12 Light.pdf

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Reflection of Light

Light

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1. What is Light?

Learning Outcomes

In this section, you’ll be able to:

 Understand how light travels and how we are able to see.  Understand the terms used for reflection:

 Normal

 Angle of incidence  Angle of reflection

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1. What is Light?



Light is a form of

energy

.



Our eyes detect light in a range of colours – red,

orange, yellow, green, blue, indigo and violet.

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1. What is Light?

How does light travel?

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1. What is Light?

What are luminous and non-luminous objects?



Luminous objects are objects which

give off

light of

their own

e.g. Lighted lamp, images on a TV screen



Non-luminous objects are objects that do not give off

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1. What is Light?

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Blind Creatures

Mexican Tetra or

Blind Cave Fish

Blind Albino Cave Crab

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 _{The paths along which the light travels are called rays.}

 _{A beam of light consists of a bundle of light rays}

 _{A light ray is represented by an arrow, pointing in the} direction that the light is travelling

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1. What is Light?

Key Ideas

1.

Light is a form of

energy

.

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1. What is Light?

Question:

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1. What is Light?

Answer:

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2. Reflection of Light

Laws of Reflection:

1. 1st Law of Reflection

The incident ray, the reflected ray and the normal to the reflecting surface all lie in the same plane.

2. 2nd Law of Reflection

The angle of incidence is equal to the angle of

reflection

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2. Reflection of Light

Question:

Draw a diagram to show clearly the incident ray, normal, reflected ray, angle of incidence and the angle of reflection. What is the relationship between angle of incidence and angle of reflection.

incident ray

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2. Reflection of Light

Question:

1. Draw a diagram to show clearly the incident ray, normal, reflected

ray, angle of incidence and the angle of reflection. What is the relationship between angle of incidence and angle of reflection. Answer:

i

r

incident ray normal mirror

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2. Reflection of Light

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2. Reflection of Light

Characteristics of the image formed by a

plane

mirror:

1.

The image is of the

same size

as the object.

2.

It undergoes

lateral

inversion.

3.

It is

upright

.

4.

It is

virtual

.

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Lateral Inversion

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Reflection of Light

Light

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3. Constructing Ray Diagrams

Learning Outcome



Apply the principle of reflection in constructing ray

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Ray Diagrams for plane mirrors

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3. Constructing Ray Diagrams

Step 1: Image distance from mirror = object distance

from mirror

i) Measure accurately the perpendicular distance between object O and the

mirror surface.

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Step 2: Draw the light rays from the image to the eye

i) Use dotted lines behind the mirror

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Step 3: Draw the incident rays from object O to the point of incidence on the mirror surface.

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4. Applications of Mirrors

1. Optical testing

Plane mirrors are used to

reduce the distance

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4. Applications of Mirrors

2. Mirrors at blind corners

 Such mirrors can be installed in shops to allow

shopkeepers to watch out for shoplifters

 Can also be installed at blind corners to allow drivers to

make a turn safely

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4. Applications of Mirrors

3. Instrument Scales

By placing a mirror below the pointer of a scale, parallax error can be avoided when taking readings. This is done by ensuring that the image of the pointer is in line with the pointer itself.

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4. Applications of Mirrors

4. Periscope

Using two plane mirrors inclined at 45°_{, a periscope allows}

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4. Applications of Mirrors

5. Other uses

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5. Key ideas

1.

Ray diagrams are used to

locate the position

of a

mirror image.

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Exercise

1. Figure below shows an extended object. Using ray diagrams, locate the image of the arrow. Describe the image formed.

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Exercise

1. Figure below shows an extended object. Using ray diagrams, locate the image of the arrow. Describe the image formed. Answer:

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2. An optician’s eye chart is fixed 0.5 m behind the eyes of a patient looking into a mirror placed 3.0 m in front of him. Find the

distance of the chart as seen by his eyes.

