Transcript Light Revision

Light
Light is a form of energy
Crooke’s Radiometer
proves light has energy
Turns in
sunlight as
the light
heats the
black side
Can you think of another example to demonstrate
that light is a form of energy?
Light travels in straight lines
How are shadows formed?
Reflection
Reflection is the bouncing of light off
an object.
When light bounces off objects it
scatters in all directions – diffuse
reflection.
Highly polished surfaces (mirror)
behave in a more predictable way.
Reflection
Angle of incidence = Angle of reflection
Normal
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.
Reflection
Laws of Reflection Animation 1
Laws of Reflection Animation 2
How is an image formed in a plane
mirror?
Virtual Image
An image that is formed by the
apparent intersection of light rays
Can not appear on a screen
d
d
Curved Mirrors
Curved mirrors consist of a series of
small mirrors combined together.
Each individual mirror must obey the
laws of reflection.
Real Image
An image that is formed by the actual
intersection of light rays.
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
Ray Diagram Example
An object 4 cm high is placed at right
angles to the axis of a concave mirror
and at a distance of 30 cm from the
mirror. If the focal length of the
mirror is 10 cm find the position, size
and nature of the image.
This can be done using a diagram or by
calculation.
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
MEASUREMENT OF THE FOCAL
LENGTH OF A CONCAVE MIRROR
Concave
mirror
Crosswire
Lamp-box
Screen
u
v
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
i
r
Repeat for different angles of incidence
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
Refractive Index(n) in terms of
relative speeds

Refractive Index
 Ratio of speeds
cair
300000000m / s
n

 1.5
cwater 200000000m / s
Refraction out of glass or water
Light stays in denser medium
Reflected like a mirror
Angle i = angle r
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
Critical Angle
 Varies according to
refractive index
1
sin C 
n
1
sin 45 
n
1
0.7071 
n
1
n
0.7071
n  1.141
Refractive Index and Critical Angle
Refractive Index is defined in relation
to light going from air into that medium
(i.e. air to glass or air to water)
Ex 1: The critical angle for a certain
medium is 500 . Find its refractive index
Ex 2: The refractive index of glass is
1.5. What is the critical angle for glass?
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”)
Mirages
Lenses
Two types of lenses
Focal Point
Focal Point
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
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.
The Eye
Power of Accommodation
-
ability to focus a real image of an object on the retina
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.
Why is not a good idea to water
plants on a sunny day?
The water forms droplets on the leaves.
 These droplets act as converging
lenses and focus the sun onto the
leaves, burning them.
 As a result the leaves will have brown
spots.
Why can’t we focus clearly under
water yet swimming goggles will
restore clear focus?
 Hint: your cornea and water have a similar
refractive index
 Light refracts when travelling from air
through the cornea of your eye, but water
and the cornea have the same refractive
index , so light does not refract.
 By wearing goggles however light which hits
your eye is coming from air, so the usual
focusing applies and objects appear normal.
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 -20
v
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
Myopia (Short Sighted)
Image is formed in front of the retina.
Correct with diverging lens.
Hyper-Myopia (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