LENSES
There are two types of lenses
1. Convex or Converging
2. Concave or Diverging
A ray of light is refracted twice by a lens, once when it
passes into the lens (air to glass) and once when it
emerges from the lens (glass to air).
Convex lens causes the rays of light to form a converging
beam and a concave lens forms a diverging beam.
The centre of the lens is called the optical centre and the
direction through the optical centre and perpendicular to
the lens is called the principal axis.
A beam parallel to the principal axis will form a
converging beam with a convex lens and a diverging
beam with a concave lens.
The focus is the point on the principal axis to which all
rays originally parallel and close to the principal axis
converge to or from which the diverge from after passing
through the lens.
The focal length is the distance from the focus to the
optical centre.
For a Convex lens the focus is real and so the focal
length f is positive
For a concave lens the focus is virtual and so the focal
length f is negative.
Also since the light may pass through a lens in either
direction there are two focus points equidistant from the
optical centre.
Ray Diagrams used to locate the image in a lens.
Three classes of rays
1. Rays parallel to the principal axis will pass
through the focus after refraction through the
lens.
2. Rays through the principal focus will emerge
parallel to the principal axis after refraction
through the lens (reversibility of light)
3. Rays through the optical centre are undeviated.
Convex Lens
1. Object between F1 and Optical Centre
Image is
(i)
(ii)
(iii)
(iv)
Behind the object
Virtual
Upright
Larger than object.
2. Object at either F1 or F2
Image is at infinity
3. Object between F1 and 2F1
Image is
(i)
(ii)
(iii)
(iv)
Beyond 2F2
Real
Inverted
Magnified
4. Object is at 2F1
Image is
(i)
(ii)
(iii)
(iv)
At 2F2
Real
Inverted
Same size as object
5. Object beyond 2F1
Image is
(i) Between F2 and 2F2
(ii) Real
(iii) Inverted
(iv) Smaller than object
6. Object at infinity
Image is
(i)
(ii)
(iii)
(iv)
At F
Real
Inverted
Smaller than object
For a concave lens the image is always
(i)
(ii)
(iii)
(iv)
Between the object and the lens
Virtual
Upright
Diminished
So again Real image implies a convex lens
Virtual and magnified implies convex
Virtual and diminished implies convex
LENS / MIRROR EQUATION
All distance values are measured with the optical centre
as the origin
We use the following symbols
u = distance from the object to the lens
v = distance from the image to the lens
f = focal length distance from focus to lens
For any lens it can be proved that
1 1 1
 
u v f
and also that the magnification
v
height of image
m

height of object u
And again using the REAL IS POSITIVE convention
give
Convex lens : Real focus so f is +ve
Concave lens : Virtual focus so f is –ve
Lenses in combination
When two lenses are placed in combination together
another lens is produced. This produces a lens with a
focal length which we will call fcombination
If f1 is the focal length of the first lens and
f2 is the focal length of the second lens
It can be shown that
1
1
1


fC
f1 f 2
POWER.
The power of a lens is defined as the inverse of its focal
length
Power of a lens 
1
1

focal length in metres f
S.I. Unit Basic m-1 Called Dioptre
Experimental Method to calculate the focal length of
mirrors / Lens
The diagram below shows where the focus may be found
for a convex lens and a concave mirror.
Convex Lens
The focus is the point through which light rays striking
the lens/mirror along a line close to and parallel to the
principal axis pass.
The focal length , f, is determined by the curvature of the
surfaces involved and is a fixed property of a lens or
mirror.
If an object is placed at a distance u from a lens/mirror of
focal length f , then an image of the object will be
focussed in the lens/mirror at a distance v from it
accord
ing to the equation:
1 1 1
 
u v f
General Lens/Mirror Equation
Method:
1. Using the illuminated grid or the light box as
your object set up a number of the arrangements
shown in the ray diagrams. Try to find the image
on a screen where possible and if this is not
possible explain why.
2.
For each case measure u and v.
3. Determine the focal length ,f , using the above
formula.
Method 2 . Using no Parallax.
The apparent relative movement of two objects owing to
the movement on the part of the observer is called
parallax.
When two objects coincide in position there is NO
PARALLAX.
No parallax is used to locate the position of a virtual
image as it is not real it can not be produced on the
screen.
An object pin is placed in front of the mirror and a search
pin is placed behind the mirror.
The pins are set up in such a way so as the image of the
object pin in the mirror and the search pin are in line
Search pin
I image of object pin in the
mirror
OBJECT PIN
The eye is moved from side to side while viewing a
search pin behind the mirror.
A position is found for which both pin and image appear
to coincide in the same straight line. When this
condition of no parallax holds the search pin gives the
position of the image.
The procedure can also be used where there is a virtual
image and v is a negative value in the equation.
PROBLEM SHEET.
Question 1. A convex lens has a focal length
20 cm and an object is placed (a) 25 cm and
(b) 15 cm from it. Calculate the image if the object is 10
cm high. For both positions produce a ray diagram
Question 2. A concave lens has a focal length of 12 cm.
An object is placed 60 cm from the lens. Find the
position and size of the image if the object 6 cm high.
(b) If a convex lens of focal length 20 cm is placed in
combination with the concave lens where is the new
image formed. (c) Calculate the power of the
combination lens.
Question 3. A lens placed 25.4 cm from an object
produces a virtual image four times as large as the object.
What type of lens is this, and what is its focal length.
Question 4. An object O forms a sharp image on a
screen when a lens is placed 10 cm from the screen.
When the lens is moved 2 cm further from O, the screen
has to be moved
2 cm closer to the lens to retain the focussed image.
Calculate the power of the lens.
Question 5. A convex lens has a focal length of 20 cm. An
object 8 cm high is placed 30 cm in front of the lens.
(i)
Draw ray diagram to find the position,
height and nature of the image.
(ii) Using the lens formula calculate the
position, height and nature of the image.
(iii) Calculate the power of the lens.
(iv) If a second lens, concave of focal length
100 cm, is placed in combination with the
first lens calculate the height of the new
image.
Question 6. A convex lens has a focal length of 20 cm.
An object of height 10cm is placed 50 cm in front of the
lens.
(i) Calculate the position, height and nature
of the image using lens/ mirror equations.
(ii) Verify the above values using a ray
diagram.
(iii) Calculate the power of the lens.
(iv) Assuming that the object position is fixed
calculate the type and focal length of a lens
that needs to be put in combination with the
first lens so that the image becomes
magnified by 5.