LIGHT
INTRODUCTION:
The simplest kind or types of mirrors are those which are of spheres. If the polished surface bulges outside, it is concave and if it bulges inwards it is a convex.
Concave mirrors
A concave mirror is formed by the silvering the outside part of a piece of glass which would form part of a hollow sphere.
NOTE: a concave mirror is sometimes referred to as converging mirror.
Useful terms on concave mirrors
In order to locate the position, size and nature of the image formed on the concave mirror, the following terms are very useful.
The center of the curvature
The center of the curvature
- The centre of the curvature is the centre of the sphere on which the convex mirror is part of.
- Pole p
The pole is the centre of the arc of the concave mirror.
- The principle axis
The principle axis is the line joining the centre of curvature C, and the pole P of the concave mirror.
- Principle focus(F)
The principles focus is the point where a beam of rays of light parallel to the principle axis falls on the concave mirror appears to convert after the reflection.
- Focal length(f)
This is the distance between the principle focus (F) and the pole P
- Radius of curvature.
This is the distance between the centre of curvature C and the pole P of the concave mirror
Such that
R= 2f
f = (1⁄2) R
The formation of images on the concave mirror
The following rules are used to draw raw diagrams for concave mirrors.
- Any ray parallel to the principle of axis is reflected through the principle focus (F)
2. Any ray passing through the principle focus f is reflected parallel to the principle axis.
3. Any ray passing through the centre of curvature C hits the mirror at right angles and is reflected straight back through the centre of curvature C.
Note: the position, the size and the nature of the image formed on a concave mirror depends very much on the p
osition of the object.
3. Any ray passing through the centre of curvature C hits the mirror at right angles and is reflected straight back through the centre of curvature C.
Note: the position, the size and the nature of the image formed on a concave mirror depends very much on the p
osition of the object.
Example
1. When an object is placed beyond the centre of curvature
- The image I forms between C and F
- The image I is diminished.
- The image I is real
- The image I is inverted ( upside down)
2. When an object is placed exactly at the centre of curvature C.
- The image I form at C.
- The image I is exactly the same size.
3. When an object is placed between the centre of curvature C and the principle focus F
- The image I form beyond C
- The image I is magnified.
- The image I is inverted that means it’s a real object.
4. When an object is placed exactly at F
- The image I is formed beyond the centre of curvature at infinity.
- The image I is highly magnified.
- The image I is real.
- The image I is inverted (upside down).
5. When an object is placed between F and P.
- The image I form behind the mirror.
- It is vertical (erect upright)
- It is magnified
- It is real.
NOTE: in most cases the images formed on the concave mirror are real except that when objects are placed between the principles of the focus (F) and the pole (P), the image formed on the concave mirror are virtual.
THE FORMATION OF IMAGES BY GRAPHICAL METHOD
Questions
- An object 2cm is high erected 8cm in front of a concave mirror of radius of curvature 10cm. by graphical method, find the position, size and nature of the image.
Therefore, the image I forms at 13.3 cm in front of a concave mirror and the size of the image is 3.3 cm. it is magnified because the object was 2cm but the image is 3.4 cm. The image is real.
2. A small spring is 4 cm long is kept at 10cm in front of the converging mirror of radius of curvature 12 cm. By scale drawi
ng, determine the position, size and state the nature of the image formed.
ng, determine the position, size and state the nature of the image formed.
This image is 15 cm in front of the mirror, height of 3 cm and it is magnified and is real.
THE CONVEX MIRROR (Diverging mirror)
They are mirrors formed by silvering or painting the inside parts of the pieces of the glasses which would form parts of the hollow spheres.
Convex mirror
Useful terms in a convex mirrors.
- The centre of curvature.
The centre of curvature of the convex mirror is the centre of the hollow sphere in which the convex mirror is part of.
- The pole of the convex mirror, P
This is the centre of the arc of the mirror.
- The principle axis of the convex mirror.
This is the line joining the centre of curvature C and the pole P of the convex mirror.
- The principle focus f
The principle focus of the convex mirror is a point where a beam of rays of light parallel to the principle axis falls on the convex mirror appears to diverge after deflection.
- Focal length.
Focal length is the distance between the pole P and the principle focus f.
6. Radius of curvature.
Is the distance between the centre of curvature C and the pole P.
Is the distance between the centre of curvature C and the pole P.
Such that; 2f = R
RULES FOR THE RAY DIAGRAM CONSTRUCTION
- Any ray passing parallel to the principle axis appears to diverge from the principle of focus F after reflection.
- Any ray passing through the principle of focus f appears to diverge parallel to the principle of axis.
- Any ray passing through the centre of curvature appears to come straight back.
Formation of images on the convex mirror
Image formed in a convex mirror are independent of the position always have the following characteristic.
- Is smaller than object (diminished)
- Formed between f and the mirror
- Image is erect
- Virtual image are formed (behind the mirror)
EXAMPLE
- An object of 4 cm high is placed 8cm in front of a convex mirror of radius of curvature 12cm. By scale drawing, find the position, the size and the nature of the image formed.
