ScienceScience

CENTRAL BOARD OF SECONDARY EDUCATION

Shiksha Kendra, 2, Community Centre, Preet Vihar,Delhi-110 092 India

PHYSICSPHYSICSLight Reflection and RefractionLight Reflection and Refraction

CHEMISTRYCHEMISTRYCarbon and its CompoundsCarbon and its Compounds

BIOLOGYBIOLOGYHeredity and EvolutionHeredity and Evolution

CBSE - i

Class-X Unit - 4

ScienceUnit - 4

PHYSICSPHYSICSLight Reflection and Refraction

CHEMISTRYCHEMISTRYCarbon and its Compounds

BIOLOGYBIOLOGYHeredity and Evolution

CBSE - iCBSE - i

CENTRAL BOARD OF SECONDARY EDUCATION

Shiksha Kendra, 2, Community Centre, Preet Vihar,Delhi-110 092 India

CLASS-X

The CBSE-International is grateful for permission to reproduce

and/or translate copyright material used in this publication. The

acknowledgements have been included wherever appropriate and

sources from where the material may be taken are duly mentioned. In

case any thing has been missed out, the Board will be pleased to rectify

the error at the earliest possible opportunity.

All Rights of these documents are reserved. No part of this publication

may be reproduced, printed or transmitted in any form without the

prior permission of the CBSE-i. This material is meant for the use of

schools who are a part of the CBSE-International only.

The Curriculum initiated by Central Board of Secondary Education -International (CBSE-i) is a progressive step in making the educational content and methodology more sensitive and responsive to the global needs. It signifies the emergence of a fresh thought process in imparting a curriculum which would restore the independence of the learner to pursue the learning process in harmony with the existing personal, social and cultural ethos.

The Central Board of Secondary Education has been providing support to the academic needs of the learners worldwide. It has about 11500 schools affiliated to it and over 158 schools situated in more than 23 countries. The Board has always been conscious of the varying needs of the learners in countries abroad and has been working towards contextualizing certain elements of the learning process to the physical, geographical, social and cultural environment in which they are engaged. The International Curriculum being designed by CBSE-i, has been visualized and developed with these requirements in view.

The nucleus of the entire process of constructing the curricular structure is the learner. The objective of the curriculum is to nurture the independence of the learner, given the fact that every learner is unique. The learner has to understand, appreciate, protect and build on values, beliefs and traditional wisdom, make the necessary modifications, improvisations and additions wherever and whenever necessary.

The recent scientific and technological advances have thrown open the gateways of knowledge at an astonishing pace. The speed and methods of assimilating knowledge have put forth many challenges to the educators, forcing them to rethink their approaches for knowledge processing by their learners. In this context, it has become imperative for them to incorporate those skills which will enable the young learners to become 'life long learners'. The ability to stay current, to upgrade skills with emerging technologies, to understand the nuances involved in change management and the relevant life skills have to be a part of the learning domains of the global learners. The CBSE-i curriculum has taken cognizance of these requirements.

The CBSE-i aims to carry forward the basic strength of the Indian system of education while promoting critical and creative thinking skills, effective communication skills, interpersonal and collaborative skills along with information and media skills. There is an inbuilt flexibility in the curriculum, as it provides a foundation and an extension curriculum, in all subject areas to cater to the different pace of learners.

The CBSE has introduced the CBSE-i curriculum in schools affiliated to CBSE at the international level in 2010 and is now introducing it to other affiliated schools who meet the requirements for introducing this curriculum. The focus of CBSE-i is to ensure that the learner is stress-free and committed to active learning. The learner would be evaluated on a continuous and comprehensive basis consequent to the mutual interactions between the teacher and the learner. There are some non-evaluative components in the curriculum which would be commented upon by the teachers and the school. The objective of this part or the core of the curriculum is to scaffold the learning experiences and to relate tacit knowledge with formal knowledge. This would involve trans-disciplinary linkages that would form the core of the learning process. Perspectives, SEWA (Social Empowerment through Work and Action), Life Skills and Research would be the constituents of this 'Core'. The Core skills are the most significant aspects of a learner's holistic growth and learning curve.

The International Curriculum has been designed keeping in view the foundations of the National Curricular Framework (NCF 2005) NCERT and the experience gathered by the Board over the last seven decades in imparting effective learning to millions of learners, many of whom are now global citizens.

The Board does not interpret this development as an alternative to other curricula existing at the international level, but as an exercise in providing the much needed Indian leadership for global education at the school level. The International Curriculum would evolve on its own, building on learning experiences inside the classroom over a period of time. The Board while addressing the issues of empowerment with the help of the schools' administering this system strongly recommends that practicing teachers become skillful learners on their own and also transfer their learning experiences to their peers through the interactive platforms provided by the Board.

I profusely thank Shri G. Balasubramanian, former Director (Academics), CBSE, Ms. Abha Adams and her team and Dr. Sadhana Parashar, Head (Innovations and Research) CBSE along with other Education Officers involved in the development and implementation of this material.

The CBSE-i website has already started enabling all stakeholders to participate in this initiative through the discussion forums provided on the portal. Any further suggestions are welcome.

