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Physics Factsheet September 2001

Number 24

Physics of the Eye The eye and sight Again when a scale drawing is constructed it is possible to determine the exact nature of the usually virtual, upright and diminished image formed by a diverging lens.

The eye can be modelled by an optical refracting system consisting of a series of lenses, seperated by media of different densities. There are broadly two types of lens: (see Fig 1) a) convex – converging lenses which are thicker at the centre than at the edge; b) concave – diverging lenses which are thicker at the edge than at the centre.

Fig 2. Graphical construction for a converging lens f'

ray 1 ray

The images that are formed by convex and concave lenses can be illustrated by graphical construction of the rays as they traverse the lens.

Object

F r ay

2

optical axis or

F'

O

principal axis 3

f

a) Graphical construction of the image from a converging lens (Fig 2.) • Ray 1 is parallel to the optical or principal axis and is refracted by the lens through the second principal focal point (F'). • Ray 2 passes through the optical centre of the lens and is undeviated. • Ray 3, which passes through the first principle focal point (F), emerges parallel to the optical axis. The intersection of these rays indicates the point at which the image is formed and the distance from the optic axis to this intersection indicates the size of the image. When the graphical construction is drawn to scale, it is therefore possible to determine whether the image is real or virtual, erect or inverted, diminished or magnified.

Image F ' = second principal focal point

F = first principal focal point f = first principal focal length O = optical centre

f ' = second principal focal length

Fig 3. Graphical construction for a diverging lens

ray 1 ray 2

b) Graphical construction of the image from a diverging lens (Fig 3.) • Ray 1 is parallel to the optical axis and diverges following refraction at the lens. The refracted ray appears to be incident from the second focal point (F'). • Ray 2 passes through the optical centre and is undeviated. The intersection between ray 2 and the dotted line indicates the point at which the image is formed. The distance from the optical axis to this intersection indicates the size of the image. This image is always virtual (this is always the case if a ray intersects the dotted construction line).

O object

optical centre

F' image second focal point

F first focal point

Fig 1. Examples of different convex and concave lenses Convex

symbol for converging lens rays from

F

O

infinity equiconvex

bi-convex

has two equally curved surfaces

has two unequally curved surfaces

plano-convex has one flat surface

f

meniscus

f = focal length F= focal point

has one convex and one concave surface

symbol for diverging lens

Concave

rays from infinity equiconcave

bi-concave

plano-concave

meniscus

1

O

Physics Factsheet

Physics of the eye Calculating the position of images - the lens formula

Basic structure and function of the eye

In addition to graphical construction, the position of an image can be calculated using the lens formula, provided the object position and the focal length of the lens are known. lens formula:

The focusing ability of the eye and accommodation The image that is formed at the retina is both inverted and laterally reversed. Light rays from an object are first refracted by the cornea where in fact most refraction occurs. In order to produce a focused image at the retina, the crystalline lens changes thickness and therefore its refracting ability.

1 1 1 + = u v f

u = object distance v = image distance f = focal length

Magnification m can be calculated using: m =

This ability to change thickness is essential to maintain focus of objects at different distances through a mechanism called accommodation. This can be illustrated by two examples:

v u

Distant object

Sign Convention

rays from distant

Most exam boards use the “real is positive” sign convention (you should check that yours does). In the real is positive sign convention all the distances to the real objects and images are taken as being positive and the distances to the virtual objects and images are taken as negative. Biconvex lenses are considered to have a positive focal length (when the image is on the opposite side to the real object) and biconcave lenses are always considered to have a negative focal length.

object parallel to visual axis For a distant object the ciliary muscle is relaxed, the suspensory ligaments are taut and this tension results in the crystalline lens being pulled into a thinner structure with a weaker refracting power but still enabling a focused image to be produced on the retina.

The alternative is the New Cartesian sign convention, which works by taking any distance to the left of the lens as being negative and any distance to the right as being positive.