The image formed is diminished by 1.8cm
The image is virtual (erect)
The image is formed between F and P by 3.4 cm.
THE MIRROR FORMULA
To express the qualitative relationship between the object distance (U) the image distance (V) and the focal length of the mirror shown by the letter (f). The following equation is appropriate.
1/f=1/U+1/V
1/f=1/U+1/V
REAL IS POSITIVE SIGN CONVENTION
FOR A CONCAVE MIRROR
- The focal length of the mirror is positive.
- The object distance is positive.
- The image distance may be positive or negative depending on the position of the object from the mirror. Why?
Because a concave mirror can produce both real and virtual images
FOR A CONVEX MIRROR
- The focal length of the mirror is negative.
- The object distance is positive.
- The image distance may be negative.
THE MAGNIFICATION OF THE IMAGE SHOWN BY “M”
The magnification of the image is the ratio of the image distance to the object distance.
Mathematically;
OR
Magnification of an image is the ratio of the image height to the object height.
Mathematically;
Lateral magnification.
EXAMPLES
- A small pin 3cm high is placed 30 cm away from a concave mirror of focal length 12cm. By using the mirror formula, find the position, the height and the nature of the image formed.
Solution
Data; – using real is positive sign.
- Focal length f = + 12 cm
- Object distance = 30cm
- Image height Hi =
- Object height Ho = 3cm
- Image distance V =?
The position of the image is 20cm in front of the mirror and is 2cm high and is real, inverted and diminished.
2. An object 3cm high is placed 20cm in front of a concave mirror of focal length
-25cm. Describe the image formed.
- Solution
Real is positive sign convention
Object distance U = 20cm
Image distance V =?
Height of object = 3cm
Focal length = -25cm
Image height Hi=?
The image is formed 11.1 cm behind the mirror.
The image is upright, diminished and virtual, plus it is formed behind the mirror.
3. An object 2cm high is situated on and perpendicularly to the axis of the concave mirror of radius of curvature 30cm and is 10cm from the mirror, Find by calculation the position and size.
- Solution
DATA: real is positive sign convention.
Focal length = 15
Object distance V=
Image distance U =
Height of image Hi=
Height of object Ho=
Image height = 2cm
APPLICATIONS OF REFLECTION IN LIFE
1. Driving Mirrors
The convex mirrors produce a very wide field of views but the images are diminished. The convex mirrors therefore are used as driving mirrors (side mirrors) in the cars.
2. Shaving Mirrors
The concave mirrors are used as shaving mirrors
3. Reflectors of Light
The concave mirrors are used as reflectors of lights such as in car head lamps or in the torch.
In the parabolic mirror/ reflector is required in order to produce a wide and strong parallel beam of light.
The filament of the torch or bulb is placed at the principle focus of the parabola reflector.
The concave mirror curved parabolic reflects are used for amplification purposes such as satellite dishes.
Satellite dishes
EXPERIMENT
To determine the focal length of the concave mirror by non parallax method
1. Place a concave mirror on a table with its reflecting surface facing upwards.
2. Clamp an optical pin OB above the mirror so that the tip of the pin is directly above the pole P of the mirror.
3. Looking down at the arrangement from above move the optical pin OB up until an inverted image IM of the object is seen.
4. Move the pin up and down until it coincides with its sharp image.
5. Looking from above, move your head from side to side to check if there is any relative motion (parallax) between the optical pin and its image.
6. If there is no relative motion (no parallax) between the pin and its image, then the pin and its image are both at the centre of the curvature of the mirror.
7. Measure the distance from the tip of the pin to the pole P of the mirror, this is the radius of curvature of the mirror.
Focal Length
In this position, the object and its image are of the same size.
The focal length of the concave mirror is obtained by dividing the distance r by 2. Thus focal length f= r/2.
2. Clamp an optical pin OB above the mirror so that the tip of the pin is directly above the pole P of the mirror.
3. Looking down at the arrangement from above move the optical pin OB up until an inverted image IM of the object is seen.
4. Move the pin up and down until it coincides with its sharp image.
5. Looking from above, move your head from side to side to check if there is any relative motion (parallax) between the optical pin and its image.
6. If there is no relative motion (no parallax) between the pin and its image, then the pin and its image are both at the centre of the curvature of the mirror.
7. Measure the distance from the tip of the pin to the pole P of the mirror, this is the radius of curvature of the mirror.
Focal Length
In this position, the object and its image are of the same size.
The focal length of the concave mirror is obtained by dividing the distance r by 2. Thus focal length f= r/2.
THE REFRACTION OF LIGHT
Introduction
Light travels in a straight line if and only if the medium through which the light travels does not alter/ change. If the light travels from one medium to another, the direction of travel changes at the point through which it crosses the first medium to the second. Light bends when traveling in different medium due to the change in speed caused by the densities of different media.
