chapter 2 refracting light - jacaranda | shop...the reason is not apparent. when we set up a special...

$: Chapter 2 Refracting light - Jacaranda | Shop...the reason is not apparent. When we set up a special situation, such as in ﬁgure 2.2, where a straight rod is placed in a beaker of$
Chapter

2Refracting

light

Figure 2.1

he path of a light ray bends when travelling from one material into

another, such as through water. A person’s legs appear shorter.

T

Remember Before beginning this chapter, you should be able to:• associate changes in direction of light rays with

refraction• describe some everyday uses of lenses.

Key ideasAfter completing this chapter, you should be able to:• describe the bending of light as it passes from one

medium into another• use the ray model to describe the refraction of light• mathematically model refraction using Snell’s Law• use the ray model of light to describe and explain

total internal reflection and mirages• describe the path of light through an optical fibre• discuss the limitations of the ray model in

describing and explaining refraction • demonstrate how a particle model and a wave

model can be used to describe and explain the reflection and refraction of light

• use the ray model to locate and describe images formed by convex and concave lenses.

Jac Phys 1 2E - 02 Page 23 Tuesday, October 21, 2003 1:43 PM

WAVE-LIKE PROPERTIES OF LIGHT24

Experience shows that when you arespearing for fish in the shallows youmust aim the spear below where thefish appears to be in the water. At thebeach or in a pool, people standing inthe shallows appear to have shorterlegs. Our perception is distorted, butthe reason is not apparent.

When we set up a special situation,such as in figure 2.2, where a straightrod is placed in a beaker of liquids thatdo not mix, the idea of change ofdirection is apparent. This change indirection of the light is called refraction.

Figure 2.2 An example ofrefraction

BENDING OF LIGHTThe ray model can help explain our observations of light. If a fish seemscloser to the surface of the water, the ray of light from the fish must havebent. To our eye, the ray seems to be coming from another direction (see

figure 2.3). Given that light can travel bothways along a light path, the fish will see

the spear thrower further towards thevertical than he actually is.

Figure 2.3 The rays from the fish bend when they enter the air. To the eye, the rays appear to come from a point closer to the surface.

The ray model not only gives us a way of describing our observations ofthe bending of light, but also of taking measurements. The angle that aray of light makes with the normal, angle of incidence and angle ofrefraction can be measured and investigated.

Snell’s LawIn 1621, the Dutch physicist Willebrand Snell investigated the refractionof light and found that the ratio of the sines of the angles of incidenceand refraction was constant for all angles of incidence.

The diagram on page 25 shows how an incident ray is affected when itmeets the boundary between air and water. The normal is a line at rightangles to the boundary, and all angles are measured from the normal.Some of the light from the incident ray is reflected back into air. The restis transmitted into the water. The following ratio is a constant for allangles for light travelling from air to water:

= constant.

Inve

stigations

Investigation 2.1Seeing is believing

Refraction is the bending of light as it passes from one medium into another.

The normal is a line that is perpendicular to a surface or a boundary between two surfaces.The angle of incidence is the angle between the incident ray and the normal. The angle of refraction is the angle between the refracted ray and the normal.

θisin θrsin

---------------------

air

water

Fish appears to be here.Fish is here.


CHAPTER 2 REFRACTING LIGHT 25

Figure 2.4 The ratio is constant for all angles for light

travelling from air to water.

Snell repeated his experiments with different substances and foundthat the ratio was still constant, but it had a different value. This sug-gested that different substances bend light by different amounts.(Remember that some light is always reflected.)

In fact, there is a different ratio for each pair of substances (forexample air and glass, air and water). A different ratio is obtained forlight travelling from water into glass. The value of the ratio is called therelative refractive index because it depends on the properties of two dif-ferent substances.

The bending of light always involves light travelling from one sub-stance to another. It is not possible to find the effect of a particularsubstance on the deflection of light without adopting one substance as areference standard. Once you have a standard, every substance can becompared with it. A natural standard is a vacuum — the absence of anysubstance. The absolute refractive index of a vacuum is given the value ofone. From this, the absolute refractive index of all other substances canbe determined. Some examples are given in table 2.1. (The word ‘abso-lute’ is commonly omitted and the term ‘refractive index’ usually refersto the absolute refractive index.)

Table 2.1 Values for absolute refractive index

incident ray

boundary

angle ofincidence

angle of reflection

normal

reflected ray

refracted rayangle ofrefraction

�r

�i�iair

water

sin θi

sin θr-----------------------

Relative refractive index is a measure of how much light bends when it travels from any one substance into any other substance.

The absolute refractive index of a substance is the relative refractive index for light travelling from a vacuum into the substance. It is commonly referred to as the refractive index.

MATERIAL VALUE

Vacuum 1.000 0

Air at 20°C and normal atmospheric pressure 1.000 28

Water 1.33

Perspex 1.49

Quartz 1.46

Crown glass 1.52

Flint glass 1.65

Carbon disulfide 1.63

Diamond 2.42



The refractive index is given the symbol n because it is a pure numberwithout any units. This enables a more useful restatement of Snell’s Law,for example:

nair sin θair = nwater sin θwater.

More generally this would be expressed as follows:n1 sin θ1 = n2 sin θ2.

This formula is more useful because the two values for a substance —its refractive index and the angle the light ray makes — are together onthe same side and are combined in the same way on both sides of theequation.

A ray of light strikes a glass block of refractive index 1.45 at an angle ofincidence of 30°. What is the angle of refraction?

nair = 1.0; θair = 30°; nglass = 1.45; θglass = ?

1.0 × sin 30° = 1.45 × sin θglass (substitute values into Snell’s Law)

sin θglass = (divide both sides by 1.45, the refractiveindex of glass)

= 0.3448 (calculate value of expression)⇒ θglass = 20.17° (use inverse sine to find angle whose sine

is 0.3448)⇒ θglass = 20°. (round off to two significant figures)

A ray of light enters a plastic block at an angle of incidence of 40°. Theangle of refraction is 30°. What is the refractive index of the plastic?

nair = 1.0; θair = 40°; nplastic = ?; θplastic = 30°

1.0 × sin 40° = nplastic × sin 30° (substitute values into Snell’s Law)

⇒ nplastic = (rearrange formula to get the unknownby itself)

= 1.286 (calculate value of expression)⇒ nplastic = 1.3. (round off to two significant figures)

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links

Refraction applet

medium 1refractive index n1

medium 2 refractive index n2

boundary

normal

�2

�1

n1 sin �1 = n2 sin �2

Figure 2.5 A graphical depiction of Snell’s Law for any two substances. Note that the light ray has no arrow, because the relationship is true for the ray travelling in either direction.

