should be at rest.

This can be achieved by voluntary relaxation of accommodation by the patient, by noncycloplegic method of inhibiting accommodation by Fogging or by using cycloplegics to prevent accommodation.

Relaxation of accommodation is difficult in younger age group and in cases with spasm of ciliary muscle. Fogging is also not effective in these situations. So, cycloplegic refraction is the answer.

Cycloplegia is used up to the age of 10 – 15 years (may be up to 20 years).Also in cases with over action of accommodation and in all cases of Convergent strabismus.

Commonly used Cycloplegics

atropine Sulphate (0.5 – 1%)
(lower strength below 5 years of age)
Used for cycloplegia in children below 7 – 10 years of age,
in cases with ciliary spasm and in Convergent Strabismus.
Ointment or drops applied thrice daily for at least three days. Amount of cycloplegia (change of refractive power) achieved is +1.0 Dioptre (more than the actual static refraction of the eye).Effect lasts for 2 weeks.

Homatropine Hydrobromide (1 – 2%)
Drops applied at 15 minutes interval for 6-8 times.
Retinoscopy is done after 1 hour.
Amount of cycloplegia (change of refractive power) achieved is +0.50 Dioptre (more than the actual static refraction of the eye).Effect lasts for 2 days.

Cyclopentolate Hydrochloride (0.5-1%)
Drops applied 2-3 times at 5 minutes interval.
Retinoscopy is done after 30 minutes.
Amount of cycloplegia (change of refractive power) achieved is +0.50 Dioptre (more than the actual static refraction of the eye).Effect lasts for 8 hours.

Tropicamide (0.5-1%)
Drops applied 2-3 times at 5 minutes interval.
Retinoscopy is done after 30 minutes.
Amount of cycloplegia (change of refractive power) achieved is +0.50 Dioptre (more than the actual static refraction of the eye).Effect lasts for 8 hours.
http://en.wikipedia.org/wiki/Tropicamide
http://en.wikipedia.org/wiki/Phenylephrine
Post Cycloplegic Subjective Refraction (Post Mydriatic Test) is to be done in possible cases.

X. Effect of Cycloplegia

2006-06-05T04:38:00.000+05:30

Latent Hypermetropia

Normal tone of ciliary muscle is affected by the cycloplegics. When the tone is lost (Cycloplegia) the lens zonules are taut flattening the lens.

This leads to less convergence of light rays by the lens. So there is a shift of refraction of the eye to the hypermetropic side (light rays focused posterior to the previous focus).This value of hypermetropia, made manifest by cycloplegia is the latent hypermetropia.

So an emmetropic eye will become hypermetropic, myopic eye becomes less myopic and hypermetropic eye will become more hypermetropic.

As a rule the latent hypermetropia
amounts to only one Dioptre.

So when we do retinoscopy under cycloplegia the values we get are to be corrected for this hypermetropic factor. I call it ‘Cycloplegic Factor’. This extra power of converging (+) lens (cycloplegic factor) used to reach the point of reversal must also be subtracted from the total power.Now the Correction Factor includes Observer Factor and Cycloplegic Factor

Observer Factor + Cycloplegic Factor = Correction Factor

Total Power – Correction Factor =Patient Factor(Refractive status of the Subject’s eye).

When the observer is at 1 metre from the subject and
the cycloplegia is with atropine then the Observer Factor
is +1.0 D and the Cycloplegic Factor is +1.0 D. So the Correction Factor is +2.0 D.

Total Power – Correction Factor = Patient Factor
Total Power – (+1 + +1) = Patient Factor
Total Power – (+2) = Patient Factor (Ref.Status)

Eg. (+2) - (+2) = 0 (zero) Eye is Emmetropic
(+3) - (+2) = +1.0 Eye is 1D Hypermetropic
(+1) - (+2) = - 1.0 Eye is 1D Myopic
( 0 ) - (+2) = -2.0 Eye is 2D Myopic
(-1) - (+2) = - 3.0 Eye is 3D Myopic

The value of Total Power can be plus or minus. So for the ease of calculation instead of subtracting +2 we can add -2 with the Total Power.

