Physics form three topic three

 Light is a form of energy which controls the sense of vision.Reflection of Light from Curved Mirrors Difference between Concave and Convex MirrorsDistinguish between concave and convex mirrorsConcave mirror is a spherical mirror whose reflecting surface is curved inwards. A Good example is the driving mirror of a car.Convex

mirror is a spherical mirror whose reflective surface is curved
outwards. A good example of a convex mirror is a shaving mirror.General demonstrations of convex and concave mirrors (curved mirrors:

The Terms Principle, Axis, Pole, Principle Focus and Radius of Curvature as Applied to Curved MirrorsExplain the terms principle, axis, pole, principle focus and radius of curvature as applied to curved mirrorsTerms used in studying curved mirrors

• Centre of curvature (C):the centre of the sphere of which a mirror is a part of.
• Radius of curvature (R): the radius of sphere of which a mirror is a part of.
• Pole (P): the central point of the reflecting surface of spherical mirror (curved or convex mirror).

• Principal axis:the straight line joining the centre of curvature (C) and the pole (P).
• Principal focus (F):the point o the principal axis where light rays tend to intersect. This point is between centre of curvature and the pole.
• Principal axis:the straight line joining the centre of curvature (C) and the pole (P).
• Principal focus (F):the point on the principal axis where light rays tend to intersect. This point is between centre of curvature and the pole.

The Images Formed by a Curved MirrorLocate the images formed by a curved mirrorCase (1)When
a beam of light parallel and very close to the principal axis, CL, is
reflected from a concave mirror, it converges to a point, F, on the
principal axis called the principal focus.

Case 2When a ray passes through the principal focus, F, it is reflected parallel to the principal axis.

Case 3When
a ray passes through the centre of curvature, C, which therefore
strikes the mirror at normal incidence, it is reflected back along its
original path.

Note: Concave mirrors have a real focus because light passes through the focus.The formation of images by concave mirror tends to change as the position of object changes.Case 1: Image (I) formed by a concave mirror when the object is beyond C.

Properties of images formed:

1. The image is between C and F
2. The image is smaller than the object
3. The image is inverted (upside down)
4. The image is real

Case 2: The object is placed at C

Properties of image

1. The image is formed at C
2. The image has the same size as object
3. The image is inverted (upside down)
4. The image is real.

Case 3: The object is placed between C and F

Properties of image formed

1. The image is real
2. The image is large than object
3. The image is formed beyond C
4. The image is inverted (upside down)
5. The image is real
6. The image is large than object
7. The image is formed beyond
8. The image is inverted (upside down)

Case 4:The object is placed at F

Properties of image:

1. The image is formed at infinity (x)
2. The image is formed beyond C
3. The image is large than object
4. The image is Real

Case 5:The object is placed between F and P.

Properties of image formed:

1. The image is virtual
2. The images is upright
3. The image is formed behind the mirror
4. The image is large than the object

Formation of images in a convex mirror:Obviously,there isonly one kind of image formed when an object is placed at any position.Properties of image formed by convex mirror:

1. the image is virtual
2. the image is upright
3. The image is smaller than object (diminished)
4. The image is formed behind the mirror.

Example 1An
object 2cm long is erected 8cm infront of a concave mirror of radius of
curvature 10cm. By using a scale drawing, determine the position, size
and nature of image formed.Data given

• Height of object, Ho = 2cm
• Object distance, U= 8cm
• Radius of curvature, r = 10cm
• Focal length,f =8cm
• Choose suitable scale.
• Say 1cm represents 5cm

From this scale then

• Height of object, Ho = 2cm
• Object distance, U= 2cm
• Focal length, F = 2.5cm

Thus,Image distance, V = XImage Height, H1=YThe Focal Length of a Concave MirrorDetermine practically the focal length of a concave mirrorFocal length (f) is the distance between the principal focus and the pole.

Convex and Concave Mirrors in Daily LifeUse Convex and concave mirrors in daily lifeCurved mirrors are used as:

1. Driving mirrors
2. Shaving mirrors
3. Reflectors

Question Time 1Why is convex mirror used as driving mirror?The convex mirror is used as driving mirror because it provides the wider field of view.Question Time 2Why concave mirror used as shaving mirror?Concave mirrors are used as shaving mirrors because they form an enlarged image when held close up.Refraction of LightThe Concept of Refraction of LightExplain the concept of refraction of lightRefraction
of light refers tothe bending of light as it passes through two
different medium because the speed of light tends to change when
travelling from one medium to another.Figure showing refraction of light as it passes from air to glass.The Angle of Incidence and Angle of RefractionMeasure the angle of incidence and angle of refractionThe angle of incidence (i)is the angle between the incident ray of light and the normal at the point of incidence.The angle of Refraction (r)is the angle between the refracted ray and the normal at the point of incidence.The Laws of RefractionState the laws of refractionFirst law of refractionThe
First Law of refraction states that “the incident ray, the refracted
ray and the normal at the point of incident are located in the same
plane.”Second law of refractionSecond
Law of refraction states that “when a light ray passes from one medium
into another medium, the angle of incidence (i) and corresponding angle
of refraction( r) are such that the ratio of sine of the angle of
incidence to the sine of the angle of refraction (sini/sinr) is a
constant value called the refractive index.”Note:
The Second Law of Refraction is called Snell’s Law in honour of a Dutch
scientist named Snell (1591 – 1626) who first described it.The Refraction Index of a MaterialDetermine the refraction index of a materialRefractive index (n) is the ratio of the sine of the angle of incidence to the sine of the angle of refraction.n = Sini/Sinr ORRefractive index (n) is the ratio of the velocity of light in air to the velocity of light in glass.n = Velocity of light in air (Va)/Velocity of light in glass (Vg)OrRefractive
index, n is the constant number which expresses how many times or to
what extent a light ray bends when passing through different medium.Absolute refractive index (na) is the refractive index between vacuum or air and any other medium.The refractive indices between air and some common media is given below:

Medium	Refractive index (n)
Diamond	2.417
Ethanol	1.360
Glass (Crown)	1.520
Quartz	1.553
Water (at 20ºC0	1.333
Air (at stp)	1.00029

Example 2The
refractive index for light passing from air to water is equal to 1.333
find the refractive index for light travelling from water to air.Data given:Refractive index anw of air to water = 1.333Required: To find refractive index from water to airSinceanw = 1.333wna = (1/anw)= (i/1.333): wna = 0.75Real and Apparent DepthReal depthis the actual height measured without taking account any refraction of lightApparent depth is the virtual height measured when viewed by observer.The Concept of Critical Angle and Total Internal Reflection of LightExplain the concept of critical angle and total internal reflection of lightCritical angleCritical
angleis the angle of incidence (i) for which the angle of refraction
(r) is equal to 90º . It is obtained when light rays moves from a dense
medium to a less dense medium.For refractive indexn=Sini/sinrBut i= Critical angle, Cr = 90ºThus n= sinC/sin 90ºn=SinC/1n = Sin Cc= Sin -1 (n)Total Internal RefractionThis
occurs when a light ray from a less dense medium is reflected into the
denser medium at the boundary separating the two media.Conditions for total internal reflection to occur include the following:

1. Light must be travelling from a more dense to less dense medium.
2. Light must incident at the boundary at an angle greater than the critical angle (C).

Optical fibresThese
are very thin tubes of plastic or glass and because they are so thin
they can bend without breaking, so they can carry light around the
corners.Uses of optical fibres

• Used in telecommunications to carry telephone calls over vast distance, without loss of intensity and without interference.
• Used
  in endoscope to view inside a patient body for example inside stomach.
  Light is carried into the stomach through a bunch of fibres and is
  reflected into small camera, which then displays a picture on a screen.

The Occurance of MirageExplain the occurrence of mirageThis
is the phenomenon inwhich an object appears to be at an incorrect
position due to the bending of light rays from the object.Mirages occur during hot days.

Refraction of Light by Rectangular PrismThe Passage of Light through a Triangular PrismTrace the passage of light through a triangular prismDeviation of light in a prism is the changing in direction of the incident ray when it enters/hits a triangular glass prism.Where iis the angle of incidences is the angle of deviationThe minimum angle of deviation ( qm)In order to determine the minimum angle deviating (Qm) then we must set triangular Glass prism as follows.

The Dispersion of White LightDemonstrate the dispersion of white lightDispersion of light is the splitting up of light beam (white light) into its seven components of colour by a prism.

Spectrum is the patch or band of colours which comprise / constitute seven component of white light.Pure section is the patch or band of colours in which the colours are clearly separated.In order to produce pure spectrum then we must use two converging lenses (convex lenses).

When colours of spectrum are combined, they form white light.In order to combine colours of the spectrum, weneed two triangular glass prisms and one lens.Impure spectrum:the band/patch of colours which overlap and are not seen clearly.The rainbow:a
bow-shaped spectrum of seven colours of white light formed when white
light undergoes dispersion within the rain drops because water is denser
than air, so has a large refractive index.Activity 1A rainbow can be demonstrated as follows:Spray some water into the air in a direction opposite to that of the sun.Look
at the water shower while you face away from the sun. You will see the
colour of the spectrum of white light in the falling drops of water. The
spectrum so formed hasthe shape like a bow. So it is called rainbow.There are two main types of rainbow:

1. Primary rainbow
2. Secondary rainbow

Primary rainbowThis
is formed when light undergoes one or single total internal reflection
in the water droplets. In this type of the rainbow the violet colour is
on the inside of the bow while the red colour is on the outside.