SAMPLEPROBLEM 2.1

Solution:

sin 30°1.45

-----------------

SAMPLEPROBLEM 2.2

Solution:

sin 40°sin 30°-----------------

ight can be bent by a strong gravitational field, such as that nearthe Sun. The gravitational field can act like a convex lens. Light

from a distant star that is behind and blocked by the Sun bendsaround the Sun so that astronomers on Earth see an image of the starto the side of the Sun.

LAS A MATTER OF FACT



Speed of light in glassDuring the seventeenth century, it was agreed that light changed speedwhen it travelled from air into glass, but how? One explanation or modelof light proposed that light travelled faster in glass. Another model oflight proposed that light travelled slower in glass. Scientists at the timedid not have the technology to measure the speed of light through mat-erials such as water or glass.

It was only in the nineteenth century that Augustin-Jean Fresnel andJean Bernard Léon Foucault were able to measure the speed of light inwater as being less than the speed in air. The second explanation abovewas accepted and the other rejected.

This means that the refractive index now has a physical meaning:

absolute refractive index of water =

wherespeed of light in a vacuum = 3.0 × 108 m s−1.

The refractive index of glass is 1.5. How fast does light travel in glass?

1.5 =

⇒ speed of light in glass = (rearrange formula to get the unknown by itself)

= 2.0 × 108 m s−1.

Apparent depthAt the beginning of the chapter, the experience of the spear thrower wasdescribed and explained. A similar phenomenon occurs when a spearthrower is directly above a fish — the fish appears to be closer to the sur-face than it actually is. This observation is known as apparent depth.Swimming pools provide another example of apparent depth — theylook shallower than they actually are. The refraction of light combinedwith our two-eyed vision makes the pool appear shallower.

The relationship is illustrated in figure 2.6 and can be expressed asfollows:

Figure 2.6 The phenomenon of apparent depth

speed of light in a vacuumspeed of light in water

-----------------------------------------------------------------------------------------------

SAMPLEPROBLEM 2.3

Solution: 3.0 108×(speed of light in glass)-------------------------------------------------------------

3.0 108×1.5

-----------------------

real depthapparent depth-------------------------------------------------------- refractive index.=

air

apparentdepth

realdepth

water



A swimming pool is 2.00 m deep. How deep does it appear as you standon the edge of the pool?

real depth = 2.00 m; apparent depth = ?; refractive index of water = 1.33

= 1.33 (substitute values into relationship)

⇒ apparent depth = (rearrange formula to get the unknown by itself)

= 1.50 m.

TOTAL INTERNAL REFLECTION AND CRITICAL ANGLELight can play some strange tricks. Many of these involve refraction awayfrom the normal and the effect on light of a large increase in the angleof incidence.

Figure 2.7 There are no mirrors in a fish tank but strange reflections can be seen. It appears that light is being reflected off the side of the fish tank and the water surface.

It has already been mentioned that some light is reflected off a trans-parent surface, while the rest is transmitted into the next medium. Thisapplies whether the refracted ray is bent towards or away from thenormal. However, a special situation applies when the refracted ray isbent away from the normal. This is illustrated in figure 2.8. As the angleof incidence increases, the angle of refraction also increases. Eventuallythe refracted ray becomes parallel to the surface and the angle of refrac-tion reaches a maximum value of 90° (see figure 2.8(b)). The corre-sponding angle of incidence is called the critical angle. If the angle ofincidence is increased beyond the critical angle, all the light is reflectedback into the water, with the angles being the same. This phenomenon iscalled total internal reflection (see figure 2.8(c)).

SAMPLEPROBLEM 2.4

Solution:

2.00apparent depth----------------------------------------

Inve

stigations

Investigation 2.2Using apparent depth

to determine therefractive index

2.001.33-----------

The critical angle is the angle of incidence for which the angle of refraction is 90°. The critical angle exists only when light passes from one substance into a second substance with a lower refractive index.Total internal reflection is the total reflection of light from a boundary between two substances. It occurs when the angle of incidence is greater than the critical angle.



Figure 2.8 Three stages of refraction leading to total internal reflection

The critical angle can be calculated using Snell’s Law (see the followingsample problem).

What is the critical angle for water given that the refractive index of wateris 1.3?

nair = 1.0; θair = 90°; nwater = 1.3; θwater = ?

1.0 × sin 90° = 1.3 × sin θwater (substitute data into Snell’s Law)

⇒ sin θwater = (rearrange formula to get the unknown by itself)

= 0.7692 (determine sine values and calculate expression)

⇒ θwater = 50.28° (use inverse sine to find angle)

θwater = 50°. (round off to two significant figures)

Total internal reflection is a relatively common atmospheric phenom-enon (as in mirages) and it has technological uses (for example, inoptical fibres).

MiragesThere are several types of mirage that can be seen when certain atmos-pheric conditions enable total internal reflection to occur. These miragesappear because the refractive indexof air decreases with temperature.

A common type of mirage occursin the desert or above a road on asunny day. As displayed in figure2.10, at ground level the air is hot(A) with a refractive index close to1 (B). As height increases, the tem-perature of the air decreases (C) andits refractive index increases (D).

Figure 2.9 Mirages such asthis are common on hot, sunny days.

air

water

(a) Before critical angle (b) At critical angle (c) After critical angle (total internal reflection)

�i�i

�r

�c�c

SAMPLEPROBLEM 2.5

Solution:

sin 90°1.3

-----------------



Figure 2.10 Temperature and refractive index profiles for the mirage phenomenon

Rays of light from a car, for example, go in all directions. The air abovethe ground can be considered as layers of air. The closer to the ground,the higher the temperature and the lower the refractive index. As a raymoves into hotter air, it bends away from the normal. After successivedeflections, the angle of incidence exceeds the critical angle for air atthat temperature and the ray is totally internally reflected. As the rayemerges, it follows a similar path, refracting towards the normal as itenters cooler air. An image of the car can be seen below street level (seefigures 2.9 and 2.11). The mirage is upside down because light from thecar has been totally internally reflected by the hot air close to the roadsurface.

Another mirage that depends on layers of air at different temperaturesis known as the ‘Fata Morgana’ in which vertical streaks, like towers or walls,appear. This occurs where there is a temperature inversion — very cold atground level and warmer above — and very stable weather conditions.