Vineet Joshi

Chairman

PREFACEPREFACE

ACKNOWLEDGEMENTSACKNOWLEDGEMENTSAdvisory Conceptual Framework

Ideators

Shri Vineet Joshi, Chairman, CBSE Shri G. Balasubramanian, Former Director (Acad), CBSE

Shri N. Nagaraju, Director(Academic), CBSE Ms. Abha Adams, Consultant, Step-by-Step School, Noida

Dr. Sadhana Parashar, Director (Training),CBSE

Ms. Aditi Misra Ms. Anuradha Sen Ms. Jaishree Srivastava Dr. Rajesh Hassija

Ms. Amita Mishra Ms. Archana Sagar Dr. Kamla Menon Ms. Rupa Chakravarty

Ms. Anita Sharma Ms. Geeta Varshney Dr. Meena Dhami Ms. Sarita Manuja

Ms. Anita Makkar Ms. Guneet Ohri Ms. Neelima Sharma Ms. Himani Asija

Dr. Anju Srivastava Dr. Indu Khetrapal Dr. N. K. Sehgal Dr. Uma Chaudhry

Coordinators:

Dr. Sadhana Parashar, Ms. Sugandh Sharma, Dr. Srijata Das, Dr. Rashmi Sethi, Head (I and R) E O (Com) E O (Maths) E O (Science)

Shri R. P. Sharma, Consultant Ms. Ritu Narang, RO (Innovation) Ms. Sindhu Saxena, R O (Tech) Shri Al Hilal Ahmed, AEO

Ms. Seema Lakra, S O Ms. Preeti Hans, Proof Reader

Material Production Group: Classes I-V

Dr. Indu Khetarpal Ms. Rupa Chakravarty Ms. Anita Makkar Ms. Nandita Mathur

Ms. Vandana Kumar Ms. Anuradha Mathur Ms. Kalpana Mattoo Ms. Seema Chowdhary

Ms. Anju Chauhan Ms. Savinder Kaur Rooprai Ms. Monika Thakur Ms. Ruba Chakarvarty

Ms. Deepti Verma Ms. Seema Choudhary Mr. Bijo Thomas Ms. Mahua Bhattacharya

Ms. Ritu Batra Ms. Kalyani Voleti

English :

Geography:

Ms. Sarita Manuja

Ms. Renu Anand

Ms. Gayatri Khanna

Ms. P. Rajeshwary

Ms. Neha Sharma

Ms. Sarabjit Kaur

Ms. Ruchika Sachdev

Ms. Deepa Kapoor

Ms. Bharti Dave Ms. Bhagirathi

Ms. Archana Sagar

Ms. Manjari Rattan

Mathematics :

Political Science:

Dr. K.P. Chinda

Mr. J.C. Nijhawan

Ms. Rashmi Kathuria

Ms. Reemu Verma

Dr. Ram Avtar

Mr. Mahendra Shankar

Ms. Sharmila Bakshi

Ms. Archana Soni

Ms. Srilekha

Science :

Economics:

Ms. Charu Maini

Ms. S. Anjum

Ms. Meenambika Menon

Ms. Novita Chopra

Ms. Neeta Rastogi

Ms. Pooja Sareen

Ms. Mridula Pant

Mr. Pankaj Bhanwani

Ms. Ambica Gulati

History :

Ms. Jayshree Srivastava

Ms. M. Bose

Ms. A. Venkatachalam

Ms. Smita Bhattacharya

Material Production Groups: Classes IX-X

English :

Ms. Rachna Pandit

Ms. Neha Sharma

Ms. Sonia Jain

Ms. Dipinder Kaur

Ms. Sarita Ahuja

Science :

Dr. Meena Dhami

Mr. Saroj Kumar

Ms. Rashmi Ramsinghaney

Ms. Seema kapoor

Ms. Priyanka Sen

Dr. Kavita Khanna

Ms. Keya Gupta

Mathematics :

Political Science:

Ms. Seema Rawat

Ms. N. Vidya

Ms. Mamta Goyal

Ms. Chhavi Raheja

Ms. Kanu Chopra

Ms. Shilpi Anand

Geography:

History :

