Date: 12/11/2024
Class: #19
Syllabus Topic: Lenses
Title: Lens Diagrams, Magnification, Human Eye, Past Paper Questions
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Types of Lenses in Ray Optics
A lens is a piece of specially shaped transparent material that can form focused images of
objects. There are two types of lens:
1. Convex lenses
2. Concave lenses
Convex Lenses:
A convex or converging lens is one that is thicker at its centre. It can converge parallel rays
of light to produce a real image.
Concave Lenses:
A concave or diverging lens is one that is thinner at its centre. It can diverge parallel rays of
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light to produce a virtual image.
Manipulative
Link:
https://phet.colorado.edu/sims/html/geometric-optics/latest/geometric-optics_all.html
This simulation will help you in drawing ray diagrams.
Terms used with lenses
Here are some terms to familiarize yourself with:
• The optical centre, 𝑂, of a lens is the point at the centre of the lens through which all rays
pass without deviation.
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• The principal axis of a lens is the line that passes through its optical centre and is
perpendicular to the faces of the lens.
• The principal focus, 𝐹, of a lens is the point on the principal axis through which all rays
parallel and close to the axis converge, or from which they appear to diverge, after passing
through the lens.
• The focal length, 𝑓, of a lens is the distance between its optical centre and its principal
focus.
• The focal plane of a lens is the surface perpendicular to its principal axis and containing
its principal focus.
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Consider the diagram below:
From the diagram above,
𝐴 – principal axis
𝐵 – focal plane
𝐶 – optical centre
𝐷 and 𝐸 – principal focus
The focal length will be the distance between 𝐶 and 𝐷.
The symbol 𝐹 is used to represent the focal length.
The points 𝐺 and 𝐻 represent 2𝐹 or in other words, twice the focal length.
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Diagram for convex lens:
Diagram for concave lens:
Important rays for lens diagrams
Parallel rays focus on the focal plane even if they are not parallel to the principal axis. In
order to determine where the rays focus, draw an incident ray straight through the optical
centre to a point on the focal plane. Then connect the other parallel rays, after passing
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through the lens, to the same point.
The diagram below summarises the rays used to construct image positions:
Constructing Scale Diagrams for Convex and Concave Lenses
1. Draw two perpendicular lines to represent the principal axis and the lens.
2. Place points, F, to scale in position, to represent the principal foci.
3. Draw the object to scale, in size and position, to stand on the principal axis.
4. Draw lines to represent the following rays from the top of the object:
•
parallel to the principal axis, and then through F after passing through the
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lens
•
straight through the optical centre.
5. Where the rays cross represents the top of the image. Draw the image from the
principal axis to this point.
1. Object is between 𝐹 and optical centre:
Object
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Image
This is the only case where the image is virtual and erect. The image is also magnified and is
on the same side of lens object. The distance of the object from the lens must be less than
the focal length.
Uses for this type of lens:
1. Magnifying glass
2. Instrument eyepieces
3. Spectacles for long-sightedness
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2. Object is at 𝐹
In this situation, no localised image is formed as the rays of light do not converge. The
image of the object will be located at infinity.
Uses for this type of lens:
1. Spot light
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3. Object is between 2𝐹 and 𝐹:
The image has been magnified, but has been inverted. The image in this situation is real and
located on the other side of the lens than the object.
Uses for this type of lens:
1. Projector
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4. Object is at 2𝐹:
The image is real, inverted and located on the other side of the object. The image falls on
the focal point and is the same size as the object.
Uses for this type of lens:
1. Telescope
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5. Object is located beyond 2𝐹:
The image is real, inverted and located on the other side of the object. However, the image
has been diminished, meaning, it is smaller than the object.
Uses for this type of lens:
1. Camera
2. Eyes
Magnification
Definition:
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Magnification, 𝑚, is the ratio of the size of the image to the size of the object.
Formula:
𝐼𝑚𝑎𝑔𝑒 𝐻𝑒𝑖𝑔ℎ𝑡
𝐼𝑚𝑎𝑔𝑒 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
Magnification = 𝑂𝑏𝑗𝑒𝑐𝑡
= 𝑂𝑏𝑗𝑒𝑐𝑡
𝐻𝑒𝑖𝑔ℎ𝑡
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑣
𝑚= 𝑢
Lens Formula
Formula:
1
𝑓
1
1
= 𝑢 + 𝑣
where 𝑓 = focal length: + for convex, − for concave
𝑢 = object distance: + if real, − if virtual
𝑣 = image distance: + if real, − if virtual
Example:
You are given the following information:
focal length = 67 𝑐𝑚
object distance = 106 𝑐𝑚
Find the image distance.
