In Grade 10, we studied how light is reflected and refracted. This chapter builds on what you have learnt in Grade 10. You will learn about lenses, how the human eye works as well as how telescopes and microscopes work.
In this section we will discuss properties of thin lenses. In Grade 10, you learnt about two kinds of mirrors: concave mirrors which were also known as converging mirrors and convex mirrors which were also known as diverging mirrors. Similarly, there are two types of lenses: converging and diverging lenses.
We have learnt how light travels in different materials, and we are now ready to learn how we can control the direction of light rays. We use lenses to control the direction of light. When light enters a lens, the light rays bend or change direction as shown in Figure 1.
- Definition 1: Lens
A lens is any transparent material (e.g. glass) of an appropriate shape that can take parallel rays of incident light and either converge the rays to a point or diverge the rays from a point.
Some lenses will focus light rays to a single point. These lenses are called converging or convex lenses. Other lenses spread out the light rays so that it looks like they all come from the same point. These lenses are called diverging or concave lenses. Lenses change the direction of light rays by refraction. They are designed so that the image appears in a certain place or as a certain size. Lenses are used in eyeglasses, cameras, microscopes, and telescopes. You also have lenses in your eyes!
- Definition 2: Converging Lenses
Converging lenses converge parallel rays of light and are thicker in the middle than at the edges.
- Definition 3: Diverging Lenses
Diverging lenses diverge parallel rays of light and are thicker at the edges than in the middle.
Examples of converging and diverging lenses are shown in Figure 2.
Before we study lenses in detail, there are a few important terms that must be defined. Figure 3 shows important lens properties:
- The principal axis is the line which runs horizontally straight through the optical centre of the lens. It is also sometimes called the optic axis.
- The optical centre (O) of a convex lens is usually the centre point of the lens. The direction of all light rays which pass through the optical centre, remains unchanged.
- The focus or focal point of the lens is
the position on the principal axis
where all light rays which run parallel to the principal axis through the lens converge (come together) at a point. Since light can pass through the lens either from right to left or left to right, there is a focal point on each side of the lens (F1F1 and F2F2), at the same distance from the optical centre in each direction. (Note: the plural form of the word focus is foci.)
- The focal length (ff) is the distance between the optical centre and the focal point.
We will only discuss double convex converging lenses as shown in Figure 4. Converging lenses are thinner on the outside and thicker on the inside.
Figure 5 shows a convex lens. Light rays traveling through a convex lens are bent towards the principal axis. For this reason, convex lenses are called converging lenses.
When an object is placed in front of a lens, the light rays coming from the object are refracted by the lens. An image of the object is produced at the point where the light rays intersect.
The type of images created by a convex lens is dependent on the position of the object. We will examine the following cases:
- the object is placed at a distance greater than 2f2f from the lens
- the object is placed at a distance equal to 2f2f from the lens
- the object is placed at a distance between 2f2f and ff from the lens
- the object is placed at a distance less than ff from the lens
We examine the properties of the image in each of these cases by drawing ray diagrams. We can find the image by tracing the path of three light rays through the lens. Any two of these rays will show us the location of the image. The third ray is used to check that the location is correct.
Aim:
To determine the focal length of a convex lens.
Method:
- Using a distant object from outside, adjust the position of the convex lens so that it gives the smallest possible focus on a sheet of paper that is held parallel to the lens.
- Measure the distance between the lens and the sheet of paper as accurately as possible.
Results:
The focal length of the lens is cm
Aim:
To investigate the position, size and nature of the image formed by a convex lens.
Method:
- Set up a candle, and the lens from Experiment Lenses A in its holder and the screen in a straight line on the metre rule. Make sure the lens holder is on the 50 cm mark.
From your knowledge of the focal length of your lens, note where ff and 2f2f are on both sides of the lens.
- Using the position indicated on the table below, start with the candle at a position that is greater than 2f2f and adjust the position of the screen until a sharp focused image is obtained. Note that there are two positions for which a sharp focused image will not be obtained on the screen. When this is so, remove the screen and look at the candle through the lens.