1

Near object rays from a near

Typical Exam Question An object is placed 0.10 m from a converging lens. It forms a real image 0.20 m from the lens. Find the focal length of the lens. [2]

object diverge from the visual axis

Exam Hint: It helps to sketch a ray diagram of the lens problems – this makes sure that your calculated answer is sensible. If you haven’t enough space to do a sketch make use of any blank areas on the exam paper and make a reference to it so that the examiner knows what you have done.

1 with accommodation

10 + 5 =

2 without accommodation

As the object becomes closer, without accommodation the image is focused behind the retina as the total refracting power is too weak. However, with accommodation the ciliary muscle contracts, the tension in the suspensory ligaments and therefore the lens is reduced. The elastic lens returns to its more bulbous state and this its refracting power is increased, therefore allowing the image to be focused on the retina.

Answer Focal length is positive since it is a converging lens: u = 0.10 m, v = 0.20 m, f = ? 11 11 11 1 1 1 + = ⇒ + = f 0.1 0.2 u v f

2

✓

1 ⇒ 1 = 0.067 ✓ f = 15 f

Fig 4. Basic structure and function of the eye aqueous humour provides nutrition for the lens and cornea cornea - transparent structure with ~ 2/3 of the refracting power of the eye pupil iris - circular, muscular diaphragm which controls the diameter / amount of light that enters the inner eye through the pupil. It is the coloured part of the eye

lens - transparent and elastic, enclosed in a capsule. Flexible structure that enables objects at different distancest to be focused vitreous humour jelly-like substance that maintains the spheroidal shape of the eye ~ 1/3 of the refracting power of the eye

yellow spot - 1mm in diameter, situated on the optical axis. At its centre is a tiny pit called the fovea - high concentration of cone cells and no rods, it is where objects can be clearly seen. optic nerve transports sensory information from retina along the visual pathway to the occupital lobes

suspensory ligaments maintain position of lens circular cillary muscle and involved in its change muscle that contracts /relaxes of thickness and changes the thickness of the lens

blind spot - where optic nerve leaves the eye, no light-sensitive cells are found at this point.

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Physics of the eye

Physics Factsheet Spherical and astigmatic refractive errors A sphere has the same surface curvature and therefore the same power of refraction in different meridians. Corresponding light rays incident along different meridians will be focused at the same point. (Fig 6.)

Other terms used in relation to accommodation Near point of accommodation is the nearest point in space relative to the fovea that an object can be seen clearly when the maximum amount of accommodation is exerted.

Fig 6. Spherical surface

Far point of accommodation is the farthest point in space relative to the fovea that an object can be seen clearly when the accommodation is fully relaxed.

A

Depth of field. For a given amount of accommodative effort, the depth of field is the distance over which an object can be displaced without causing a reduction in the sharpness of an image. The smaller the pupil diameter, the greater the depth of field. (Fig 5)

B

B'

Fig 5. Depth of field A

O

B

Refracting Surface

Image A'

depth of field A' = focal point of object A B' = focal point of object B

A' B' depth of focus

AA' is the 90o meridian BB' is the 180o meridian

In contrast, an astigmatic surface is one where the surface curvature in different meridians is different. Therefore light rays that are incident along different meridians are focused at different points. (Fig 7.)

With the eye focusing an object at point O, there is a fixed level of accommodation. Moving the object to points A or B produces the same sized blurred image at the retina therefore the sharpness of the image of objects A and B are the same.

Fig 7. Astigmatic surface A FA FB

Refractive errors and corrections

B

The following table illustrates different refractive errors and the lenses that are used to correct them.

B' fB

Table 1. Refractive errors and corrections Refractive Error

Image Position

Lenses Used

A'

Normal vision

FA = focal point of meridian AA' FB = focal point of meridian BB' AA' is the weaker 90o meridian and produces a longer focal length fA BB' is the stronger 180o meridian and produces a shorter focal length fB

No correction required Image of object at infinity is focused at the fovea

Spectacle correction in astigmatism

Myopia (Short Sightedness)

The astigmatic eye has two focal points which have to be individually corrected. A cylindrical lens has power in one meridian only. It is used to produce a single point of focus before a spherical lens is used to produce a focus of the object at the fovea (Fig 8).

