Form 3 Physics Notes Msomi Maktaba All Notes Physics Study Notes

PHYSICS FORM THREE TOPIC 3: LIGHT

 

Light is aform of energy which controls the sense of vision.

Reflection of Light from Curved Mirrors

Difference between Concave and Convex Mirrors

Distinguish between concave and convex mirrors

Concave 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:

www.msomimaktaba.com

The Terms Principle, Axis, Pole, Principle Focus and Radius of Curvature as Applied to Curved Mirrors

Explain the terms principle, axis, pole, principle focus and radius of curvature as applied to curved mirrors

Terms 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).

www.msomimaktaba.com

  • 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 Mirror

Locate the images formed by a curved mirror

Case (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.

www.msomimaktaba.com

Case 2

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

www.msomimaktaba.com

Case 3

When 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.

www.msomimaktaba.com

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.

www.msomimaktaba.com

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

www.msomimaktaba.com

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

www.msomimaktaba.com

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

www.msomimaktaba.com

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.

www.msomimaktaba.com

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 1

An 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 = X

Image Height, H1=Y

The Focal Length of a Concave Mirror

Determine practically the focal length of a concave mirror

Focal length (f) is the distance between the principal focus and the pole.

www.msomimaktaba.com

Convex and Concave Mirrors in Daily Life

Use Convex and concave mirrors in daily life

Curved mirrors are used as:

  1. Driving mirrors
  2. Shaving mirrors
  3. Reflectors

Question Time 1

Why 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 2

Why 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 Light

The Concept of Refraction of Light

Explain the concept of refraction of light

Refraction 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.

www.msomimaktaba.com

Figure showing refraction of light as it passes from air to glass.

The Angle of Incidence and Angle of Refraction

Measure the angle of incidence and angle of refraction

The 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 Refraction

State the laws of refraction

First law of refraction

The 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 refraction

Second 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 Material

Determine the refraction index of a material

Refractive index (n) is the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

n = Sini/Sinr OR

Refractive 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)

Or

Refractive 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 2

The 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 = 1.333

Required: To find refractive index from water to air

Since

anw = 1.333

wna = (1/anw)

= (i/1.333)

: wna = 0.75

Real and Apparent Depth

Real depthis the actual height measured without taking account any refraction of light

Apparent depth is the virtual height measured when viewed by observer.

The Concept of Critical Angle and Total Internal Reflection of Light

Explain the concept of critical angle and total internal reflection of light

Critical angle

Critical 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 index

n=Sini/sinr

But i= Critical angle, C

r = 90º

Thus n= sinC/sin 90º

n=SinC/1

n = Sin C

c= Sin -1 (n)

Total Internal Refraction

This 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 fibres

These 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 Mirage

Explain the occurrence of mirage

This 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.

www.msomimaktaba.com

 
 

Refraction of Light by Rectangular Prism

The Passage of Light through a Triangular Prism

Trace the passage of light through a triangular prism

Deviation of light in a prism is the changing in direction of the incident ray when it enters/hits a triangular glass prism.

Where i

is the angle of incidence

s is the angle of deviation

The minimum angle of deviation ( qm)

In order to determine the minimum angle deviating (Qm) then we must set triangular Glass prism as follows.

www.msomimaktaba.com

The Dispersion of White Light

Demonstrate the dispersion of white light

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

www.msomimaktaba.com

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).

www.msomimaktaba.com

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 1

A 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 rainbow

This 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.

www.msomimaktaba.com

The Angles of Deviation and Minimum Deviation

Determine the angles of deviation and minimum deviation

Finding the refractive index (n) of glass by using the deviation of light in a prism:

Refracting angle of prism is A

Snell’s law

S in i/Sin r = N

Sin i= n sin r

Sin e’ = nsin i

From Geometry of figure

I = A- r

The 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’- A

When 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)/2

Sin (A/2)

Where

A = Apex angle ( angle of prism)

Smin – The angle of minimum deviation

A Simple Prism Binocular

Construct a simple prism binocular

Simple prism binocular

www.msomimaktaba.com

 
 

Colours of Light

The Component of White Light

Explain the component of white light

There are two types of colour of light

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

Primary colour of light

These 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 light

These are colour of light obtained after mixing primary colours of light

Colour mixing by Addition

This 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 Light

Recombine colours of white light

When 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 Light

Explain the appearances of coloured object under white light

There 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 Light

Identify primary, secondary and complementary colours of light

Primary Coloured pigments

These are basic coloured pigments which form secondary coloured pigment when combined.

The primary coloured pigments include:Yellow, Cyan and Magenta

Secondary colour pigments

These 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 Colours

Distinguish between additive and subtractive combination of colours

Colour Mixing by Substration

Is 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 = Blue

Example 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.

Hence

Magenta + Yellow = Red

Example 5

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

The colour which is common to both green will appearwhile Blue and Red will disappear

Hence

Cyan + Yellow = Green

 
 

Refraction of Light by Lenses

Difference between Convex and Concave Lenses

Distinguish between convex and concave lenses

A 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 Lenses

Explain the terms focal length, principle focus, principle axis and optical centre as applied to lenses

Optical 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 Lens

Determine practically the focal length of a lens

Focal 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 6

An object is 2 cm high and placed 24cm from a convex lens. An image formed 72 cm. find the focal length of the lens.

Solution

i/f = 1/u + 1/v

1/f =1/24 + 1/72

1/f = 4/72

f = 18cm.

The Immage Formed by a Lens

Locate the image formed by a lens

Rays 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)

www.msomimaktaba.com

The position, Size and Nature of the Image formed by Lens

Determine the position, size and nature of the image formed by lens

The 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 Camera

Determine the magnification of the lens camera

As 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 7

A 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.

Solution

A 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 / f

or, 1 / u = 1 / v – 1 / f

1 / u = 1 / -10 – 1 / (-15) = – 1 / 10 + 1 / 15

1 / u = (-3+2) / 30 = 1 / (-30)

or, u = – 30 cm.

Thus, the object-distance is 30 cm.

Magnification m = v/ u

m = -10 cm / -30 cm = 1 / 3 = +0.33

The 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 Lenses

Determine the relationship between focal length (f) object distance (u) and image distance (v) as applied to Lenses

The 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 8

An object is placed 12 cm from converging lens of focal length 18 cm. Find the position of the image.

Solution

Since the lens is converging f = +18 cm. 1/v = 1/18 -1/12, v = -36.

The image is virtual.

 
 

Leave a Comment