- Fill in the relevant information on the table below
Results:
Table 1
| Relative position of object |
Relative position of image |
Image upright or inverted |
Relative size of image |
Nature of image |
| Beyond 2f2f
cm
|
|
|
|
|
| At 2f2f |
|
|
|
|
|
cm
|
|
|
|
|
| Between 2f2f and ff |
|
|
|
|
|
cm
|
|
|
|
|
| At ff |
|
|
|
|
|
cm
|
|
|
|
|
| Between ff and the lens
|
|
|
|
|
|
cm
|
|
|
|
|
QUESTIONS:
- When a convex lens is being used:
- A real inverted image is formed when an object is placed
- No image is formed when an object is placed
- An upright, enlarged, virtual image is formed when an object is placed
- Write a conclusion for this investigation.
Aim:
To determine the mathematical relationship between d0,did0,di and ff for a lens.
Method:
- Using the same arrangement as in Experiment Lenses B, place the object (candle) at the distance indicated from the lens.
- Move the screen until a clear sharp image is obtained. Record the results on the table below.
Results:
ff = focal length of lens
d0d0 = object distance
didi = image distance
Table 2
| Object distance |
Image distance |
1
d
0
1
d
0
|
1
d
i
1
d
i
|
1
d
0
+
1
d
i
1
d
0
+
1
d
i
|
| d0d0 (cm) |
didi (cm) |
(cm-1-1) |
(cm-1-1) |
(cm-1-1) |
| 25,0 |
|
|
|
|
| 20,0 |
|
|
|
|
| 18,0 |
|
|
|
|
| 15,0 |
|
|
|
|
| |
|
|
|
Average = |
Reciprocal
of
average
=
1
1
d
0
+
1
d
i
=
̲
(
a
)
Focal
length
of
lens
=
̲
(
b
)
Reciprocal
of
average
=
1
1
d
0
+
1
d
i
=
̲
(
a
)
Focal
length
of
lens
=
̲
(
b
)
(1)
QUESTIONS:
- Compare the values for (a) and (b) above and explain any similarities or differences
- What is the name of the mathematical relationship between d0d0, didi and ff?
- Write a conclusion for this part of the investigation.
Drawing Ray Diagrams for Converging Lenses
Ray diagrams are normally drawn using three rays. The three rays are labelled R1R1, R2R2 and R3R3. The ray diagrams that follow will use this naming convention.
- The first ray (R1R1) travels from the object to the lens parallel to the principal axis. This ray is bent by the lens and travels through the focal point.
- Any ray travelling parallel to the principal axis is bent through the focal point.
- If a light ray passes through a focal point before it enters the lens, then it will leave the lens parallel to the principal axis. The second ray (R2R2) is therefore drawn to pass through the focal point before it enters the lens.
- A ray that travels through the centre of the lens does not change direction. The third ray (R3R3) is drawn through the centre of the lens.
- The point where all three of the rays (R1R1, R2R2 and R3R3) intersect is the image of the point where they all started. The image will form at this point.
Convex lens:
Concave lens:
We can locate the position of the image by drawing our three rays. R1R1 travels from the object to the lens parallel to the principal axis, is bent by the lens and then travels through the focal point. R2R2 passes through the focal point before it enters the lens and therefore must leave the lens parallel to the principal axis. R3R3 travels through the center of the lens and does not change direction. The point where R1R1, R2R2 and R3R3 intersect is the image of the point where they all started.
The image of an object placed at a distance greater than 2f2f from the lens is upside down or inverted. This is because the rays which began at the top of the object, above the principal axis, after passing through the lens end up below the principal axis. The image is called a real image because it is on the opposite side of the lens to the object and you can trace all the light rays directly from the image back to the object.
The image is also smaller than the object and is located closer to the lens than the object.
In reality, light rays come from all points along the length of the object. In ray diagrams we only draw three rays (all starting at the top of the object) to keep the diagram clear and simple.
We can locate the position of the image by drawing our three rays. R1R1 travels from the object to the lens parallel to the principal axis and is bent by the lens and then travels through the focal point. R2R2 passes through the focal point before it enters the lens and therefore must leave the lens parallel to the principal axis. R3R3 travels through the center of the lens and does not change direction. The point where R1R1, R2R2 and R3R3 intersect is the image of the point where they all started.