The phenomenon is named after Morgan le Fay (Fata Morgana inItalian) who was a fairy and half-sister to King Arthur of the Celticlegend. She used mirages to show her powers and, in the Italian versionof the legend, lived in a crystal palace under the sea. The mirage is oftenseen in the Strait of Messina and over Arctic ice. As shown in figure 2.12,the light rays from a distant point are each refracted by the differentlayers of air, arriving at different angles to the eye. The effect is that thepoint source (P) becomes a vertically extended source, like a tower orwall (see figure 2.13).

temperature

A D

B C

refractive index

refractiveindexincreasing

temperatureincreasing

high

low 1.0 (vacuum)

1.000 28(normal air)

groundlevel

treelevel

height

warm air

hot air

road

Figure 2.11 The mirage of the car appears upside down due to total internal reflection in the hot air close to the ground.

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Mirages and more

Figure 2.12 Ray paths for the Fata Morgana

roughseaice P



Figure 2.13 An example of the Fata Morgana. The conditions that encourage the Fata Morgana are particularly common in the polar regions over ice.

Optical fibresAnother example of total internal reflection is in the important techno-logical application of optical fibres. Optical fibres have become a featureof modern life. A thin, flexible cable containing an optical fibre can beplaced inside a person’s body to transmit pictures of the condition oforgans and arteries, without the need for invasive surgery. The same canbe done in industry when there is a problem with complex machinery.

Optical fibres are also the basis of the important telecommunicationsindustry. They allow high quality transmission of many channels of infor-mation in a small cable over very long distances and with negligiblesignal loss (see figure 2.14).

Figure 2.14 A bundle of optical fibres. Each fibre in the bundle carries its signal along its length. If the individual fibres remain in the same arrangement, the bundle will emit an image of the original object.

An optical fibre is like a pipe with a light being shone in one end andcoming out of the other. An optical fibre is made of glass which is about10 micrometres (10 × 10−6 m) thick. Light travels along it as glass istransparent, but the fibre needs to be able to turn and bend aroundcorners. The optical fibre is designed so that any ray meeting the outersurface of the glass fibre is totally internally reflected back into the glass.As shown in figure 2.15, the light ray meets the edge of the fibre at an

An optical fibre is a thin tube of transparent material that allows light to pass through without being refracted into the air or another external medium.

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links

Fibre optics



angle of incidence greater than the critical angle and is reflected backinto the fibre. In this way, nearly all of the light that enters the fibreemerges at the other end.

Figure 2.15 A light ray travels along an optical fibre through total internal reflection.

If the glass fibre is exposed to the air, the critical angle for light travel-ling from glass to air is 42°, which is quite small. Any angle of incidencegreater than this angle will produce total internal reflection. If the fibreis very narrow, this angle is easily achieved.

However, in both medical and telecommunication uses, fibres arejoined in bundles with edges touching. The touching would enable lightrays to pass from fibre to fibre, confusing the signal. To overcome this, aplastic coating is put around the glass to separate the glass fibres. Thetotal internal reflection occurs between the glass and the plastic. Thecritical angle for light travelling from glass to plastic is 82°. This valuepresents a problem because light meeting the edge of the glass at anyangle less than 82° will pass out of the fibre (see figure 2.16).

This has implications for the design of the optical fibre and the beamof light that enters the fibre. The fibre needs to be very narrow and thelight entering the fibre has to be a thin beam with all the rays parallel.

Figure 2.16 Light rays entering the fibre at too sharp an angle are refracted out of the fibre.

LIMITATIONS OF THE RAY MODELSo far in this chapter we have used the ray model to describe how light isrefracted. Ray diagrams illustrate Snell’s Law and have allowed us to visu-alise a range of optical phenomena such as mirages and to developtechnologies such as optical fibres. However, the ray model, which viewslight as a pencil thin beam, does not offer an explanation of why lightrefracts. More sophisticated models are needed to provide an expla-nation for refraction, and in doing so they suggest further experimentsto investigate the properties of light more deeply, and to develop newtechnologies.

optical fibre

light ray

82°

optical fibre

light rays



Two very different models of light were devel-oped in the seventeenth century one by SirIsaac Newton (1642–1727) in England and theother by Christiaan Huygens (1627–1695) inHolland.

Newton’s model was described as a ‘particlemodel’. In his model, light consists of a stream oftiny, mass-less particles he called corpuscles. Theparticles stream from a light source like waterfrom a sprinkler.

Huygens proposed a wave model of light, wherelight travels in a similar way to sound and waterwaves. Light leaves a source in the same way thatwater ripples move out from a dropped stone. Thedisturbance of the water surface travels outwardsfrom the source.

Figure 2.17 Huygens proposed that light travelled outwards from a source like circular ripples on a pond.

How light travelsNewton’s particle model Once ejected from alight source the particles continue in a straightline until they hit a surface.Huygens’s wave model Huygens proposed abasic principle: ‘Every point in the wavefront is asource of a small wavelet. The new wavefront isthe envelope of all the wavelets.’

Reflection of lightNewton’s particle model As particles approach a surface they are repelled bya force at the surface that slows down and reverses the normal component ofthe particle’s velocity, but does not change its tangential component. The par-ticle is then reflected from the surface at an angle equal to its angle ofapproach. The same process happens when a billiard ball hits the cushion.

Huygens’s wave model As each part of the wavefront arrives at the surface,it produces a reflected wavelet. The new wavelets overlap to produce the nextwavefront, which is travelling away from the surface at an angle equal to itsangle of approach.

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Huygens’s principleapplet

Figure 2.19Newton’s particle model of reflection

mirror

i i´

Figure 2.20 The wave model of reflection. C and D are parallel, incoming rays. AB is the wavefront. When A hits the mirror a circular wavelet is produced. By the time B has reached the mirror at E, the reflected wavelet has travelled out to F. The line EF is the reflected wavefront.

HOW DO THE TWO MODELS EXPLAIN

Source

S

Figure 2.18 Every point in the wavefront is a source of a small wavelet. The new wavefront is the envelope of all the wavelets.

(continued )mirror AE

B F

D C

THE PROPERTIES OF LIGHT?



Refraction of lightNewton’s particle model In approaching a denser medium, the particlesexperience an attractive force which increases the normal component of theparticle’s velocity, but does not affect the tangential component. This has theeffect of changing the direction of the particles, bending them towards thenormal where they are now travelling faster in the denser medium. Snell’sLaw can be explained by this model.

