Reflection
Reflection is the bouncing off of light from a surface.
A ray of light heading towards an object is called an incident ray. If it reflects off the
object, it is called a reflected ray. A perpendicular line drawn at any point on a surface is
called a normal (just like with normal force). The angle between the incident ray and
normal is called the angle of incidence, i, and the angle between the reflected ray and the
normal ray is called the angle of reflection, r.
Laws of reflection
 The incident ray, the reflected ray, and the normal to the surface of the mirror all
lie in the same plane
 The angle of incidence i is equal to the angle of reflection r
Types of reflection
There are two types of reflection
 Regular reflection/ specular reflection - Bouncing of light from a smooth surface
 Diffuse reflection- Bouncing of light from a rough surface.
Properties of Images formed by plane mirrors
 laterally inverted


same size as the object
virtual
 Same distance behind the mirror as the object is in front of the mirror.
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Real vs. Virtual Images
Real images are formed by mirrors or lenses when light rays actually converge and pass
through the image. Real images will be located in front of the mirror forming them. A real
image can be projected onto a piece of paper or a screen. If photographic film were placed
here, a photo could be created.
Virtual images occur where light rays only appear to have originated. For example,
sometimes rays appear to be coming from a point behind the mirror. Virtual images can’t
be projected on paper, screens, or films since the light rays do not really converge there.
Concave and Convex Mirrors
Spherical mirrors
Concave and convex mirrors are curved mirrors similar to portions of a sphere.
Concave mirrors reflect light from their inner surface, like the inside of a spoon
Convex mirrors reflect light from their outer surface, like the outside of a spoon.
Concave Mirrors
•
•
•
•
•
Concave mirrors are approximately spherical and have a principal axis that goes
through the center, C, of the imagined sphere and ends at the point at the center
of the mirror, A. The principal axis is perpendicular to the surface of the mirror at
A.
CA is the radius of the sphere, or the radius of curvature of the mirror, R .
Halfway between C and A is the focal point of the mirror, F. This is the point
where rays parallel to the principal axis will converge when reflected off the
mirror.
The length of FA is the focal length, f.
The focal length is half of the radius of the sphere.
2
r = 2f
Concave mirrors can form both real and virtual images, depending on where the
object is located, as will be shown in upcoming slides.
•
•
•
•
Convex Mirrors
A convex mirror has a negative focal length (used later in the mirror
equation).
Light rays reflected from convex mirrors always diverge, so only virtual
images will be formed.
Rays parallel to the principal axis will reflect as if coming from the focus
behind the mirror.
Rays approaching the mirror on a path toward F will reflect parallel to the
principal axis.
Mirror Formula
The mirror formula is expressed as;
=
+
Where
U=object distance
V=image distance
f= focal length
We use the real is positive sign convention is when doing calculations using
the mirror formula.
In real-is-positive sign convention;
 The focal length of a concave mirror is positive
 The focal length of a convex mirror is negative
 V for a convex mirror is negative
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Magnification
 The ratio of the image size to the object size.
 Denoted by M
 No units
Another formula for magnification
M=
M=Height of the image/Height of the object
Examples
1. Suppose AllStar, who is 3.5 cm tall, stands 27 cm in front of a concave mirror
with a radius of curvature of 20 cm. Where will his image be reflected and
what will its size be?
Solution
1/f=1/u+1/v
1/v=1/f-1/u
1/v=1/10-1/27
Evaluating and taking the reciprocal we get
V=15.88
Image size=2.06
2. Casey decides to join in the fun and she finds a convex mirror to stand in
front of. She sees her image reflected 7 cm behind the mirror which has a
focal length of 11 cm. Her image is 1 foot tall. Where is she standing and
how tall is she?
Solution
Since it is a convex mirror f is negative and v is also negative.