Definition
Refraction:Is the tendency in which light bends or changes its direction of travel when crossing from one medium to another.
NB: Even though the light bends when crossing from one medium to another, it still travels in a straight line in the same medium.
KEY CONCEPTS
Consider the diagram below.
- The beam of light IC is called incident ray which is always in the first medium.
- The beam of light CR is called refracted ray which should always be in the second medium.
- The line ACB is called a plane of incidence or boundary or line which separates the two media.
- The line perpendicular to AB is the line NB which is called the normal ray ( this intersects with line AB at right angles at point C)
- The angle between the incident and the normal ray is called an angle of incidence (i) measured in degrees.
- The angle between the refracted ray and the normal ray is called the angle of refraction (r) measured in degrees.
- The Characteristics of the refraction of light
1. The beam of light is refracted towards the normal if and only if it travels from a less dense medium to a high dense medium due to the fact that the light travels faster in less dense medium than in high dense medium.
2. The beam of light is refracted away from the normal if and only if it travels from high dense medium to low dense medium due to the fact that light travels faster in less dense media than in high dense media.
LAWS OF REFRACTION OF LIGHT
The first law states that; “the incident ray, normal ray and refracted ray all are located on the same plane of incidence.”
The first law states that; “the incident ray, normal ray and refracted ray all are located on the same plane of incidence.”
NOTE: conduct a simple experiment where by a beam of light is allowed to travel from one medium(less dense) to another medium (high dense). Record different angles of incidence and refraction in the table below
r | Sin i | Sin r | Sin i Sin r | |
i1 | r1 | Sin i1 | Sin r1 | C |
i2 | r2 | Sin i2 | Sin r2 | C |
i3 | r3 | Sin i3 | Sin r3 | C |
i4 | r4 | Si n i4 | Sin r4 | C |
i5 | r5 | Sin i5 | Sin r5 | C |
i6 | r6 | Sin i6 | Sin r6 | C |
i7 | r7 | Sin i7 | Sin r7 | C |
OBSERVATION
It will be noted that the ratio of to for values of i and r for a given pair of medium is constant.
NOTE; when the graph of against is plotted, the line graph passes through the origin.
i.e. slope is constant.
i.e. slope is constant.
The second law states that (Snell’s law) “the ratio of the sin of an angle of incidence to the sin of an angle of refraction is a constant for a given pair of medium”
NB: the physical meaning of constant is the refractive index denoted by µ
Refractive index
This is the measure to what extent the light bends or changes direction from one medium to another.
Hence
Or
Suppose light travels from our medium to glass.
NOTE: the refracting index can be given in terms of the ration of the velocity of light in air to the velocity of light in a different medium.
Mathematically;
The light from air to glass.
The light from water to glass.
NB: the refractive index between vacuum or air and any other material is called absolute refractive index.
Or simply b
Refractive index of second medium.
E.g. the light from air to glass
aµ g = µ g
Generally
aµ medium = µ medium
NB: if none of the two media is air or vacuum, the refractive index is called relative refractive index.
Suppose light travels from water to glass, the refractive index of glass with respect to water is
Some refractive index of mediums
Medium | Refractive index |
Ethyl alcohol | 1.36 |
Glass | 1.52 |
Quartz | 1.55 |
Water | 1.33 or 4/3 |
Air | 1.00 |
NB: the refractive index for light traveling from a dense medium to a less dense medium is equal to the reciprocal of the refractive index for light from the less dense to the denser medium.
Example 1
The refractive for light traveling from air to water is 1.3. fin the refractive index of light travelling from water to air.
Solution
Data
wµ a = 0.7693
Example 2
If the light has a velocity of 3×108 and has a velocity 1.97 X 108 m/s in the glass.
- What is the refractive index of the glass?
- Calculate the refractive index for light traveling from glass to air.
Solution
2.
THE IMAGE FORMED BY REFRACTION OF LIGHT
- The apparent upward bending of a stick when placed in water.
See the diagram below;
Where H = real depth
= apparent depth
The rays of light from end B of the stick pass from water to air and are bent away from the normal since they are passing through to less dense medium.
On entering the eye, the rays appear to be coming from Bi above B. Bi is the image of B as a result of the refraction of light.
The shape of the stick is AEB this is true for any point inside the container.
Real and apparent depth
Definitions;
Real depth of an object is the actual height measured without taking into account any refraction of light denoted by H.
Apparent depth of an object is the depth at which an object is the depth at which an object appears to be raised denoted by letter h.
NB: an apparent depth inside a liquid is usually less than the corresponding real depth when the point is observed oblique. That is why the water pond appears shallow than it really is.
MEASUREMENT OF THE REFRACTIVE INDEX OF WATER BY APPARENT DEPTH METHOD
APPARATUS
- Optical pin
- Container
- Water
- Ruler
- Slider with a wire.
Setup
- Place an optical pin at the bottom of the beaker and then fill the beaker with water ( 2/3 full)
- Keep a straight ruler close to the beaker
- Move a straight ruler of thin wire horizontally up and down along the ruler.(slider)
- Place your eye at “0” with respect to the beaker.
