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Dark Room Procedures for Ophthalmolgy Practicals: Part 1

Author: rxpg, Posted on Monday, August 04 @ 15:10:40 IST by rxpg  

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RxPG thanks Dr Sanjay Dhawan, Maulana Azad Medical College and GNEC, for providing these high yield notes. DARKROOM PROCEDURES : ENLIGHTENED


Dark-room procedures (DRP) constitute an essential part of the examination of the eye beside being an important part of the undergraduate professional examinations. The main DRPs are:

  1. Preliminary examination at 1 m (including Retinoscopy).

  2. Distant Direct Ophthalmoscopy at 22/25 cm.

  3. Indirect Ophthalmoscopy.

  4. Direct Ophthalmoscopy.

All DRP are performed in very low ambient illumination (not necessarily pitch dark) and with the examiner preferably dark adapted.

Before describing DRP it is important to understand the principle of coaxial illumination which is applicable to all DRP.

Coaxial Illumination

Why does the pupil appear black no matter how bright light we use to examine the eye? And why does it appear red when seen through ophthalmoscope or retinoscope even with very faint light? A very basic principle of optics is that in any ideal optical system the light rays retrace their own path. Therefore the light rays from the torch-light falling into the eye get reflected back in the same direction and since the observer’s eye is not a source of light, no light comes in that direction and therefore the pupil appears black (in eyes with high refractive errors the pupil may appear red because the eye is no longer an ideal optical system). However the ophthalmoscope or retinocsope have an ingenious optical system of coaxial illumination i.e. the axis of illumination and the axis of observation are in same line making the eye of the observer a virtual source of light, as depicted in the diagram below:

The light reflected from the patient’s eye gets transmitted through the partially mirrored glass and falls into the observer’s eye, therefore the pupil appears red (the fundal glow).

What is the color of the retina? Theoretically purple because of the pigment visual purple (as deciphered from its absorption/emission spectrum) but in practice when any amount of light falls on the retina the pigment gets bleached making it almost transparent. Then why does the fundal glow appear red? Because of the light reflected from the choroidal vasculature.

I. Preliminary Examination at 1m (including Retinoscopy)

Patient is seated in a dimly lit room with light source above and behind patient’s left shoulder. Observer sits at 1m and using plane mirror of the Priestley-Smith Retinoscope (described later) light is reflected onto patient’s eye while looking through its hole. If the ocular media are clear the pupil appears red due to fundal glow.

Retinoscopy (or Skiascopy) is an objective method of determining the refraction of the eye (not to be confused with ophthalmoscopy which means visualization of the retina).

To understand the methodology of retinoscopy we will make following assumptions:

  • Done at 1m.

  • Using plane mirror.

  • We move the light (or the mirror) from our left to our right.

The reflected light from plane mirror is moved across the patient’s eye by turning the mirror from left to right while the movement of the glow (light) in the pupil is observed. There are 3 possibilities:

1. Glow in the pupil moves in the same direction as the light outside. This means the patient can be any of the following:

  • Hypermetropic

  • Emetropic

  • Myopic less than -1.0 D

2. Glow in the pupil moves in the opposite direction. This means patient is myopic by more than -1.0 D.

3. Glow does not move (and the pupil is either uniformly lit or uniformly dark). This means patient has myopia of 1.0D.

Above assertions can be easily remembered by studying the following diagram:

It may be seen from the retinoscopy line that if the glow moves in the + (same) direction patient is anywhere on the + side and if the glow moves in the - (opposite) direction the patient is anywhere in the - side; but in both these situations we do not know where exactly the patient is. However, when the glow in the pupil does not move that we know for certain that the patient has myopia of -1.0 D. So what do we do if the glow moves in the same or the opposite direction? We neutralize the movement i.e. bring the optical system to the point where the glow stops moving, by putting in front of the eye increasing power of + or - lenses depending on + or - direction of movement of glow, respectively. If the glow moves in the + (same) direction we neutralize it by + lenses and if it moves in the - (opposite) direction then by - lenses. When the glow stops moving then the ‘optical system’ viz. the eye of the patient and the lens in front of it, has come to the point of neutralization and therefore has myopia of -1.0 D. Now if we prescribe -1.0 D lens to this ‘optical system’ it will become emetropic, which is same as adding -1.0 to the power of lens in front of the eye (the retinoscopic value). Thus, we get the refractive error of the patient.