Applying the mirror formula gives
U=19.25 cm
Height of Casey =2.75 cm
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Applications of mirrors





Shaving mirrors
In optical instruments
By dentists when examining teeth
Used in supermarkets
As car side mirrors
REFRACTION
This is the bending of light rays as it travels from one medium to another e. g air-glass
media
Air
N
I
Glass
O
OI – Incident ray
W
i
r
R
OR – Reflected ray
I – Angle of incidence
r – Angle of refraction
LAWS OF REFRACTION
1. The incident ray, the refracted ray and the normal at the point of incidence all lie on the
same plane
2. For a given pair of media, the ratio of sine of the angle of incidence to the sine of the
angle of refraction is a constant i.e
sin i
 constant; This is called Snell’s law
sin r
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REFRACTIVE INDEX
The
sin i
is called the refractive index for light travelling from the first to the second
sin r
medium.
If the medium containing the incident ray is denoted by 1 and that containing the
refracted ray by 2, then the refractive index is denoted as
n2
1
=
sin i
sin r
n2
1
Medium 1
Medium 2
The absolute refractive index for crown glass = 1.50 and that for water = 1.33. The
greater the refractive index of a medium, the greater the change in the direction
suffered by light travelling from air to the medium e. g refraction for air-glass boundary
has a bigger change in direction rather than refraction in air-water boundary. In both
cases however, the reflected ray is bent towards the normal and light is said to be
travelling to an optically denser medium.
A ray of light travelling from glass or water to air would thus be bent away from the
normal.
Refractive index of air = 1.00
Refractive index can also be defined in terms of velocity of light in the two medium as
follows
n2 =
1
velocity of light in medium 1
velocity of light in medium 2
For absolute refractive index =
velocity of light in air / vacuum
c
=
v
velocity of light in the required medium
6
Where c is the speed of light 3x10^8 m/s
Hence refractive index may also be defined as the ratio of the speed of light in air to the
speed of light in the second medium.
The speed of light in a vacuum is 300,000,000 m/s, and the speed of light in water
is 225,000,000 m/s. What is the refractive index of water?
300000/225000
=1.33
REFRACTIVE INDEX RELATIONS
1. Air-glass or air-water media
n2 =
1
sin i
sin r
From the principle of reversibility of light, a ray RO would be refracted through OI.
n1 =
2
sin r
sin i
n2 =1/2n1
1
Refractive index in terms of real and apparent depth
n=
Re aldepth
apparaentdepth
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Example
Calculate the real depth of a swimming pool if the apparent depth is 1.2m. take
refractive index of water=1.33
Solution
n=
Re aldepth
apparaentdepth
1.33 =
R.D
1.2
real depth = 1.59
Definition:
TOTAL INTERNAL REFLECTION .
Total internal reflection is the reflection of light back into the same medium for a ray of
light travelling from a denser medium to a less dense medium.
When light travels from a material such as glass into a material of lower refractive index
e.g air, experiment shows that there is a maximum angle of incidence that will give a
refracted ray i.e the angle of incidence for which the angle of refraction is 90 0.
The refracted ray moves along the surface of the glass. The angle of incidence in the
denser medium corresponding to the angle of refraction of 900 is called the critical
angle.
At this point, the internally reflected ray is still weak but as c is slightly increased, the
reflected ray suddenly becomes brighter and the refracted ray suddenly disappears i.e
the light is reflected back into the optimally denser medium. This is called total internal
reflection since all the incident ray is reflected back into the same medium.
Conditions for total internal reflection
 The angle of incidence must be greater than the critical angle
 Light must be travelling from a denser medium to a less dense medium
Formula
Where
=
8
n=refractive index
C=critical angle
EXAMPLE
Calculate the critical angle of diamond given that its refractive index is 2.42
Solution
n=1/sinc
Sinc=1/n
Sinc=1/2.42
c=sin-1(1/2.42)
Applications

C=24.40
Optical fibres contain a glass core surrounded by cladding made from another type
of glass that has a lower index of refraction; fibres achieve total internal
reflection, which allows data to be transmitted.

Retroreflectors are small plastic prisms that reflect light directly back in the

Mirage
direction in came from. (For example, in a bicycle reflector.)
