• Nature of Light
• Movement of Light
• Simple Lens systems
• Multiple Lens systems
• Lens formula
• Ray Tracings- Lenses
• Mirrors
• Physical Optics
• Prisms
• Prentice's Rule
• Lenses
• Schematic eye
• Instruments
• Magnification
• Telescopes
• Aniseikonia

Light resembles sound in that it passes through a media; but unlike sound, it can also travel across a vacuum. This dual behavior of light (or ability to travel through a media as well as across a vacuum) has led to separate theories of its nature: wave theory and quantum theory.

Classically, light has been considered as a "stream of particles", a "stream of waves" or a "stream of quanta".

• Physical Optics examines light as energy particles that are emitted by light sources and absorbed by other substances (Wave or Quanta Theory of Light).
• Wave Theory helps to understand how light interacts with itself, different media and various surfaces. Wave theory allows us to understand the naturally occurring phenomena of interference, diffraction and polarization.
• Geometric Optics deals with the formation of images by rays of light acted on by lenses, prisms and mirrors (Particle Theory of Light).
• The concept of vergence is the unifying concept between wave theory and geometric optics.
• Quantum Optics deals with the interaction of light and matter. It considers light as having both wave and particle (photon) characteristics.

Visible light is in the very narrow portion of the electromagnetic spectrum with wavelengths roughly between 400 and 800 nanometers (380 – 760nm or 4x10^-6 m to 8x10^-6 m), which represents approximately 1% of the sun’s electromagnetic spectrum that ranges from 1x10^-16 m to 1x10⁶ m.

Yellow light is the standard wavelength for calibration. It holds mid position in the chromatic interval of the emmetropic eye and so is in best focus.

A photon of wavelength 100 nm has 12.50 eV per photon. A photon of wavelength 193 nm has 6.4 eV per photon. This shows why shorter wavelengths of light (e.g. ultraviolet) have greater potential for photic damage, due to their higher energy level.

• Newton, in 1665 stated that light was made of particles that moved in straight lines.
• One hundred years later, Kristian Huygens, a Dutch mathematician, suggested that the light was a wave form, after observing that a small amount of light was always bent onto the shadow behind an opaque object.
• Thomas Young proved the wave nature of light with a double slit refraction experiment.
• Einstein taught that the speed of light in a vacuum is always 186,000 miles per second regardless of the speed of the observer or the source. This was proven by the Michelson-Morely experiment. Einstein’s work on light concluded that light really does act as a particle, but a particle that has wave properties.
• The Heisenberg principle rationalizes that when you try to measure something too precisely, the act of measurement itself, changes the thing being measured. This has lead to light particles being called photons or quanta. Heisenberg suggested that quanta have wave properties.
• When light is considered as being composed of quanta, the results of all experiments and physical phenomenon can be predicted.
• A quanta of light’s energy is described by the equation: E = hv, where E=energy, h = Planck’s Constant, and v = frequency.
• Frequency and wavelength of light are related in the equation c = wV where c = speed of light, V = frequency and w = wavelength. Therefore the constancy of the speed of light, c, guarantees a constant relationship between frequency and wavelength.

• Movement of light by convention is from left to right. Positive numbers measure in the direction of light, negative measure against the direction of light. Therefore, a positive lens or waveform is converging and a negative lens or waveform is diverging.
• All naturally occurring wavefronts are diverging as they emerge from a source.
• As light rays approach infinity, they become parallel.
• Optical infinity is considered 20 feet or greater.

Vergence is defined as the reciprocal of the distance from a reference point (in meters) to the point of focus.

• Vergence is measured in diopters. (1 diopter = 1/1m = 100/100cm)
• Therefore, the vergence of the light rays coming from an object is directly related to the distance from the object.

In the preceding figure, the divergence of rays of light emanating from point O is, at A, 1/-.25 = –4.00D; at B, 1/-.50 = – 2.00D; at C, 1/-1 = –1.00D; at D, 1/-2 = –0.50D and at E, 1/-3 = –0.33D.

In the following figure, the convergence of the rays of light, converging to the point I is, at A, ¼ = +0.25D; at B, 1/3 = +.033D; at C, ½ = +0.50D, at D, 1/1 = +1.00D and at E, 1/.5 = +2.00D

• By convention, divergence is given in minus (-) vergence power and convergence is given in plus (+) vergence power.
• Convergent wavefronts are not found spontaneously in nature. They are the result of an alteration of a planar or divergent wavefront by a refractive or reflective medium.
• An object (O) is defined as a point or extended source that the pencil/beam of light comes from.
• The object’s distance from the object to the point of reference is designated as u.
• The object's vergence (U) is the distance from the point of reference to the object. U = 100/u (cm)
• The image (I) is defined as a point or extended source that the pencil/beam of rays go to.
• The image distance is measured from the point of reference to the image and is defined as u’.
• The image’s vergence (V) is the distance from the point of reference to the image. V = 100/u’ (cm)
• When defining an optical system, it is conventional to set the incoming rays as object rays, the outgoing rays as image rays, and light travels from left to right.

Question: Define the term Diopter

Answer: A diopter is a unit of accommodative amplitude, describes the vergence of a wave form, describes the vergence at a specific distance from the source, and is also defined as the power of the lens.