Huygens’s wave model When the wavefront meets a heavier medium thewavelets do not travel as fast as before. This causes the wavefront to changedirection. In this case the wavefront bends towards the normal when itenters a medium where the wave is slowed down. Snell’s Law can beexplained by this model.

A point of differenceNow, with these two explanations of refraction, there is a clear distinctionbetween the two models. When light bends towards the normal as it enterswater (a denser medium), the particle model says it is because light travelsfaster in water (the denser medium), whereas the wave model says it isbecause the light is travelling slower.

In the seventeenth century they did not have the technology to measurethe speed of light in water. However, the particle model became theaccepted explanation, partly because of Newton’s status, and partly becauseHuygens’s principle suggested that light should bend around corners likesound, and there was no evidence of this at the time. (Newton himselfactually thought that the particles in his model needed to have some wave-like characteristics to explain some of his other observations of light andcolour.)

New evidence emergesIn 1802, Thomas Young (1773–1829) showed that in fact light could bendaround an edge. This led to the rejection of the particle model, as it had nomechanism to explain how particles could bend around a corner, and conse-quently the wave model re-emerged. In 1856, Foucault measured the speedof light in water and showed that it was indeed slower than that in air,clinching the argument for the wave model.

Figure 2.21 The particle model of refraction. The particles are pulled towards the denser medium, resulting in a change in direction.

waterr

i

air

Inve

stigations

Investigation 2.3Refraction of particles

Inve

stigations

Investigation 2.4Refraction of waves

Figure 2.22 The wave model of refraction. C and D are parallel, incoming rays. AB is the wavefront. When A hits the surface a circular wavelet of slower speed and so smaller radius is produced. By the time B has reached the surface at E, the refracted wavelet has only gone as far as F. The line EF is the refracted wavefront, heading in a direction bent towards the normal compared to the incoming wavefront, AB.

water

airA

E

B

F

D

C



FORMING IMAGES WITH LENSESThe refraction of light at a boundary can be put to use if the boundariesare curved. There are two possibilities for curved boundaries — curvinginwards or curving outwards. A convex lens has its faces curving out-wards. A lens that curves inwards is a concave lens. The simple ray tracingin figure 2.23 illustrates what each lens does to the light rays.

As rays enter the glass they are bent towards the normal. When theyreach the air on the other side of the lens, they are bent away from thenormal. In the case of the convex lens, the emerging rays come togetheror ‘converge’ at a point called the focus (F). For the concave lens, therays move apart or ‘diverge’, so that they appear to come from a point,also called the focus, on the other side of the lens. For these reasons,convex lenses are sometimes called converging lenses, and concave lensescan be called diverging lenses.

In fact, the focus is more than just a point. It is a plane — a focal planethrough the focus point. In figure 2.24, for example, parallel light raysfrom a distant object coming in at an angle to the lens are still brought toa focus, not at the focus but elsewhere in the focal plane. This is of use indesigning telescopes.

Convex lenses — locating images using ray tracingThere are similarities in the nature of images produced by a convex lensand those produced by a concave mirror. The features of a convex lensare illustrated in figure 2.25. The convex lens has two symmetrical curvedsurfaces, which means that it has two focus points.

Figure 2.25 The convex lens has two focus points.

A convex lens is a lens that is thicker in the middle than at the edges.A concave lens is a lens that is thinner in the middle than at the edges.Converging lenses are those that bend incident parallel rays towards a focus. That is, they converge light.Diverging lenses are those that bend incident parallel rays away from each other. That is, they diverge light.

F

(a) Convex lens

F

(b) Concave lens

Figure 2.23 Refraction of rays through (a) a convex and (b) a concave lens

Inve

stigations

Investigation 2.5The convex lens as a

magnifying glass

F

focalplane

principalaxis

Figure 2.24 Rays converge at a point on the focal plane, which passes through the focus point (F).

focus focus

pole

principalaxis

convex lens



The ray model can now be used to locate an image (see figures 2.26and 2.27).

Figure 2.26 The location of the image is determined according to the point where the three rays cross. All the rays that pass through the lens pass through the image.

The ray diagram in figure 2.26 demonstrates the following:• Ray 1 leaves the head of the object parallel to the principal axis, is

refracted by the lens and passes through the focus on the other side ofthe lens.

• Ray 2 passes through the focus on the same side as the object, isrefracted by the lens and emerges travelling parallel to the principalaxis.

• Ray 3 travels towards the centre of the lens. If the angle of theincoming ray is small and the lens is considered thin, then the raywould appear to continue on in the same direction.All three rays pass through the same point. This is where the image of

the head of the object is located. Note that the image could have beenlocated with any two of the rays, but the other can be used to confirm theoriginal location of the image.

The features of the image can now be described — location, size,orientation and nature (see chapter 1, page 12).

Convex lenses are used in a variety of applications. There are differentapplications for each location of the object. A range of these applicationsis given in table 2.2 (see page 38).

Convex lenses — using a formula to locate and describe imagesThe similarities between the ray diagrams for the concave mirror and theconvex lens suggest that the same formula (see below) can be used forboth, but with a new sign convention which reads:

Real image distances are expressed as positive quantities and virtualimage distances are expressed as negative quantities.

+ =

magnification (M) = =

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links

Convex lens applet

F F

4

1

3

2

object convex lens

image

Inve

stigations

Investigation 2.6Describing images

produced by a convex lens

eMode

lling

Model of a convex lens

FF

13

2object convex

lens

image

Figure 2.27 The object is inside the focus. This makes the rays diverge after passing through the lens. ‘Backtracking’ these rays reveals that they appear to be coming from a point behind the object. The image is located at this point.

1u--- 1

v--- 1

f---

height of image Hi( )height of image Ho( )------------------------------------------------------- v

u---

FFobject

imageHo

H i

u

vf

Figure 2.28 Representation of the elements of a formula to locate images in a convex lens



Describe fully the image of a 4.0 cm object 15 cm in front of a convexlens with a focal length of 10 cm.

u = 15 cm; v = ?; f = 10 cm; M = ?; Ho = 4.0 cm; Hi = ?

+ = (substitute data into the lens formula)

⇒ = − (rearrange to get the unknown by itself)

= − = (combine fractions with lowest commondenominator)

⇒ v = 30 (invert fractions to get v as numerator)The image is 30 cm from the lens and on the opposite side of the lens tothe object.