Dispersion
A ray of white light can be split into a spectrum of colours. This is known as
dispersion. The different colours of light have different wavelengths. The different
wavelengths are refracted by different amounts.
The speed of light in vacuum is 300,000 km/s . However in other media this speed
varies,
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INTRODUCTION
LENSES
A lens is generally a transparent material having at least one curved surface.
Lenses are usually made of glass, clear plastic or Perspex. A lens works by way of
refraction of light.
TYPES OF LENSES
There are two common types of lenses namely, convex (converging) and concave
(diverging) lenses. A lens which is thicker at its centre than at its edges converges
light and is called convex or converging lens. A lens which is thicker at its edges
than at its centre diverges light and is known as concave or diverging lens.
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EFFECTS OF LENSES OF PARALLEL RAYS OF LIGHT
Biconcave lens
When a bi-convex lens is used, the rays are converged at a point, see the fig. above
But when a bi-concave lens is used, the rays diverge as if they are coming from a
point in front of the lens. See fig. below
N/B- The point of convergence/ divergence is called the principal focus.
DEFINITION OF TERMS
a) Centre of curvature, C- it is the centre of the sphere of which the surface of
the lens is part. A lens has two centres of curvature since it has two surfaces.
C
F
F
C
b) Radius of curvature, r- it is the radius of the sphere of which the lens is part.
c) Principal axis- an imaginary straight line joining the two centres of curvature.
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d) Optical centre, P- it is a point on the principal axis midway between the lens
surfaces. Any ray of light through this point passes on undeviated.
e) Principal focus, F- for a converging lens, it is the point along the principal axis
at which rays parallel and close to the principal axis converge after refraction
by the lens. For a diverging lens, it is the point along the principal axis from
which rays parallel and close to the principal axis seem to diverge from after
refraction by the lens.
(b) Principal foci of a diverging
lens
f) Focal length, f- it is the distance between the optical centre and the principal
focus. It is real for a converging lens (positive) and virtual (negative) for a
diverging lens.
g) Focal plane- when parallel rays which are not parallel to the principal axis are
incident on a lens, the rays converge at or appear to diverge from a point which
is perpendicular to the principal axis and passes through the principal focus, F.
this plane is called the focal plane
For a converging lens, the rays converge at F after refraction while for a diverging
lens; the rays appear to diverge/emerge from F after refraction.
LINEAR MAGNIFICATION
The linear magnification produced by a lens defined as the ratio of the height of
the image to the height of the object, denoted by letter ‘m’.
Therefore;
Magnification, m
=
Height of the image
Height of the object
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Magnification is also given by;
M
=
M
=
V
U
Distance of the image from the lens
Distance of the object from the lens
Height of image
Magnification (m) =Height of object
Image distance from lens
== distance from lens
Object
V
U
Example 1
An object 0.05 m high is placed 0.15 m in front of a convex lens of focal length 0.1
m.
a) Find by construction, the position, nature and size of the image.
b) What is the magnification?
SOLN
Let 1 cm represent 5 cm. Hence 0.05 m = 5 cm = 1 cm – object height
0.15 m = 15 cm = 3 cm
0.1 m = 10 cm = 2 cm – focal length.
a) Image
-
formed is
Real
Magnified
image is beyond 2 F
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Inverted
-
b) Magnification M
=
v / u = 30 cm / 15 cm
=
2.
THE LENS FORMULA AND MAGNIFICATION
1
f
=
1
u
+
1
v
This equation is called the lens formula. The equation takes into account the signs
of u, v and f and holds for both the converging and diverging lens. The sign –
convention of real-is-positive is adopted here.
The ratio of the image size to the object size is called magnification of the lens.
When the magnification is less than one the image is diminished while when it is
more than one, the image is magnified. When the magnification is one, then the
object and image are of the same size.
Example 2
An object is placed 15cm in front of a convex lens of focal length 10cm. Calculate
the image size and the magnification.