• Real objects have diverging rays and are on the same side as the incoming object rays.
• Virtual objects are not naturally occurring and have converging rays.
• Real images have a focal point that can be focused on a screen and therefore are on the same side as the outgoing image rays.
• Virtual images cannot be focused on a screen and are always on the left side of the lens system.
• A virtual object may also have a virtual image.
• When light rays are about to cross, they are considered to have positive (+) vergence (convergence) and when they are receding from their crossing point, they are said to have negative (-) vergence (divergence).
• Rays that cross at the focal point of the lens are considered to have an infinite amount of vergence. Therefore rays that are parallel (have no crossing point), have a vergence of zero.
• Converging lenses have real and inverted images that are on the opposite side of the lens from the object.
• A virtual image is erect.
• Diverging lenses create virtual and erect images that are on the same side as the object.

U + D = V and 100/u (cm) + D = 100/v(cm)

Where: U = vergence of object at the lens u = object position = 1/U (m)

D = lens power

V = Vergence of image rays v = image position = 1/V (m)

Question: Define a plus lens

Answer: Plus lenses always add vergence, defines a focal point to which image rays are convergent, and produces a real imag to the right of a plus lens.

Question: Define a minus lens

Answer: A minus lens always reduces vergence, defines a focal point, and diverges all rays at this point to produce a virtual image to the left of the minus lens.

When working with a multiple lens system, it is essential to first calculate the position of the image formed by the first lens. Only after locating the first image is it possible to calculate the vergence of light as it reaches the second lens. This is the method by which any number of lenses can be analyzed. Always remember to locate the image formed by the first lens and use it as the object for the second lens to calculate the vergence of light as it reaches the second lens. Repeat the process for each subsequent lens.

M_T = I/O=U/V = v/u

The ratio of the image size to the object size or image vergence to object vergence is called transverse or linear magnification.

• If the absolute number is greater than 1, the image is larger. If the absolute number is less than 1, the image is smaller. If the absolute number is equal to 1, it is the same size.
• If the magnification number is negative (-), the image is inverted, if the magnification number is positive (+), the image is erect.
• If the object and image are on the same side of the lens, the image is erect, if not, the image is inverted.
• Generally, if the image is located farther from the lens than the object is, the image is larger than the object, if the image is closer to the lens than the object, the image is smaller than the object.

• The primary focal point F₁, is a point that is conjugate to infinity(object point). If you place an object here, the image rays will emerge parallel.
• The primary focal length, f₁, is the distance from the optical surface to the primary focal point. If F₁ is a real object, f₁ is negative. If F₁ is a virtual object, f₁ is positive.
• Secondary focal point F₂ is a point which is conjugate to infinity (image point). When parallel rays enter the optical surface, they will focus at the secondary focal point. If F₂ is a real image, f₂ is positive. If F₂ is a virtual image, f₂ is negative.
• For a plus (+)/convergent lens, the focal point is to the right of the lens.
• For a minus (-)/divergent lens, the focal point is to the left of the lens.

For lenses, convex lenses form real images on the opposite side from the object and concave lenses form virtual images on the same side as the object.

When doing ray tracing, there are three rays that follow simple known paths before and after their refraction by the lens. Use two of the three construction rays to find the image.

1. Draw a line from the object to the lens, parallel with the direction of light. At the lens, draw the line through the primary focal point of the lens.
2. Draw a line from the object, through the center of the lens (no deviation at the lens).
3. Draw a line from F₁ to the lens. At the lens, draw the line parallel with the direction of light.

Where the lines cross is where the image will be. By doing this, you can estimate the approximate location of the image, tell whether the image is erect or inverted as well as real or virtual.

The following figure shows the construction of images produced by a plus lens:

1. Object farther from the lens than F₁ , image is real
2. Object lies in F₁ plane, the image therefore is at infinity
3. Object closer to lens than F₁ , image is virtual
4. Object at infinity, image at F₂ .
5. Object lies to the right of the lens, (a virtual image projected by another optical system), image is real.

The following figure shows the construction of images produced by a minus lens:

1. Object real, image virtual
2. Object at infinity, image at F₂ plane
3. Virtual object in F₁ plane, image at infinity
4. Virtual object closer to lens than F₁, image real
5. Virtual object farther than F₁ , image virtual

• The focal length of a curved mirror is always ½ its radius of curvature (f= r/2)
• f = focal length of the mirror in meters
• r = radius of curvature of the mirror in meters.
• For mirrors or reflecting surfaces: U + 2/r_m = V, (r_m is in meters)
• If the mirror is convergent or plus, the focal point is to the left of the mirror.
• If the focal point is to the right of the mirror, the mirror is divergent or minus.
• Convex mirrors form virtual images on the opposite side from the object
• Concave mirrors form real images on the same side as the object.
• A plus (concave) mirror adds positive vergence, while a minus (convex) mirror adds minus vergence.
• Convex mirrors add negative vergence like minus lenses.
• Concave mirrors add positive vergences like plus lenses.
• Plain mirrors add no vergence.
• The field of view of a plane mirror is 2 times its size.
• The reflecting power of a mirror in diopters D_M = 1/f (m)

When the object is located closer to a converging lens or a converging mirror than its focal distance, the image will be virtual and erect, not real and inverted. These are the principles used with magnifying glasses used to read small print and a concave mirror, used as a shaving mirror.

Consider a concave mirror whose radius of curvature is 50cm. Therefore, the focal length of the mirror is f = r/2 = .5/2 = .25m, and the reflecting power of the mirror is 1/f = 1/.25 = +4.00D.

1. If an object lies 1m in front of the mirror, the image is (-1 +4 = +3) and therefore is real and lies .33cm in front of the mirror.
2. If an object point is 50cm in front of the mirror (coincides with C – the center of curvature) the image point (-2 + 4 = +2) also coincides with C.
3. If an object point coincides with F, the focal point of the mirror, the image point (-4 + 4 = 0) is at infinity.
4. If an object point lies 20cm in front of the mirror, the image point (-5 + 4 = -1) is virtual (reflected rays are divergent) and lies 1m in back of the mirror.