Magnification =

=

⇒ = (substitute values into magnification formula)

= 2.0

⇒ Hi = 2.0 × 4.0 (rearrange to get the unknown by itself)= 8.0 (calculate expression)

The image is 8.0 cm high, real, inverted and located 30 cm from the lens.

Concave lenses — locating images using ray tracingConcave lenses curve inwards so they are thinner in the middle, asopposed to the convex lens which is thicker in the middle. This meansthat the rays of light are bent away from the principal axis at both thefront and the back surfaces of the lens (see figure 2.23(b), page 35. Forthis reason they are also called diverging lenses.

Concave lenses have a principal focus, likeconvex lenses, but it is located on the sameside of the lens as the object and is calleda virtual focus. Light rays parallel tothe principal axis spread out afterpassing through the concave lens as ifthey are coming from the virtual focus.

Figure 2.29 The rays diverge for allobject positions. An upright, diminished, virtual image

is located between the focus and the lens.

Concave lenses — using a formula to locate and describe imagesAs with the formula method with the convex mirror, the focal length of aconcave lens can be represented as a negative number because it is avirtual focus.

SAMPLEPROBLEM 2.6

Solution:1

15------ 1

v--- 1

10------

1v--- 1

10------ 1

15------

330------ 2

30------ 1

30------

Skillc

hecks

Significant figures(p. 493)

Skillc

hecks

Using spreadsheets(p. 504)

Hi

Ho-------

vu---

Hi

4.0------- 30

15------

Hi

4.0-------

eMode

lling

Using a spreadsheetto locate the image

from a slide projector

Web

links

Concave lens applet

Inve

stigations

Investigation 2.7Locating an imagefor a concave lens FF

object

image concave lens



Locate and describe the image formed by a 6.0 cm high object 15 cm infront of a concave lens with a focal length of 10 cm.

Ho = 6.0 cm; Hi = ?; u = 15 cm; v = ?; f = −10 cm

Using + =

⇒ + = (substitute data into the lens formula)

⇒ = − − (rearrange to get the unknown by itself)

= (combine fractions with lowest commondenominator)

= (simplify expression)

⇒ v = (invert to get v in the numerator)

= −6.0.The image is located 6.0 cm in front of the lens. It is virtual and upright.

Magnification = =

= = 0.4 (substitute data into magnification formula)

Hi = 0.4 × 6.0 = 2.4. (rearrange to get the unknown by itselfand calculate expression)

The image is 2.4 cm high.

Because they spread light rays further apart, concave lenses do notform real images. Therefore, they are usually used in combination with aconvex lens.

Table 2.2 Simple applications of convex lenses

SAMPLEPROBLEM 2.7

Solution:1u--- 1

v--- 1

f---

115------ 1

v--- 1

−10----------

110------ 1

15------

(−3 − 2)30

----------------------

5–30------

30−5------

Hi

Ho------- v

u---

Hi

6.0------- 6.0

15-------

LOCATION OF OBJECT USES DESCRIPTION OF IMAGE

Very large distance away from lens

Objective lens of refracting telescope

Real, inverted, diminished and located near the opposite focus

Beyond twice the focal length from lens

Human eye; camera Real, inverted, diminished and located on other side between one and two focal lengths from lens

At twice the focal length from lens

Correction lens for terrestrial telescope

Real, inverted, same size and located two focal lengths from lens

Between twice the focal length from lens and the focus

Slide projector; objective lens of microscope

Real, inverted, magnified and located on other side of lens beyond two focal lengths

At the focus Searchlight; eyepiece of refracting telescope

No image. The emerging parallel rays do not meet.

Between focus and lens Magnifying glass; eyepiece lens of microscope; spectacles for long-sightedness

Virtual, upright, magnified and located on same side of the lens and further away



MICROSCOPES: AN APPLICATION OF LENSESA convex lens can be used as a simple magnifying glass, but to obtaingreater magnification another lens is needed.

Compound microscopeIn its simplest form, a compound microscope has two convex lenses,one fatter and with a shorter focal length than the other. The fatterone is located near the object to be magnified and is called the objec-tive lens. The other lens is close to the eye and is called the eyepiece(see figure 2.31).

An object placed just outside the focus of the objective lensproduces a real, inverted, magnified image. The eyepiece lens isadjusted so that this image is formed inside the focus of the eyepiecelens.

The real, inverted image now becomes the object for the eyepiecelens. Since it is inside the focus of the eyepiece, it produces a finalmagnified image which is virtual and inverted compared with theoriginal object.

PHYSICS IN FOCUS

AFlat lenses?

lens works by changing the direction of the light ray at the frontsurface and then again at the back surface. The glass in the

middle is there to keep the two surfaces apart. Augustin-Jean Fresneldevised a way of making a lens without the need for all the glass inthe middle.

The glass surface of the lens is a series of concentric rings. Eachring has the slope of the corresponding section of the full lens,but its base is flat. The slopes of the rings get flatter towards thecentre.

Figure 2.30 A side view of a convex Fresnel lens showing how it is constructed

This design substantially reduces the weight of the lens, so lensesof this type are used in lighthouses. Their relative thinness meansthey are also used where space is at a premium, such as in overheadprojectors, and as a lens to be used with the ground-glass screens incamera viewfinders.

Flat lenses, or Fresnel lenses as they are called, are now attached tothe rear windows of vans and station wagons to assist the driver whenreversing or parking.

Inve

stigations

Investigation 2.8Examining a flat lense

Skillc

hecks

Using spreadsheets(p. 506)

eMode

lling

Using a spreadsheetto model amicroscope

Jac Phys 1 2E - 02 Page 39 Friday, October 24, 2003 2:08 PM


HUMAN AND ANIMAL VISIONThe human eye is an extraordinary device. It is able to respond to anenormous range of light brightness. The strongest light that the eye cansafely detect is 10 000 million times as bright as the weakest light it candetect. It is able to focus on objects from many billions of kilometres awayto objects a few centimetres away. It can also detect colour. The parts ofthe human eye are shown in figure 2.32.

The purpose of the eye is to produce a sharp, real image on a screen— the retina. Light passes through many refracting elements in thehuman eye on its way to the retina (see table 2.3). It achieves a focus intwo stages. The curved cornea at the front of the eye does two-thirds ofthe focusing, due to the large difference in refractive index between thecornea and the air. The lens does the fine focusing as there is little differ-ence in the refractive index of the lens and that of the liquid on eitherside of it.

