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Light
By Neil Bronks
Light is a form of energy
Crooke’s Radiometer
proves light has energy
Turns in
sunlight as
the light
heats the
black side
Light travels in straight lines
Reflection-Light bouncing off object
Angle of incidenceNormal
= Angle of reflection
Reflected ray
Incident ray
Angle of
incidence
Angle of
reflection
Mirror
Laws of Reflection
 The angle of incidence ,i, is
always equal to the angle
of reflection, r.
 The incident ray, reflected
ray and the normal all lie
on the same plane.
Virtual Image
An image that is formed by the eye
Can not appear on a screen
d
d
Real Image
Rays really meet
Can be formed on a screen
2F
F
All ray diagrams in curved
mirrors and lens are drawn
using the same set of rays.
Concave Mirror
Object
Principal
Axis
F
Pole
F
You can draw any ray diagram by
combining 2 of these rays
The only difference is where the
object is based.
F
Ray Diagrams- Object outside 2F
1/. Inverted
2/. Smaller
2F
F
3/. Real
The images can be formed on a screen so they are real.
Object at 2F
1/. Inverted
2/. Same Size
3/. Real
2F
F
The image is at 2F
Object between 2F and F
1/. Inverted
2/. Magnified
2F
F
The image is outside 2F
3/. Real
Object at F
2F
F
The image is at infinity
Object inside F
F
1/. Upright
2/. Magnified
3/. Virtual
The image is behind the mirror
Convex Mirror
The image is behind the mirror
1/. Upright
2/. Smaller
3/. Virtual
F
Convex Mirror – only one ray diagram
F
The image is behind the mirror
Uses of curved mirrors
Concave Mirrors
Dentists Mirrors
Make –up mirrors
•Convex Mirror
Security Mirrors
Rear view mirrors
Calculations
 Use the formula
1 1 1
 
f u v
u
F
v
f=focal length
u=object distance
v=image distance
Example
An object is placed 20cm from a concave mirror
of focal length 10cm find the position of the
image formed. What is the nature of the
image?
Collect info f=10 and u=20
Using the
formula
1 1 1
 
10
f 20
u v
1 1 1
 
f u v
1 1
1
 
v 10 20
1 1 V=20cm real

v 20
Magnification
What is the magnification in the last
question?
Well u=20 and v=20
As
v 20
v
m m 
u 20
u
m
2v
2u
• m=1
• Image is same size
Example
An object is placed 20cm from a concave
mirror of focal length 30cm find the position
of the image formed. What is the nature of
the image?
Collect info f=30 and u=20
Using the
formula
1 1 1
 
f u v
1
1 1 1 1
1
1


 

30 20 v v 30 20  60
V=60cm
Virtual
Example
An object is placed 30cm from a convex mirror
of focal length 20cm find the position of the
image formed. What is the nature of the
image?
Collect info f=-20 and u=30
Using the
formula
1
1 1


 20 30 v
1 1 1
 
f u v
V=60/5cm =12cm
The minus is Virtual
Because the
1Mirror1is 1
5
convex
 

v
30
20
60
Questions
An object 2cm high is placed 40cm in
front of a concave mirror of focal
length 10cm find the image position and
height.
An image in a concave mirror focal
length 25cm is 10cm high if the object
is 2cm high find the distance the object
is from the mirror.
MEASUREMENT OF THE FOCAL
LENGTH OF A CONCAVE MIRROR
Concave
mirror
Crosswire
u
Lamp-box
Screen
Approximate focal length by focusing image of window
onto sheet of paper.
Place the lamp-box well outside the approximate focal
length
Move the screen until a clear inverted image of the
crosswire is obtained.
Measure the distance u from the crosswire to the mirror,
using the metre stick.
Measure the distance v from the screen to the mirror.
Repeat this procedure for different values of u.
Calculate f each time and then find an average value.
Precautions The largest errors are in measuring with the
meter rule and finding the exact position of the
sharpest image.
Refraction
The fisherman sees
the fish and tries to
spear it
Fisherman use a trident
as light is bent at the
surface
Refraction into glass or water
AIR
WATER
Light bends towards the
normal due to entering a more
dense medium
Refraction out of glass or water
Light bends away from the
normal due to entering a less
dense medium
Refraction through a glass block
Light bends towards the
normal due to entering a more
dense medium
Light slows down but is
not bent, due to entering
along the normal
Light bends away from the
normal due to entering a less
dense medium
Laws of REFRACTION
The incident ray, refracted ray and
normal all lie on the same plane
SNELLS LAW the ratio of the sine of
the angle of incidence to the sine of
the angle of refraction is constant
for 2 given media.
sin i = n (Refractive Index)
sin r
Proving Snell’s Law
Sin i
i
r
Sin r
A straight line though the origin proves Snell’s law.
The slope is the refractive index.
Proving Snell’s Law
Sin i
i
Glass
Block
Protractor
r
Sin r
A straight line though the origin proves Snell’s law.
The slope is the refractive index.
H/W
LC Ord 2006 Q2
Refractive Index
 Ratio of speeds
cair
300000000m / s
n