When doing ray tracing, there are three rays that follow simple known paths before and after their refraction by the mirror, just as there was with lenses. Use two of the three construction rays to find the image

1. Draw a line from the object to the mirror, parallel with the direction of light. At the lens, draw the line through the primary focal point of the mirror.
2. Draw a line from the object, through the center of curvature of the mirror.
3. Draw a line from F to the mirror and then back parallel to the axis of the mirror.

Where the lines cross is where the image will be formed. By doing this, you can estimate the approximate location of the image, tell whether the image is erect or inverted as well as real or virtual.

The following drawing shows the construction of mirror images produced by a concave mirror:

1. Object farther from mirror than C, image real and inverted
2. Object at C, image also at C, real and inverted
3. Object at F, image at infinity
4. Object closer to mirror than F, image virtual and erect

The following drawing shows the construction of mirror images produced by a convex mirror :

• Object real, image virtual, erect and minified

This mirror has a power of zero. Therefore U + V and m = +1. This indicates that any real object has a virtual erect image of the same size and any virtual object has a real, erect image of the same size.

Indexes of refraction of various mediums

n = speed of light in a vacuum (c)/speed of light in a particular medium (v).

The index tells us how much light has slowed down. Denser media have higher n values; rarer media have smaller n values.

• Vacuum - 1.00
• Air is assumed to be 1.00
• Water - 1.33
• Crown glass - 1.523
• Plastic (CR-39) - 1.498
• Polycarbonate (higher index than glass or plastic) - 1.586
• High index glasses - 1.6/1.7/1.8
• Titanium glass is now available with an index of 1.806. However, it is 2 ½ times heavier than CR-39.

With higher index lenses, chromatic aberration becomes a factor (chromatic aberration will be discussed later). Although glass higher index lenses are thinner, they have a higher specific gravity and so are considerably heavier than plastic or crown glass. Polycarbonate lenses continue to be the lens of choice, because of its greater safety and lighter weight.

As light goes from a vacuum to a medium, the light waves slow down slightly. The denser the medium, the slower they move.

Object vergence V = n/u

Image vergence V’ =n’/u’

Where: n = index of refraction light is coming from

n’ = index of refraction light is going to

u = object distance

u’ = image distance

If light hits the surface of a media at less than a 90° angle, the angle formed between the line representing the path of light and a line that is perpendicular to the surface (the so called normal line), is called the angle of incidence. The line representing the light that emerges on the other side of the interface, measured from the normal line, is called the angle of refraction.

n sin i = n’ sin r where: i = angle of incidence as measured from the normal

r = angle refracted as measured from the normal

When light enters a denser media at an angle, it slows down so that the path becomes a bit more perpendicular. Therefore light assumes a more nearly perpendicular path when passing from a less dense, into a denser medium, but assumes a less perpendicular path when passing from a denser medium into a less dense one.

• When moving from rarer to denser medium, light is bent towards the normal.
• When moving from a denser to a rarer medium, light is bent away from normal.

When light enters a medium, it may be—reflected off the surface, refracted (bending of light due to a change in velocity when it hits the medium) or absorbed (where it is changed into a different type of energy).

Apparent Thickness Formula: n/u = n’/u’

Question: A butterfly is embedded 10cm deep in a piece of CR-39 (n = 1.498) lens material. How far into the lens material does the butterfly appear to be?

Answer: Using the formula n/u = n’/u’, 1.498/10cm = 1/u’, u’ = 10cm/1.498 = 6.68cm.

Question: If a fisherman is going to spear a fish that is 50cm below the surface of the water, that he sees at an angle of 40^o from the surface of the water, where should he aim to spear the fish?

Answer: The image that is viewed, as 40^o from the surface of the water will be 50^o from the normal (a line perpendicular with the surface of the water). Using the formula n sin i = n’ sin r = 1 sin 50^o = 1.33 sin r and r = 35.17^o.

Using the formula n/u = n’/u’ = 1/u’ = 1.33/50cm, u’ = 50cm/1.33 = 37.59cm.

Therefore, the fisherman should aim behind and below the fish because light from the fish is passing from a more dense to a less dense medium and will be refracted away from the normal. The fisherman sees a virtual image in front of and above the actual fish.

Law of reflection A = a’ or A = angle of incidence and a’ equals the angle of reflection. The angle of incidence and the angle of reflection are both measured from a normal to the surface. The normal is at 90° to the surface the light is hitting.

The Critical Angle occurs when going from a denser to a rarer medium. There is a point where a’ = 90° and all light is therefore internally reflected. The angle of incidence (a) that produces this condition is termed the Critical Angle (AC). Total internal reflection occurs when the angle exceeds AC. As n increases, AC decreases. The angle can be determined using Snell’s law as follows:

n sin i_c = n’ sin 90^o where: i_c = the critical angle and the refracted angle is 90^o,

Sin i_c = n’/n x 1

Question: What types of ophthalmic instruments is the application of the Critical Angle the basis for?

Answer: Fiberoptics, gonioscopy, reflecting prisms, and Goldmann lens funduscopy. It is not related to retinoscopy. Total internal reflection occurs when light is trapped in the incident medium.

Lensmaker equation: The surface power of a lens = D_s = (n’- n)/r, where r is in meters, n = the index of the object space and n’ = the index of the image space. This is also called the refractive power or simply the power of a spherical refracting surface.

Prisms are defined as a transparent medium that is bound by two plane sides that are inclined at an angle to each other. Prisms are used to deviate light, but do not change the vergence and for this reason, they do not focus light. With prisms, light is bent towards the base, and the image appears bent towards the apex.

