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CHAPTER 5 – LIGHT 5.1 – Reflection of light Light rays trave in a straight line . A ray is a narrow beam of light. An object can only be seen when light rays from the object enter our eyes. Reflection of light on a PLANE MIRROR Mirror works because it reflects light. The light ray that strikes the surface of the mirror is called incident ray . The light ray that bounces off from the surface of the mirror is called reflected ray . The normal is a line perpendicular to the mirror surface where the reflection occurs. The angle between the incident ray and the normal is called the angle of incidence , i. The angle between the reflected ray and the normal is called the angle of reflection , r. Laws of reflection 1. The incident ray, the reflected ray and the normal lie on the same plane. 2. The angle of incidence, i, is equal to the angle of reflection, r. Example 1: Two plane mirrors are placed at right angles to each other. A light ray is incident on one of the mirrors at 45˚. Draw the path of light rays. What can you say about the path of incident ray and the final reflected ray? Example 2: Find the angle of reflection.

SPM Form 4 Physics - Light

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Page 1: SPM Form 4 Physics - Light

CHAPTER 5 – LIGHT

5.1 – Reflection of light

Light rays trave in a straight line. A ray is a narrow beam of light. An object can only be seen when light rays from the object enter our eyes.

Reflection of light on a PLANE MIRROR Mirror works because it reflects light.The light ray that strikes the surface of the mirror is called incident ray.The light ray that bounces off from the surface of the mirror is called reflected ray.The normal is a line perpendicular to the mirror surface where the reflection occurs.The angle between the incident ray and the normal is called the angle of incidence, i.The angle between the reflected ray and the normal is called the angle of reflection, r.

Laws of reflection 1. The incident ray, the reflected ray and the normal lie on the same plane.

2. The angle of incidence, i, is equal to the angle of reflection, r.

Example 1:Two plane mirrors are placed at right angles to each other. A light ray is incident on one of the mirrors at 45˚. Draw the path of light rays. What can you say about the path of incident ray and the final reflected ray?

Example 2:Find the angle of reflection.

Characteristics of image formed by a plane mirror

1. Upright2. Same size as the object3. Virtual4. Laterally inverted5. Image is at the same distance as you are

from the mirror.

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Object distance = Image distance

Example 3:A woman of height 1.5m stands 3m in front of a plane mirror. (Draw the diagram)

a) What is the height for her image?b) How for is she from her image?c) Is she walks 2m towards the mirror, how

far is she from her image now?

Example 4:Chayah’s height is 160cm. She stands facing a mirror mounted on a wall at a distance of 160cm. The distance between her eyes and the floor is 150cm and she could see her feet at the bottom edge of the mirror. (Draw the diagram)

a) What is the distance between Chayah and her image in the mirror?

b) Draw light rays to show how she could see her head and her feet.

c) How high above the floor is the bottom edge of the mirror?

d) What is the minimum height of the top edge of the mirror from the floor if she is able to see the top of her head?

Example 5:A girl stands 4m in front of a mirror and a boy

Example 6:A boy stands 5m in front of a plane mirror. If the

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stands 2m behind her. What is the distance between the girl and the image of the boy. (Draw the diagram)

boy moves 2m forward. What is the distance between the boy and his image now? (Draw the diagram)

Curved mirrors- a mirror with curved reflective surface

Principal axis A line passing through the vertex and the centre of curvature.Centre of curvature, C The centre of the sphere that forms the curved mirror.Radius of curvature, CP

The radius of the sphere.

Focal point, F Concave mirror:The point on the principal axis where all reflected rays converge and meet.Convex mirror:The point on the principal axis where all the reflected rays appear to diverge from behind the mirror.

Focal length, f Distance between the focal point and the surface of the mirror.Image distance, i Distance between the image and the surface of the mirror.Object distance, u Distance between the object and the surface of the mirror.

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Convex mirror Concave mirror

Drawing a ray diagram:1. Draw a straight line as principal axis.2. Draw a curved line as the curved mirror and dotted vertical line at the vertex, P.3. Mark position of F and C on the axis. CP=FP4. Draw an arrow as the object standing upright on the axis at a given distance.5. Draw ray P, ray F and ray C.6. The top of the image is where any two of the reflected rays meet after reflection.7. Draw image.