Ms. Suparna Sharma

Ms. Leela Grewal

Ms. Leeza Dutta

Ms. Kalpana Pant

Material Production Groups: Classes VI-VIII

C o n t e n tC o n t e n t

Physics

1. Syllabus Coverage 1

Unit 4 -

Core and Extension

3. Scope Document 5

Learning Objectives

Cross Curricular Links

Suggested Activities

4. Teachers' Notes (TN) 6

Warm up : Nature of Light

Reflection of Light Regular and Diffused

Plane Mirror Properties

Refraction of Light

Spherical Lenses and their Image formation

Total Internal Reflection and its applications

Applications of Refraction of light in nature

Dispersion

Structure and Function of Human eye

Eye defects and their correction

Extension

5. Teacher Student Support Material (TSSM) 11

6. Rubrics of Assessment For Learning - Physics 81

Light Reflection and Refraction

2. Matrix 2

Light Reflection and Refraction

Chemistry

1. Syllabus Coverage 83

Carbon and its Compounds

3. Scope Document

Learning Objectives

Cross Curricular Links

Suggested Activities

4. Teachers' Notes (TN) 91

Warm up

Pre content

Bonding in Carbon Compounds

Nomenclature of Organic Compounds

Homologous series

Fractional distillation of petroleum

The Physical and Chemical Properties of Hydrocarbons

Alcohols

Carboxylic Acid

Polymes-Nylon and Terylene

5. Teacher Student Support Material (TSSM) 115

6. Formative Assessment

7. Rubrics of Assessment 173

8. Suggested Videos and Resources 175

2. Matrix 84

Carbon and its Compounds

Biology

1. Syllabus Coverage 176

Heredity and Evolution

3. Scope Document 182

Learning Objectives

Cross Curricular Links

Suggested Activities

4. Teachers' Notes (TN) 184

Structure of DNA

Traits

Gene, genotype, phenotype, homologous chromosomes

Rules for Inheritance

Genetic Variation

Chromosomal basis of sex determination in human beings

our solar system and life

origin of life

Evolution

A Long Time- a time line

evidences that help in tracing evolutionary relationships

Evolution and Classification

the steps in species formation

human evolutionz

5. Teacher Student Support Material (TSSM) 199

6. Formative Assessment

7. Rubrics of Assessment 276

8. Suggested Videos and Resources 277

2. Matrix 177

Heredity and Evolution

1

PHYSICS

LIGHT Reflection and Refraction

UNIT 4

SYLLABUS COVERAGE

Unit 4 Light Reflection and Refraction

S

Y

L

L

A

B

U

S

CORE

Nature of Light

Regular and Diffused reflection

Laws of reflection

Image formation by plane mirror

Refraction

Refractive Index

Spherical lenses

Image formation by lenses

Practical Application of reflection and refraction

Dispersion

Scattering of light

Structure and function of human eye

Eye defects and their correction

EXTENSION

Electromagnetic Waves

2

Matrix

CONTENT/

CORE

INTENDED LEARNING SKILL

Nature of light Students will be able to understand the

nature/properties of light by

observing/performing a few activities.

Observation,

Understanding

and analysis

Reflection of light Students will be able to understand the

phenomenon of reflection of light and its

reasons.

Observation, To

Differentiate

Types of

Reflection

They will be able to differentiate between

regular and diffused reflection and state the

two laws of reflection.

Identification,

Application,

Learning by

doing.

Illustration.

Laws of reflection

Image formed by

a plane mirror

and its

characteristics.

Student will understand that a plane mirror

is the most suitable example of regular

reflection.

Understanding,

Application,

Observation,

Learning by

doing Use of plane

mirrors in making

kaleidoscope or

periscope etc.

They will be able to explain the

characteristics of the image formed by a

plane mirror. They will be able to construct

kaleidoscope, periscope etc.

http://www.goo

d-science-fair-

projects.com/

3

Refraction Students will understand the phenomenon of

refraction and will be able to explain/ reason

out the difference in real and apparent

positions of objects in water.

Observation.

Refraction index They may do some activities and watch

videos from the link provided to strengthen

the concepts.

Learning by

doing,

Application and

reasoning.

www.youtube.co

m/watch?v=kc2o

73FyN3

http://hyperphys

ics-astro.gsu.edu

Spherical lenses

and the images

formed by them

Student will be able to understand and

illustrate the images formed by convex and

concave lenses with the help of ray

diagrams. They will understand the Lens

formula and its use through numerical They

will know the use of these lenses on the basis

of their properties.

Observation,

Illustration,

Reasoning

Numerical and

analytical skills

Practical

application of

reflection and

refraction

Students will be able to apply the

phenomenon of reflection and refraction in

day to day activities.

Observation,

Application

Dispersion Students will be able to understand the

phenomenon of dispersion and its cause.

Observation,

Learning by

http://www.youtube.com/watch?v=kc2o73FyN3http://www.youtube.com/watch?v=kc2o73FyN3http://www.youtube.com/watch?v=kc2o73FyN3http://hyperphysics-astro.gsu.edu/http://hyperphysics-astro.gsu.edu/

4

They will be able to depict the formation of

rainbow with the help of activities.

doing,

Illustration

Scattering of light Students will be able to explain the

appearance of blue sky and reddish orange

sun(during the sunrise and the sunset) with

the help of the phenomenon of scattering.

Observation,

reasoning and

application

http://hyperphys

ics-astro.gsu.edu

Structure and

Function of Human eye

Students will be able to gain the knowledge

about the structure of the human eye. They

will also understand the function of various

parts of the eye.

Observation,

Analysis

The concept can be explained with the help

of the Ppt provided herewith.

http://www.enc

hantedlearning.co

m/subjects/anato

my/eye/label/la

beleye.shtml

Eye defects and

their correction

Students will come to know the various

types of defects of vision and the causes and

correction fot them. They will be able to

illustrate The defective and corrective eyes

with the help of ray diagrams.

Observation,

reasoning,

application and

analysis.

Extension Students will learn about properties and

types of Electromagnetic waves.

http://hyperphysics-astro.gsu.edu/http://hyperphysics-astro.gsu.edu/http://www.enchantedlearning.com/subjects/anatomy/eye/label/labeleye.shtmlhttp://www.enchantedlearning.com/subjects/anatomy/eye/label/labeleye.shtmlhttp://www.enchantedlearning.com/subjects/anatomy/eye/label/labeleye.shtmlhttp://www.enchantedlearning.com/subjects/anatomy/eye/label/labeleye.shtmlhttp://www.enchantedlearning.com/subjects/anatomy/eye/label/labeleye.shtml

5

Scope Document

Intended Learning outcomes

At the end of this unit, students should be able to

Describe the phenomenon of reflection, refraction and dispersion and scattering.

State, explain and apply the laws of reflection and explain the practical

application of reflection, refraction, dispersion and scattering.

Illustrate the formation of real and virtual images by convex and concave lenses.

Describe and understand the structure and function of human eye.

Describe the causes and correction of defects of vision.

