# Physics Tutorial: Lenses. Equation of Lenses. Image Formation of Lenses

In this Physics tutorial, you will learn:

• What are lenses?
• How many types of lenses are there?
• What are the main features of lenses?
• How the image is formed in converging and diverging lenses?
• What is the equation of lenses?
• How to find the magnification produced by lenses?
• What are aberrations in lenses?
• How to combine various optical tools in an optical system?

## Introduction to Lenses. Equation of Lenses. Image Formation of Lenses

Why do people use optical glasses? What is the shape of these glasses?

Why normal glasses such as window glasses cannot be used to see better?

What happens when you direct a magnifying glass towards sunlight?

What happens to sunlight when it passes through a magnifying glass?

This tutorial is entirely focuses on lenses as the most important tools in optical systems. Since their mathematical apparatus is not very different from curved mirrors, we will deal more with some technical details that were not discussed in the previous tutorial Mirrors, Equation of Curved Mirrors and Image Formation in Plane and Curved Mirrors. Therefore, the tutorial we will explain now is an extension of the previous tutorial mentioned above.

## What Are Lenses? Types of Lenses

Lenses are optical tools used to enlarge or reduce the size of images. This is made possible through refraction occurring when light passes through them.

In a certain sense, lens is an extension of the concept of curved mirrors because unlike the latter, a lens can be used in both sides, as it is a piece of transparent material, usually circular in shape, with two polished surfaces, either or both of which is/are curved.

There are two main categories of lenses: converging (concave) and diverging (convex). They are shown in the figure below. From the figure, you can see that a converging lens is thicker at middle and thinner in extremities, while diverging lenses are thinner at middle and thicker in extremities.

The lenses shown in the figure are both regular, i.e. they behave equally in both directions. However, there is a variety of lenses with non-identical sides. Some of them are shown in the figure below. All lenses except plane-concave and plane-convex ones have two focîbecause they are formed by joining two spherical parts. As a result, they have two centres as well (one in each side). They can be at different distances from the lens depending on their curvature but for simplicity, we will only consider lenses that have identical sides, i.e. which have the same value for focîand centres in both directions, as shown in the figure below. ## Special Rays in Lenses

Like in spherical mirrors, we have to use the special rays to build up the image in lenses. However, in this case, two special rays are enough to build the image. They are:

1. The ray that originates from the top of object, is incident to the lens in parallel to the principal axis and after refracting through lens, it passes through focus in converging lenses while in diverging lenses the extension passes through focus.
2. The ray that originates from the top of object, passes through the middle of lens (point O) then moves away without changing direction, as is normally incident to the lens surface.

The two special rays used in lenses are shown in the figure below. ## Image Formation in Converging Lenses

Converging lenses are very similar in concept to concave mirrors. Therefore, we have again six possible cases of image formation in converging lenses depending to the position of object in respect to the lens.

1. The object is beyond the centre of curvature. In this case, the image is formed at the other side of lens, between focus and centre of curvature; it is diminished and inverted. The image is also real because it is produced from the two reflected rays and not from their extensions. 2. The object is at centre of curvature. In this case, the image will form at the other centre of curvature; it has the same size as the object and is inverted. The image is real because it is produced from the two reflected rays and not from their extensions. 3. The object is between centre of curvature and focus. The image is formed at the other side of lens, beyond the other centre of curvature. The image is magnified and inverted; it is real because is produced from the two reflected rays and not from their extensions. 4. The object is at focus. In this case, there is no image as the refracted rays are parallel. 5. The object is close to the lens than focus. In this case, the reflected rays diverge and therefore, we take their extension to build up the image. As a result, the image is at the same side to the object; it is enlarged, erect and virtual, as it is obtained by considering the rays' extensions. 6. The object is at infinity. In this case, the image has no dimensions. It is a bright point at focus as all parallel rays coming from the object converge at focus. ## Image Formation in Diverging Lenses

Like in convex mirrors, the image formation in diverging lenses has only one case. The image is formed closer to the mirror than focus. It is erect and diminished. Since the image is obtained from rays extensions, it is virtual. ### Example 1

The image produced by a lens is shown in the figure below. If the incident rays coming from the object refract to the lens from left to right, determine:

1. The type of lens used to produce the image shown.
2. The approximate position of the object.

### Solution 1

1. Since the object is on the left of the lens and the image on its right, this is a converging lens, as in diverging lenses the image is on the same side to the object.
2. Since the image is beyond the centre of curvature and it is on the other side of object position in respect to the lens, we can easily deduce this is the third case of image formation in converging lenses. This means the object is located between C and F on the left side. Yet, it is smaller than the image. ## Equation of Lenses

The equation of lenses is identical to that of curved mirrors. Thus, if we denote by do the position of object in respect to the lens, by dîthe position of image and by F the focus (focal length), we obtain

1/do + 1/dî = 1/F

The sign rules are identical to those used in spherical mirrors, i.e.