If the movement in the same direction gets neutralized by +5.0D lens then the refractive error is +4.0D (-1.0 added algebrically to +5.0).

If movement in the same direction is neutralized by +1.0D then the error is 0.0 i.e. the patient is emetropic[-1+(+1)].

If movement in same direction is neutralized by +0.25D then the error is -0.75D[-1+(+0.25)].

If the movement in the opposite direction is neutralized by -1.0D then the error is -2.0D[-1+(-1)].


The higher the refractive error the fainter the glow in the pupil and slower does it move. But as one approaches the point of neutralization the glow gets brighter and moves faster. And at the point of neutralization the glow is the brightest and completely fills the pupil.

In some cases direction of movement of glow is not clearly defined, instead there is scissoring of the glow; here the point of neutralization is reached when two limbs of the scissors start from the center of the pupil and move equally in opposite directions.

If two light reflexes are seen, one central and the other peripheral, then one should neutralize the central glow because the central parts of cornea and lens are more important in forming image on the retina.

Theoretically the ideal distance for doing retinoscopy is infinity (Ą ) because the retinoscopy directly gives the refractive error. The neutralization point correspond to myopia of -1¸ distance in meter which is also the amount to be added to the retinoscopy value.

At neutralization point the patient’s and the observer’s eyes become conjugate foci of the optical system (as the image of the illuminated points on the patient’s retina are formed at the observer’s pupil).

The method described above gives refractive error only in the horizontal meridian, whereas the error may not be the same in all meridians as is seen in astigmatism. However, as most patients have regular astigmatism in which two principal meridians disposed at right angle to each other can be defined. Also, these meridians are most commonly aligned vertically and horizontally. Therefore, it is customary to do retinoscopy both vertically and horizontally, and note the values separately as follows:

where x denotes retinoscopy value along horizontal meridian and y denotes the value along the vertical meridian (obtained by moving the mirror vertically). If these two values are equal then there is no astigmatism and a spherical lens alone will correct the error. But if these two values are not equal then it denotes presence of astigmatism which needs a cylindrical lens (alone or in combination with a spherical lens) for its correction, as explained next.

A cylinder is a lens which has refractive power only in one meridian (i.e. at right angle to its axis) and no power at right angle to it (i.e. along the axis). Now if the retinoscopy values are e.g.:

Now a +3.0 D spherical lens would completely correct the vertical meridian and would partly correct the horizontal meridian leaving a residual error of +1.0 D [+4.0-(+3.0) = +1.0]. This is corrected by +1.0 D cylinder whose axis is placed at 90° (vertically) because the power is required to act at 180° (horizontally). Thus the prescription would be:

+3.0 D sphere / +1.0 D cylinder at 90°

Transposition means an equivalent prescription with the cylinder of opposite sign. While transposing a prescription the spherical-equivalent (and not the sphere) of the lens is kept constant. Following are the steps to transpose a prescription:

Algebraic sum of the sphere and the cylinder gives the new sphere.

Same cylinder with opposite sign.

Axis is placed at right angle to the previous axis.

Spherical Equivalent of a spherocylindrical lens (combination of a sphere and cylinder) is a spherical lens with same average refractive power obtained by algebraically adding half the value of the cylinder to the sphere. Note the focal point of the spherical equivalent coincides with the circle of minimal blur of the spherocylindrical lens.

This article is part 1 of the three article series. For other parts see the ophthalmology section of this site.

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