The image of an object placed at a distance equal to 2f2f from the lens is upside down or inverted. This is because the rays which began at the top of the object, above the principal axis, after passing through the lens end up below the principal axis. The image is called a real image because it is on the opposite side of the lens to the object and you can trace all the light rays directly from the image back to the object.
The image is the same size as the object and is located at a distance 2f2f away from the lens.
We can locate the position of the image by drawing our three rays. R1R1 travels from the object to the lens parallel to the principal axis and is bent by the lens and then travels through the focal point. R2R2 passes through the focal point before it enters the lens and therefore must leave the lens parallel to the principal axis. R3R3 travels through the center of the lens and does not change direction. The point where R1R1, R2R2 and R3R3 intersect is the image of the point where they all started.
The image of an object placed at a distance between 2f2f and ff from the lens is upside down or inverted. This is because the rays which began at the top of the object, above the principal axis, after passing through the lens end up below the principal axis. The image is called a real image because it is on the opposite side of the lens to the object and you can trace all the light rays directly from the image back to the object.
The image is larger than the object and is located at a distance greater than 2f2f away from the lens.
We can locate the position of the image by drawing our three rays. R1R1 travels from the object to the lens parallel to the principal axis and is bent by the lens and then travels through the focal point. R2R2 passes through the focal point before it enters the lens and therefore must leave the lens parallel to the principal axis. R3R3 travels through the center of the lens and does not change direction. The point where R1R1, R2R2 and R3R3 intersect is the image of the point where they all started.
The image of an object placed at a distance less than ff from the lens is upright. The image is called a virtual image because it is on the same side of the lens as the object and you cannot trace all the light rays directly from the image back to the object.
The image is larger than the object and is located further away from the lens than the object.
The Thin Lens Equation
We can find the position of the image of a lens mathematically as there is a mathematical relation between the object distance, image distance, and focal length. The equation is:
1
f
=
1
d
o
+
1
d
i
1
f
=
1
d
o
+
1
d
i
(2)
where ff is the focal length, dodo is the object distance and didi is the image distance.
The object distance dodo is the distance from the object to the lens. dodo is positive if the object is on the same side of the lens as the light rays enter the lens. This should make sense: we expect the light rays to travel from the object to the lens. The image distance didi is the distance from the lens to the image. Unlike mirrors, which reflect light back, lenses refract light through them. We expect to find the image on the same side of the lens as the light leaves the lens. If this is the case, then didi is positive and the image is real (see Figure 11). Sometimes the image will be on the same side of the lens as the light rays enter the lens. Then didi is negative and the image is virtual (Figure 12). If we know any two of the three quantities above, then we can use the Thin Lens Equation to solve for the third quantity.
Magnification
It is possible to calculate the magnification of an image. The magnification is how much bigger or smaller
the image is than the object.
m
=
-
d
i
d
o
m
=
-
d
i
d
o
(3)
where mm is the magnification, dodo is the object distance and didi is the image distance.
If didi and dodo are both positive, the magnification is negative. This means the image is inverted, or upside down. If didi is negative and dodo is positive, then the image is not inverted, or right side up. If the absolute value of the magnification is greater than one, the image is larger than the object. For example, a magnification of -2 means the image is inverted and twice as big as the object.
An object is placed 6 cm from a converging lens with a focal point of 4 cm.
- Calculate the position of the image
- Calculate the magnification of the lens
- Identify three properties of the image
- Step 1. Identify what is given and what is being asked :
f
=
4
cm
d
o
=
6
cm
d
i
=
?
m
=
?
f
=
4
cm
d
o
=
6
cm
d
i
=
?
m
=
?
(4)
Properties of the image are required.