Total Power + (-2) = Refraction of Patient’s eye
Eg. (+2) + (-2) = 0 (zero) Eye is Emmetropic
(+3) + (-2) = +1.0 Eye is 1D Hypermetropic
(+1) + (-2) = - 1.0 Eye is 1D Myopic
( 0 ) + (-2) = -2.0 Eye is 2D Myopic
(-1) + (-2) = - 3.0 Eye is 3D Myopic

IX. Practice of Retinoscopy

2006-06-01T04:50:00.000+05:30

New visitors please go through previous posts - I to VIII
Practice of Retinoscopy

Observer sitting at a distance of 1 metre from the subject (in a darkened room) reflects light to subject's eye with the mirror (Reflecting Retinoscope).

Observe the behaviour of the light reflex at the pupil of the subject when the mirror is titled (rotated) along one meridian.(movement of the reflex is noted in relation to that of the mirror or external movement of light across the subject's eye)..
When plane mirror is used......
If we get a 'with movement' of the reflex, cancel the movement (to reach the end point) by using Convex lens. Note down the value (Total Power) on the corresponding axis of the 'cross diagram'. Repeat the procedure for the opposite axis.If we get an 'against movement' cancel it with Concave lenses.
At Neutral Point (end point) we get a uniformly illuminated or uniformly dark pupil.
Repeat the procedure for the other eye.

(Concave mirror gives opposite results – reason already explained. Movements are cancelled with opposite type of lens compared to the use of Plane mirror, ie. in with movement with - lens and in against movement with + lens)

Movement of reflex is decided by the movement of immediate source of illumination and the refractive state of the eye. Movement of immediate source of illumination is decided by the type of mirror used.

Now we can find out the Refractive status of subject's eye.
Total Power = Observer Factor + Patient's Ref.Status
Total Power - Observer Factor = Ref.Status
Total Power - (+1.0 ) = Ref .Status (observer at 1 metre.)

You can get Total power with plus and minus values depending on the refraction of the eye . So to avoid confusion in the calculation instead of subtracting +1.0 from the Total power you always add -1.0 with the Total Power.

Now we get the actual power along each axis. If both the values are same, there is no astigmatism. If we get a + value the eye is that much Hypermetropic. If it is minus the eye is Myopic. If it is zero the eye is emmetropic.

Effect of using cycloplegics on the Retinoscopy values will be discussed later.

VIII.Optics of Retinoscopy-Contd.

2006-05-29T15:11:00.000+05:30

Light rays entering subject's eye. (Illumination Stage)

Consider only the immediate source of illumination in front of the subject's eye . This may be a virtual image or real image of an original source of illumination behind the subject. When the immediate source is towards one side, the other side of the retina will be illuminated. When the source is shifted to the other side, illumination will be shifted to the opposite side of the retina.
This will be the situation in all states of refraction of the eye. (but rays coming out of the eye behave differently in different types of refraction).Type of mirror we are using to reflect the rays from the original source of illumination is not the primary factor deciding the optics.

If we are using a Plane mirror, the virtual image of the original source of illumination is formed as far behind the mirror as the original source is in front of it. So, the tilt of the mirror to one side will shift the image (immediate source of illumination) to the other side. Tilt of the plane mirror and the shift of illuminated patch on the retina are in the same direction.

If we are using a Concave mirror, the real image of the original source of illumination is formed in front of the mirror ( position depends on the focal length of the mirror). So, the tilt of the mirror to one side will shift the image (immediate source of illumination) to the same side itself. Tilt of the concave mirror and the shift of illuminated patch on the retina are in opposite directions.