CONCAVE MIRROR

Object distance, u

Ray diagram Image distance, v

Characteristics

Slightly behind C

At C

Between C

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and F

At F

Between F and P

At infinity

CONVEX MIRROR

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Object distance, u

Ray diagram Image distance, v

Characteristics

At infinity

At a distance from the mirror

At close distance to the mirror

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APPLICATIONS OF REFLECTION OF LIGHT

1. Anti-parallax mirror in ammeters and voltmeters.A parallax error occur when you can see both the pointer and its image.Our eyes must be normal to the pointer when the image of the pointer cannot be seen.

2. PeriscopeUsed to see over the top of high obstacles such as a wall.Also used in submarines to observe the surrounding above the water.Consist of 2 plane mirror inclined at 45˚. Final image appears upright.

3. AmbulanceThe word “AMBULANCE” is purposely inverted laterally on an ambulance car.Images seen throught the rear mirror of a car is laterally inverted.

4. Make up mirrorConcave mirrors with long focal lengths can be used.Images appear upright and magnified.

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5. Reflector of torchlightLight bulb is fixed in position at the focal point of the concave mirror to produce a beam of parallel light rays.The beam of parallel light rays will maintain a uniform intensity for a greater distance.Other examples of applications:Headlights of motor vehicles and the lamp of slide projectors.

6. Widening of vision fieldWhen a convex mirror is used, the vision field is larger than a plane mirror.Examples:Rear view mirrors in cars, security mirrors.

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5.2 – Refraction of light

Definition: The bending of light ray at the boundary as it travels from one medium to another with different optical densities.

It is due to a change in the velocity of light as it passes from one medium to another. The more optically dense a medium is, the slower light travels through it. A light ray travels much slower in a denser medium. The change in speed of light ray causes it to

change its direction.

When light ray travels from less dense to a denser medium.

When light ray travels from a denser to a less dense medium.

When light ray is incident normally on the boundary between the two medium.

Laws of refraction 1. Incident ray, refracted ray and the normal lie in the same plane.

2. The ratio of the sine of the angle of incidence, I to the sine of angle of refraction, r is a constant.

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Refractive index, n The value of the

constant for a light ray passing through a vacuum into a given medium is called refractive index.

The is known as Snell’s law.

Refractive index has no unit.

An indication of light-bending ability of a medium as light ray enters its surface from air/vacuum.

Formula:

A material with higher refractive index has a greater bending effect on light as it slows down the speed more. It causes a larger angle of deviation of ray of light, dending the ray of light more towards the normal.

Example 7:Draw a figure showing the angle of incidence on an air-water boundary of 50˚.Calculate the angle of refraction, r if the refrative index of water is 1.33.

Example 8:Draw the diagram accordingly and determine the refractive index of the glass.

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Example 9:Draw the diagram accordingly.Refractive index of glass is 1.62.What is the angle of incidence?

Example 10:a) The speed of light passing through a

medium is 1.8x108 ms-1. Calculate the refractive index of the medium.

b) What is the speed of light in a medium with a refractive index of 2.4?

Example 11:A light ray is incidently normal on a glass prism that has refractive index of 1.50.

a) Complete the ray diagram.b) Find the incident angle and the

refractive angle.

Example 12:A light ray travels from glass to air. Refractive index of glass is 1.54 and speed of light in air is 3x108 ms-1.Calculate:

a) The angle of refractionb) Speed of light in glass

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Phenomena due to refraction of light

Shallow swimming pool 1. Swimming pool appears to be shallower than it really is.

2. Light rays from bottom of the pool is refracted away from the normal as they enter the air.

3. When lgiht rays reach the eye, they appear to come from I, which is higher up.

4. Therefore, the pool appears to be shallower.

Bent objects in water 1. The pencil appears to be bent at the air-water boundary.

2. Light rays from the end of pencil are refracted at the air-water boundary and bent away from the normal.

3. The image of the pencil under the water appears to be higher than it really is.

4. Thus, the pencil appears bent at the surface.

Fish-eye view 1. A fish or a diver can see objects above the water surface, but the objects on the shore would appear to be up in the air at different positions.