The Angles of Deviation and Minimum DeviationDetermine the angles of deviation and minimum deviationFinding the refractive index (n) of glass by using the deviation of light in a prism:Refracting angle of prism is ASnell’s lawS in i/Sin r = NSin i= n sin rSin e’ = nsin iFrom Geometry of figureI = A- rThe total angle of deviation (s) is the angle between the direction orf the incident ray and the emergent ray.Again from the Geometry Q is given by:S= i+r’- AWhen the deviation is a minimum (Sm) the passage of light through the prism will be symmetrical so:I = r ‘and r = I’This means that;A + Smin = 2i = 2r’Therefore;Refractive index, n = Sin (A + Smin)/2Sin (A/2)WhereA = Apex angle ( angle of prism)Smin – The angle of minimum deviationA Simple Prism BinocularConstruct a simple prism binocularSimple prism binocular

Colours of LightThe Component of White LightExplain the component of white lightThere are two types of colour of light

1. Primary colour of light
2. Secondary colour of light

Primary colour of lightThese are basic (fundamental ) Colour of light to which the eye is most sensitive.Primary Colour of light Include the following

1. Red
2. Green
3. Blue

Secondary colours of lightThese are colour of light obtained after mixing primary colours of lightColour mixing by AdditionThis is the process of combining primary colours of light without loss any colour to form secondary colours of light.

	Primary color	Secondary color
	Red + Blue	Magenta
	Red + Green	Yellow
	Blue + Green	Cyan

Colours of White LightRecombine colours of white lightWhen all white light ( Red , Blue and Green)Combineforms WHITE LIGHT.Complementary colours of light: These are the colours which produce white light when combined.

• Red + Blue+ Green – White light
• Red + Cyan – White light
• Blue + Yellow – White light
• Green + Magenta – White light

The Appearances of Coloured Object under White LightExplain the appearances of coloured object under white lightThere are two types of coloured paints ( pigments) which Include the following

1. Primary coloured pigment (paints)
2. Secondary coloured pigment (paints)

Primary, Secondary and Complementary Colours of LightIdentify primary, secondary and complementary colours of lightPrimary Coloured pigmentsThese are basic coloured pigments which form secondary coloured pigment when combined.The primary coloured pigments include:Yellow, Cyan and MagentaSecondary colour pigmentsThese
are coloured pigments which are formed when two primary colours
combine, whichis always accompanied with the removal of other colours.Difference between Additive and Subtractive Combination of ColoursDistinguish between additive and subtractive combination of coloursColour Mixing by SubstrationIs the process of mixing two primary coloured paints ( pigments) to form secondary colour white.Example 3

• Magenta + Cyan
• Magenta = ( Blue) + ( Red)
• Cyan = (Blue) + (green)

The colour which is common to Blue will appear while red and green disappear.Magenta + Cyan = BlueExample 4

• Magenta + yellow
• Magenta = (Blue) + (Red)
• Yellow = (Green) + (Red)

The colour which is common to both red will appear while blue and green will disappear.HenceMagenta + Yellow = RedExample 5

• Cyan + yellow
• Cyan = (Blue) + (Green)
• Yellow = (Red) + (Green)

The colour which is common to both green will appearwhile Blue and Red will disappearHenceCyan + Yellow = GreenRefraction of Light by LensesDifference between Convex and Concave LensesDistinguish between convex and concave lensesA lens is a transparent medium bounded by two surfaces of regular shape. There are two major categories of lenses which include:The Terms Focal Length, Principle Focus, Principle Axis and Optical Centre as Applied to LensesExplain the terms focal length, principle focus, principle axis and optical centre as applied to lensesOptical center is a geometric center of a lens. Center of curvature is the center of the sphere in which a lens is a part. Principal axis is an imaginary line which passes through the optical center of the lens at right angle to the lens. Principle focus is a point through which all rays traveling close and parallel to the principal axis pass through.The Focal Length of a LensDetermine practically the focal length of a lensFocal
length is a distance between between optical centre and the principal
focus. It is important to note that the the principal focus is not the
halfway between the optical centre and the centre of curvature in lenses
as it is in mirrors. The plane through the principal focus which is at
right angles with the principal axis is called the focal plane.Example 6An object is 2 cm high and placed 24cm from a convex lens. An image formed 72 cm. find the focal length of the lens.Solutioni/f = 1/u + 1/v1/f =1/24 + 1/721/f = 4/72f = 18cm.The Immage Formed by a LensLocate the image formed by a lensRays diagrams are normally used toillustratesthe formation of images by lenses.