 1.5
cwater 200000000m / s
Real and Apparent Depth
 A pool appears
shallower
Re al
n
Apparent
MEASUREMENT OF THE
REFRACTIVE INDEX OF A LIQUID
Cork
Pin
Apparent depth
Mirror
Real depth
Water
Image
Pin
Finding No Parallax – Looking Down
Pin at
bottom
Pin
reflection
in mirror
Parallax
No Parallax
Set up the apparatus as shown.
Adjust the height of the pin in the cork above the mirror
until there is no parallax between its image in the mirror
and the image of the pin in the water.
Measure the distance from the pin in the cork to the
back of the mirror – this is the apparent depth.
Measure the depth of the container – this is the real
depth.
Calculate the refractive index n= Real/Apparent
Repeat using different size containers and get an average
value for n.
Refraction out of glass or water
Light stays in denser medium
Reflected like a mirror
Angle i = angle r
Snell’s Window
Finding the Critical Angle…
1) Ray gets refracted
3) Ray still gets refracted (just!)
THE CRITICAL
ANGLE
2) Ray still gets refracted
4) Total Internal
Reflection
Semi-Circular Block Expt and on
the internet click here
Mirages
Critical Angle
 Varies according to
refractive index
1
sin C 
n
1
sin 45 
n
1
0.7071 
n
1
n
0.7071
n  1.41
Uses of Total Internal Reflection
Optical fibres:
An optical fibre is a long, thin, transparent rod made of glass
or plastic. Light is internally reflected from one end to the
other, making it possible to send large chunks of
information
Optical fibres can be used for communications by sending e-m
signals through the cable. The main advantage of this is a
reduced signal loss. Also no magnetic interference.
Practical Fibre Optics
It is important to coat the strand in a material of low n.
This increases Total Internal Reflection
The light can not leak into the next strand.
1) Endoscopes (a medical device used to see inside the body):
2) Binoculars and periscopes (using “reflecting prisms”)
Now is a good time to get out the
light demo kit
H/W
LC Ord 2003 Q7
Lenses
Two types of lenses
Focal Point
Focal Point
Focal Length=f
Focal Length=f
Converging Lens
Diverging Lens
Ray Diagrams
Optical Centre
F
F
2F
F
F
2F
Converging LensObject outside 2F
Image is
1/. Real
2/. Inverted
3/. Smaller
2F
F
F
2F
Object at 2F
Image is
1/. Real
2/. Inverted
3/. Same size
2F
F
F
2F
Object between 2F
and F
Image is
1/. Real
2/. Inverted
3/. Magnified
2F
F
F
2F
Object at F
Image is at infinity
F
F
Object inside F
Image is
1/. Virtual
2/. Erect
3/. Magnified
F
F
H/W
Draw the 5 ray diagrams for the
converging lens and the diagram for the
diverging lens .
Write 3 characteristics of each image.
Calculations
 Use the formula
1 1 1
 
f u v
f=focal length
u=object distance
v=image distance
u
2
F
F
F
v
2F
Example
An object is placed 30cm from a converging
lens of focal length 40cm find the position
of the image formed. What is the nature of
the image?
Collect info f=40 and u=30
Using the
formula
1 1 1
 
40
f 30
u v
1 1 1
 
f u v
1 11 11 1
 
- =
vu -120
v
vf fu40 30
V=120cm
virtual
Magnification
What is the magnification in the last
question?
Well u=30 and v=120
As
120
v
v
mm 
30
u
u
m
4v
1u
• Image is larger
MEASUREMENT OF THE FOCAL LENGTH
OF A CONVERGING LENS
Show on OPTICAL BENCH
Lamp-box with
crosswire
Screen
Lens
u
v
1.
Place the lamp-box well outside the approximate focal
length
2. Move the screen until a clear inverted image of the
crosswire is obtained.
3. Measure the distance u from the crosswire to the lens,
using the metre stick.
4. Measure the distance v from the screen to the lens.
5. Calculate the focal length of the lens using
1 1 1
 
f u v
6. Repeat this procedure for different values of u.
7. Calculate f each time and then find the average value.
H/W
LC Ord 2002 Q3
Accommodation
The width of the lens
is controlled by the
ciliary muscles.
For distant objects
the lens is
stretched.
For close up
objects the
muscles relax.
Diverging Lens
Image is
1/. Virtual
2/. Upright
3/. Smaller
F
F
Example
An object is placed 30cm from a diverging lens
of focal length 20cm find the position of the
image formed. What is the nature of the
image?
Collect info f=-20 and u=30
Using the
formula
1
1 1


 20 30 v
1 1 1
 
f u v
V=60/5cm =12cm
The minus is Virtual
Because the
1
1 1
5
Diverging
  lens


v
30
20
60
Example
An object is placed 30cm from a diverging lens
of focal length 60cm find the position of the
image formed. What is the nature of the
image? (Remember f must be negative)
Collect info f=-60 and u=30
Using the
formula
1
1 1
 
-60
30
f
u v
1 1 1
 
f u v
1
11 11 1
 - =
-60
uf 30vu -20v
vf
V=20cm
virtual
Magnification
What is the magnification in the last
question?
Well u=30 and v=20
As
v 20
v
m m 
u 30
u
m
2v
3u
• Image is smaller
Sign Convention
1 1 1
 
f u v
f
Positive
V
either
f
negative
V
negative
f
Positive
V
either
f
negative
V
negative
Myopia (Short Sighted)
Image is formed in front of the retina.
Correct with diverging lens.
Hyperopia (Long-Sighted)
Image is formed behind the retina.
Correct with a converging lens
Power of Lens
Opticians use power to describe lenses.
1
P=
f
So a focal length of 10cm= 0.1m is written as
P=10m-1
A diverging lens with a negative focal length
f=-40cm=-0.4m
Has a power of P = -2.5m-1
Lens in Contact
Most camera lens are made up of two lens
joined to prevent dispersion of the light.
The power of the total lens is
Ptotal=P1+ P2
H/W
LC Higher 2002 Q12 (b)
LC Higher 2003 Q3