Twinkling stars 1. Stars do not twinkle.2. Light from the stars travel in straight

lines thorugh the vacuum of space.3. When these light rays enter the earth’s

atmosphere, they are refracted by layers of atmosphere of different densities.

4. Densities of atmosphere layers are constanly changing (due to wind and temperature), thus altering the path of the light rays.

5. One moment the light rays are refracted in such a way that it reaches your eyes and the next moment they are not.

6. Therefore, we see the stars twinkle.

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REAL DEPTHThe distance of the real object, O from the

surface of water.

APPARENT DEPTHThe distance of the virtual imgae, I from the

surface of water.

Example 13:A fish at the bottom of a pond appears to be 1.2m from the water surface, What is the depth of the pond? (Refractive index of water=1.33) Draw and answer.

Example 15:A boy is standing inside a swimming pool. His legs appead shorter when his friend observes him from the pool side.

a) Explain how this phenomena happen.

b) Draw a ray diagram to show how his legs appear shorter.

c) If the depth of the pool is 0.8m, calculate the distance of the image of his feet as seen from the surface of water. (n of water=1.33)

Example 14:Draw accordingly, and find the refractive index of water.

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Example 16:An object is placed under a glass block. The eye of an observer is positioned at a distance of 10cm above the glass block. The image seen by the observer is 12cm from the eye. If the refractive index of the glass block is 1.50, what is the thickness of the glass block?

Example 17:A glass block with length 18cm has refractive index of 1.50. There is a bubble in the glass block. When the bubble is seen from one side, it appears at a distance of 8cm from that side.When the bubble is seen from the opposite side, it appears at a distance, d2. Determine the value of d2.

5.3 – Total internal reflection of light

Light ray bends away from normal when it travels through a denser medium to a less dense medium. Angle of refraction is larger than angle of incidence.

When the angle of incidence increases, the angle of refraction also increases.

The refracted ray travels along the glass-air boundary.

This is the limit of the light ray that can be refracted in air as the angle in air cannot be any larger than 90˚.

The angle of incidence in the denser medium at this limit is called the critical angle, c.

If the angle of incidence is increased further so that it is greater than the critical angle, the light ray is not refracted anymore but is internally reflected.

This is called TOTAL INTERNAL REFLECTION.

Critical angle, c:The angle of incidence in the denser medium when the angle of refraction, r in the less dense medium is

90˚.

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Two conditions:1. Light ray must travel from a denser medium to a less dense medium.

2. Angle of incidence must be grater than the critical angle of the medium.

Formula:

Example 18:i. A glass block has a refractive index of 1.52.

Calculate the critical angle, c for glass.

ii. The critical angle for water is 49˚, Determine the refractive index of water.

Example 19:Draw accordingly and find the refractive index of the glass prism.

Natural phenomena involving total internal reflection of light.

Mirages

Mirage is caused by refraction and total internal reflection. Mirage normally occur in the daytime when the weather is hot.

The air above the road surface consists of many layers.

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The layers of air nearest the road are hot and the layers get cooler and denser towards the upper layers.

The refractive index of air depends on its density. The lower or hotter layers have a lower refractive index than the layers above them.

Rainbow

A rainbow is formed by refraction, dispersion and total internal reflection of light within water droplets.

When sunlight shines on millions of water droplets in the air during rain and after rain, we see a multicoloured arc.

The colours run from violet along the lower part of the arc to red along upper part of the arc.

Light that forms the primary rainbow is first refracted and dispersed in water droplets. Then, the light is reflected once inside each water droplets. Finally the light is refracted and dispersed again upon exiting the water droplet.

This results in the light being spread out into a spectrum of colour.

Applications of total internal reflection of light.Prism periscope Periscope is built using two right-angled prisms.

The critical angle of the glass prism is 42˚. Roral internal reflection occurs when the light

rays strike the inside face of a 45˚ angles with an angle of incidence greater than the critical angle.

Image produced is upright and has the same size as the object.