Construct devices like kaleidoscope etc which work on the phenomenon of

reflection and refraction.

Cross curricular links

Mathematics Measuring angles with the help of protractor, measuring the size

of the cardboard for making kaleidoscope etc.

Biology Explaining the structure of human eye.

6

Teachers Notes (TN)

The teacher would go through the links, video clips and PPT prior to teaching in the

class. She is expected to frame thought provoking questions based on the web links.

1. Teacher will demonstrate few warm up Activities to explain properties of light.

2. Using daily life examples teacher will define Reflection of light and differentiate

between Regular and Diffused Reflections.

3. Teacher may demonstrate Mirror Magic Activities to explain properties of plain

mirror and organize hands-on activity in the classroom to build a

kaleidoscope/Periscope in order to teach characteristics of image formation. Teacher

may find the link http://www.good-science-fair-projects.com useful.

4. Following video links and Activities mentioned in TSSM can be used by the teacher

to explain the phenomenon of Refraction of light.

www.youtube.com/watch?v=kc2o73FyN3 and

http://hyperphysics-astro.gsu.edu

5. Teacher may remember following points while teaching the application of

Refraction in forming images by Spherical lenses.

Refraction may be defined not as the bending of a ray of light in going from one

medium to another but as the collection of phenomenon associated with the

change in the speed of a ray of light as it goes from one medium to another. We

usually (but not always) observe the effect of this change in speed of light

through the bending of a ray of light.

All ray diagrams, associated with refraction phenomenon, are always drawn on

the basis of the law of refraction. This is so both for convex and concave lenses

and it would be interesting to show that a ray incident on a convex lens, bends

http://www.good-science-fair-projects.com/http://www.youtube.com/watch?v=kc2o73FyN3http://hyperphysics-astro.gsu.edu/

7

towards the principal axis on both the surfaces of the lens. For a concave lens, the

bending is away from the principal axis at both the surfaces. Application of

Snell's law may also be done to show the reversal in the behavior of the two

types of lenses when the surrounding medium has refractive index more than

that of the material of the lens. All the time, it has to be remembered that the

normal to a spherical surface (for a lens there are two such surfaces) is drawn by

joining the point on the lens surface to the centre of curvature of THAT surface. It

would be pertinent to point out the three special rays, used for drawing ray

diagrams, are all drawn on the basis of the law of refraction.

The importance of the sign convention has to be emphasized.(We would

otherwise get different lens formulae for the convex and concave lenses) The

terms u, v, and f in the(single) lens formula, are all algebraic terms and we have

to put both the sign and the magnitude when substituting their values for any

numerical calculation. It would be interesting to work out the power of the

normal eye lens when viewing objects at Infinity and at the normal distance of

distinct vision I.e.,25 cm. The image distance for both the cases would be the size

of the eyeball which may be roughly taken as 2.3 cm.The power values are quite

high ( of the order of 45 D) and the difference in the power values, for the two

cases,(of the order of 4 D) is an indicator of the power of accommodation of the

normal eye.

The students may be given sufficient practice in drawing ray diagrams for

different situations. Some students have difficulty understanding how the entire

image of an object can be deduced once a single point on the image has been

determined. If the object is merely a vertical object (such as the arrow object used

in the example below), then the process is easy. The image is merely a vertical

line. In theory, it would be necessary to pick each point on the object and draw a

separate ray diagram to determine the location of the image of that point. That

would require a lot of ray diagrams as illustrated in the diagram below.

8

If the object is a vertical line, then the image is also a vertical line. For our

purposes, we will only deal with the simpler situations in which the object is a

vertical line that has its bottom located upon the principal axis. For such

simplified situations, the image is a vertical line with the lower extremity located

upon the principal axis.)

The teacher may interest the students by telling them the relationship between

the object distance and object size and the image distance and image size of a

convex lens. Starting from a large value, as the object distance decreases (i.e., the

object is moved closer to the lens), the image distance increases; meanwhile, the

image height increases. At the 2F point, the object distance equals the image

distance and the object height equals the image height. As the object is brought

closer than 2F the size and distance of the image starts increasing reaching to

highly magnified image formed at infinity when the object is kept at the focus F.

In case of a concave lens, as the object distance

is decreased, the image distance is decreased

and the image size is increased. So as an object

approaches the lens, its virtual image on the

same side of the lens approaches the lens as

well; and at the same time, the image becomes

9

larger thoughthe size of the image will always remain smaller than the object.

Method used for convex lens can be used again to explain the students, how the

entire image of an object can be deduced once a single point on the image has

been determined. If the object is merely a vertical object (such as the arrow object

used in the example below), then the process is easy. The image is merely a

vertical line. This is illustrated in the diagram below. In theory, it would be

necessary to pick each point on the object and draw a separate ray diagram to

determine the location of the image of that point. That would require a lot of ray

diagrams as illustrated in the diagram below. Fortunately, a shortcut exists. If the

object is a vertical line, then the image is also a vertical line. For our purposes, we

will only deal with the simpler situations in which the object is a vertical line that

has its bottom located upon the principal axis. For such simplified situations, the

image is a vertical line with the lower extremity located upon the principal axis.

The ray diagram above illustrates that the image of an object in front of a double

concave lens will be located at a position behind the double concave lens.

6. Teaching practical applications of Reflection and Refraction of light can be made

very interesting by taking examples from daily life. Hands on session for

building Pin-hole camera may be arranged in the classroom as part of the activity

session. Concepts of Dispersion and Scattering of light may be taught by

explaining formation of rainbow and blue color of sky. Teacher may use the link

http://hyperphysics-astro.gsu.edu to explain these phenomenons.

