Download Today`s Objectives

Lenses
Astrophysics Lessons 1
Homework
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No homework except to revise for the mock
exam next Friday!
Textbooks!
(Astrophysics is not covered in the
specification book)
Plan for today…
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P2: Peer mark the open book exam please ask
any questions you wish (~30 mins).
Then we start astrophysics!
P4: Draw ray diagrams and play with some
lenses.
Today’s Objectives
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Explain what is meant by the principal focus and
focal length of a converging lens.
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Calculate the power of a lens using power = 1/f
Objectives
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In the exam you are expected to know about:
Principal focus, focal length of converging lens
power = 1/f
Formation of images by a converging lens
Ray diagrams
Lens formula: 1/f = 1/u + 1/v
Astronomical telescope consisting of two converging lenses
Ray diagram to show the image formation in normal adjustment
Angular magnification in normal adjustment
M=
angle subtended by image at eye
angle subtended by object at unaided eye
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Focal lengths of the lenses
M = fo/fe
Convex Lens
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Lenses work by refracting light at a glass-air
boundary. Although refraction occurs at the
boundary, we will treat all lenses as bending the
rays at the lens axis.
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The lens in the eye is a convex or converging
lens. This means that the lens makes rays of
light come together, or converge.
Convex Lens
Principal Focus/Focal Length
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The rays parallel to the principal axis are
converged onto the principal focus. The focal
length is the distance between the lens axis and
the principal focus (strictly speaking, the focal
plane).
The power of a lens
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Thicker lenses bend light more, and are
therefore described as more powerful. Powerful
lenses have short focal lengths. The power of a
lens is measured in dioptres (D) and is given by
the formula:
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Power =
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1
focal length (m)
Quick Question
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The power a lens is +0.2 D What is the focal
length in metres?
Real
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The principal focus of a convex lens is called
real. The images made by convex lenses are in
most cases real. This means that the image can
be projected onto a screen. We will see now
how images are made with ray diagrams.
Ray Diagrams
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We can determine where an image lies in
relation to the objects by using a ray
diagram. We can do this by using two simple
rules:
Draw a ray from the top of the image parallel to
the principal axis. This ray bends at the lens axis
and goes through the principal focus.
Draw a ray from the top of the object through
the centre of the lens.
Ray Diagrams
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Where the two rays meet, that is where the
image is found. The following diagrams shows
how we do a ray diagram step-by-step:
A Worked Example
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Step 1: Draw the ray parallel to the principal axis.
A Worked Example
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Step 2: Draw the refracted ray so that it passes through
the principal focus.
A Worked Example
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Step 3 Draw a ray from the top of the object through
the middle of the lens. This ray is undeviated.
A Worked Example
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Step 4 Where the rays meet, that is where the image is.
Ray Diagrams
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It is a good idea to draw your ray diagrams on
graph paper if possible.
Be careful with your drawing; a small change in
the angle of the undeviated ray can lead to quite
a big change in the final position of the image.
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Use a sharp pencil.
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Now lets try some…(F = 3cm)
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1) What is the image like if the object is beyond
2F?
2) What is the image like if the object is at 2F?
3) What is the image like if the object is between
2F and F, (1.5F)?
4) What is the image like if the object is at F?
5) What is the image like if the object is less
than F?
Object at > 2F
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Image is real, inverted and diminished
Object at 2F
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Image is real, inverted, and the same size.
Object between 2F and F
Image is real, inverted, and magnified.
Object at F
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Image is real, inverted, at infinity.
Object at < F
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Image is virtual, right way up, and magnified.
The Lens Formula
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Lens diagrams have the main disadvantage that there is
uncertainty in precisely where the image is. Therefore
the use of the lens formula is better. The lens formula
is:
1 1 1
 
f u v
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where f is the focal length, u is the object distance and v
is the image distance.
Question
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An object of height 1.6 cm is placed 50 cm from
a converging (convex) lens of focal length 10
cm.
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What is the position of the image?
What is the magnification?
What is the size of the image?
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Question
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An object of height 1.6 cm is placed 50 cm from
a converging (convex) lens of focal length 10
cm. What is the position of the image?
v = 1/0.08 = 12.5 cm
Question
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An object of height 1.6 cm is placed 50 cm from a
converging (convex) lens of focal length 10 cm.
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M = 12.5 ÷ 50 = 0.25
Image 1.6 × 0.25 = 0.4 cm = 4 mm
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The convention for the equation is that real is
positive. For a concave lens, the focal length is
negative, because the principal focus is virtual. If the
image position gives a negative value, then the image is
virtual.
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It does not matter if you work in cm, as long as
you are consistent. However if you are going to
use dioptres you must work in metres.
Magnification
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The magnification is worked out using this simple
formula:
v
M
u
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Since v is in metres, and u is in metres, M has no units.
Question
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Find the position and size of a pound coin, 2.2
cm in diameter placed 20 cm from a converging
lens of focal length 40 cm
Question
Find the position and size of a pound coin, 2.2 cm in
diameter placed 20 cm from a converging lens of focal
length 40 cm
 1/f = 1/u + 1/v
 1/40 = 1/20 + 1/v
 1/v = 0.025 - 0.05 = -0.025
 v = - 40 cm
 magnification = v/u = -40/20 = (-)2
 Therefore the image is 4.4 cm across