Prism binoculars A 45˚/45˚ prism can cause light rays to bend through 180˚ when the light is incident at right angles onto the hypotenuse of the prism.

Light rays will be totally reflected internally two times in a pair of binoculars.

Benefits:o Image produced is upright.o Distance between the objective lens and

the eyepiece is reduced. This makes the binoculars shorter as compared to a teloscope.

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Fibre Optics An optical fibre optic consists of an inner core of high refractive index glass and is surrounded by an outer cladding of lower refractive index.

When light in introduces into the inner core at one end, it will propagate along the fibre in a zigzag path undergoing a series of total internal reflections.

Optical fibres are useful for getting light to inaccesible places. For examples: as an endoscope in medical field, as fibreoptic cables in telecommunications.

Advantage of fibreoptic cables over copper cables:o Much thinner and lighter.o Large number of signals can be sent

through at one time.o Transmit signals with very little loss over

long distance.o Signals are safe and free from electical

interference.o Can carry large data for computers and

telecommunications.

5.4 – Lenses

Convex lenses Concave lenses

Convex lens Concave lens

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Power of a lens:

Power = 1/f

Unit: Dioptre (D)The shorter the focal length, the greater the power.

Power for convex lens in positive while power for concave lens in negative.

CONVEX LENS

Object distance, u

Ray diagram Image distance, v

Characteristics

At infinity

Beyond 2Fu>2f

At 2Fu=2f

Between 2F and F

At F

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Between F and C

CONCAVE LENS

Image formed by concave lens is always virtual, upright and diminished.

Magnification 1. Size of image formed by a lens varies with the position of object.2. Simplest way to compare the image witht the object is by the ratio of

their sizes.3. Formula:

Example 20:Draw accordingly and find the height of image. Height of object is 5cm.

Example 21:A convex lens forms an image of an object. If the height of object is 4cm, what is the height of the image?

Lens formula:

f = focal lengthu = object distancev = image distance

Sign convention:Distance Positive value

(+)Negative value (-)

u Real Virtualv Real Virtualf Convex lens Concave lens

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Example 22:An object is placed in front of a convex lens with a focal length, f of 10cm. What are the characteristics of the image formed if the object distance is 15cm?

Example 23:An object of height 6cm is places at a distance of 20cm from a concave lens. Its focal length is 10cm. Find the position and size of the image.

Example 24:A convex lens with focal length 15cm formed an image which is real, inverted and same size with the object. What is the object distance from the lens?

Example 25:When an object of height 3cm is placed 20cm from a concave lens of focal length 30cm, what is the height of the image formed?

Simple microscope 1. A magnifying glass is the simplest microscope.

2. It consists of a single convex lens with short focal length.

3. When the magnifying glass is held near to the eye and the object is placed inside its focal length (u<f), a virtual, magnified and upright image is formed.

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Compound microscope 1. To view very small objects like microorganisms.

2. 2 powerful conves lenses of short focal lengths:

a. Objective lensb. Eyepiece lens

3. Focal length for objective lens is shorter than focal length for eyepiece lens. (fo<fe)

4. Object to be observed must be plaved between Fo and 2Fo.

5. 1st image: Real, inverted and magnified6. Eyepiece lens is used as magnifying

glass to magnify the first image.7. Eyepiece lens must be positioned so

that the first image is between the lens and Fe

8. 2nd image: Virtual, upright, magnified.9. In normal adjustment, distance

between the lenses is greater than the sum of their individual focal length (L>fo+fe)

10. Magnification of a compound microscope:

m = mo x me

Teloscope 1. To view distant objects like stars.2. 2 convex lenses:

a. Objective lensb. Eyepiece lens

3. fo > fe

4. Objective lens converges the parallel rays from a distant object and forms a real, inverted and diminised image at its focal point.

5. Eyepiece lens is used as a magnifying glass to form a virtual, upright and magnifies image.

6. At normal adjustment, final image is formed at infinity.

7. This is done ny adjusting the position of the eyepiece lens so that the first real image becomes the object at the focal point of the eyepiece lens.

8. Normal adjustment: Distance between lenses = fo + fe .