A prism diopter is defined as a deviation of 1 cm at 1 meter. For angles under 45 prism diopters, one prism diopter deviates light by approximately .50°. Therefore, 1 degree of deviation equals approximately 2 prism diopters (Approximation Formula).

Tells how much deviation you get by looking off center of a lens. Deviation in prism diopters equals h (cm) x F where F = power of the lens and h = distance from the optical center of the lens.

Image jump is the sudden shift in the image position when the visual axis descends from the distance optical center to the bifocal segment. Flat top, or executive style bifocals have the least amount of image jump of any bifocal segment.

Question: What is the induced prism for an individual wearing +5.00D OU, when reading at the usual reading position of 2mm in and 8mm down from the optical center of his lenses?

Answer: PD = hF, therefore, vertically +5.00D x .8 = 4PD BU per eye

horizontally +5.00D x .2 = 1PD BO per eye

Question: What vertically compensating prism is needed for an individual wearing +5.00D OD and +2.00D OS when they are viewed in the normal reading position of 8mm down from the optical center of the lens?

Answer: +5.00D x .8 = 4PD BU and +2.00 x .8 = 1.6PD BU, 4-1.6 = 2.4PD BU needed for the left eye.

Spherical - power and radius is the same in all meridians

Aspheric - radius changes from the center to the outside (becomes less curved usually)

Cylindrical- different powers in different meridians

Astigmatism- difference in power between the two meridians.

The power meridian is always 90 degrees to the axis. Therefore, if the axis is 45 degrees, the power meridian is at 135 degrees.

Plano + 5.00 x 45 = +5.00 @ 135/Plano @ 45

• A cylinder is specified by its axis
• The power of a cylinder in its axis meridian is zero
• Maximum power is at 90 degrees to its axis, in the power meridian
• The image formed by the power meridian is a focal line parallel to the axis

The image formed by the axis meridian is unchanged, because the axis meridian has no power

The Interval or Conoid of Sturm is a conical image space bound by the two focal lines of a spherocylinder lens. At the center of the Conoid of Sturm is the Circle of Least Confusion. The Circle of Least Confusion is the dioptric midpoint between the two focal lines, where the horizontal and vertical dimensions of the blurred image are approximately equal. The Circle of Least Confusion is the dioptric midpoint of a cylindrical lens and is defined as the Spherical Equivalent of the cylindrical lens.

• For single vision spherical lenses it is the weaker of the two curves.
• For astigmatic single vision lenses, it is the lesser of the two curves on the side in which the cylinder is ground. Because almost all lenses are designed in a minus cylinder form, manufacturers identify their lenses in terms of the front curve.
• For multifocal lenses, it is the side that has the segment.

• Chromatic (color) Aberrations: is the change in light direction in materials with different refractive index due to the different wavelengths of light.
• Shorter wavelengths (blue) deviate the most in materials with higher index of refraction. This is called Chromatic Dispersion of refracted rays.
• Non-paraxial rays of light cause monochromatic aberrations. Monochromatic aberrations include spherical aberration, coma, oblique astigmatism, curvature of field and distortion.

Aberration can be modified by

• Changing the shape of the lens
• Changing the refractive index of the lens
• Changing the aperture size (results in fewer marginal rays)
• Changing the position of the aperture

In general, it is not possible to eliminate all aberration at once. Minimize one may worsen the other, therefore we need to prioritize and minimize the most irritating aberrations.

• Aberrations are all object/image distance dependent.

Monochromatic Aberration includes

1. Spherical aberration: is shape dependent. Spherical aberration normally increases as you move towards the peripheral portion of the lens. This is because the deviating power of the lens increases towards the periphery of the lens. To minimize spherical aberration, a biconvex lens is used. Aspheric lenses, lenses where the radius of curvature gets flatter in the periphery (have less power at the edge of the lens) also help minimize aberrations. The cornea is an aspheric surface that gradually flattens towards the periphery. Aperture size: The larger the aperture, the more spherical aberration from marginal rays occurs.
2. Coma: is an off axis spherical aberration. Peripheral rays produce coma. The image is a series of circles that form a comet shape. This is primarily a problem for large aperture optical systems and can be ignored in spectacles because of the limited affect of the pupil. If we increase the aperture size, we have more coma. Shorter objects have less coma. Objects off axis have more coma. Lens shape will minimize spherical aberration, but not totally eliminate coma. When the aperture is closer to the lens, greater coma occurs. Aplanatic system is free of spherical aberration and coma.

Increasing pupil diameter causes greater spherical aberration. This is due to off axis points or extended objects that result in light rays passing through the marginal surface of the lens. This results in the lens not focusing the image at the same point due to para-axial rays. The difference in angles causes the aberration. The pupil of the eye corrects spherical aberration and coma.

3. Oblique astigmatism is corrected with curved lenses and is proportional to (object size)².
4. Curvature of field is corrected by the curvature of the retina. Curvature of field is an advantageous aberration in the human eye because it produces a curved image on the retina, as opposed to a flat image.
5. Distortion: is another aberration of thick lenses. It concerns the distortion of straight edges of square objects. There are two types of distortion resulting from lateral magnification of the image that results in a lateral displacement of the image.

• Barrel distortion - where the rays in the center are more magnified than the rays’ further off axis. Due to minification of the corners of a square object, more then the sides, from minus lenses.
• Pincushion - where the central rays are less magnified than the rays off axis. Due to magnification of the corners of a square object, more then the sides, from plus lenses.

Assumes an eye power at the corneal surface of +60.00D (actual power is +59.00D)

Average index of refraction of all ocular media = 1.333

Anterior focal point is approximately 17mm (1/-60 = -16.67mm in front of the cornea)

Eye is 22.6mm in length with the nodal point 5.6mm behind the cornea.