1. The object's distance do is always taken as positive.
2. The image's distance dîis taken as positive when the image is real, otherwise it is negative.
3. Focal length F is positive for converging lenses and negative for diverging ones.

### Example 2

An object is placed 12 cm on the left of a converging lens of focal length equal to 8 cm.

1. What type of image is produced?
2. What is the position of image in respect to the lens?

### Solution 2

1. From the description it is clear that this is an example of 3rd case of image formation in converging lenses as the object is between F and 2F (12 is between 8 and 16). Therefore, the image is real, inverted and magnified. It is formed beyond C = 2F on the right of the lens.
2. We use the lenses equation
1/do + 1/dî = 1/F

to find the position of image in respect to the lens. Thus, substituting the values (given that do = + 12 cm and F = + 8 cm) we obtain

1/12 + 1/dî = 1/8
1/dî = 1/8 - 1/12
= 3/24 - 2/24
= 1/24

Therefore dî= 24 cm. This result means the image is produced at 24 cm on the right of the lens as shown in the figure below. ## Magnification Produced by Lenses

The most important feature of lenses (for which they are produced) is the magnification they provide. The approach is the same as magnification in curved mirrors. This means we can use two formulae for the calculation of magnification:

M = himage/hobject

where h stands for height, and

M = dimage/dobject

For example, if we take the height of the object in the previous example equal to 3 cm, we can work out the image's height by combining the two formulae of magnification:

himage/hobject = dimage/dobject

Substituting the known values, we obtain for the image's height:

himage/3 cm = 24 cm/12 cm

Thus,

himage = 6 cm

## Aberration in Lenses

When we use lenses, light passes from air to another medium (usually glass) and refracts through it. It is a known fact that all lenses have their own thickness, which is different in various part of them. We have stated earlier that converging lenses are thicker at middle and diverging lenses are thicker at edges.

Aberration is the non-regular deviation of light rays through lenses due to non-uniform thickness, causing images of objects to be blurred.

In an ideal system, every point on the object will focus to a point of zero size on the image. However, in reality this does not occur, because lenses are not ideal optical tools. As a result, parallel rays do not converge at a single dimensionless point as assumed earlier, but in a zone around focus, as shown in the figure. In curved mirrors, aberrations are less visible as light does not enter inside the glass but it is reflected by the mirroring surface. Therefore, curved mirrors are more preferable than lenses to be used in powerful optical systems such as telescopes and microscopes.

## Combined Systems of Lenses and Curved Mirrors

We can combine optical tools such as plane and curved mirror with lenses to produce new optical systems, like we did in the previous tutorial where two curved mirrors were combined to produce new images. Let's see an example in this regard.

### Example 3

An optical magnifying device is composed by a converging lens and a concave mirror as shown in the figure. What is the total magnification produced by this optical system if the object is placed at the distance shown?

### Solution 3

First, let's calculate the position of the image produced by the converging lens. From the figure, we can extract the following clues:

do = 22 cm - 16 cm = 6 cm
FL = 8 cm
dî = ?

Applying the equation

1/do + 1/dî = 1/FL

we obtain

1/6 + 1/dî = 1/8
1/dî = 1/8 - 1/6
= 3/24 - 4/24
= - 1/24

This means the image produced by the converging lens is 24 cm on its left. Thus, the magnification produced by the converging lens is

ML = di/do
= - 24/6
= - 4

This means the image produced by the converging lens is 4 times larger than the object.

This image acts as an object for the concave mirror. Its distance from the mirror is

do = 16 cm + 24 cm = 40 cm
F = 20 cm
dî = ?

We have:

1/do + 1/dî = 1/FM
1/40 + 1/dî = 1/20
1/dî = 1/20 - 1/40
= 2/40 - 1/40
= 1/40

Thus, the second image is 40 cm on the left of the converging lens. In this case, there is no magnification as

MM di/do
= 40/40
= 1

Thus, the mirror is used only to turn the image upright.

Therefore, the total magnification produced by this system is

Mtotal = ML × MM
= - 4 × 1
= - 4

This means the final image is 4 times larger than the object.

## Physics Revision: Lenses. Equation of Lenses. Image Formation of Lenses Summary

Lenses are optical tools used to enlarge or reduce the size of images by means of refraction of light. A lens is a piece of transparent material, usually circular in shape, with two polished surfaces, either or both of which is/are curved.