Solution:
1
𝑓
= 𝑢 + 𝑣
1
1
1
67
= 106 + 𝑣
1
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1
1
𝑣
= 67 − 106
1
1
1
𝑣
= 7102
39
𝑣=
7102
39
𝑣 = 182 𝑐𝑚 (to the nearest 𝑐𝑚)
Question:
Determine the distance of the image from the lens if the focal length is 10 𝑐𝑚 and the object
was placed a distance of 12 𝑐𝑚 in front of the lens.
Solution:
𝑓 = 10 𝑐𝑚
𝑢 = 12 𝑐𝑚
1
𝑓
= 𝑣 + 𝑢
1
1
1
10
= 𝑣 + 12
1
1
1
𝑣
= 10 − 12
1
1
1
𝑣
= 60
1
𝑣 = 60 𝑐𝑚
June 2015 – Question 6
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(a) (i) Copy Figure 3 and draw the ray drawn as it emerges on the other side of the lens and
its relation to the focus. Show the principal axis and the focal length.
[5]
(ii) Write, in words or symbols, the formula for the magnification of an object.
[1]
(b) An object 𝐴𝐵 is placed 20 𝑐𝑚 in front of a converging lens of focal length, 𝐹, 10 𝑐𝑚 as
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seen in Figure 4.
(i)
Calculate the position of the image formed and state on what side of the lens it is
located.
[5]
(ii)
Calculate the magnification of the image formed.
[3]
(iii)
Is the image formed real or virtual?
[1]
Total: 15 marks
June 2015 – Question 6 – Solution
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(a) (i) Copy Figure 3 and draw the ray drawn as it emerges on the other side of the lens and
its relation to the focus. Show the principal axis and the focal length.
The completed figure is as follows:
[5]
[1]
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(ii) Write, in words or symbols, the formula for the magnification of an object.
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑚𝑎𝑔𝑒
Magnification = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡
or
𝑖𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
Magnification = 𝑜𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
(b) An object 𝐴𝐵 is placed 20 𝑐𝑚 in front of a converging lens of focal length, 𝐹, 10 𝑐𝑚 as
seen in Figure 4.
(i)
Calculate the position of the image formed and state on what side of the lens it is
located.
Object distance, 𝑢 = 20 𝑐𝑚
[5]
Focal length, 𝑓 = 10 𝑐𝑚
Image distance, 𝑣 = ?
Using the Lens formula:
= 𝑢 + 𝑣
1
1
1
10
= 20 + 𝑣
1
1
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1
𝑓
1
𝑣
= 10 − 20
1
1
1
𝑣
= 20
1
𝑣 = 20 𝑐𝑚
Note that the image is at 2𝐹on the opposite side of the lens to the object and is
20 𝑐𝑚 from the lens.
(ii)
Calculate the magnification of the image formed.
[3]
𝐼𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
Magnification = 𝑂𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
20
Magnification = 20
Magnification = 1
(iii)
Is the image formed real or virtual?
The image is real.
[1]
Total: 15 marks
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June 2017 – Question 6
(a) Using fully labelled diagrams, define the principal focus of
(i) a converging lens
[3]
(ii) a diverging lens
[3]
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(b) An object of height 7. 2 𝑐𝑚 is placed 18. 0 𝑐𝑚 from a converging lens of focal length
12. 0 𝑐𝑚. Determine
(i)
the image distance
[3]
(ii)
the magnification
[3]
(iii)
the height of the image formed
[2]
(iv)
whether the image formed is real or virtual
[1]
Total: 15 marks
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June 2017 – Question 6 – Solution
(a) Using fully labelled diagrams, define the principal focus of
(i) a converging lens
[3]
For a converging lens,
The principal focus of a converging lens is a point on the principal axis which all rays
initially parallel to the principal axis will converge on after refraction by the lens.
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(ii) a diverging lens
[3]
For a diverging lens,
The principal focus of a diverging lens is a point on the principal axis which all rays
initially parallel to the principal axis appear to diverge from after refraction by the
lens.
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(b) An object of height 7. 2 𝑐𝑚 is placed 18. 0 𝑐𝑚 from a converging lens of focal length
12. 0 𝑐𝑚. Determine
(i)
the image distance
[3]
Using the lens formula,
1
𝑓
= 𝑢 + 𝑣
1
1
1
12
= 18 + 𝑣
1
𝑣
= 12 − 18
1
𝑣
= 36
1
1
1
1
1
𝑣 = 36 𝑐𝑚
∴ The image distance is 36 𝑐𝑚.
(ii)
the magnification
𝐼𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
Magnification = 𝑂𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
[3]
36
𝑀𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 = 18
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Magnification = 2
(iii)
the height of the image formed
[2]
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑚𝑎𝑔𝑒
Magnification = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡
∴ Image height = magnification × object height
∴ Image height = 2×7. 2
∴ Image height = 14. 4 𝑐𝑚
(iv)
whether the image formed is real or virtual
[1]
The image is real.
Total: 15 marks