Procedures
- Measure the real depth.
- Measure the apparent depth (h) of the optical pin by using a ruler by moving your eye from side to side; adjust the vertical position of the thin horizontal wire until there is no relative motion between the image of the optical pin and the wire.
Data analysis/ relative motion calculations
- finding the ratio of H to h
- Change the amount of water 3 to four times and then repeat all the procedures above.
- Calculate the average value of H/h
- When the graph of real depth against apparent depth is plotted, a line is produced.
The slope represents the refractive index of water.
Observation
The average value of H⁄h is 1.333 and hence the refractive index of water aµw = 1.333
Example
- The apparent depth of a certain point at the bottom of water pond is 25cm. find the real depth of this point given that the refractive index is 4⁄3
Solution
Apparent depth, h= 25cm.
aµw = 4⁄3
Real depth =?
From
aµw = H⁄h
The real depth of the point is 33.3*
2. Given that the refractive index of ethyl alcohol is 1.36*
Find the apparent depth in the beaker if the real depth of the optical pin is 52cm.
Solution
From
aµw = H⁄h
1.36= 52/h
h=52/1.36
∴ the apparent depth is 38.23 cm
3. A fish is located 10m deep in the liquid when viewed from the top. The depth of the fish is 8m. Find the refractive index of the liquid.
Data given
Real depth (H) = 10 CM
Apparent depth (h) = 8 m
Refractive index =?
Solution:
THE CRITICAL ANGLE AND TOTAL INTERNAL REFLECTION
Consider the diagram below where by a beam of light is traveling from high dense medium to low dense medium.
Where AO = incident ray
OC = refractive ray
i = angle of incidence.
r = angle of refraction.
Definition
Critical angle
This is an angle of incidence whose corresponding angle of refraction is equal to right angle (900) when the light travels from a high dense medium to a low dense medium.
What are the necessary conditions for a critical angle to form?
The critical angle forms when the light travels from a high dense medium to a low dense medium such that r= 900, its corresponding angle of incidence is called critical angle denoted by “c” or “ic”
NOTE: When the critical angle is reached, the refracted ray travels along the line of interface, the line separating the two medium.
Suppose that
The angle of incidence is made bigger than the critical angle; the refraction of light doesn’t take place. Because the whole incident ray reflects back into high dense medium
NOTE: this phenomenon is called total internal reflection. A situation where by a beam of light travelling from high dense medium to low dense medium reflecting fully into high dense medium at the (line of separating medium) is called total internal reflection.
Total internal reflection is a situation where by a beam of light is travelling from high to low medium reflected fully into high dense medium at the line of separating medium.
NB: consider the diagram below
Snell’s law
Since; sin 900 = 1
Then sin c = iµii condition; sin at any angle should be -1 ≤0≤1
EXAMPLES
- Calculate the critical angle for air and water medium if the refractive index of water is 4⁄3.
Data
Critical angle C =?
Angle of refraction = 90
aµw = 4⁄3
By Snell’s law
aµ ii = (sin i)/(sin r)
sin C = 1/ aµw
sin C = 3⁄4
C = sin-1 ()
C = 480 38’
The critical angle is 48 0 38’
- Given that the refractive index of glass is 1.5, what is the value of the critical angle?
Solution
Data
Refractive index = 1.5
Critical angle =?
Refractive angle = 900
By Snell’s law
Sin C = 0.6667
C = sin-1 (0.6667)
C = 41049l
The critical angle is 410491
MIRAGE
This is a form of total internal reflection. This is the appearance of water on a high way during a clear hot summer day. The air cools rapidly with height. The hotter air is optically less dense medium as has less dense.
The rays of light passing from high dense to less dense medium i.e. from colder air to warmer air bend gradually from the incident direction until it enters a layer of hot air at certain height where the total internal refraction occurs.
Its reflection, that is the image is also blue, this interpretation that the light is from the blue sky as if it is coming from the high way, hence looks like blue water.
The same type of mirage is often seen at desert areas and has given false hope to many thirsty travelers.
ASSIGNMENT
Pools of water which dry up when approached appear to be on a surface on the tarmac road on a Sunday. Explain this phenomenon.
This is due to the total internal reflection of light. The rays of light passing from high dense to low dense medium that i
s from colder air to warmer air bend as they pass the layer of hot air where total internal reflection occurs.
s from colder air to warmer air bend as they pass the layer of hot air where total internal reflection occurs.
The refraction by triangular prism
Consider the ray of light incident on the prism from one side. It gets refracted on entering the prism. When it reaches the other side the light bends away from the normal.
When a beam of light travels from air to prism, it bends towards the normal and bends away from the normal on leaving the prism.
THE DISPERSION OF WHITE LIGHT
Introduction
The dispersion of white light is the separation of color from a mixture.Consider a white light PO being dispersed at O b y the first surface of the prism into its various colors. The coloured rays are then refracted without further color change at the second surface of the prism but the amount of dispersion is increased. The emerging light is deviated towards the base of the prism.