1. A ray parallel to the principal axis passes through or appears to diverge from the principal focus after refraction.
2. A ray of light passing through the principal focus of a lens is refracted parallel to the principal axis of the lens.
3. A ray of light through the optical center of the lens continues throughundeviated(Not change direction)

The position, Size and Nature of the Image formed by LensDetermine the position, size and nature of the image formed by lensThe
nature, position and size of the image formed by a lens depends on the
position of the object in relation to the type of lens. For example in
converging lens when the object is between the lens and principal focus
the image will be formed at the same side as the object but further from
the lens. It is virtual, erect, and magnified. The image by concave
lens is erect, virtual and reduced.Activity 2

1. Take a convex lens. Find its approximate focal length in a way described in Activity 11.
2. Draw
   five parallel straight lines, using chalk, on a long Table such that
   the distance between the successive lines is equal to the focal length
   of the lens.
3. Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
4. The
   two lines on either side of the lens correspond to F and 2F of the lens
   respectively. Mark them with appropriate letters such as 2F1, F1, F2and 2F2, respectively.
5. Place a burning candle, far beyond 2F1to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
6. Note down the nature, position and relative size of the image.
7. Repeat this Activity by placing object just behind 2F1, between F1and 2F1at F1, between F1and O. Note down and tabulate your observations.

The
nature, position and relative size of the image formed by convex lens
for various positions of the object is summarized in the table below:

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F2	Highly diminished, point-sized	Real and inverted
Beyond 2F1	Between F2and 2F2	Diminished	Real and inverted
At 2F1	At 2F2	Same size	Real and inverted
Between F1and 2F1	Beyond 2F2	Enlarged	Real and inverted
At focus F1	At infinity	Infinitely large or highly enlarged	Real and inverted
Between focus F1and optical centre O	On the same side of the lens as the object	Enlarged	Virtual and erect

Activity 3

1. Take a concave lens. Place it on a lens stand.
2. Place a burning candle on one side of the lens.
3. Look
   through the lens from the other side and observe the image. Try to get
   the image on a screen, if possible. If not, observe the image directly
   through the lens.
4. Note down the nature, relative size and approximate position of the image.
5. Move
   the candle away from the lens. Note the change in the size of the
   image. What happens to the size of the image when the candle is placed
   too far away from the lens.

Nature, position and relative size of the image formed by a concave lens for various positions of the object

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F1	Highly diminished, point-sized	Virtual and erect
Between infinity and optical centre O of the lens	Between focus F1and optical centre O	Diminished	Virtual and erect

The Magnification of the Lens CameraDetermine the magnification of the lens cameraAs
we have a formula for spherical mirrors, we also have formula for
spherical lenses. This formula gives the relationship between object
distance (u), image-distance (ν) and the focal length (f ). The lens
formula is expressed as1/ν – 1/u = 1/f(8)The
lens formula given above is general and is valid in all situations for
any spherical lens. Take proper care of the signs of different
quantities, while putting numerical values for solving problems relating
to lenses.The
magnification produced by a lens, similar to that for spherical
mirrors, is defined as the ratio of the height of the image and the
height of the object. It is represented by the letter m. If h is the
height of the object and h′ is the height of the image given by a lens,
then the magnification produced by the lens is given by,m = Height of
the Image / Height of the object = h’ / h(9)Magnification
produced by a lens is also related to the object-distance u, and the
image-distance ν. This relationship is given byMagnification (m ) = h’ /
h = ν / u(10)Example 7A
concave lens has focal length of 15 cm. At what distance should the
object from the lens be placed so that it forms an image at 10 cm from
the lens? Also, find the magnification produced by the lens.SolutionA concave lens always forms a virtual, erect image on the same side of the object.Image-distance v = –10 cm;Focal length f = –15 cm;Object-distance u = ?Since, 1 /v – 1 / u = 1 / for, 1 / u = 1 / v – 1 / f1 / u = 1 / -10 – 1 / (-15) = – 1 / 10 + 1 / 151 / u = (-3+2) / 30 = 1 / (-30)or, u = – 30 cm.Thus, the object-distance is 30 cm.Magnification m = v/ um = -10 cm / -30 cm = 1 / 3 = +0.33The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object.The Relationship between Focal Length (f) Object Distance (u) and Image Distance (v) as Applied to LensesDetermine the relationship between focal length (f) object distance (u) and image distance (v) as applied to LensesThe
lens equation is given as 1/f =1/u + 1/v , if sign convection is used
for u, v and f the equation applies to both converging and diverging
lenses for all cases of object and image.Example 8An object is placed 12 cm from converging lens of focal length 18 cm. Find the position of the image.SolutionSince the lens is converging f = +18 cm. 1/v = 1/18 -1/12, v = -36.The image is virtual.