In the rest of the eye, the differences in refractive index are marginal,but they are crucial when the eye wishes to produce clear images of farand near objects.

objective lens eyepiece lens

convex lenses

FoFo

I1

I2

FeFeobject

• • ••

Figure 2.31 Ray diagram for a compound microscope

Inve

stigations

Investigation 2.9Calculating with a

microscope

retinasclera

choroid

opticnerve

iris(coloured)

ciliarymuscle

foveacentralis

macula

blindspot

cornea

pupil

lens

aqueoushumour

vitreoushumour

Figure 2.32Cross-section of a human eye

The retina takes the role of the screen of the eye. It is covered with nerve cells that detect the brightness and colour of the light falling on it.The cornea is the curved front surface of the eye. It refracts light towards the pupil so that it can pass through towards the lens.



Table 2.3 Refractive index of the parts of the eye

Accommodation mechanismsConsider the light from a light globe that passes through a convex lensonto a screen to produce an image of the globe’s filament. If the lens ismoved to a new position, the screen needs to be moved to obtain a sharpimage. The human eye can produce a sharp image on its ‘screen’ (theretina) of objects at various distances. But the eye’s screen stays put, sosomething else has to change to achieve a sharp image. The only thingthat can change is the focal length of the lens.

Convex lenses form images of very distant objects at their focus. Asobjects get closer to the lens, the image is formed behind the focus. Forthe human eye with its fixed ‘screen’, this means that the focal lengthmust shorten to keep a sharp image on the retina. This adjustment iscalled accommodation.

How short can the focal length of the lens in an eye become? Whenyou bring an object closer to your eye from an arm’s length away, there isa point at which the object becomes fuzzy in appearance. This point iscalled your near point. As you age, your near point becomes further away(see table 2.4). That is why some older people hold the newspaper fur-ther away to read it.

Table 2.4 Average near points by age

PART OF THE EYE REFRACTIVE INDEX

Tears 1.33

Cornea 1.37

Aqueous humour 1.33

Lens cover 1.38

Lens centre 1.41

Vitreous humour 1.33

he eye is an amazing instrument. It has an automatic apertureadjustment called the iris. The act of blinking operates the

cornea’s built-in scratch remover, lens cleaner and lubricator. In dimlight, the eye operates as a supersensitive, black and white televisioncamera. It allows us to see objects with less than 0.1 per cent, or one-thousandth, of the light we need for colour vision.

The optic nerve packages the visual information from the retina sothat the brain receives about 30 discrete ‘frames’ per second in asimilar way to television and cinema.

TAS A MATTER OF FACT

Web

links

Modelling the eyeapplet

Accommodation is the adjustment to the thickness of the lens in the eye to ensure that the image on the retina is sharp. When the thickness of the lens changes, so does its focal length.The near point of your eye is the closest an object can be to your eye so that your eye can produce a sharp image of the object on its retina.

Inve

stigations

Investigation 2.10Measuring your near

point

AGE (YEARS) NEAR POINT (CM) AGE (YEARS) NEAR POINT (CM)

10 7 40 22

15 8.5 50 40

20 10 60 65

30 15 70 200



To enable the focal length of the eye to change, the lens has tochange its shape. To achieve a shorter focal length, the lens needs tobecome fatter. This raises important questions. How does the lenschange shape? Which is the natural rest position of the lens in the eye?Is the lens relaxed in the short focal length position and needs to be‘stretched’ to see distant objects? Or is it relaxed in the long focallength position and needs to be ‘squashed’ to see near objects? Musclescan act only by contraction, so how are the muscles arranged in the eyeto squash the lens?

In fact, the relaxed human eye has a long focal length located at theretina to form images of very distant objects. Ligaments are attached tothe outside edge of the lens which are continually pulling outwards; how-ever, around the lens itself is a circular muscle. When this circular musclecontracts, it produces a smaller circle, making a lens that is smaller in dia-meter, but fatter front to back.

Do the eyes of animals work in the same way as human eyes? What lensmechanism would be appropriate for an animal in the wild? In fact, mostanimals’ eyes are relaxed in the long focal length position because, if theanimal is roused from sleep by the sound of a distant predator, the eye isready to focus on it.

While many animals use an adjustable focal length lens like humans,some use other strategies. One strategy is to move the lens backwards andforwards. In most fish and snakes, the lens moves within the eyeball in thesame way that a camera lens is moved to produce a sharp image on film.

Correcting eye defectsSometimes our eyes do not work properly and corrective lenses need tobe prescribed. Corrective lenses were the first prosthetic devices to beinvented (in about 1300) but, until early in the twentieth century, theywere available only to the wealthy. The two main eye defects are hyper-metropia and myopia.

Myopia, or short-sightedness, means that the person’s vision of nearbyobjects is good, but distant objects are unclear. Parallel rays of light fromdistant objects are brought to a focus in front of rather than on theretina. The eyeball is too large for the ‘relaxed focal length’ of the eye(see figure 2.33). This defect is not usually noticed until the eyeballapproaches its final size in adolescence. It can be overcome by placing aconcave lens in front of the eye that spreads the rays apart slightly beforethey enter the eye (see figure 2.34).

Web

links

Animal eyes

PHYSICS IN FOCUS

The power of the lensptometrists and opticians describe the focusing ability of a lensin terms of its power. The power of a simple lens is defined as:

Power = .

The unit of the power of a lens is the dioptre (D). For example, aconcave lens of a focal length of 25 cm has a power given by:

Power = = = −4.0 D.

O1f---

1f--- 1

0.25–--------------



Figure 2.33 Myopic vision causes light rays to focus in front of the retina.

Figure 2.34 Corrective concave lenses are used in spectacles and contact lenses.

Sometimes myopia can be caused by excessive curvature of the cornea,the principal focusing agent in the eye. Recent medical developmentsuse low temperature ultraviolet lasers to ‘shave off’ some tissue from thefront of the cornea, to flatten the curvature.

Hypermetropia, or long-sightedness, is the reverse of myopia. The lensproduces an image of near objects behind instead of on the retina(see figure 2.36). Near objects are not clear, while distant objects are infocus. The muscles around the lens cannot contract enough to shortenthe focal length to bring the image onto the retina. The eye needs helpto achieve this, and an extra convex lens can help (see figure 2.37).

Figure 2.36 Hypermetropia causes light rays to focus behind the retina. This produces a ‘fuzzy’ image.