The dispersion of white light is due to the different speeds in glass of various colors.
Note; red color is deviated the least by the refraction and violet is deviated the most by refraction.
The light consisting of only one color is called monochromatic light such as red light, green light, yellow light.
The light consistin
g of several colors is called polychromatic light such as sunlight.
g of several colors is called polychromatic light such as sunlight.
Pure spectrum
Pure spectrum is a spectrum in which each part has only one color. There are no overlapping colors.
NO OVERLAPPING COLORS
Impure spectrum
This is a spectrum in which different colors of white light are overlapping each other and only the edges are distinct. The spectrum gives a continuous band of colors.
The middle band colors overlap.
The recombination of colors of light
When another prism is placed with its apex angle pointing in the opposite direction to the first one and the two facing prisms sides are parallel, the colors from an incident white light(W), tend to recombine. The dispersion due to one prism is now neutralized by another.
NOTE: the recombination of colors may also be demonstrated by the use of Newton’s color disk, this consist of sections painted with different colors of the spectrum. When the disk is allowed to spin about its axis at very high speed, all the color of the spectrum recombine to form white light. When it is slowed down, the individual colors of the spectrum are seen again.
Newton’s color disk
THE RAINBOW
The rainbow is an arc formed out of a spectral band of colors with the violet on the inside and the red on the outside, this is a natural phenomenon of the dispersion of light.
Formation of the rainbow
Whenever there are water droplets in the air and the sunshine’s from behind, the observer at low altitude or at an angle can observe the rainbow.
This is caused by the dispersion of sunlight as it is refracted and reflected by the rain drops which are roughly in shape.
NB all rain drops refract and reflect the sunlight in the same way but only the light from some drops reach the observers eyes.
PRIMARY AND SECONDARY RAINBOW
The primary rainbow is formed as a result of light that has undergone one total internal reflection water droplets, violent color is inside and red color is outside.
The secondary rainbow is due to light that has undergone two total internal reflections in the water droplets.
VISIBLE SPECTRUM AND COLORS
Visible spectrum (ROYGBIV)
Color is the property of light that reaches our eyes. The objects can absorb certain colors of light that fall on them and reflect other colors. The color of an object depends on the light falling on it and the color it absorbs or reflects. This phenomenon is called SELECTIVE ABSORPTION AND REFLECTION.
Example
- A blue flower or coat absorbs all colors of light except blue which it reflects. If all the colors falling on an object are reflected, then the entire object will appear white. The objects appear black if and only if it absorbs the colors and doesn’t reflect any color.
NOTE; when a colored object is viewed under a colored, it appears blue in blue light and red in red light, but if an object is blue in color and is viewed in red light, it appears black
PRIMARY AND SECONDARY COLOUR OF LIGHT
Primary colors of light
They are colors like red, blue and green because they can be used successfully in the color matching.
Definition;
A primary color of light is a color that cannot be created by mixing other colors.
NB: a primary filter is the one that transmits light of one of the primary colors when white color falls on it.
Examples; traffic light and car indicators etc
Primary color
SECONDARY COLORS
Mixing 2 primary colors produces a secondary color such as magenta cyan and yellow.
- Blue + green = cyan
Primary secondary
Colors color
- Red + green = yellow
Primary secondary
Colors color
- red + blue = magenta
Primary secondary
Colors color
COMPLEMENTARY COLORS OF LIGHT
These are any 2 colors that when mixed together (primary color and secondary color) produce a white color.
I.e. primary color + secondary color = white light
When all 3 primary colors of light are mixed together, the result draws to white.
ADDITIVE AND SUBTRACTING MIXING OF COLORS
ADDITIVE COLOR MIXING (COLOUR OF LIGHT)
This is the mixing of colored light. Increasing the number of colours can be obtained by adding different colors of light together. More colors of light added together, the closer the result draws to white.
PIGMENTS
These are substances which give color to paints by reflecting light of certain colors only and absorbing all other colors. Most pigments are impure; they reflect more than one colour when they are mixed. The color which is reflected is common to all.
NOTE:
The primary colors of pigments are the same as the secondary spectrum colors of light.
The primary colors of pigments are
– Magenta pigment
– Cyan pigment
– Yellow pigment
The secondary colors of pigments are
– Red pigment
– Blue pigment
– Green pigment.
NB: the mixing of all 3 primary colors of pigments produces BLACK. As the entire beam of white light is complementary dissolved.
SUBTRACTIVE COLOR MIXING (PIGMENT OF COLORS)
– When subtracting process of color mixing is the one in which when mixing different colour pigments most of the colour pigments are absorbed because each colored
pigments absorbs (subtracts) certain colors of light. More subtraction gives black (no light).
pigments absorbs (subtracts) certain colors of light. More subtraction gives black (no light).