In Practice we are considering the movement of the mirror .
When we use plane mirror we get movement of the mirror and movement of the retinal illumination in the same direction.( because of the movement of immediate source of illumination - ie virtual image formed behind the mirror - in the opposite direction of movement of the mirror).
When we use concave mirror we get movement of the mirror and movement of retinal illumination in opposite directions (because of the movement of immediate source of illumination - ie real image formed in front of the mirror – in the same direction of movement of the mirror).

VII.Optics of Retinoscopy

2006-05-29T05:17:00.000+05:30

Rays coming out of Subject's Eye. (Reflex & Projection Stages)

If the illuminated patch on Patient's retina is away from the principal axis, the rays coming out will not enter Observer's eye. When the illumination is moved across the fundus towards the principal axis the rays coming out enters observer's eye.

VII.a) In Finding 1.The first ray of light entering observer's eye is from the same edge of the pupil as the first position of retinal illumination. As the illumination on the retina moves towards the principal axis the light reflex in the pupil also moves in the same direction. The last ray of light entering the observer's eye is from the other edge of the pupil. (see diagrams). This gives a with movement reflex in the pupil.( seen in emmetropia, hypermetropia and myopia less than 1D - observer at 1 metre from patient.)

VII.b) In Finding 2. The first ray of light entering observer's eye is from the opposite edge of the pupil when the position of the retinal illumination is considered. (Light rays cross at FP and the diverging rays are entering observer's eye). As the retinal illumination moves towards the principal axis the light reflex in the pupil moves in the opposite direction. The last ray of light entering the observer's eye is from the other edge of the pupil. (see diagrams). This gives an against movement reflex in the pupil. (seen in myopia more than 1D - observer at 1 metre)

VII.c) In Finding 3. As the retinal illumination moves towards the principal axis, the rays coming out will focus on the nodal point of observer's eye. This makes the pupil of the patient uniformly illuminated. (The rays coming out through all the parts of the pupil are focused at the nodal point of observer's eye). By a slight shift of retinal illumination this focal point is displaced away from observer's eye making patient's pupil uniformly dark. (see diagrams) This finding you get as end point of retinoscopy and in eyes with 1 D myopia. (observer at 1 metre from the patient.)

From the description and the diagrams you know the position of the FP in each finding. (behind your position or in front of you or at your nodal point). Now it is easy to decide which type of lens you have to use to reach the point of reversal. (convex lens if FP is behind the observer, ie in 'with movement' and concave lens if FP is in front, ie in 'against movement') Then by changing the power of lens you reach the end point.

Here movements are described in relation to movements of the retinal illumination. Not related to the movement of the mirror. They differ with type of mirror (plane or concave).

(Details discussed along with discussion of the illumination stage of retinoscopy.)
nodal point

VI.Practical Aspects – Contd.

2006-05-28T15:10:00.000+05:30

How do you know that you are at the FP of the Subject's eye ?.

When you are sitting at 1 metre from the patient, how do you know that you have brought the FP of the patient's eye (by using lenses) to the nodal point of your eye at 1 metre ?.

The technique of Objective method of Refraction (Retinoscopy) will answer these questions.

Principle : Behaviour of the luminous reflex in the pupil of the patient is studied by moving the illumination across the fundus. This behaviour depends on the vergence of the light rays coming out of the pupil. It also depends on the position of the observer. (optics discussed later)

VI.a) Finding 1.When the observer is at 1 metre from the patient and if the rays coming out from the patient's eye form a focus (FP of the patient) behind the observer at a finite distance or at infinity or 'beyond infinity' (virtual point behind patient's retina), then the luminous reflex in the pupil will move in the same direction of movement of the illumination across the retina ('With Movement'). (This finding we get in myopia less than 1 D, in emmetropia and in hypermetropia)

VI.b) Finding 2.When the observer is at 1 metre from the patient and if the rays coming out from the patient's eye form a focus (FP) in front of the observer (between patient and observer) the luminous reflex in the pupil will move in the opposite direction of movement of the illumination across the retina ('Against Movement'). (This finding we get in myopia more than 1 D.)