- Step 2. Calculate the image distance (didi) :
1
f
=
1
d
o
+
1
d
i
1
4
=
1
6
+
1
d
i
1
4
-
1
6
=
1
d
i
3
-
2
12
=
1
d
i
d
i
=
12
cm
1
f
=
1
d
o
+
1
d
i
1
4
=
1
6
+
1
d
i
1
4
-
1
6
=
1
d
i
3
-
2
12
=
1
d
i
d
i
=
12
cm
(5)
- Step 3. Calculate the magnification :
m
=
-
d
i
d
o
=
-
12
6
=
-
2
m
=
-
d
i
d
o
=
-
12
6
=
-
2
(6)
- Step 4. Write down the properties of the image :
The image is real, didi is positive, inverted (because the magnification is negative) and enlarged (magnification is >> 1)
An object is placed 5 cm to the left of a converging lens which has a focal length of 2,5 cm.
- What is the position of the image?
- Is the image real or virtual?
- Step 1. Set up the ray diagram :
Draw the lens, the object and mark the focal points.
- Step 2. Draw the three rays :
- R1R1 goes from the top of the object parallel to the principal axis, through the lens and through the focal point F2F2 on the other side of the lens.
- R2R2 goes from the top of the object through the focal point F1F1, through the lens and out parallel to the principal axis.
- R3R3 goes from the top of the object through the optical centre with its direction unchanged.
- Step 3. Find the image :
The image is at the place where all the rays intersect. Draw the image.
- Step 4. Measure the distance between the lens and the image :
The image is 5 cm away from the lens, on the opposite side of the lens to the object.
- Step 5. Is the image virtual or real? :
Since the image is on the opposite side of the lens to the object, the image is real.
An object, 1 cm high, is placed 2 cm to the left of a converging lens which has a focal length of 3,0 cm. The image is found also on the left side of the lens.
- Is the image real or virtual?
- What is the position and height of the image?
- Step 1. Draw the picture to set up the problem :
Draw the lens, principal axis, focal points and the object.
- Step 2. Draw the three rays to locate image :
- R1R1 goes from the top of the object parallel to the principal axis, through the lens and through the focal point F2F2 on the other side of the lens.
- R2R2 is the light ray which should go through the focal point F1F1 but the object is placed after the focal point! This is not a problem, just trace the line from the focal point F1F1, through the top of the object, to the lens. This ray then leaves the lens parallel to the principal axis.
- R3R3 goes from the top of the object through the optical centre with its direction unchanged.
- Do not write R1R1, R2R2 and R3R3 on your diagram, otherwise it becomes too cluttered.
- Since the rays do not intersect on the right side of the lens, we need to trace them backwards to find the place where they do come together (these are the light gray lines). Again, this is the position of the image.
- Step 3. Draw the image :
- Step 4. Measure distance to image :
The image is 6 cm away from the lens, on the same side as the object.
- Step 5. Measure the height of the image :
The image is 3 cm high.
- Step 6. Is image real or virtual? :
Since the image is on the same side of the lens as the object, the image is virtual.
- Which type of lens can be used as a magnifying glass? Draw a diagram to show how it works. An image of the sun is formed at the principal focus of a magnifying glass.
- In each case state whether a real or virtual image is formed:
- Much further than 2f2f
- Just further than 2f2f
- At 2f2f
- Between 2f2f and ff
- At ff
- Between ff and 0
- An object stands 50 mm from a lens (focal length 40 mm). Draw an accurate sketch to determine the position of the image. Is it enlarged or shrunk; upright or inverted?
- Draw a scale diagram (scale: 1 cm = 50 mm) to find the position of the image formed by a convex lens with a focal length of 200 mm. The distance of the object is 100 mm and the size of the object is 50 mm. Determine whether the image is enlarged or shrunk. What is the height of the image? What is the magnification?
- An object, 20 mm high, is 80 mm from a convex lens with focal length 50 mm. Draw an accurate scale diagram and find the position and size of the image, and hence the ratio between the image size and object size.
- An object, 50 mm high, is placed 100 mm from a convex lens with a focal length of 150 mm. Construct an accurate ray diagram to determine the nature of the image, the size of the image and the magnification. Check your answer for the magnification by using a calculation.
- What would happen if you placed the object right at the focus of a converging lens? Hint: Draw the picture.
We will only discuss double concave diverging lenses as shown in Figure 20. Concave lenses are thicker on the outside and thinner on the inside.
Figure 21 shows a concave lens with light rays travelling through it. You can see that concave lenses have the opposite curvature to convex lenses. This causes light rays passing through a concave lens to diverge or be spread out away from the principal axis. For this reason, concave lenses are called diverging lenses. Images formed by concave lenses are always virtual.