Solution
/f = 1/u + 1/v
1
/10 = 1/15 + 1/v
1
/v = 1/10 - 1/15
1
/v = 1/30
1
V=30cm
M= v/u =
30
/15 =2
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DETERMINATION OF THE FOCAL LENGTH OF A CONVERGING LENS
Different methods are used in determination of the focal lengths of the lenses.
Some of three methods include;
(i)
(ii)
(iii)
(iv)
By focusing a Distant object
By plane mirror method; i.e., non-parallax method and using an
illuminated object
By use of lens formula
By displacement method.
OTHER POSSIBLE GRAPHS FROM THE LENS FORMULA
1. From the lens formula; 1/f=1/u +1/v
1/f = (v+u) / uv
uv = (u+v) f
Hence a graph of uv against (u+v) is a straight line through the origin and
whose slope equal to the focal length, f of the lens.
2. Also, from the lens formula; 1/f= 1/u + 1/v
Multiplying through by v, we obtain v/f= v/u + 1
But v/u= m
Therefore, m=v/f – 1
Hence a graph of m against v is a straight line whose slope equal to 1/f and
the y-intercept= -1.
POWER OF A LENS
It is the measure of the refracting ability of the lens. It is expressed as the
reciprocal of the focal length i.e.
Power of a lens= 1/focal length
It is measured in dioptre (D).
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The shorter the focal length the higher its refracting ability. Thus, a lens of small
focal length is more powerful than that of large focal length. The power of a
converging lens is positive while that of a diverging lens is negative.
For example, the power of a converging lens of focal length 10cm is given by;
Power
=
=
1/ 0.1
+10D or simply 10D
That of diverging lens of the same focal length is given by;
Power
=
=
1/-0.1
-10D
APPLICATIONS OF LENSES
Lenses have many uses, especially in optical instruments. Some optical instruments
that use lenses are the magnifying glass, compound microscope and the human eye.
1) A SIMPLE MICROSCOPE
It is also known as a magnifying glass. When an object is placed between a convex
lens and its principal focus, the image formed is it virtual, upright/erect and
magnified. When used this way it serves as a simple microscope or magnifying glass.
Usually, a lens of short focal length (high power) is preferred.
2) A COMPOUND MICROSCOPE
There are two casa under which a converging lens can produce magnified images,
namely;
(i)
When the object is between F and 2F
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(ii)
When the object is between the lens and F
A compound microscope combines the two cases above. It consists of two
converging lenses, objective lens and eyepiece lens both of short focal lengths.
The lens closer to the object is called the objective lens while that closer to
the eye is called the eyepiece lens. The focal length of the eyepiece lens is
longer than that of the objective lens. The object is placed between F and C of
the objective lens.
The first image formed by the objective lens is real, inverted and magnified. This
image then acts as the object for the eyepiece lens. The eyepiece lens forms a
final image which is greatly magnified. The eyepiece acts as a magnifying glass and
produces a final image that is greatly magnified.
A compound microscope overcomes the limitations of a simple microscope by use of
an objective lens with many lenses and an eyepiece with more than one lens.
Total Magnification Produced By Compound Microscope
A compound microscope has two lenses and each magnifies.
Assuming the magnification of the objective lens is m o =v/u; where v is the
distance of first image from the L1 and u the distance of object. Taking f 0 as the
focal length of the objective lens, we have, from lens formula;
1
/u + 1/v = 1/fo
Multiplying through by v;
1 + v/u = v/fo
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Substituting m0 = /u we obtain;
v
Like wise, me =
where D is the distance of I1 from the eyepiece.
The total magnification of the compound microscope m= m o x
Example
me.
In a compound microscope, the focal length of the objective lens is 2.0cm and that
of the eyepiece lens is 2.2cm and they are placed at a distance of 8.0cm. a real
object of size 1.00mm is placed 3.0cm from the objective lens.
a) Use the lens formula in turn for each lens to find the position of the final
image formed.
b) Calculate the magnification produced by the arrangement of these lenses
and the size of the final image viewed by the eye.
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