F

(a) Spectacle lens

F

(b) Contact lens

Figure 2.35 Contact lenses can be a solution to eye defects. They alter the curvature of the front of the eye.



Figure 2.37 Convex lenses are used in spectacles and contact lenses to correct hypermetropia.

As people age, the muscles around the lens slowly weaken and the lensbecomes thicker and less flexible. The loss in flexibility means that thenear point moves further away and the newspaper needs to be held outto be read. The thickening of the lens, even when the muscles arerelaxed, means that distant objects start becoming fuzzy. This means thatthe eye is showing signs of myopia for distant objects and hypermetropiafor near objects, which require different corrective solutions. Hence theneed for bifocal lenses, which have a convex surface at the bottom tolook down to read the newspaper and a concave surface at the top tolook up to drive the car (see figure 2.38).

Figure 2.38 Example of a bifocal lens design

(a) Spectacles (b) Contact lens

lens for watchingTV and driving

lens for reading



CHAPTER REVIEW

• Light bends as it travels from one medium toanother. A measure of a medium’s capacity tobend light is given by its refractive index.

• If light travels into a medium of a higher refrac-tive index, the light is bent towards the normal.If light travels into a medium of a lower refrac-tive index, the light is bent away from thenormal. This change in direction is summarisedin Snell’s Law. Snell’s Law can be expressed asn1 sin θ1 = n2 sin θ2.

• When light travels into a medium of a lowerrefractive index, there will be an angle of inci-dence for which the angle of refraction is 90°.This angle of incidence is called the criticalangle. For angles of incidence greater than thecritical angle, all the light is reflected back intothe medium. This phenomenon is called totalinternal reflection.

• Two improvements on the ray model of lightare the particle model and the wave model.Both of these models explain various lightphenomena, but they differ in their predictionof the speed of light in a medium. The wavemodel’s prediction was found to be correct.

• When refractive materials are shaped intoconvex and concave lenses, both real and vir-tual images can be formed.

• The ray model explains the formation and prop-erties of images in convex and concave lenses.

• The formation of images in convex and con-cave lenses can be mathematically modelled

with the equations + = and M = .

• The formation and properties of images inoptical devices such as optical fibres, micro-scopes and spectacles can be modelled mathe-matically or by ray tracing.

Understanding1. What is the angle of refraction in water

(n = 1.33) for an angle of incidence of 40°? Ifthe angle of incidence is increased by 10°, byhow much does the angle of refraction increase?

2. A ray of light enters a plastic block at an angleof incidence of 55° with an angle of refractionof 33°. What is the refractive index of the plastic?

3. A ray of light passes through a rectangularglass block with a refractive index of 1.55.

The angle of incidence as the ray enters theblock is 65°. Calculate the angle of refractionat the first face of the block, then calculatethe angle of refraction as the ray emerges onthe other side of the block. Comment onyour answers.

4. Immiscible liquids are liquids that do not mix.Immiscible liquids will settle on top of eachother, in the order of their density, with thedensest liquid at the bottom. Some immiscibleliquids are also transparent.(a) Calculate the angles of refraction as a ray

passes down through immiscible layers asshown in the figure below.

(b) If a plane mirror was placed at the bottomof the beaker, calculate the angles ofrefraction as the ray reflects back to thesurface. Comment on your answers.

5. Calculate the critical angle for light travellingthrough a diamond (n = 2.5) towards the surface.

6. A ray travelling through water (n = 1.33)approaches the surface at an angle of inci-dence of 55°. What will happen to the ray?Support your answer with calculations.

7. (a) Calculate the refractive index of the glassprism shown in the figure below so that thelight ray meets the faces at the critical angle.Is this value of the refrac-tive index the minimumor maximum value forsuch a reflection?

(b) Draw two parallelrays entering the block.How do they emerge?

8. Calculate the refractiveindex of the plastic coatingon an optical fibre if thecritical angle for glass toplastic is 82.0° and the refrac-tive index of glass is 1.500.

SUMMARY

1u--- 1

v--- 1

f--- v

u---

QUESTIONS

air

25°

acetone

glycerol

carbontetrachloride

glass beaker

lightray

n = 1.00

n = 1.357

n = 1.4746

n = 1.4601

n = 1.53

45°


WAVE-LIKE PROPERTIES OF LIGHT46CHAPTER REVIEW

9. When light passes from air into a glass block,some light is reflected off the surface as well.(a) Describe how the wave model of light

explains this observation.(b) Why can’t the particle model explain this

observation?

10. Use ray tracing to determine the full descrip-tion of the following objects:(a) a 4.0 cm high object, 20 cm in front of a

convex lens with a focal length of 15 cm(b) a 3.0 mm high object, 10 cm in front of a

convex lens with a focal length of 12 cm(c) a 5.0 cm high object, 200 cm in front of a

convex lens with a focal length of 10 cm.

11. A bull’s eye has a diameter of about 5.0 cm.The bull needs a clear image of the food it iseating, so its near point is about 30 cm. Calcu-late the minimum focal length of the lens inthe bull’s eye.

12. What does ‘accommodation mechanism’ mean?Give an example.

Application13. A ray of light enters a parallel-sided glass block

(n = 1.55) at an angle of incidence of 35°, asshown in the figure below.

(a) Calculate the angle of refraction at the topsurface and the angles of incidence andrefraction as the ray emerges at the bottomsurface.

(b) If the initial ray (at the same angle)entered the block further to the right, theemerging ray would hit the corner of theblock. If the block is 5.0 cm wide, how farfrom the edge should the initial ray enterthe block for this to occur?

(c) If the initial ray is moved even closer to theedge, the emerging ray will meet theadjacent side edge. What angle of inci-dence does the light ray make with the adja-cent face? Will the ray pass out of the blockor will it be totally internally reflected?

(d) Some or all of the light will be reflectedfrom the adjacent face. Draw the sub-sequent path of this light, showing angles.

(e) Would your answer to (c) be different ifthe initial angle of incidence was larger orsmaller than 35°C? (Hint : Assume the raymeets the adjacent face at the criticalangle and then work back to the initialangle of incidence.)

(f) Does the answer to (e) depend on therefractive index of the material? Set up thespreadsheet (shown at the base of the page)to investigate how the initial angle of inci-dence varies with the refractive index if theray emerges from the adjacent face at 90°.

Increase the refractive index in smallsteps to find the value for which the angleof incidence becomes 90°. What is unusualabout this value for the refractive index?