NOTE:
Recall complementary colors
1p. Blue (B) + yellow (R+G) = white.
2. Red + Cyan = white
THE REFRACTION OF LIGHT THROUGH THIN LENSES
Lens
This is a transparent or translucent material or medium that alters the direction of light passing through it.
Types of lenses
- Convex lenses
- Concave lenses
- Convex lenses (converging lenses)
Are thicker in the middle than near the edges
2. Concave lenses (diverging lenses)
These are thinner in the middle than near the edges.
Definition of the terminologies
- Optical centre
This is a geometric centre of a lens.
2. Centre of curvature.
2. Centre of curvature.
This is a geometric centre of the sphere of which the lens surface is a part of. Since the lens has 2 surfaces, it has two centers of curvatures.
IMAGE FORMATION BY THIN LENSES
The way in which lenses form images of the object can be shown by means of ray diagram constructions.
The following are 3 important rules which are used in the construction of ray diagrams.
- Any ray parallel to the principle axis passes through or appears to diverge from the principle focus after the reflection through the lens.
2. Any ray of light incident on the principle focus of the length is refracted parallel to the principle axis.
3. Any ray through the optical centre of the lens continues through undeviated.
NB: The location of the image formed by the convex lens depends on the position of the object relative to the principle focus of the object.
The image formed by the concave lens are virtual erect and diminished in size.
CONVEX LENSES
- Case A
When an object is beyond centre of curvature
2. Case B
2. Case B
When an object is at c
3. Case C
3. Case C
When an object is between c and f
4. Case D
When an object is at f
- The image is real and inverted
- The image is highly magnified
- The image is formed beyond infinity.
5. Case E
When an object is between f and O
- The image is virtual
- The image is erect
- The imago is magnified
- The image forms between
c and f.
CONCAVE LENSES (DIVERGING LENSES)
Image which formed in concave lens have the following properties independent of position of an object
- The image is virtual
- The image is diminished
- The image is erect
- The image is formed between f and O
- As the object distance increases from pole to infinity the image distance increases from the pole to f.
THIN LENS FORMULA
1/F=1/U+1/V
Real image is positive sign convention
For convex lenses
U=+ve
F=+ve
V=+ve or-ve
F=+ve
V=+ve or-ve
For concave lens
U=+ve
F=-ve
V=-ve (always)
F=-ve
V=-ve (always)
Examples
- The object is placed 20 cm from a converging lens for focal length 15cm . fin the position, the magnification and the nature of the image
Solution:
u= 20cm
f= 15cm
v=?
m=?
From;
1/F=1/U+1/V
1/15=1/20+1/V
1/V=(4-3)/60
1/V=1/60
V=60cm
m=v/u
m=60/20
m=3
m=3
∴ The position is 60 cm from the lens, the magnification is 3 and the nature is a real image
2. Find the nature and position of the image of an object, placed 10cm from a diverging leans of focal length 15cm (concave).
Solution:
U= 10cm
V =?
F= 15cm
M=?
Using:
1/F=1/U+1/V
1/(-15)=1/10+1/V
1/V=(-3-2)/30
1/V=(-6)/30
V=-6cm
M=V/U
M=(-6)/10
M=0.6
The position is -6 cm from the lens, the magnification is 0.6
M=(-6)/10
M=0.6
The position is -6 cm from the lens, the magnification is 0.6
- An object stands vertically on the principle axis of a converging lens of focal length 10mm and at a distance of 17mm from the lens. Find the position, size and nature of the image.
Solution:
U= 17mm
F= 10mm
V=?
H1=?
H0= 10mm
M=?
By Using:
1/F=1/U+1/V
1/F=1/U+1/V
1/10=1/17+1/V
1/v=1/10-1/17
1/v=1/10-1/17
1/v=(17-10)/170
V=170/7
V=170/7
V=24.3
We get V = 24.14
M=V/U
M=24.4/7
M = 3.5 mm
M = 3.5 mm
M=HI/HO
3.5= 10/HO
3.5HO=1O
3.5HO=1O
HO=10/3.5
HO=2.9
∴The position is 24.14 cm from the lens, the size is 2.9 mm and the image is real.
OPTICAL INSTRUMENTS
These are the instruments which are used to help human eye to view small objects or distant objects more clearly.
These include instruments like:-
- Simple microscope
- Compound microscope
- Telescope
- Lens camera
- Projector
- Human eye
NB: The instruments consist of convex lens (the converging lens)
- SIMPLE MICROSCOPE (MAGNIFYING LENS)
The simple microscope consists of conveying lens of short focal length. The object is nearer to the lens than its focal length. The image is viewed directly by the eye is erect virtual.
-Image I is erect
-Image I is virtual
-Image I is highly magnified
NB: The thin lens formula obeys
LINEAR MAGNIFICATION
The ratio of the image distance to the object distance is called linear magnification (M).
LATERAL MAGNIFICATION
The ratio of the image height or size to the object height or size is called lateral magnification.