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2. An optician’s eye chart is fixed 0.5 m behind the eyes of a patient looking into a mirror placed 3.0 m in front of him. Find the

distance of the chart as seen by his eyes.

Answer:

The distance of the eye chart is 3.5 m from the mirror.

Hence the image of the eye chart is 3.5 m behind the mirror. Therefore chart’s image is 6.5 m away from the patient’s eyes.

mirror 3.0 m _Patient’s Chart

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Chart

In the absence of mirror,



need longer space

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Refraction

LESSON-3

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Learning Outcomes



Understand the terms used for refraction:

 Normal

 Angle of incidence  Angle of refraction

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 Light can travel through transparent materials such as glass, water or plastic.  Figure below shows a light ray traveling from air into glass, and then into air again.

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Refraction

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What causes refraction?

 Light travels at different speed in different media.

 For example, in air its speed is 3.0 × 108 m s-1, while in glass it

is 2.0 × 108 _{m s}-1_.

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Refractive Index,

n

speed of light in vacuum or air Refractive index of medium =

speed of light in medium

c

n =

v

For light passing from air or vacuum into a medium, then

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(a) State the speed of light in air.

(b) Calculate the speed of light in a glass block which has a refractive index of 1.53.

(a) Speed of light in air = 3 x 108 _m/s

speed of light in vacuum or air Refractive index of medium =

speed of light in medium (b)

1.53 = 3 x 108 _{/ v}

v = 3 x 108 _{/ 1.53}

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Misperception of Depth (Apparent depth)

The effect of refraction can make the depth seems shallower

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Draw a diagram to show how the eye sees a coin at the bottom of a bucket of water.

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Draw a diagram to show how the eye sees a coin at the bottom of a bucket of water.

water

air

Exercise 2

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Laws of Refraction

1.

The

incident

ray, the

normal

and the

refracted

ray all

lie in the

same plane

.

2.

For two particular media, the

ratio

of the

sine

of the

angle of incidence

to the

sine

of the

angle of

refraction

is a

constant

, i.e.

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Snell’s Law

n

_i

sin i = n

_r

sin r

where n_i = refractive index of medium i

n_r = refractive index of medium r

i

r

incident ray

n

_i

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Refractive index

Glass (n_glass = 1.50) a higher refractive index than water (n_water= 1.33). When a light ray enters glass, it will bend towards the normal more than compared to when it enters water.

air

i

air

i

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At what angle of incidence in air will light pass through another transparent medium without being refracted?

Answer:

When the angle of incidence,

i = 0° from the normal, the

air

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Key Ideas

1. Refraction occurs because the speed of light changes when travelling through different optical media.

2. The two laws of refraction are:

a. The incident ray, the normal and the refracted ray all lie in the same plane.

b. For two particular media, _{sin i}

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3. The refractive index n of a transparent medium is

c

n =

v

where c is the speed of light in vacuum or air,

v is the speed of light in the medium

4.

sin i

= n

sin r

where _{in air or vacuum}i is the angle of incidence of light

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Total Internal Reflection

Light

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6. Total Internal Reflection

Learning Outcomes



Define the terms critical angle and total internal

reflection.



Identify the main ideas in total internal reflection and

apply them to the use of optical fibres in

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6. Total Internal Reflection

Total internal reflection can occur when light passes

from an

optically denser

to a

less dense

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What is the critical angle?

1. When a ray of light is directed perpendicularly through the semicircular glass block, the light ray will pass through

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6. Total Internal Reflection

What is the critical angle?

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6. Total Internal Reflection

What is the critical angle?

3. As the incident angle is increased, the refracted ray is seen to bend further away from the normal. It comes a point when the refracted

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6. Total Internal Reflection

What happens if the angle of incidence is greater than the critical angle?

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6. Total Internal Reflection

Total Internal Reflection takes place only when

1. a ray of light travels from an optical denser medium to a

less dense medium

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How to determine the critical angle of a medium?