App’s Law: One problem in treating refractive errors is that the corrective lens usually changes the size of the retinal image. Many patients can tolerate the change in image size, but problems arise if the difference in image size between the two eyes is too large. According to App’s Law, the retinal image size will not be different between the two eyes when the spectacle lens is placed in the eye’s front focal plane. The front focal plane of the eye is about 17 mm in front of the cornea (see schematic eye information). Although it is possible to wear glasses so that the spectacle lens is 17 mm in front of the eye, most people prefer to wear them from 10-15 mm in front of the cornea.

Refractive Myopia: occurs when the power of the eye exceeds 60D and the length of the eye is 22.6mm.

Axial Myopia: occurs when the power of the eye is 60D but the eye is longer than 22.6mm.

Refractive Hyperopia: occurs when the power of the eye is less than 60D and the length of the eye is 22.6mm.

Axial Hyperopia: occurs when the power of the eye is 60D but the eye is shorter than 22.6mm.

Far point of the eye: is the object point imaged by the eye onto the retina in an unaccommodated eye. If a corrective lens is used to correct for myopia, the lens has its secondary focal point coincident with the far point of the eye.

• The far point of the emmetropic eye is at infinity.
• Myopia exists if, without accommodation, the point at infinity is imaged in front of the retina (in the vitreous). The stimulus on the retina is therefore not a point, but a blur circle. Moving the object closer to the myopic eye, until the image is a point focus on the retina, establishes the far point of the eye.
• Hyperopia exists, if without accommodation, an object point at infinity is imaged neither in the vitreous nor on the retina, but theoretically, behind it.
• When doing retinoscopy, the examiner is attempting to put the far point of the patient’s eye at the plane of the examiner’s pupil.
• When the reflex shows "against" motion, the far point plane lies between the patient’s eye and the examiner’s eye, indicating myopia.
• When the reflex shows "with" motion, the far point lies outside the interval between the patient’s eye and the observer’s eye, indicating hyperopia, emmetropia or mild myopia.

Near point of the eye: is found when the uncorrected refractive error of the eye is added to the accommodative ability of the eye.

If the amplitude of accommodation is 10D, the near point is 10cm in front of the eye (specifically, 10cm in front of the vertex of the cornea which is used as a convenient reference point)

Question: What is the interval of clear vision for an uncorrected 5.00D myope with 10D of accommodative amplitude?

Answer: Far point = 100/5 = 20cm

Near point = 100/(5 + 10) = 6.67cm

Interval of clear vision = 6.67cm to 20cm

Question: What is the interval of clear vision for an uncorrected 2.00D hyperope with 4.00D of accommodative amplitude?

Answer: Far point is 50 cm behind the eye

With +2.00D of accommodation, the far point is at infinity

Near point = 100/(4-2) = 50cm

Interval of clear vision is 50cm to infinity

• The lens clock physically measures the sagital height/depth.
• The reading is in power (diopters)
• The lens clock assumes that n is in air and n’ = 1.53

To calculate for the lens radius (assumes that s is very small) r = y²/2s

To calculate true power of a single refracting surface (SRS)

F _true = F _{lens clock}((n’ _true – n)/(n’ _{lens clock} – n)

For lenses made with indexes of refraction greater than crown glass, the lens clock will underestimate the true lens power.

The lensometer measures the vertex power of the lens. The vertex power is the reciprocal of the distance between the back surface of the lens and its secondary focal point. This is also known as the back focal length. For this reason, a lensometer does not really measure the focal length of a lens.

A lensometer is really an optical bench consisting of an illuminated moveable target, a powerful fixed lens, and a telescopic eyepiece focused at infinity. The key element is the field lens that is fixed in place so that its focal point is on the back surface of the lens being analyzed.

Direct – image is upright. Magnification is based on the total refractive power of the eye. Using the basic magnification formula of M = F/4, an emmetropic eye of +60.00D would provide +60/4 = +15X. An aphakic eye of +40.00D would provide +40/4 = +10X.

Indirect – the image of the fundus becomes the object of the condensing lens, which then forms an aerial image that is larger and inverted. The two plus lenses (the eye and the condensing lens) determine the magnification of the aerial image. For the emmetropic eye, using the formula M_A = (-)D_E/D_O = (-)60/D (condensing lens), we find that a 20D condensing lens results in (-)60/20 = -3X.

As the power of the condensing lens decreases, the magnification increases. Axial magnification increases exponentially, based on the formula Axial magnification: M_A = (M)². (Additional information on axial magnification is found latter in this review).

Traditionally, 3 types of magnification are discussed :

The easiest way to magnify an object is to bring the object closer to the eye. Children with visual impairments do this naturally. Adults will require reading glasses to have the object in focus.

With reading glasses, as the lens power increases - working distance decreases. The reading glasses do not magnify by their power alone when worn in the spectacle plane. Magnification occurs because the lens strength requires the individual using them to hold things closer to have the object in focus.

Enlarge the object while maintaining the same working distance

Example: Large print

Occurs when the object is not changed in position or size, but has an optical system interposed between the object and the eye to make the object appear larger.

Objects appear larger because the device presents an image to the eye that has a larger angular subtense than the original object had when viewed by the eye alone.

Angular magnification indicates how large an image will appear to an observer relative to the appearance of the original object.

Examples: Telescopes and hand magnifiers

The optical system produces a virtual image smaller than the original object but much closer to the eye. The image has a larger angular subtense than the original object; therefore, the objects appear larger when seen through this optical system even though the virtual image is smaller than the object.