NEAR POINT
It is the least distance whereby an image appears clearest (sharpest). The image appears clearest when it is about 25 cm from the eye. The distance is called near point.
When the image is at the near point; v = 25.From thin lens formula:
Real is positive sign conventionSince image I is virtual
Multiplying by v through out
Example:
- If the focal distance of the converging lens is 5 cm the object distance is 4 cm. Find the magnification of the image.
Solution
M = 6
The magnification of the object is 6 cm
- Given that the focal length of the simple microscope is 12cm.Find the magnification of the image of the object distances.
Solution
APPLICATION OF SIMPLE MICROSCOPES
1. The simple microscope are used to view specimens in laboratory
2. The simple microscopes are used to reads small prints
- COMPOUND MICROSCOPE
It produces a much greater magnification more than the simple microscope. It has an objective lens and eye piece lens which are both convex lenses and of short focal length.
CONSTRUCTION
An object A is placed just beyond the principle focus (FO) of the objective lens which forms magnified image B. The eye piece is used as a simple magnified glass to increase the magnification further that a large inverted image is seen as point C.
The image C is magnified
The image C is virtual
The image C is inverted as the object B
MAGNIFICATION
The magnification produced by the objective lens is given by
-The magnified image acts as the object for eye piece lens then let L is the distance between 2 le
nses.
nses.
-The distance VO for the objective lens is approximately equal to
where L is the distance between the 2 lenses i.e.
where L is the distance between the 2 lenses i.e.
Hence
The eye piece lens acts as a simple microscope hence
-The total magnification M is given by the product of the magnification of the objective lens and magnification of the eye force lens.
Hence
Note: For very large magnification F0 and Fe must be very small as compound to L.
Applications of a compound microscope
i) It is used to magnify small objects
ii) The compound microscope it is used to observe the Brownian motion in science
iii) It is used to study the characteristics of the micro-organism and cells in biology
iii) It is used to study the characteristics of the micro-organism and cells in biology
iv) It is used to analyze the laboratory samples in hospitals such as tissues or body fluids to check for infections caused by micro-organism
Example:
A compound microscope has an objective lens of focal length 2 cm and eye piece lens of fo cal length 6cm.An object is placed 2.4 cm from an objective lens. If the distance between the objective lens and eye piece lens is 19 cm. Find
a) The distance of the final image from the eyepiece lens
b) Compound magnification
Solution
a)
Ve = ∞
The final image from the eye piece lens is at infinity.
(b)
: . M = ∞
Example:
A certain microscope consists of 2 converging lenses of focal length 4 cm and 10 cm for objective 3 eyepiece lenses respectively. The 2 lenses are separated by the distance of 30cm.The instrument is focused so that the image is at infinity. Calculate the position of the object and the magnification of the objective lens.
Solution:
a)
uo = 5 cm
Mo = 4
:. The magnification produced by the eye piece lens is 4
E.g. The following diagram represents a compound microscope
(a). Identity A and B are objective and eyepiece lenses
(b). What is the relationship between fo and fe?
__ fo > fe
- TELESCOPE
A telescope is an optical instrument which is used to view distant object(far object). It uses two convex lenses,the same as compound microscope but it has a large focal length on the objective lens and short focal length on the eyepiece lens.
Constructions
- The image I is virtual
- The image I is magnified
- A looks larger close to the observers’ eye
- The image I is inverted
Magnification
Magnification of telescope is the ratio of the focal length of objective lens to focal length of the eye piece lens.
Note: If L is the distance of separation between the objective and eyepiece lens for normal adjustment is equal to fo + fe
L = fo +f e
Applications of telescope
- They are used to view distant objects example stars and other heavenly bodies in the sky
- They are used in military bases to see enemies
Example: A telescope is consisting of two converging lens of lens at focal length 25 and 4 respectively. The final image is found at distinct vision that is 25 cm in front of the eyepiece lens. Find the position of the first image from the eyepiece.
Solution
= 3.4cm
Example: An astronomer telescope has its 2 lens 78 cm apart. If the objective lens has a focal length of 75.5 cm, what is the magnification produced by the telescope under normal vision.
Solution
L= fo + fe
78 = 75.5 + fe
Fe = 2.5
M = 30.2
:. The magnification is 30.2
4. PROJECTORS (PROJECTION LANTERNS)
A projector is an optical instrument which is used to display large images on the screen with a slide projector. This consists of a powerful small source of light (O). It also consists of condensing lens (2 Plano convex lenses to collect the light from (O) and send it through the slide 5 which is then illuminated powerfully).It has a converging projection lens. A also has a distant wide screen.
CONSTRUCTION
Where :
u = distance from slides (object) to lens L
u = distance from slides (object) to lens L
V = distance from lens L to the distance white screen A
f = Focal length of the lens L
- The image I is highly magnified
- The image I is real
- The image I erect
Note: The thin lens formula is applicable
MAGNIFICATION
The magnification produced by the projector is given by the ratio of the image distance to the object distance.
or
-The magnification produced by the projector is given by the ratio of the image height to the object height.