7. Teacher may use the PPT to teach structure, function and defects of human eye.

Some more Teaching Suggestions, Activities and Demonstrations

1. Construct a kaleidoscope which allows the angle between the mirrors to be

adjusted. Describe what happens to the patterns observed when the angle

between the mirrors changes.

http://hyperphysics-astro.gsu.edu/

10

2. Experimentally investigate the characteristics of images formed in a plane mirror

using a optical bench, mirror and a candle. Illustrate using ray diagrams or

explain why a plane mirror produces a virtual image.

3. Suggest practical applications which illustrate the lateral inversion of an image in

a plane mirror and differentiate between lateral and vertical inversions.

4. Straddle a full-length mirror sideways, so that students see one leg in front of the

mirror, while the other leg is behind the mirror out of their view. The mirror

should be tall enough so that it reaches from the floor to your crotch. While

balancing on the rear leg, out of the view of the class, slowly raise the leg that is

visible to them and lean forward slightly. The reflection in the mirror creates the

illusion that the rear leg is also being lifted, allowing you to "levitate" before their

very eyes. Wearing a cape, having a fan blowing over the cape to produce a

breeze, and holding your arms outstretched may create an illusion of flight. This

demonstration is likely to be a success if it is set up carefully beforehand.

A follow-up might include a discussion of how mirrors are used to produce theatrical

effects and optical illusions.

5. Show students how to use ray diagrams as a means of verification for numerical

solutions to problems.

6. Perform an activity to observe the number of images formed by two mirrors

placed at 60, 45, and 30 angles to one another. Confirm that the number of

images observed and calculated agree with one another.

11

TEACHER STUDENT SUPPORT

MATERIAL (TSSM)

12

Nature of Light

We think of light as an agent that enables us to see and observe the world around us.

This is just one of the functions of what we now refer to as visible light. In general, we

can say that

Light is a form of energy and it can be converted to other forms

Light travels in straight lines

Light forms shadows.

Activity 1

We know that energy is the ability to do work. Hence, to show

that light is a form of energy we must be able to show that it can

do work. Work is done when an object get moved. So, if we can

show that light can move something, we can demonstrate that it

is a form of energy

Demonstration

Connect a motor to a solar panel. It is observed that when light shines on the solar

panel, the motor turns. This is because the solar panel convert light energy into

electrical energy and this then gets converted to kinetic energy in the motor.

Activity 2

To Show that Light Travels in Straight Lines.

Demonstration

Take a bulb and three symmetric pieces of cardboards

with a hole in the centre of each one of them. Align them

using a piece of cord. Notice that you can only see light

from the bulb when the holes in the cards are lined up.

13

Activity 3

We all know that shadows get formed when light coming from a

source is blocked. The shadow is then similar in outline to the object

blocking the light. We can use different objects, and our imagination,

to form very many interesting shadows. Some of them are shown

below.

In nature also, we understand the phenomenon of Lunar and Solar eclipses through the

formation of shadows.

The size and nature of the shadow observed, depends on the nature and size of both the

source of light and the obstacle put in its way. When we place our fingers between a

candle and a screen we notice that as we bring our finger closer to the candle, more

light gets blocked out and the shadow gets bigger.

Activity 4

Objective: Make simple observations of light and shadows to demonstrate and

understand that

Light travels in straight lines.

Write your observations and answer questions for each of the following:

1. Go to a dark room and turn on a light kept in one corner of the room. Observe

the places, if any, in the room which gets lighted up or stay dark.

2. Stand in the room with your back towards the light. Observe the change in your

shadow as you move around the room.

14

3. Make a small hole of diameter about 1 cm in an opaque sheet. Use this sheet to

block the light from a bulb by keeping it about 10 cm from the light source.

Position your eye at a place where you can see the light passing through the hole.

Record the details about the position of your eye when you see the light from the

lamp.

4. Show the relationship between the positions of your eye, the hole, and the source

of light on a top view diagram. Hence state your conclusion, if any, about the

condition necessary for you to see the light through the hole in the paper.

5. We often speak of ray of light or a beam of light to describe the propagation of

light. Explain how we are justified in using these words to describe light.

Activity 5

Objective: Observe the nature of the changes in the shadow with a change in distance

between the light source and the object.

Procedure:

1. Go to a dark room with an opaque linear object (say a pencil) and keep it about

20 cm from the light source on a blank sheet of paper.

2. Turn on the lamp and observe the shadow on the table cast by the pencil.

3. Observe the shape of the shadow carefully and see whether the shadow is lighter

and darker at different places. On the white paper draw a picture of the shadow

and give the details of the description of your observations.

4. Take the object 40 cm and then to 80 cm away from the light source. Do you

observe any changes in the shadow?

15

5. Analyse your observations to predict what the shadow would look like when the

object is moved to 120 cm from the light source.

Use a paper with a small hole in the center between the light and your standing

object. The size of the light source can now be thought of as being closer to a

point source of light.

Repeat the above steps in the procedure and again above observe the effect of

this nearly point light source on the size, shape, brightness, and sharpness of the

shadow cast by the standing opaque object.

6. Write out the details of the changes you observe in the shadows with this point

light source and try to analyse and explain your observations.