Unlike converging lenses, the type of images created by a concave lens is not dependent on the position of the object. The image is always upright, smaller than the object, and located closer to the lens than the object.
We examine the properties of the image by drawing ray diagrams. We can find the image by tracing the path of three light rays through the lens. Any two of these rays will show us the location of the image. You can use the third ray to check the location, but it is not necessary to show it on your diagram.
Drawing Ray Diagrams for Diverging Lenses
Draw the three rays starting at the top of the object.
- Ray R1R1 travels parallel to the principal axis. The ray bends and lines up with a focal point. However, the concave lens is a diverging lens, so the ray must line up with the focal point on the same side of the lens where light rays enter it. This means that we must project an imaginary line backwards through that focal point (F1F1) (shown by the dashed line extending from R1R1).
- Ray R2R2 points towards the focal point F2F2 on the opposite side of the lens. When it hits the lens, it is bent parallel to the principal axis.
- Ray R3R3 passes through the optical center of the lens. Like for the convex lens, this ray passes through with its direction unchanged.
- We find the image by locating the point where the rays meet. Since the rays diverge, they will only meet if projected backward to a point on the same side of the lens as the object. This is why concave lenses always have virtual images. (Since the light rays do not actually meet at the image, the image cannot be real.)
Figure 22 shows an object placed at an arbitrary distance from the diverging lens.
We can locate the position of the image by drawing our three rays for a diverging lens.
Figure 22 shows that the image of an object is upright. The image is called a virtual image because it is on the same side of the lens as the object.
The image is smaller than the object and is closer to the lens than the object.
An object is placed 4 cm to the left of a diverging lens which has
a focal length of 6 cm.
- What is the position of the image?
- Is the image real or virtual?
- Step 1. Set up the problem :
Draw the lens, object, principal axis and focal points.
- Step 2. Draw the three light rays to locate the image :
- R1R1 goes from the top of the object parallel to the principal axis.
To determine the angle it has when it leaves the lens on the other side, we
draw the dashed line from the focus F1F1 through the point where R1R1
hits the lens. (Remember: for a diverging lens, the light ray on the opposite
side of the lens to the object has to bend away from the principal axis.)
- R2R2 goes from the top of the object in the direction of the other focal
point F2F2. After it passes through the lens, it travels parallel to the
principal axis.
- R3R3 goes from the top of the lens, straight through the optical centre
with its direction unchanged.
- Just like for converging lenses, the image is found at the position where
all the light rays intersect.
- Step 3. Draw the image :
Draw the image at the point where all three rays intersect.
- Step 4. Measure the distance to the object :
The distance to the object is 2,4 cm.
- Step 5. Determine type of object :
The image is on the same side of the lens as the object, and is upright. Therefore it is virtual. (Remember: The image from a diverging lens is always virtual.)
The properties of the images formed by converging and diverging lenses depend on the position of the object. The properties are summarised in the Table 3.
Table 3: Summary of image properties for converging and diverging lenses
| |
Image Properties
|
|
Lens type
|
Object Position
|
Position
|
Orientation
|
Size
|
Type
|
| Converging |
>
2
f
>
2
f
|
<
2
f
<
2
f
|
inverted |
smaller |
real |
| Converging |
2
f
2
f
|
2
f
2
f
|
inverted |
same size |
real |
| Converging |
>
f
,
<
2
f
>
f
,
<
2
f
|
>
2
f
>
2
f
|
inverted |
larger |
real |
| Converging |
f
f
|
no image formed |
| Converging |
<
f
<
f
|
>
f
>
f
|
upright |
larger |
virtual |
| Diverging |
any position |
<
f
<
f
|
upright |
smaller |
virtual |
- An object 3 cm high is at right angles to the principal axis of a concave lens of focal length 15 cm. If the distance from the object to the lens is 30 cm, find the distance of the image from the lens, and its height. Is it real or virtual?
- The image formed by a concave lens of focal length 10 cm is 7,5 cm from the lens and is 1,5 cm high. Find the distance of the object from the lens, and its height.