VI.c) Finding 3.When the observer is at 1 metre from the patient and if the rays coming out from the patient's eye form a focus (FP) at the nodal point of the observer then the pupil of the patient will appear uniformly illuminated. By slight tilt of illumination across retina the pupil will appear uniformly dark. (This finding we get in myopia of 1D and also as the end point of retinoscopy with the observer at 1 metre from the patient.)

Reason for the differences in behaviour of the reflex in the pupil is explained along with the optics of retinoscopy.

Neutral Point or End Point or Point of Reversal in Retinoscopy is reached by bringing the FP of patient's eye to the Nodal Point of observer's eye by using lenses. (except in myopia of 1 D, where it is reached without any lens - already explained).

V.Practical Aspects

2006-05-27T05:29:00.000+05:30

If the observer is at a distance of one metre from the subject the Observer Factor is +1.0 Dioptre. (+1 Dioptre is the power of the lens which can focus parallel rays at 1 metre.) Then...............

V.a) In emmetropia the parallel rays coming out of the subject's eye can be brought to a focus at 1 metre by using converging (convex) lens of 1 Dioptre kept close to subject's eye. In other words, if the power of the lens used to bring the FP to 1 metre is +1 D then we know that the rays coming out are parallel (which normally meet at infinity) and the eye we are examining is emmetropic.

Total Power = +1 D,ObserverFactor +1 D.

So Patient Factor is 0 (zero)

Total Power - Observer Factor = Patient Factor (Ref.Error)

(+1) - (+1) = 0 (zero) Subject is emmetropic

V.b) In hypermetropia the diverging rays coming out of the eye are brought to a focus at 1 metre by using converging lenses of power more than +1 D. (+1 required for parallel rays in emmetropia). The power used in excess of +1 D is the measure of hypermetropia of that eye.

eg. Total Power = +3 D, Observer Factor = +1 D.
So Patient Factor is +2 D
Total Power - Observer Factor = Patient Factor.
(+ 3) - (+1) = +2. This means the patient is 2 Dioptres Hypermetropic

V.c) In myopia of less than 1 Dioptre the converging rays coming out are focused at 1 metre by using converging lens of power less than +1 D .This difference in power compared with the emmetropic eye will give the exact measure of myopia.

eg. Total Power= +0.50 D, Observer Factor= +1 D.
So Patient Factor (ref.error) is -0.50D
Total Power - Observer Factor = Patient Factor (Ref. error)
(+0.50) - (+1.0) = - 0.50 (Patient is 0.50 Dioptre myopic).

V.d) In myopia of 1 Dioptre the rays coming out of the eye are convergent and meet at 1 metre with out using any lens. In other words if we are not using any lens to bring the FP to 1 metre we know that we are dealing with an eye which is 1 D myopic.

Total Power = 0 (zero), Observer Factor = +1 D,
so Patient Factor (error) is - 1.0 D.
Total Power - Observer Factor = Patient factor (ref.error)
( 0 ) - ( + 1.0 ) = - 1.0 ( Patient is 1 Dioptre myopic)

V.e) In myopia of more than 1 Dioptre the converging rays coming out of the eye will focus at the FP which is less than 1 metre from the eye. (between the patient and the observer). So this focus (FP) can be brought to 1 metre by using diverging (concave) lens. So this much of diverging (minus) power is in excess when compared with an eye with FP at 1 metre (ie.eye with 1 D myopia).

eg. Total Power = - 2.0 D, Observer Factor = +1D,
so Patient Factor (error) is - 3.0D.
Total Power - Observer Factor = Patient Factor (ref.error)

( - 2.0 ) - ( + 1.0 ) = - 3.0 ( Patient is 3 Dioptre myopic )

IV.Theoretical Approach

2006-05-26T17:26:00.000+05:30

If we can locate the Far Point of subject's eye, then we can calculate the refraction of that eye. But observer moving towards FP is not practical. (In emmetropia FP is at infinity and in hypermetropia FP is virtual point behind patient's eye).