There are two main categories of lenses: converging (concave) and diverging (convex). Converging lenses are thicker at middle and thinner in extremities, while diverging lenses are thinner at middle and thicker in extremities.

All lenses except plane-concave and plane-convex ones have two foci, because they are formed by joining two spherical parts. As a result, they have two centres (one in each side) as well.

Like in spherical mirrors, we have to use the special rays to build up the image in lenses. However, unlike in curved mirrors, two special rays are enough to build the image in lenses. They are:

1. The ray that originates from the top of object, is incident to the lens in parallel to the principal axis and after refracting through lens, it passes through focus in converging lenses while in diverging lenses the extension passes through focus.
2. The ray that originates from the top of object, passes through the middle of lens (point O) then moves away without changing direction, as is normally incident to the lens surface.

Converging lenses are very similar in concept to concave mirrors. Therefore, we have again six possible cases of image formation in converging lenses depending to the position of object in respect to the lens.

1. The object is beyond the centre of curvature. In this case, the image is formed at the other side of lens, between focus and centre of curvature; it is diminished and inverted. The image is also real because it is produced from the two reflected rays and not from their extensions.
2. The object is at centre of curvature. In this case, the image will form at the other centre of curvature; it has the same size as the object and is inverted. The image is real because it is produced from the two reflected rays and not from their extensions.
3. The object is between centre of curvature and focus. The image is formed at the other side of lens, beyond the other centre of curvature. The image is magnified and inverted; it is real because is produced from the two reflected rays and not from their extensions.
4. The object is at focus. In this case, there is no image as the refracted rays are parallel.
5. The object is close to the lens than focus. In this case, the reflected rays diverge and therefore, we take their extension to build up the image. As a result, the image is at the same side to the object; it is enlarged, erect and virtual, as it is obtained by considering the rays' extensions.
6. The object is at infinity. In this case, the image has no dimensions. It is a bright point at focus as all parallel rays coming from the object converge at focus.

Like in convex mirrors, the image formation in diverging lenses has only one case. The image is formed closer to the mirror than focus. It is erect and diminished. Since the image is obtained from rays extensions, it is virtual.

The equation of lenses is identical to that of curved mirrors. Thus, if we denote by do the position of object in respect to the lens, by dîthe position of image and by F the focus (focal length), we obtain

1/do + 1/dî = 1/F

The sign rules are identical to those used in spherical mirrors, i.e.

1. The object's distance do is always taken as positive.
2. The image's distance dîis taken as positive when the image is real, otherwise it is negative.
3. Focal length F is positive for converging lenses and negative for diverging ones.

The most important feature of lenses (for which they are produced) is the magnification they provide. The approach is the same as magnification in curved mirrors. This means we can use two formulae for the calculation of magnification:

M = himage/hobject

where h stands for height, and

M = dimage/dobject

Aberration is the non-regular deviation of light rays through lenses due to non-uniform thickness, causing images of objects to be blurred.

In an ideal system, every point on the object will focus to a point of zero size on the image. However, in reality this does not occur, because lenses are not ideal optical tools. As a result, parallel rays do not converge at a single dimensionless point as assumed earlier, but in a zone around focus.

In curved mirrors, aberrations are less visible as light does not enter inside the glass but it is reflected by the mirroring surface. Therefore, curved mirrors are more preferable than lenses to be used in powerful optical systems such as telescopes and microscopes.

We can combine optical tools such as plane and curved mirror with lenses to produce new optical systems.

## Physics Revision Questions for Lenses. Equation of Lenses. Image Formation of Lenses

1. The image produced by a converging lens is shown in the figure below. What are the image features if the object is on the left of the lens?

1. Real, inverted and magnified
2. Virtual, inverted and magnified
3. Real, inverted and diminished
4. Virtual, erect and magnified

2. An object is placed 12 cm on the left of a diverging lens. If the image is produced 8 cm on the left of the lens, what is the height of the image if the object is 6 cm high?

1. 4.0 cm
2. 3.0 cm
3. 2.5 cm
4. 2.0 cm

3. An optical system composed by a diverging lens and a plane mirror is shown in the figure below. How far from the plane mirror is the second image formed in it (the image formed by the first image produced in the lens) if the plane mirror is placed at the centre of curvature of the diverging lens? Express the result in terms of F.

1. 2F/3 due left of the mirror
2. 8F/3 due left of the mirror
3. 2F due right of the mirror
4. 8F/3 due right of the mirror