14. Describe what a diver would see when lookingup at a still water surface.

15. A right-angled glass prism (n = 1.55) is placedunder water (n = 1.33), as shown in the figurebelow. A ray of light enters the longest sidealong the normal. What happens to the ray oflight?

16. A fish looking up at the surface of the watersees a circle, inside which it sees the ‘airworld’. Outside the circle it sees the reflectionof the ‘water world’. If the fish is 40 cm belowthe surface, calculate the radius of the circle(nwater = 1.33).

17. Light enters an optical fibre 1.0 µm in diameter,as shown in the figure on page 47. Some lightgoes straight down the centre. Another ray isangled, leaving the central line and meeting theoutside edge at slightly more than the criticalangle of 82° then reflects back to the central line.

(a) How much further did thisray travel?

(b) Calculate the speed of light inthe glass and determine the

5 cm n = 1.55

35°

45°

air

water



CHAPTER REVIEW

(i) (ii) (iii) (iv) (v) (vi)

(a) (b) (c) (d) (e) (f) (g)

time delay between the two rays after oneinternal reflection. Do you think this couldbe a problem in an optical fibre? If so,when? How could the problem be over-come?

18. (a) You are carrying out a convex lens investi-gation at a bench near the classroom windowand you obtain a sharp image of the windowon your screen. A teacher walks past outsidethe window. What do you see on the screen?

(b) The trees outside the classroom areunclear on the screen. What can you do tobring the trees into focus?

19. Use the ray tracing or the formula method todetermine the magnification of an objectplaced under the following two-lens micro-scope. The object is placed 5.2 mm from anobjective lens of focal length 5.0 mm. The eye-piece lens has a focal length of 40 mm. Thepoles of the lenses are 150 mm apart.

20. Light rays are shown passing through boxes inthe figure at the base of the page. Identify thecontents of each box from the options (a)–(g)given below. Option (b) is a mirror. All othersare solid glass. Note : There are more optionsthan boxes.

21. A convex lens with a focal length of 5.0 cm isused as a magnifying glass. Calculate the sizeand location of the image of text on this pageif the centre of the lens was placed:(a) 4.0 cm above the page(b) 3.0 cm above the page.

22. A 35 mm slide is placed in a slide projector. Asharp image is produced on a screen 4.0 m away.The focal length of the lens system is 5.0 cm.(a) How far is the slide from the centre of the

lens?(b) What is the size of the image?

(c) Looking from the back of the slide pro-jector, the slide contains a letter ‘L’. Whatshape will appear on the screen?

(d) The slide projector is moved closer to thescreen. The image becomes unclear.Should the lens system be moved closer toor further away from the slide?

23. A teacher is using a slide projector but theimage on the screen is smaller than the screen.What needs to be done to produce a clearimage on the full screen?

24. Calculate the distance the photographic filmneeds to be from the centre of a camera lensof focal length 5.0 cm in order to take a sharpphotograph of a family group 15.0 m away.

25. (a) To appear invisible you need to becometransparent. What must your refractive indexbe if your movement is not to be detected?

(b) The retina of your eye is a light-absorbingscreen. What does that imply about yourown vision if you are to remain invisible?(Hint: If you are invisible all light passesthrough you.)

More of a challenge26. Calculate the angle of deviation at a glass–air

interface for an angle of incidence of 65° andrefractive index of glass of 1.55.

27. Calculate the sideways deflection as a ray oflight goes through a parallel-sided plasticblock (n = 1.4) with sides 5.0 cm apart, as inthe figure below.

82°

optical fibrelight rays

1 m µ

30°

n = 1.4 5 cm


WAVE-LIKE PROPERTIES OF LIGHT48CHAPTER REVIEW

28. Calculate the angle of deviation as the light raygoes through the triangular prism shown inthe figure below.

29. A ray of light enters a glass sphere (n = 1.5), asin the figure below. What happens to the ray?

30. (a) Model a heat mirage on a spreadsheet (seethe example at the base of the page) as aset of air layers whose refractive indiceseach decrease by 0.000 02, starting at1.0003. As the light ray travels througheach layer, the angle of refraction at thelayer’s top surface becomes the angle ofincidence at its bottom surface. Totalinternal reflection will occur when thespreadsheet tries to find an angle whosesine is greater than 1 and displays #NUM.

(b) If the initial angle of incidence is 89°, inwhat layer does total internal reflectionoccur? Experiment with different initialangles of incidence, initial refractiveindices and different rates of change inthe refractive index from layer to layer.

31. In the diagram of Fata Morgana (figure 2.12,page 30), describe how the temperature andrefractive index vary with height.

32. A hair 1.0 mm in diameter is looked atthrough a microscope. The objective lens has afocal length of 3.0 mm, the eyepiece has a

focal length of 25 mm and the lenses are18.0 cm apart. The hair is placed just outsidethe focus of the objective lens at 3.5 mm.(a) Calculate the location of the image pro-

duced by this lens. Using this image as theobject for the eyepiece lens, calculate thelocation of the final image.

(b) Calculate the size of the image producedby the objective lens, then calculate thesize of the final image produced by themicroscope.

33. A screen captures the image of a small globefrom a convex lens. The screen is then takenaway. Where should you place your eye andwhere should you look to see the image?

34. A hollow 15 cm tube contains a convex lens ofunknown focal length but greater than 15 cm.However, you do not know where the lens is inthe tube. You have a light source, a screen anda ruler. How would you determine the focallength of the lens and its position in the tube?

35. Ray tracing can be used to locate images pro-duced by a concave lens. The only differencefrom ray tracing with convex lenses is that theprincipal focus is on the opposite side of theconcave lens from the object. Draw ray tracingdiagrams for objects at different distances anddescribe the properties of images produced bya concave lens.

36. The closer an object is to your eye, the moredetail you can see. The object casts a largervisual angle at the eye. But this is only true upto your near point, which is the limit of youreye’s focusing ability. This is an important con-sideration in the design of magnifying glassesand microscopes. For the most detailed image,the device should produce an image located atthe user’s near point.(a) A slide viewer is a small magnifying glass in

a frame which holds a slide at a distance of5.0 cm from the magnifying lens. Whatshould be the focal length of the lens ifthe average near point is 25 cm and innormal use the slide viewer is held up tothe eye?

(b) A person with a near point of 50 cm asksyou to design a magnifying glassto enable the person to read abook. What focal length convexlens should you use?(Remember to estimate objectand image distances.)

40°

60°n = 1.5

?

30°

glass sphere

centre ofcircle


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