- An object 6 cm high is 10 cm from a concave lens. The image formed is 3 cm high. Find the focal length of the lens and the distance of the image from the lens.
This demonstration shows that:
- The eyeball has a spherical shape.
- The pupil is a small hole in the front and middle of the eye that lets light into the eye.
- The retina is at the back of the eyeball.
- The images that we see are formed on the retina.
- The images on the retina are upside down. The brain inverts the images so that what we see is the right way up.
You will need:
- a round, clear glass bowl
- water
- a sheet of cardboard covered with black paper
- a sheet of cardboard covered with white paper
- a small desk lamp with an incandescent light-bulb or a candle and a match
You will have to:
- Fill the glass bowl with water.
- Make a small hole in the middle of the black cardboard.
- Place the black cardboard against one side of the bowl and the white cardboard on the other side of the bowl so that it is opposite the black cardboard.
- Turn on the lamp (or light the candle).
- Place the lamp so it is shining through the hole in the black cardboard.
- Make the room as dark as possible.
- Move the white cardboard until an image of the light bulb or candle appears on it.
You now have a working model of the human eye.
- The hole in the black cardboard represents the pupil. The pupil is a small hole in the front of the eyeball that lets light into the eye.
- The round bowl of water represents the eyeball.
- The white cardboard represents the retina. Images are projected onto the retina and are then sent to the brain via the optic nerve.
Tasks
- Is the image on the retina right-side up or upside down? Explain why.
- Draw a simple labelled diagram of the model of the eye showing which part of the eye each part of the model represents.
Eyesight begins with lenses. As light rays enter your eye, they pass first through the cornea and then through the crystalline lens. These form a double lens system and focus light rays onto the back wall of the eye, called the retina. Rods and cones are nerve cells on the retina that transform light into electrical signals. These signals are sent to the brain via the optic nerve. A cross-section of the eye is shown in Figure 26.
For clear vision, the image must be formed right on the retina, not in front of or behind it. To accomplish this, you may need a long or short focal length, depending on the object distance. How do we get the exact right focal length we need? Remember that the lens system has two parts. The cornea is fixed in place but the crystalline lens is flexible – it can change shape. When the shape of the lens changes, its focal length also changes. You have muscles in your eye called ciliary muscles that control the shape of the crystalline lens. When you focus your gaze on something, you are squeezing (or relaxing) these muscles. This process of accommodation changes the focal length of the lens and allows you to see an image clearly.
The lens in the eye creates a real image that is smaller than the object and is inverted
(Figure 27).
In a normal eye the image is focused on the retina.
If the muscles in the eye are unable to accommodate adequately, the image will not be in focus. This leads to problems with vision. There are three basic conditions that arise:
- short-sightedness
- long-sightedness
- astigmatism
Short-sightedness or myopia is a defect of vision which means that the image is focused in front of the retina. Close objects are seen clearly but distant objects appear blurry. This condition can be corrected by placing a diverging lens in front of the eye. The diverging lens spreads out light rays before they enter the eye. The situation for short-sightedness and how to correct it is shown in Figure 29.
Long-sightedness or hyperopia is a defect of vision which means that the image is focused in behind the retina. People with this condition can see distant objects clearly, but not close ones. A converging lens in front of the eye corrects long-sightedness by converging the light rays slightly before they enter the eye. Reading glasses are an example of a converging lens used to correct long-sightedness.
Astigmatism is characterised by a cornea or lens that is not spherical, but is more curved in one plane compared to another. This means that horizontal lines may be focused at a different point to vertical lines. Astigmatism causes blurred vision and is corrected by a special lens, which has different focal lengths in the vertical and horizontal planes.
We have seen how a simple lens can be used to correct eyesight. Lenses and mirrors are also combined to magnify (or make bigger) objects that are far away.
Telescopes use combinations of lenses to gather and focus light. Telescopes collect light from objects that are large but far away, like planets and galaxies. For this reason, telescopes are the tools of astronomers. Astronomy is the study of objects outside the Earth, like stars, planets, galaxies, comets, and asteroids.
