Geometrical optics: Converging lenses (Grade 11)

Introduction

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.

Lenses

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 [link].

The behaviour of parallel light rays entering either a converging or diverging lens.
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!

Converging Lenses

Converging lenses converge parallel rays of light and are thicker in the middle than at the edges.

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 [link].

Types of lenses

Before we study lenses in detail, there are a few important terms that must be defined. [link] shows important lens properties:

Properties of lenses.

Converging Lenses

We will only discuss double convex converging lenses as shown in [link]. Converging lenses are thinner on the outside and thicker on the inside.

A double convex lens is a converging lens.

[link] 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.

Light rays bend towards each other or converge when they travel through a convex lens. F1F1 and F2F2 are the foci of the lens.

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:

  1. the object is placed at a distance greater than 2f2f from the lens
  2. the object is placed at a distance equal to 2f2f from the lens
  3. the object is placed at a distance between 2f2f and ff from the lens
  4. 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.

Experiment : Lenses A

Aim:

To determine the focal length of a convex lens.

Method:

  1. 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.
  2. Measure the distance between the lens and the sheet of paper as accurately as possible.

Results:

The focal length of the lens is                       cm

Experiment : Lenses B

Aim:

To investigate the position, size and nature of the image formed by a convex lens.

Method:

  1. 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.
  2. 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.
  3. Fill in the relevant information on the table below
Experimental setup for investigation.

Results:

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:

  1. When a convex lens is being used:
    1. A real inverted image is formed when an object is placed                      
    2. No image is formed when an object is placed                      
    3. An upright, enlarged, virtual image is formed when an object is placed                      
  2. Write a conclusion for this investigation.
Experiment : Lenses C

Aim:

To determine the mathematical relationship between d0,did0,di and ff for a lens.

Method:

  1. Using the same arrangement as in Experiment Lenses B, place the object (candle) at the distance indicated from the lens.
  2. 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

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 )

QUESTIONS:

  1. Compare the values for (a) and (b) above and explain any similarities or differences
  2. What is the name of the mathematical relationship between d0d0, didi and ff?
  3. 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.

  1. 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.
  2. Any ray travelling parallel to the principal axis is bent through the focal point.
  3. 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.
  4. 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.
  5. 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.
In ray diagrams, lenses are drawn like this:

Convex lens:

Concave lens:

CASE 1: Object placed at a distance greater than 2f2f from the lens
An object is placed at a distance greater than 2f2f away from the converging lens. Three rays are drawn to locate the image, which is real, and smaller than the object and inverted.

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.
CASE 2: Object placed at a distance equal to 2f2f from the lens
An object is placed at a distance equal to 2f2f away from the converging lens. Three rays are drawn to locate the image, which is real, the same size as the object and inverted.

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.

CASE 3: Object placed at a distance between 2f2f and ff from the lens
An object is placed at a distance between 2f2f and ff away from the converging lens. Three rays are drawn to locate the image, which is real, larger than the object and inverted.

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.

CASE 4: Object placed at a distance less than ff from the lens
An object is placed at a distance less than ff away from the converging lens. Three rays are drawn to locate the image, which is virtual, larger than the object and upright.

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 and magnification

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

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 [link]). 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 ([link]). 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

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.

Using the lens equation

An object is placed 6 cm from a converging lens with a focal point of 4 cm.

  1. Calculate the position of the image
  2. Calculate the magnification of the lens
  3. Identify three properties of the image
  1. f = 4 cm d o = 6 cm d i = ? m = ? f = 4 cm d o = 6 cm d i = ? m = ?

    Properties of the image are required.

  2. 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
  3. m = - d i d o = - 12 6 = - 2 m = - d i d o = - 12 6 = - 2
  4. The image is real, didi is positive, inverted (because the magnification is negative) and enlarged (magnification is >> 1)

Locating the image position of a convex lens: I

An object is placed 5 cm to the left of a converging lens which has a focal length of 2,5 cm.

  1. What is the position of the image?
  2. Is the image real or virtual?
  1. Draw the lens, the object and mark the focal points.

    • 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.
  2. The image is at the place where all the rays intersect. Draw the image.

  3. The image is 5 cm away from the lens, on the opposite side of the lens to the object.

  4. Since the image is on the opposite side of the lens to the object, the image is real.

Locating the image position of a convex lens: II

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.

  1. Is the image real or virtual?
  2. What is the position and height of the image?
  1. Draw the lens, principal axis, focal points and the object.

    • 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.
  2. The image is 6 cm away from the lens, on the same side as the object.

  3. The image is 3 cm high.

  4. Since the image is on the same side of the lens as the object, the image is virtual.

Phet simulation for Geometrical Optics
Converging Lenses
  1. 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.
  2. In each case state whether a real or virtual image is formed:
    1. Much further than 2f2f
    2. Just further than 2f2f
    3. At 2f2f
    4. Between 2f2f and ff
    5. At ff
    6. Between ff and 0
    Is a virtual image always inverted?
  3. 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?
  4. 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?
  5. 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.
  6. 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.
  7. What would happen if you placed the object right at the focus of a converging lens? Hint: Draw the picture.