Angular magnification is the ratio of the angular subtense of the image produced by a device, divided by the angular subtense of the original object. Angular magnification takes into account not only the size of an image, but also its distance from the observer.

• Perceived size is proportional to the size of the object’s image on the retina.
• Retinal image size is proportional to the objects angular subtense.
• Angular subtense is directly proportional to the object size and inversely proportional to the object’s distance from the observer.

Magnification looks at the ratio of object size (Y) to the image size (Y’) or the ratio of the angular subtense of the image viewed with the optical system to the angular subtense of the object viewed without the optical system

• Plus (+) indicates the image is upright
• Minus (-) indicates the image is inverted.
• When the image is smaller than the object, magnification is numerically between 0-1.
• If the absolute number is greater than 1, the image is larger.
• If the absolute number is equal to 1, it is the same size.
• If the image is inverted with respect to the object, the magnification is negative.
• A magnification of –4 implies that the image is inverted and 4x larger than the object
• A magnification of 0.2 implies that the image is erect and 1/5 the size of the object.

Question: An object 15mm long is placed 20cm in front of a +10.00 diopter lens. The resultant linear magnification will be?

Answer: 1

Explanation: In order to calculate linear magnification for a single lens system, one must know only the object distance and the image distance and/or the object vergence and the image vergence.

The formula is Magnification = image distance (v)/object distance (u) = U/V.

If the object distance is 20cm, the rays incident on the lens have a vergence of 100/-20 = -5.00D. After refraction, through the +10.00 diopter lens, the rays have a vergence of

-5.00 + (+10.00D) = +5.00D. Therefore, a real image is formed 100/+5.00D = +20cm behind the lens.

As it turns out, the object distance of 20cm and the image distance of 20cm, are equal, so the magnification is 1. Also the object vergence is –5.00D and the image vergence is +5.00D giving a magnification of –5.00/+5.00 of –1, indicating the image is inverted.

• Indicates the physical size of the image relative to the object.
• Is typically used to characterize devices that produce images on film or screens such as cameras or projectors. Transverse magnification is the ratio of image height/object height.

Axial magnification is used when talking about objects that do not occupy a single plane (3D objects). Axial magnification is the distance, along the optical axis, between the two image planes divided by the distance between the two object planes (extreme anterior and posterior points on the object with their conjugate image points). Axial magnification is proportional to the product of the transverse magnifications for the pair of conjugate planes at the front and back of the object.

For objects with axial dimensions that are relatively small, M₁ and M₂ are usually very close in numerical value, which leads to the approximate formula of:

Axial Magnification = M²

Where M is the transverse magnification for any pair of the object’s conjugates.

Question: An example of the importance of axial magnification is the evaluation of optic nerve cupping using indirect ophthalmoscopy. The cup can be evaluated using a +20.00 diopter lens, but a 14.00 diopter lens markedly improves the evaluation. What is the axial magnification of a 20D versus a 14D-condensing lens?

Answer: Lateral magnification produced through the indirect ophthalmoscope is the ratio of the total refracting power of the eye (60D) to the power of the condensing lens. The 14.00 diopter lens gives a slightly larger transverse magnification (60/20 = 3X versus 60/14 = 4.286X), but a significantly larger axial magnification since axial magnification increases as the square of transverse magnification (3X² = 9X versus 4.286X² = 18.37X). Larger axial magnification increases the distance between the optic nerve rim and the base of the cup in the aerial image, improving assessment of the cup.

Where d = reference distance in meters to the object (image is formed at infinity)

If d = 25cm than M_e = F₁/4

If d = 40cm than M_e = F₁/2.5

Question: A +24.00 diopter lens is used as a hand held magnifier with the patient viewing an object that is 50cm from the eye and at the focal point of the lens. How much larger do things appear to the patient?

Answer: d = .50m, F= +24.00D, M_e=dF_e = .50(24) = 12X

Rated Magnification = M_r = F/4

Assumes that the individual can accommodate up to 4.00 diopters when doing close work which gives d=25cm (25cm is the standard reference distance used when talking about magnification).

Question: A simple lens magnifier to be used as a low vision device is marked 5X (reference plane at 25cm). What would you expect to find when you measure the lens on a lensometer?

Answer: M = F/4 = 5 = F/4, F = 20D

Question: A view of the retina is obtained through an indirect ophthalmoscope, using a 30-diopter lens. The observer is 40cm from the arial image. What is the perceived lateral magnification?

Answer: 1.25x

Explanation: Lateral magnification produced through the indirect ophthalmoscope is the ratio of the total refracting power of the eye (60D) to the power of the condensing lens (30D), assuming the standard reference distance for magnification of 25cm from the observer to the arial image. If the distance is greater than 25cm, the lateral magnification is multiplied by the ratio of the standard reference distance, 25cm, to the distance in question, 40cm.

60/30D = 2x magnification at 25cm (2 x 25cm/40cm) = 1.25x magnification

Angular/Apparent/Perceived Magnification = M_a=1+hF₁

Gives the ratio of image size to object size at whatever distance the object is located (h equals the eye to lens distance).

M_a increases with increased object distance while the effective magnification (M_e=dF₁) remains constant for a specific reference distance d.

The underlying assumption in this equation is that the patient is "supplying" one unit (1X) of magnification

Some companies use F/4 (Rated Magnification) while others use F/4+1 (Conventional Magnification) to determine magnification strength for their magnifiers.

• Magnification needs are based on the initial reference value and the desired final value. Clinically, it is the acuity divided by the goal acuity (h/h^’).
• Diopter power, on the other hand, is an absolute value and is the same under all conditions.