Usually the object viewed with a telescope is very far away. Objects closer to Earth, such as the moon, appear larger, and with a powerful enough telescope, we are able to see craters on the Moon's surface. Objects which are much further, such as stars, appear as points of light. Even with the most powerful telescopes currently built, we are unable to see details on the surfaces of stars.
There are many kinds of telescopes, but we will look at two basic types: reflecting and refracting.
A refracting telescope like the one pictured in Figure 32 uses two convex lenses to enlarge an image. The refracting telescope has a large primary lens with a long focal length to gather a lot of light. The lenses of a refracting telescope share a focal point. This ensures that parallel rays entering the telescope are again parallel when they reach your eye.
Some telescopes use mirrors as well as lenses and are called reflecting telescopes. Specifically, a reflecting telescope uses a convex lens and two mirrors to make an object appear larger. (Figure 33.)
Light is collected by the primary mirror, which is large and concave. Parallel rays traveling toward this mirror are reflected and focused to a point. The secondary plane mirror is placed within the focal length of the primary mirror. This changes the direction of the light. A final eyepiece lens diverges the rays so that they are parallel when they reach your eye.
The Southern African Large Telescope (SALT) is the largest single optical telescope in the southern hemisphere, with a hexagonal mirror array 11 metres across. SALT is located in Sutherland in the Northern Cape. SALT is able to record images of distant stars, galaxies and quasars a billion times too faint to be seen with the unaided eye. This is equivalent to a person being able to see a candle flame on the moon.
SALT was completed in 2005 and is a truly international initiative, because the money to build it came from South Africa, the United States, Germany, Poland, the United Kingdom and New Zealand.
SALT is part of SAAO. Write your investigation as a short 5-page report. Include images of the instrumentation used.
We have seen how lenses and mirrors are combined to magnify objects that are far away using a telescope. Lenses can also be used to make very small objects appear bigger.
Figure 12 shows that when an object is placed at a distance less than ff from the lens, the image formed is virtual, upright and is larger than the object. This set-up is a simple magnifier.
If you want to look at something very small, two lenses may work better than one. Microscopes and telescopes often use two lenses to make an image large enough to see.
A compound microscope uses two lenses to achieve high magnification (Figure 34). Both lenses are convex, or converging. Light from the object first passes through the objective lens. The lens that you look through is called the eyepiece. The focus of the system can be adjusted by changing the length of the tube between the lenses.
Drawing a Ray Diagram for a Two-Lens System
You already have all the tools to analyze a two-lens system. Just consider one lens at a time.
- Use ray tracing or the lens equation to find the image for the first lens.
- Use the image of the first lens as the object of the second lens.
- To find the magnification, multiply: mtotal=m1×m2×m3×...mtotal=m1×m2×m3×...
A compound microscope consists of two convex lenses. The eyepiece has a focal length of 10 cm. The objective lens has a focal length of 6 cm. The two lenses are 30 cm apart. A 2 cm-tall object is placed 8 cm from the objective lens.
- Where is the final image?
- Is the final image real or virtual?
We can use ray tracing to follow light rays through the microscope, one lens at a time.
- Step 1. Set up the system :
To prepare to trace the light rays, make a diagram. In the diagram here, we place the image on the left side of the microscope. Since the light will pass through the objective lens first, we'll call this Lens 1. The eyepiece will be called Lens 2. Be sure to include the focal points of both lenses in your diagram.
- Step 2. Find the image for the objective lens. :
- Step 3. Find the image for the eyepiece. :
The image we just found becomes the object for the second lens.
- A lens is any transparent material that is shaped in such a way that it will converge parallel incident rays to a point or diverge incident rays from a point.
- Converging lenses are thicker in the middle than on the edge and will bend incoming light rays towards the principal axis.
- Diverging lenses are thinner in the middle than on the edge and will bend incoming light rays away from the principal axis.
- The principal axis of a lens is the horizontal line through the centre of the lens.
- The centre of the lens is called the optical centre.
- The focus or focal point is a point on the principal axis where parallel rays converge through or diverge from.
- The focal length is the distance between the focus and the optical centre.
- Ray diagrams are used to determine the position and height of an image formed by a lens. The properties of images formed by converging and diverging lenses are summarised in Table 3.