Image of object at Diverging lens is used to infinity is focused focus the image at the in front of the fovea fovea

Fig 8. Correction in astigmatism

Hypermetropia (long sightedness)

Presbyopia - with increasing age the level of accommodation decreases. Beyond the age of 40 years this decreases to a level where objects at near distances e.g. less than 1 m cannot be focused

fA

Image of object at infinity is focused behind the fovea

Converging lens is used to focus the image at the fovea

Image of object at near distance is focused beyond the fovea

Converging lens is used to focus the image at the fovea

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Physics Factsheet

Physics of the eye

Scotopic and photopic vision Rod cells react over a wide range of wavelengths (see Fig 10) and respond to lower levels of light (low luminance) than the cones. So in dim light we detect shape but not colour. This is known as scotopic vision. At high light levels (high luminance) cones respond to stimulation by light. This is known as photopic vision. When the luminance level changes from photopic to scotopic levels of luminance there is a shift of the spectral sensitivity curve. This is called the Purkinje shift. Therefore a red light appears less luminous in scotopic conditions compared to photopic conditions.

Calculating the power of the correcting lens The unit of power of a lens is called the dioptre, D. It is given by the reciprocal of the focal length f in metres. For example, a lens with a focal length of 25 cm has power: 1 1 = +4.00D D= = f 0.25 The sign convention is important in this equation.

Fig 10. Relative sensivity of a light and dark adapted eye f = −0.5 m

f = +0.333 m 1 1 = +3.00 D = 0.333 f

light-adapted eye

1 1 _ = 2.00 D = _ 0.50 f In order to calculate the power of a spectacle lens used to correct a refractive error, the lens formula has to be used. In these calculations it is assumed that the distance between the spectacle lens and the eye compared with the distance between the object and the lens is negligible. D=

0.8

Physical principles of the retina The retina consists of rod cells which are most sensitive to light of low luminance levels e.g. at night. They are not colour sensitive. In contrast, cone cells are most sensitive to high luminance levels (photopic conditions) e.g. daytime. They are sensitive to different colours and responsible for providing high visual acuity. There are the three classes of cone cells which are differentiated by the wave length: red, green and blue-sensitive cone cells

Relative sensitivity

D=

dark-adapted eye

1.0

0.6

0.4

0.2

0 400

500

600 Wavelength/nm

700

Fig 9. Classes of cone cells blue-sensitive cones 100 Percentage of maximum absorbtion

Some differences between photopic vision and scotopic vision are shown in Table 2

green-sensitive cones red-sensitive cones

Table 2. Differences between photopic and scotopic vision Scotopic vision

75

uses rods night vision distinguishes only light and shadow little detail seen

50

25

Photopic vision uses cones mainly daytime vision distinguishes colours large amount of detail seen

Practice Question A 52 year old man whose eyeball is 0.025 m long suffers from presbyopia. His near point has increased to 0.5 m and his far point is 3.5 m.

0 400

450

500

550

600

650

Wavelength/nm

(a) Explain what presbyopia is and why it occurs as people age.

[2]

(b) Calculate the range of power of the man’s eye.

[4]

(c) Explaining your reasoning, calculate the power of the lenses needed to allow the man to form a clear image of: (i) an object at infinity (ii) an object at 0.20 m from the eye. [6]

The graph illustrates the approximate absorbtion of the three human cones. A – blue (short wave receptor) B – green (medium wave receptor) C – red (long wave receptor)

(d) Explain what your answers to (c) suggest about the sort of spectacles that the man requires. [1]

The proportion of different classes of cones that are stimulated by light of different wavelengths is therefore one step in the perception of colour.

Exam Hint – always be guided by the number of marks allocated to each part of question. Examiners often work on the “mark per minute” rule – but do remember that you would be expected to know and understand the theory in an examination!

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Physics of the eye

Physics Factsheet Answers

Exam Workshop This is a typical poor student’s answer to an exam question. The comments explain what is wrong with the answer and how they can be improved. The examiner’s answer is given below.