Keplerian telescopes have a weak (+) objective lens and a strong (+) eyepiece lens. The lenses are separated by the sum of their focal lengths. Keplerian (astronomical) telescopes give an inverted image so they require an erecting lens or prisms to make it a Terrestrial telescope.

Galilean telescopes have a weak (+) objective lens and a strong (-) eyepiece lens. The lenses are separated by the difference of their focal length.

The angular magnification of a telescope is equal to the power of the eyepiece divided by the power of the objective. M_{A Telescope} = (-) F_E/F_O

• The eyepiece in the Galilean telescope has a negative power. Therefore, the magnification given by the equation above is positive, indicating an upright image.
• A Keplerian telescopes have both positive objective and eyepiece lenses, the magnification is negative, indicating an inverted image.
• Galilean telescopes have several practical advantages for low vision work. The image is upright, without the need for image erecting prisms and the device is shorter. Galilean telescopes typically are 2, 3 or 4x in strength, inexpensive, light, and have a large exit pupil, which makes centering less difficult.
• 4x telescopes and stronger are usually Keplerian in design which gives an optically superior image, but are more expensive with a smaller exit pupil requiring better centering and aiming. In binoculars, they contain prisms to erect the otherwise inverted image.
• Galilean telescopes used as surgical loupes, require an add to be combined with the objective lens. The field size is far smaller than that obtained with bifocal spectacles.
• Telescopic loupes can produce asthenopia with any type of refractive error. If binocular loupes are not aligned properly, vertical or horizontal phorias can be induced. Adopting a working distance too far inside the focal distance of the "add" can require excessive accommodation, even for a myope.
• When viewing a near object through an afocal telescope, the telescope acts as a vergence multiplier. The approximate accommodation required is given by A_oc = M²U where A_oc = vergence at eyepiece = accommodation, U = object vergence at objective = 1/u, M is the magnification of the telescope. M is (+) for Galilean and (-) for Keplerian when used in this equation. U is also a negative number for these equations.
• With any telescope, the secondary focal point of the first lens must coincide with the primary focal point of the second lens. With Galilean telescopes, the second lens is minus and so the primary focal point is virtual.

"May be due to differences in the size of the optical images on the retina or may be anatomically determined by a different distribution in spacing of the retinal elements" (Duke-Elder), 1963.

Aniseikonia is a term coined by Dr. Walter Lancaster in 1932. It means literally "not equal images (either size, shape or both)" from the two eyes, as perceived by the patient and is one of the conditions most frequently associated with the correction of anisometropia with spectacles. It is an anomaly of the binocular visual process that effects the patient’s perceptual judgment. The most common cause is the differential magnification inherent in the spectacle correction of anisometropia, producing different sized retinal images. Approximately 1/3 of the cases of aniseikonia are predicted from anisometropia. Aniseikonia is more commonly caused by unequal refractive errors such as monocular aphakia or pseudophakic surprises. However, it is also found with retinal problems and occipital lobe lesions.

Aniseikonia occurs in 5-10% of the population with only 1-3% having symptoms.

The perception of an image size disparity between the two eyes is due to the image on the retina not falling on corresponding retinal points. The ocular image is the final impression received in the higher cortical centers, involving the retinal image with modifications imposed by anatomical, physiologic, and perhaps psychological properties of the entire binocular visual apparatus. This is why there are cases of aniseikonia in individuals with emmetropia and isometropia.

In general aniseikonia is associated with a false stereoscopic localization and an apparent distortion of objects in space. Aniseikonia can be the cause of asthenopia, diplopia, suppression, poor fusion, headaches, vertigo, photophobia, amblyopia, and strabismus. The differences in size may be overall, that is, the same in all meridians, or meridional, in which the difference is greatest in one meridian and least in the meridian 90° away.

Clinically, aniseikonia usually occurs when the difference in image size between the two eyes approaches 0.75%. Individuals with greater than 4-5% image size difference, have such a large disparity in image size, that they generally do not have binocularity. It is usually assumed that patients can comfortably tolerate up to 1% of aniseikonia in non-astigmatic cases.

A change in refractive correction is always accompanied by some change in the retinal image size and in the conditions under which the patient sees. The magnitude of these changes and the patient’s tolerance determines whether these changes will produce symptoms of discomfort or inefficiency. Persons with normal binocular vision can readily discriminate differences in image size as low as .25 to .5 percent. For persons with normal binocular vision, a deviation of 4-5x the threshold of discrimination is usually considered significant.

Aniseikonia can be noted when a patient, for the sake of comfort, prefers to use one eye for reading or watching moving objects. If an individual can learn to rely on non-stereoscopic, rather than stereoscopic clues, they may be able to avoid irritation from aniseikonia, even when it is present.

Aniseikonic patients may see an apparent slant of level surfaces, such as tabletops and floors. The effect is more pronounced with objects on the surfaces, for instance, with an irregular pattern carpet on the floor. For high levels of cylinder correction, spherical equivalents may help reduce the aniseikonia.

• A handy rule of thumb for spectacle correction is that each diopter changes the retinal image size by about 2%. Plus lenses magnify, minus lenses minify.
• The magnification for flat trial case cylinders is approximately 1.5% per diopter.
• The uncorrected axial myopic eye will have a larger image by 1.5% per diopter and the uncorrected axial hyperopic eye will have a smaller image by 1.5% per diopter. This holds for anisometropia primarily of refractive origin. However, since anisometropia may be partially axial, an estimate of 1% per diopter is more clinically useful.
• The corrected axial myopic eye will have a smaller image by 1.5% per diopter and the corrected axial hyperopic eye will have a larger image by 1.5% per diopter. This holds for anisometropia primarily of refractive origin.