REFLECTION OF LIGHT

Reflection is the bouncing of light from a surface. We can see objects because

some of the light which leaves the objects hits our eyes. Luminous objects are a

source of light while non-luminous objects are seen as a result of light

reflected from them.

Examples of luminous objects: the sun, a light-bulb, a candle.

Examples of non-luminous objects: everything else (not emitting their own light)

Reflection of light changes the direction of a wave at an interface (surface of

separation) between two different media so that the light wave returns into the

medium from which it originated. The law of reflection state that (i) the angle at

which the wave is incident on the surface equals the angle at which it is reflected

and (ii) the incident ray, reflected ray and the normal lie on the same plane.

http://en.wiktionary.org/wiki/interfacehttp://en.wikipedia.org/wiki/Medium_(optics)

16

Types of Reflection

Reflection of light is either regular (mirror-like) or diffused (retaining the energy, but

losing the image) depending on the nature of the interface. Incoming and reflected

lights have same angle with the surface. If the surface reflects most of the light then we

call such surfaces as mirrors. Examine the given pictures below. They show regular and

diffuse reflection of light from given surfaces.

Diffused Reflections Regular Reflection

Regular Reflection

In regular reflection the reflected rays follow a set pattern as shown in the figure above.

A mirror provides the most common model for regular light reflection, and typically

consists of a glass sheet with a metallic coating where the reflection actually occurs.

Reflection also occurs at the surface of transparent media, such as water or glass.

Diagram of regular reflection

http://en.wikipedia.org/wiki/Diffuse_reflectionhttp://en.wikipedia.org/wiki/Energyhttp://en.wikipedia.org/wiki/Transparency_(optics)http://en.wikipedia.org/wiki/Waterhttp://en.wikipedia.org/wiki/Glass

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In the given diagram, a light ray PO strikes a vertical mirror at point O, and the

reflected ray is OQ. By projecting an imaginary line through point O perpendicular to

the mirror, known as the normal, we can measure the angle of incidence { The angle which

the incident ray PO makes with the normal } i and the angle of reflection { angle which

the reflected ray OQ makes with the normal } r.

Regular reflection forms images.

Diffuse/ Irregular Reflection

When light strikes the surface of a (non-metallic) material it bounces off in all directions

due to multiple reflections by the microscopic irregularities inside the material and by its

surface, if it is rough. Thus, an 'image' is not formed. This is called diffuse or irregular

reflection. The exact form of the reflection depends on the structure of the material. Most

of the objects around us reflect light irregularly. The light sent to our eyes by most of

the objects we see is due to diffuse reflection from their surface. Diffuse reflection is

the reflection of light from a surface such that an incident rays do not follow a set

pattern and is reflected at many angles rather than at just one angle as in the case of

regular reflection.

http://en.wikipedia.org/wiki/Imagehttp://en.wikipedia.org/wiki/Diffuse_reflectionhttp://en.wikipedia.org/wiki/Diffuse_reflectionhttp://en.wikipedia.org/wiki/Diffuse_reflectionhttp://en.wikipedia.org/wiki/Reflection_(physics)http://en.wikipedia.org/wiki/Lighthttp://en.wikipedia.org/wiki/Ray_(optics)http://en.wikipedia.org/wiki/Anglehttp://en.wikipedia.org/wiki/Specular_reflection

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Figure - General mechanism of diffused reflection by a solid surface

A surface may also exhibit both regular and diffuse reflection, as is the case, for

example, of glossy paints as used in home painting, which give also a fraction of regular

reflection, while matte paints give almost exclusively diffuse reflection.

Laws of reflection

The laws of reflection are as follows:

1. The incident ray, the reflected ray and the normal to the reflection surface at the

point of the incidence lie in the same plane.

2. The angle which the incident ray makes with the normal is equal to the angle

which the reflected ray makes to the same normal.

3. The reflected ray and the incident ray are on the opposite sides of the normal.

http://en.wikipedia.org/wiki/Glossyhttp://en.wikipedia.org/wiki/Painthttp://en.wikipedia.org/wiki/Matte_(surface)

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Plane Mirror and the Image Formed by it.

The plane mirror is a highly polished surface which is the best example of regular

reflection. It forms a virtual image of an object reflecting light on to it.

A virtual image is formed at the location in space where all the reflected light appears

to diverge from. Since light from the object appears to diverge from this location, a

person who sights along a line at this location will perceive a replica or likeness of the

actual object.

Characteristics of the image formed by a plane mirror

1. In the case of plane mirrors, the image is said to be a virtual image. Virtual images

are images that are formed in locations where light does not actually reach. Light

does not actually pass through the location on the other side of the mirror; it only

appears to an observer as though the light is coming from this location. Whenever a

mirror (whether a plane mirror or otherwise) creates an image that is virtual, it will

be located behind the mirror where light does not really come from.

2. Plane mirror images are laterally inverted there are several other characteristics that

are worth noting. There is an apparent left-right reversal of the image formed by a

plane mirror. That is, if you raise your left hand, you will notice that the image raises

what would seem to be its right hand. If you raise your right hand, the image raises

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what would seem to be its left hand. This is often termed left-right reversal or

lateral inversion. The word AMBULANCE is written in inverse form in front of the

vehicle. The person in a vehicle in front of the ambulance can read it properly in

his/her rear view mirror. While there is an apparent left-right reversal of the

orientation of the image, there is no top-bottom vertical reversal. The image is said

to be upright or erect.