(a) Presbyopia is the failure of the eye to accommodate for the whole range of vision; so presbyopic people cannot see objects if they are too close. It develops as the eye lens becomes less flexible as it ages so that it cannot accommodate fully as a young “normal” eye would.

(a) Explain what is meant by a defect of the eye. [2] An eye that doesn’t work properly or has some difficulty that gives a 0/2 blurred image.!!

(b)

Although this answer might be acceptable in general conversation it is simply too vague to gain marks on a physics paper – be precise with your answers.

= 42 dioptre ✓ 1 1 1 + + = u v f

(b) Explain what is meant by an eye being (i) short-sighted, (ii) long-sighted and explain how these defects can be corrected. [6] Short sight is when you cannot see things in the distance. Wearing glasses with concave lenses can correct it. ""! Long sight is when you can see distant object but not ones close to you. This time you need glasses with convex lenses. ""! 4/6

1 1 1 + + = power of eye in reading at far point ✓ = 3.5 0.025 ff = 40.3 dioptre ✓ range of power = 42.0 – 40.3 = 1.7 dioptre ✓

The basic information is correctly communicated in these answers but the “explain” part has been ignored – why do these lenses correct these defects of vision?

(c) (i)

(c) (i) What is presbyopia? (ii)Explain why a presbyopic eye is difficult to correct. [4] Presbyopia is when you can’t see things clearly when they are close up or far away. "! It is difficult to correct because you need both a convex and a concave lens so you can wear them both at the same time. "! 2/4

The effect of the lens is to make the object at infinity seem as if it were at the man’s far point (i.e. produce a virtual image 3.5 m from the eye). This virtual image behaves as a virtual object for the eye lens, which will then project a real image onto the man’s retina. ✓ 11 1 11 uu ++ vv += ff = power 1 1 1 ✓ + + = ∞ − 3 .5 f

Again an answer with much correct physics - but one that hasn’t been developed sufficiently to score all of the marks. The hint is the number of marks available and in each case its pretty clear that the student is making a single point each time.

1 1 = -0.3 dioptre ✓ = f -3.5 i.e. it must be a diverging lens.

Examiner’s Answers (a) An eye defect causes the image formed on the retina to be blurred. " It arises because the focal length of the eye lens is either too long or too short or because there are irregularities in the shape of the cornea (meaning that it is not truly spherical) " (this is a brief answer for two marks; you could write a number of books to answer this question!)

(ii) To improve the man’s near sight, an object at the improved near point (0.2 m) must form a virtual image at the man’s actual near point (0.5 m) ✓ 1 1 1 = power + + = u v f

(b) (i) Short-sightedness or myopia results in the image being formed in front of the fovea – meaning that it is blurred". The defect originates as a result of the eye lens having too short a focal length" and so needs be corrected using a diverging lens" (which makes the combined focal length of the correction lens + the eye lens longer than the eye lens alone).

1 1 1 ✓ + + = 0.2 − 0.5 f = +3 dioptre ✓

(ii) Long-sightedness or hypermetropia results in the image being formed behind the fovea – again meaning that it is blurred. " The defect originates as a result of the eye lens having too long a focal length" and so needs to be corrected using a converging lens" (which makes the combined focal length of the correction lens + the eye lens shorter than the eye lens alone). Ray diagrams would be helpful in focusing your answers here (c) (i)

1 1 1 ✓ u + v = f 1 1 1 + + = = power of eye in reading at near point ✓ 0.5 0.025 f n

i.e. it must be a converging lens. (d) The man needs two pairs of glasses for close-up and for distance viewing (or bifocal lenses). ✓

Presbyopia is the lack of ability of the eye to be able to accommodate. " This develops as we age, because the ciliary muscles become less able to change the shape of the lens in order to focus objects that are either very close or far away from the eye. "

Acknowledgements: This Physics Factsheet was researched and written by Nirinder Hunjan The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

(ii) Since a converging lens is needed to correct the presbyopic’s near vision and a diverging lens to correct his distant vision" either he needs to use two pairs of glasses or else glasses with bifocal or varifocal lenses. "

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