When considering axial versus refractive anisometropia:

If the amount of anisometropia is >2D – assume it to be axial

If the amount of anisometropia is < 2D or is in cylinder only – assume it to be refractive

Spectacle correction of astigmatism produces meridional aniseikonia with accompanying distortion of the binocular spatial sense. Anisometropia is commonly stated to be present if the difference in the refractive correction is 2.00D or more either spherical or astigmatic. However, smaller differences than 2.00D may be significant.

When prescribing aniseikonic lenses, it is important to realize that the size and shape of the final image does not matter, it is only important that the images of each eye match each other. For this reason, instead of magnifying the image of one eye, it may be easier to minify the image of another. This may allow for a more cosmetically acceptable spectacles or at least ones that are easier to manufacture, and therefore, less costly.

Cylindrical corrections in spectacle lenses produce distortion. This is a problem of aniseikonia, which may be solved by prescribing iseikonic spectacle corrections. Iseikonic spectacle corrections (Iseikonia is when perceived images are the same size)

may be complicated and expensive and the vast majority of practitioners prefer to prescribe cylinders according to cylinder judgment using guidelines that have evolved over the years. Remember the reason for intolerance of an astigmatic spectacle correction is distortion caused by meridional magnification which is more poorly tolerated. Unequal magnification of the retinal image in the various meridians produced monocular distortion manifested by tilted lines or altered shapes of objects. The monocular distortion by itself is rarely a problem. The effect is too small.

Oblique meridional aniseikonia causes a rotary deviation between fused images of vertical lines in the two eyes. The maximum tilting of vertical lines is called the Declination Error. The maximum declination error occurs when the corrected cylinder axis is at 45 or 135°, but even under these conditions, each diopter of correcting cylinder power produces only about 0.4° of tilt. Clinically significant problem begin to occur when the declination approaches 0.3%. Minor degrees of monocular distortion can produce major alterations in binocular spatial perception.

Knapp’s Rule states that the proper corrective lens placed at the anterior focal point of an eye will produce retinal images of the same size no matter what amount of axial ametropia exists. To prevent this from being strictly applied in clinical practice: ametropia is almost never purely axial, and a vertex distance of 16-17 mm for spectacle correction is impractical. Furthermore, the retina in the myopic eye of a unilaterally high myope is stretched; this increase in separation of photoreceptors can result in the effective magnification not being what you would expect.

The Total magnification of a lens (M_T) is found by adding the magnification from its power (M_P) and the magnification from its shape (M_S). Therefore, total magnification (M_{T) =} M_P + M_S.

Magnification from power (M_P) is dependent on the dioptric power of the lens (D_V) and its vertex distance (H). If H is measured in cm, the relationship is M_P = D_VH. From this formula, we see that moving a lens away from the eye increases the magnification of a plus lens and the minification of a minus lens. Moving a lens toward the eye (decreasing the vertex distance) decreases the magnification of a plus lens and the minification of a minus lens. These effects are especially notable with higher powered lenses.

+10.00D lens @ 10mm and 15mm vertex distance

@ 10mm, M_P = D_VH = +10.00 x 1.0 = +10

@ 15mm, M_P = D_VH = +10.00 x 1.5 = +15

-10.00D lens @ 10 and 15 mm vertex distance

@ 10mm, M_P = D_VH = -10.00 x 1.0 = -10

@ 15mm, M_P = D_VH = -10.00 x 1.5 = -15

For spectacle lenses remember, as you move a lens closer to the eye, you must add plus power to the lens. Therefore remember CAP = Closer Add Plus

Magnification from shape (M_S) is dependent on the curvature of the front surface of the lens D₁ and the center thickness of the lens t. The 1.5 in the following equation is the index of refraction (approximately) of glass or plastic. M_S = D₁ (t_cm/1.5). Therefore, the more curved the lens, the larger the D₁ and the more magnification from shape the lens has. Also, the thicker the lens (t) the more magnification from shape.

Front curve of a +2.00D lens is +2.00D and +6.00D, Thickness is 2mm.

M_S = D₁ (t_cm/1.5) = +2.00(.2/1.5) = +.27

M_S = D₁ (t_cm/1.5). = +6.00(.2/1.5) = +.80

Front curve of a +2.00D lens is +2.00D and +6.00D, Thickness is 4mm.

M_S = D₁ (t_cm/1.5) = +2.00(.4/1.5) = +.53

M_S = D₁ (t_cm/1.5). = +6.00(.4/1.5) = +1.60

Magnification may be reduced by making the front surface power of a lens less positive.

Decreasing center thickness also decreases magnification.

However, a change in either the front curve or the thickness of the lens will also cause the vertex distance (h) to be changed so that the magnification from the power factor (M_P) is also affected. If the front curve is changed to give the magnification or minification needed, the back curve must also be changed to maintain the same power of the lens.

If for example the front curve is increased, the back curve must also be increased, which increases the vertex distance. If the front curve is flattened, the back curve must be flattened, which causes the vertex distance to decrease. If center thickness is increased to increase the magnification of the lens, but the front curve if left the same, the increase moves the back surface closer to the eye by the amount of the increase, therefore decreasing the vertex distance. On the other hand, it the center thickness is decreased, but the front curve is left the same, the decrease causes the vertex distance to be increased by that amount.

For further review on this subject, go to the

THILL Aniseikonia Worksheet in Duane’s Clinical Ophthalmology, Volume 1, chapter 47, page 9

Contact lenses may provide a better solution than spectacles in most patients with anisometropia, particularly children, where fusion may be possible, because it gives the least change in image size from the uncorrected state in refractive ametropia.