The method used for calculation is keeping the subject and the observer at fixed places, and bringing (shifting) the FP of the subject to the position of the nodal point of observer's eye. This is done by using converging or diverging lenses.
Now we know the exact distance of the FP from patient's eye (ie. exact distance at which we are sitting) and also the power of lens used to bring the FP to this position.
( From the measurement of distance we can calculate the power required to bring the FP to this position in emmetropia - I call it observer factor - see below. The power used other than this power, gives the refractive error - patient factor - see below)

The Total Power of lenses we use contain two parts.
First one is the power used because of the position of the observer (Observer Factor or Induced Factor or False Factor). If the subject is emmetropic this is the only factor that will be there. This is the power used, only because of the position of the observer and not because of the patient's refractive error.
Second part is the power of lenses used because of the patient's refractive error (Patient Factor or True Factor). From the above three values we can calculate the refraction of the subject's eye.

Total Power = Observer Factor + Patient Factor.
Total Power - Observer Factor = Patient Factor (Refraction of Patient's Eye).

III . Far Point - contd.

2006-05-26T04:32:00.001+05:30

In Optics Direction of Light Ray is Reversible.

What happens to the light rays coming out of the eye from a point on retina?
They meet at the FP of the eye (when the accommodation is at rest).

In emmetropia light rays coming out from a point on the retina through the optical system of the eye will be parallel (when the accommodation is at rest). These rays can meet only at infinity ( ie. where the Far Point is located in emmetropia).

In myopia light rays coming out from a point on the retina through the optical system of the eye will be convergent. These rays meet at the FP of the eye (at a finite distance) when the accommodation is at rest.

In hypermetropia light rays coming out are divergent. They will only meet "beyond infinity" when the accommodation is at rest. They can meet only at the FP of the eye, which is a virtual point behind the retina. (by extrapolating the divergent rays in the reverse direction to meet behind the retina.

Optical state (refraction) of the eye decides the position of the FP of the eye.
If the position of the FP is known the refraction can be calculated.

II.Far Point of the Eye

2006-05-25T21:54:00.000+05:30

Far Point (FP) is the furthest point at which objects can be seen clearly by the eye. Object kept at FP of the eye can be seen clearly with accommodation at rest.
Position of FP depends on the optical state (static refraction) of the eye. In emmetropia it is at infinity. In myopia it is at a finite distance. In hypermetropia it is a virtual point behind the retina.

What happens to the light rays coming from the FP and entering the eye?

In emmetropia the FP is at infinity. So the rays coming from FP, and entering the eye are parallel. They are focused upon the retina (with accommodation at rest) to give a clear image of object at FP.

In myopia the FP is at a finite distance. So the rays coming from object at the far point, and entering the eye are divergent. These rays are focused upon the retina to give a clear image (with accommodation at rest).
In hypermetropia object can't be placed at the FP ( It is a virtual point behind the retina).Here the converging rays directed towards the FP behind the retina can be focused upon the retina to give a clear image by the dioptric (optical) system of the eye (with accommodation at rest).
The Far Point and the Point of Focus on the retina are Conjugate Foci.

I. Refraction of the Eye

2006-05-25T04:56:00.000+05:30

Refraction is vergence (bending) of light ray when it passes from one medium to another medium of different optical density (refractive index ).

Normal refraction (optical state) of the eye is essential for normal vision.Eye with normal optical state is Emmetropic and that with abnormality is Ametropic (with Refractive Error).

Vergence of light rays entering the eye and coming out of the eye is decided by the dioptric system (optical system) of the eye.

In emmetropia parallel rays from infinity are focused on the retina (with accommodation at rest). This will give a clear image of object at infinity (which has sufficient size).
In ametropia parallel rays from infinity are not focused on retina. So no clear image of object at infinity is formed on the retina of that eye.

Optical aids like spectacles or contact lenses are required to get clear images of objects at infinity in these eyes with abnormal optical system.
I.Refraction of the eye