3. The third characteristic of plane mirror images pertains to the relationship between

the object's distance of the object in front of the mirror to the distance of the image

inside the mirror. For plane mirrors, the object distance is equal to the image

distance. That is the image is the same distance behind the mirror as the object is in

front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must

focus at a location 2 meters behind the mirror in order to view your image.

4. The fourth and final characteristic of plane mirror images is that the dimensions of

the image are the same as the dimensions of the object. If a 1.6-meter tall person

stands in front of a mirror, he/she will see an image that is 1.6-meters tall. The ratio

of the image dimensions to the object dimensions is termed the magnification. Plane

mirrors produce images that have a magnification of 1.

To summarize, images formed by plane mirrors are virtual, erect, laterally inverted,

the same distance from the mirror as the object's distance, and the same size as the

object.

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Reflection Activities

Activity 6

Mirror Magic

Fix a comb, with its teeth pointing upwards, just in front of

a board or box-lid. Aim the comb at the sun - but do not

look at the sun. What do we observe? We will see that white

light shine through the teeth, to form parallel (side-by-side)

rays, making white lines. This activity shows that light

usually travels in a straight line.

Now hold a plane mirror, with its edge touching the board, in the path of the rays as

shown in the figure. Observe how the mirror reflects them. Angles formed with the

mirror before and after reflection are always equal. This verifies the law of reflection.

Activity 7

To Prove Laws of Reflection

Things required: a large mirror, two cardboard tubes of similar length and diameter, a

flashlight, some books.

Steps to be followed

1. Place the books to prop the mirror upright.

2. Hold one tube at an angle with one end touching the mirror.

3. Ask a friend to hold the second tube at a matching angle.

4. Shine the flashlight into the tube you are holding.

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Observation

When the tubes are at the correct angle, the light will bounce off the mirror and down to

the end of the second tube. If we hold our hand at the end of the second tube, we will

see a circle of reflected light. On a rough surface, light is not reflected like this. It is

scattered back in several different directions.

This also proves that the incident ray and the reflected ray lie on the same plane.

Activity - 8

Make a Kaleidoscope

The word "kaleidoscope" means "beautiful to look at. A lot of beautiful symmetrical

patterns can be seen through a Kaleidoscope.

Material Required: A stiff cardboard, a pencil, scissors, black paper or a thick black felt

pen, aluminum foil, glue, clear plastic, tracing paper, cellophane tape, small colored

shapes or beads, broken pieces of bangles etc.

Hoe to go about it:

1. Cut out a piece of cardboard about 22 cm by 15 cm { 9 inches by 6 inches }

2. With the pencil, divide the card into four equal strips. Each strip should be

1{inches wide }

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3. Stick foil over two of the panels as shown in the figure. Make sure it is as smooth

as possible.

4. Stick black paper over the third panel or color it black.

5. Leave the fourth panel blank.

6. Fold the cardboard to make a triangular shape and tape the side to hold it in

place.

8. Stick a piece of clear plastic over each end of the kaleidoscope.

9. Put the colored shapes or beads over one piece of plastic and stick some tracing

paper over the top. Leave enough room for the shapes to slide about.

10. Patterns can be seen when the Kaliedoscope is held towards bright light and

looked into.

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How did this happen?

The light bounces back and forth between the foil mirrors. The reflections of the colored

shapes or beads make interesting patterns. To change the pattern, shake your

kaleidoscope so the shapes or beads move into new positions

Activity 9

Alternate Method to Build a Kaleidoscope.

Material required: 3 plane mirrors, rubber bands, cardboard, tracing or greaseproof

paper, small coloured objects, scissors.

Method:

Hold the mirrors with the reflective sides pointing inwards.

Wrap the mirrors in a cardboard and fix with a rubber band.

Seal one end with tracing or greaseproof paper.

Put some pieces of coloured objects inside the tube and view.

Additional Activity:

1. Place two mirrors at 90.

2. Put a small object like a table tennis ball between the mirrors and check the

number of images Produced.

3. Slowly reduce the angle between the mirrors and check how many images are

produced.

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Activity 10

Making of a Periscope

Things required: two plane mirrors, cardboard or long box, cello tape, scissors.

Hoe to go about it?

1. Arrange the things as shown in the figure.

Questions:

1. Explain how a periscope works.

2. Suggest two uses of a periscope.

REFRACTION OF LIGHT

Refraction is usually defined as the bending of light as it passes from one medium to

another.

Take a glass slab and illuminate it with a beam of

light. Observe that it bends on the way in and

also on the way out. Light travels in a straight

line in any homogenous medium. However,

when it passes from one medium to another

medium it changes its direction. This change in

the direction of light is called refraction. Since

the densities of the media are different, light

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travels with different speed in different media. Speed of light in vacuum is 300.000.000

km per hour. The decrease in the speed of light, when it passes from a rarer medium to

a denser medium, bends the light ray toward the normal to the boundary between the

two media.

Amount of bending depends on the refractive index of the media and the angle

between the light ray and the line perpendicular (normal) to the surface separating the

two media (medium/medium interface)

Each medium has a different refractive

index. The angle between the incident

light ray and the normal to the

boundary, at the point of incidence, is

called the angle of incidence. The

angle between the refracted light ray

and the normal is called the angle of

refraction.

Index of Refraction

We find the amount of refraction by using the refractive indices of the media. Refractive

index is the ratio of the speed of light in vacuum to the speed of the light in given

medium.

Approximate values of the indices of refraction of some common substances are given

below.