When the angle of incidence is at the critical angle c, then the angle of refraction r = 90o_{. Since the light ray is from the medium into}

air, we use the principle of reversibility. Hence

o

sin 90 1

n = =

sin c sin c

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6. Total Internal Reflection

Key Ideas

1.

Total internal reflection takes place only when light

travels from optical

denser

to a

less dense

medium.

2.

The

critical angle

c

is the

angle of incidence

in

the optically

denser

medium at which the

angle of

refraction

in the

less dense

medium is

90

°

.

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Applications of total internal reflection

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Applications of total internal reflection

Optical fibres are made up of thin glass or plastic fibres

that transmit light over long distances through total

internal reflection, even when bent. An optical fibre has a

core of a high refractive index, coated with another

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Test Yourself

Determine the critical angle of a glass prism with a refractive index of 1.9.

Answer:

Given refractive index, n = 1.9

Applying the equation ₁

sin c = n

1

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Thin Converging Lens

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7. Thin Converging Lens

Learning Outcomes:



Describe the action of a thin lens (converging and diverging)

on a beam of light.



Define the term

focal length

for a converging lens.

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What is a lens?

A lens is a piece of

clear plastic or glass

with

curved

surfaces.

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How does a lens refract light?

A lens can be thought of as a set of blocks and prisms.

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Types of Lenses

–

Converging and Diverging



For converging lenses, parallel light rays are brought to

focus

at a point

. (Converge)



For diverging lenses, parallel light rays tend to be

spread out

.

(Diverge)

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Thin converging lens and its main features

1. Optical centre C – Midpoint between the lens surfaces on its principal axis.

2. Principal axis – The horizontal line passing symmetrically through C.

3. Focal point F – All parallel rays to the principal axis converge at the focal point F.

4. Focal length, f – The distance

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Tracing path of light through a thin converging

lens

3 definite paths for light rays passing through a thin converging lens:

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Key Ideas

1. Lenses are used to converge and diverge a beam of light.

2. The main features of a thin converging lens are:

 Optical centre C

 Focal length f

 Focal point F

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8. Ray Diagram for Lenses

How to locate the position of an image?

3 steps to locate the image:

Step 1: Set up the lens and the ray diagram. Step 2: Placing the object.

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Step 1: Set up the lens and the ray diagram

i. Draw principal axis ii. Draw the lens

iii. Mark optical centre C iv. Mark the focal point F

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Step 2: Placing the object.

i. Place the object O to the left of the lens. ii. Mark the object distance as u.

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Step 3: Trace the light rays and draw the image.

i. Draw 2 of the 3 definite paths e.g. Path 1 and 2.

ii. The point where the two light paths intersect is the position of the image.

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Virtual image formed by a converging lens

When the object O is placed near to the lens such that the object distance u is less that the focal length f, then a virtual image is formed. A virtual image cannot be captured on screen.

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9. Applications of Converging Lenses

Converging Lenses are used in many optical instruments:

 Magnifying glass  LCD projector  Camera

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Magnifying glass

When the object is placed such that u < f, image is magnified,

virtual and upright.

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LCD Projector

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Camera Lens

For the camera lens, the image is diminished, real and inverted.

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Visual Correction for Long-sightedness

People with long-sightedness are unable to focus a clear image of near objects on the retina. Spectacles with converging lenses can partially

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Key Ideas

1. Ray diagrams can be constructed to locate the position and type

of image formed by a thin converging lens.

2. Light paths using Path 1 and Path 2 to construct the ray diagram. 3. A real and inverted image is formed when object distance u >

focal length f. The image is formed on the other side of the lens.

4. A virtual and upright image is formed when the object distance u

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Self-Test

How far from a thin converging lens must an object be placed

to produce a magnified image?

Answer:

1.When object distance u < focal length f, the image is magnified,

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Recall

Trigo. Ratios:

sin



= opp. side / hypot. = b/c

cos



= adj. side / hypot. = a/c

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=49.0