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 [link].
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 converge parallel rays of light and are thicker in the middle than at the edges.
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].
Before we study lenses in detail, there are a few important terms that must be defined. [link] shows important lens properties:
We will only discuss double convex converging lenses as shown in [link]. Converging lenses are thinner on the outside and thicker on the inside.
[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.
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:
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:
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:
Results:
Relative position of object | Relative position of image | Image upright or inverted | Relative size of image | Nature of image |
Beyond $2f$ cm | ||||
At $2f$ | ||||
cm | ||||
Between $2f$ and $f$ | ||||
cm | ||||
At $f$ | ||||
cm | ||||
Between $f$ and the lens | ||||
cm |
QUESTIONS:
Aim:
To determine the mathematical relationship between ${d}_{0},{d}_{i}$ and $f$ for a lens.
Method:
Results:
$f$ = focal length of lens ${d}_{0}$ = object distance ${d}_{i}$ = image distance
Object distance | Image distance | $\frac{1}{{d}_{0}}$ | $\frac{1}{{d}_{i}}$ | $\frac{1}{{d}_{0}}+\frac{1}{{d}_{i}}$ |
${d}_{0}$ (cm) | ${d}_{i}$ (cm) | (cm${}^{-1}$) | (cm${}^{-1}$) | (cm${}^{-1}$) |
25,0 | ||||
20,0 | ||||
18,0 | ||||
15,0 | ||||
Average = |
QUESTIONS:
Drawing Ray Diagrams for Converging Lenses
Ray diagrams are normally drawn using three rays. The three rays are labelled ${R}_{1}$, ${R}_{2}$ and ${R}_{3}$. The ray diagrams that follow will use this naming convention.
Convex lens:
Concave lens:
We can locate the position of the image by drawing our three rays. ${R}_{1}$ travels from the object to the lens parallel to the principal axis, is bent by the lens and then travels through the focal point. ${R}_{2}$ passes through the focal point before it enters the lens and therefore must leave the lens parallel to the principal axis. ${R}_{3}$ travels through the center of the lens and does not change direction. The point where ${R}_{1}$, ${R}_{2}$ and ${R}_{3}$ intersect is the image of the point where they all started.
The image of an object placed at a distance greater than $2f$ 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.
We can locate the position of the image by drawing our three rays. ${R}_{1}$ 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. ${R}_{2}$ passes through the focal point before it enters the lens and therefore must leave the lens parallel to the principal axis. ${R}_{3}$ travels through the center of the lens and does not change direction. The point where ${R}_{1}$, ${R}_{2}$ and ${R}_{3}$ intersect is the image of the point where they all started.
The image of an object placed at a distance equal to $2f$ 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 $2f$ away from the lens.
We can locate the position of the image by drawing our three rays. ${R}_{1}$ 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. ${R}_{2}$ passes through the focal point before it enters the lens and therefore must leave the lens parallel to the principal axis. ${R}_{3}$ travels through the center of the lens and does not change direction. The point where ${R}_{1}$, ${R}_{2}$ and ${R}_{3}$ intersect is the image of the point where they all started.
The image of an object placed at a distance between $2f$ and $f$ 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 $2f$ away from the lens.
We can locate the position of the image by drawing our three rays. ${R}_{1}$ 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. ${R}_{2}$ passes through the focal point before it enters the lens and therefore must leave the lens parallel to the principal axis. ${R}_{3}$ travels through the center of the lens and does not change direction. The point where ${R}_{1}$, ${R}_{2}$ and ${R}_{3}$ intersect is the image of the point where they all started.
The image of an object placed at a distance less than $f$ 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:
where $f$ is the focal length, ${d}_{o}$ is the object distance and ${d}_{i}$ is the image distance.
The object distance ${d}_{o}$ is the distance from the object to the lens. ${d}_{o}$ 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 ${d}_{i}$ 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 ${d}_{i}$ 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 ${d}_{i}$ 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.
where $m$ is the magnification, ${d}_{o}$ is the object distance and ${d}_{i}$ is the image distance.
If ${d}_{i}$ and ${d}_{o}$ are both positive, the magnification is negative. This means the image is inverted, or upside down. If ${d}_{i}$ is negative and ${d}_{o}$ 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.
Properties of the image are required.
The image is real, ${d}_{i}$ 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.
Draw the lens, the object and mark the focal points.
The image is at the place where all the rays intersect. Draw the image.
The image is 5 cm away from the lens, on the opposite side of the lens to the object.
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.
Draw the lens, principal axis, focal points and the object.
The image is 6 cm away from the lens, on the same side as the object.
The image is 3 cm high.
Since the image is on the same side of the lens as the object, the image is virtual.