Applications of projectors
- They are used to project films, slides and transparency
- They are used in the projection of opaque objects
- Projections are used in search lights
- They are used in projector apparatus in industries for gauge and screw thread testing
- They are used in the projection of spectrum and interference
- They are used in the projection of minute objects
- The projectors are used in the projection microscope
Example: The lantern projector uses a slide off 2 cm by 2 cm, 2×2 to produce a picture 1 m by 1m on a screen 12cm from the projection lens. How far from the lens must the slide be?
Find the approximate focal length of the projection lens.
Solution:
5. LENS CAMERA
This instrument has a convex lens, diaphragm, shutter and a mounting base for the film. (A light sensitive film)
- The image forms on the film.
- The image is diminished.
- The image is inverted.
- The object is focused by moving the lens.
- The amount of light entering the lens is controlled by the diameter of the diaphragm or stop and by the speed of the shutter which is pre-set for the particular lighting conditions.
The Magnification
The magnification produced by a lens of camera is always less than one which is given by the ratio of the image distance to the object distance that is;
Or it is given by the ratio of image height and object height;
From the formula
Example: Given that an object 2m high is placed 2010cm in front of the lens camera of focal length 10cm.calculate the minimum size of the film frame.
Solution:
u = 20.1m = 2010cm
f=10cm
Ho=200cm
M = 0.0005
Applications of lens camera
- To take photographs
- Video camera are used to take motion pictures
- High speed cameras are used to record movements of pictures
- CCTV cameras are used for high security instruments, banks and other places
- Digital cameras are used to capture images that can be fed into computers.
6. THE HUMAN EYE
A human eye is sensitive to white light (ROYGBIV).It can detect colors of different objects.
STRUCTURE OF THE HUMAN EYE
The human eye has the following main parts
- Crystalline flexible convex lens (surrounded by liquid) all aqueous humour and V.H/Vitreous humour.
- It has iris (coloured and surrounding the lens)
- Pupil (hole in the middle of the night) contracting by day and widening by night to control the amount of light entering the eye.
- It has ciliary muscles to control the focal length of the eye lens by altering the radii of curvature of surfaces.(When the lens stretches the focal length increases)
- Retina (light sensitive layer of cells at the rear of the eye) Rear means backside
- Fovea (yellow spot)
-It is the most sensitive part of the retina
Note:
- V.H (Vitreous Humour)
- The vitreous humour is used to present the eye from collapsing due to change in atmospheric pressure. If focuses the images clearly on the retina (by maintaining refraction)
- A.H (Aqueous Humour)
- It is a watery fluid. When we blink our eyes, a tiny drop of the eye, it washes the eye and keeps the cornea moist in order to avoid opaque.
ACTION OF THE EYE
- The process whereby the eye can alter its focal length in order to form images at different distances is called accommodation.
The process which the ciliary muscles alter the focal length of the cry stalling flexible convex lens so as to focus near or far-off objects clearly on the retina.
SIMILARITIES BETWEEN THE HUMAN EYE AND THE LENS-CAMERA
- Both human eye and lens-camera have convex lenses which form diminished, real and inverted images
- Both human eye and the lens-camera are blackened inside in order to prevent total internal reflection.
- In the human eye the pupil controls the amount of light whole in lens camera a diaphragm controls the amount of light.
- In the human eye the image is formed on the retina while in lens on the lens-camera the image is formed on the photographic film.
- The human eye can adjust the focal length of its lens by contraction and relaxation of ciliary muscles in order to focus different objects at different distances, in lens-camera the objects can be focused by moving the lens forwards and backwards.
DIFFERENCES BETWEEN THE EYE AND LENS-CAMERA
- The human eye can alter its focal length by using ciliary muscles while the focal length of a lens-camera is fixed.
- The eye has a fluid inside which assists for refraction while a lens-camera has air inside
- Light is refracted by the cornea, lens and fluid in the eye but it is refracted by the lens only in the lens-camera
- Focusing is done in the eye by changing the shape of the lens but it is obtained in the lens-camera by adjusting the focal length relative to the film.
DEFECTS OF VISION AND THEIR CORRECTION
- SHORT-SIGHTEDNESS (MYOPIA)
For normal vision, the objects a long way off cannot be seen clearly with the short sighted person can see the objects more clearly at shorter distances. This problem is caused by the following
- Too long eye ball
- Too powerful eye-lens
CORRECTION
A short sighted person is provided with a diverging lens (concave lens) to correct his/her sight.
2. LONG-SIGNTEDNESS (HYPERMETROPIA)
2. LONG-SIGNTEDNESS (HYPERMETROPIA)
For normal vision, far objects away can be seen clearly. A person with long-sightedness cannot see near objects clearly. Objects at longer distances can be seen more clearly due to the following possible causes:
a) Too short eyeball
b) Less powerful eye lens
CORRECTION
Long sightedness person is provided with convex lens to correct its sight
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