- The human eye consists of a lens system that focuses images on the retina where the optic nerve transfers the messages to the brain.
- Defects of vision include short-sightedness, long-sightedness and astigmatism.
- Massive astronomical bodies, such as galaxies, act as gravitational lenses that can change the apparent positions of the images of stars.
- Microscopes and telescopes use systems of lenses to create magnified images of very small or very distant objects.
- Select the correct answer from the options given:
- A ........................ (convex/concave) lens is thicker in the center than on the edges.
- When used individually, a (diverging/converging) lens usually forms real images.
- When formed by a single lens, a ........................ (real/virtual) image is always inverted.
- When formed by a single lens, a ........................ (real/virtual) image is always upright.
- Virtual images formed by converging lenses are ........................ (bigger/the same size/smaller) compared to the object.
- A ........................ (real/virtual) image can be projected onto a screen.
- A ........................ (real/virtual) image is said to be "trapped" in the lens.
- A ray that starts from the top of an object and runs parallel to the axis of the lens, would then pass through the ........................ (principal focus of the lens/center of the lens/secondary focus of the lens).
- A ray that starts from the top of an object and passes through the ........................ (principal focus of the lens/center of the lens/secondary focus of the lens) would leave the lens running parallel to its axis.
- For a converging lens, its ........................ (principal focus/center/secondary focus) is located on the same side of the lens as the object.
- After passing through a lens, rays of light traveling parallel to a lens' axis are refracted to the lens' ........................ (principal focus/center/secondary focus).
- Real images are formed by ........................ (converging/parallel/diverging) rays of light that have passed through a lens.
- Virtual images are formed by ........................ (converging/parallel/diverging) rays of light that have passed through a lens.
- Images which are closer to the lens than the object are ........................ (bigger/the same size/smaller) than the object.
- ........................ (Real/Virtual) images are located on the same side of the lens as the object - that is, by looking in one direction, the observer can see both the image and the object.
- ........................ (Real/Virtual) images are located on the opposite side of the lens as the object.
- When an object is located greater than two focal lengths in front of a converging lens, the image it produces will be ........................ (real and enlarged/virtual and enlarged/real and reduced/virtual and reduced).
- An object 1 cm high is placed 1,8 cm in front of a converging lens with a focal length of 0,5 cm. Draw a ray diagram to show where the image is formed. Is the final image real or virtual?
- An object 1 cm high is placed 2,10 cm in front of a diverging lens with a focal length of 1,5 cm. Draw a ray diagram to show where the image is formed. Is the final image real or virtual?
- An object 1 cm high is placed 0,5 cm in front of a converging lens with a focal length of 0,5 cm. Draw a ray diagram to show where the image is formed. Is the final image real or virtual?
- An object is at right angles to the principal axis of a convex lens. The object is 2 cm high and is 5 cm from the centre of the lens, which has a focal length of 10 cm. Find the distance of the image from the centre of the lens, and its height. Is it real or virtual?
- A convex lens of focal length 15 cm produces a real image of height 4 cm at 45 cm from the centre of the lens. Find the distance of the object from the lens and its height.
- An object is 20 cm from a concave lens. The virtual image formed is three times smaller than the object. Find the focal length of the lens.
- A convex lens produces a virtual image which is four times larger than the object. The image is 15 cm from the lens. What is the focal length of the lens?
- A convex lens is used to project an image of a light source onto a screen. The screen is 30 cm from the light source, and the image is twice the size of the object. What focal length is required, and how far from the source must it be placed?
- An object 6 cm high is place 20 cm from a converging lens of focal length 8 cm. Find by scale drawing the position, size and nature of the image produced. (Advanced: check your answer by calculation).
- An object is placed in front of a converging lens of focal length 12 cm. By scale diagram, find the nature, position and magnification of the image when the object distance is
- A concave lens produces an image three times smaller than the object. If the object is 18 cm away from the lens, determine the focal length of the lens by means of a scale diagram. (Advanced: check your answer by calculation).
- You have seen how the human eye works, how telescopes work and how microscopes work. Using what you have learnt, describe how you think a camera works.
- Describe 3 common defects of vision and discuss the various methods that are used to correct them.