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Snell's Law

Snell's Law relates the indices of refraction n of the two media to the directions of

propagation in terms of the angles to the normal.

If the incident medium has the larger index of refraction, then the angle with the normal

is increased by refraction. The larger index medium is commonly called the "internal"

medium, since air with n=1 is usually the surrounding or "external" medium.

Lenses, spectacles, magnifying glass, microscope, binoculars, telescopes, camera lenses,

prisms, projectors, endoscope, periscope etc are some of the applications of refraction of

light.

Demonstration of Refraction of Light by Water

Refraction at the water surface gives the "broken pencil"

effect shown above. Submerged objects always appear to be

shallower than they are because the light bends downward

towards the water.

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/refr.html#c2#c2

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Activity - 11

The Aim: To use refraction to make a pencil seem like it is broken.

Material Needed: A pencil {fairly long}, Water, A drinking glass.

Method:

Put water in the glass so that it is half to three quarters full.

Stand the pencil in the glass.

Look at it from the side of the glass.

Then look at the pencil from above the glass.

Results:

From the side of the glass, the pencil seems broken (check out the photos below

the first photo was actually a photo of the same experiment using a glass bowl

with a straight side. The second photo shows the pencil in the glass). Notice the

difference in the two photos? The curved glass and water act as a magnifying

glass!

When looking from above, the pencil seems like it is bending at the surface of the

water.

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If you move your eyes up and down from the side of the glass to the top you will

get to a point where you will see what seems to be another pencil in the water.

The Conclusion

We can see all the effects mentioned in the results in the

picture below. The different effects are seen because we are

looking at the pencil from different angles. From above the

bent light will make image of the pencil above the real pencil

and so gives the impression that the pencil is bent. From the

side, the bent light will lower the image of the pencil, but

because we are looking at it from the side, it will seem to be

broken. The largeness of the pencil is because of

magnification caused by the curved glass.

Following Ray diagrams show formation of images due to Refraction of light in water.

The bent pencil...

And the broken pencil...

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Activity 12

The Aim: To create a rainbow with all its colors using refraction.

Experiment 1:

Equipment Needed: A torch, A fairly shallow glass transparent rectangular bowl, A

small mirror, A white piece of cardboard or thin tissue paper, Water

Method:

Pour the water into the glass bowl, until it is about half or three quarters full.

Balance the mirror against the side of the bowl and in the water at an angle, with

the mirror pointing up out of the water.

Shine the torch from above the water, through the water onto the mirror.

Hold the white cardboard above the mirror until you have the reflection from the

mirror on it.

Alternatively, make a frame with the cardboard and stick the tissue over the

cardboard, much like a screen. When you get the image onto the tissue screen

you will be able to see it from the top as it will shine through, rather than have to

look underneath to see the colors.

HINT: If you have a sunny window and you look at the reflection of it in the

mirror in the water, you can also see the colors of the rainbow.

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Experiment 2:

Equipment needed: A prism, A strong light source, A piece of cardboard or plastic

with a small slit on one side, A white piece of cardboard

Method:

Set up the light source and the cardboard with the slit so that the light shines

through the slit creating a thin beam of light on the surface you are working on.

Place the white cardboard so that the beam of light shines onto it.

Place the prism in-between the light source and the white cardboard so that the

beam of light passes through it.

Move the prism around while watching the white cardboard.

Results

Experiment 1: The cardboard has the colors of the rainbow shining on it.

Experiment 2: While watching the cardboard and moving the prism, we will notice

rainbow colors.

The Conclusion

In both situations above, the white light was split into its seven colors. This happened

because of refraction of light in the water in experiment 1 and in the prism in

experiment 2.

http://www.good-science-fair-projects.com/online-science-dictionary.html

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Spherical Lenses and their Image Formation

If a piece of glass or other transparent material takes an appropriate shape, it is possible

that parallel incident rays would either converge to a point or appear to be diverging

from a point. A piece of glass that has such a shape is called as a lens.

Most lenses are spherical lenses: their two surfaces are parts of the surfaces of spheres,

with the lens axis ideally perpendicular to both surfaces. Each surface can be convex

(bulging outwards from the lens), concave (depressed into the lens), or planar (flat). The

line joining the centers of the spheres making up the lens surfaces is called the principal

axis of the lens.

If the lens is biconvex or plano-convex, a beam of light travelling parallel to the lens

principal axis and passing through the lens will be converged (or focused) to a spot on

the axis, at a certain distance behind the lens (known as the focal length). In this case, the

lens is called a positive or converging lens.

http://en.wiktionary.org/wiki/convexhttp://en.wiktionary.org/wiki/concavehttp://en.wikipedia.org/wiki/Focal_length

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If the lens is biconcave or plano-concave a beam of light passing through the lens is

diverged (spread); the lens is thus called a negative or diverging lens. The beam after

passing through the lens appears to be emerging from a particular point on the axis in

front of the lens; the distance from this point to the lens is also known as the focal

length, although it is negative with respect to the focal length of a converging lens.

The Anatomy of a Lens

Lenses can be thought of as a series of tiny refracting prisms, each of which refracts

light to produce their own image. When these prisms act together, they produce a

bright image focused at a point.

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REFRACTION THROUGH LENSES

If a symmetrical lens were thought of as being a slice of a sphere, then there would be a

line passing through the center of the sphere and attaching to the mirror in the exact

center of the lens. This imaginary line is known as the principal axis. An imaginary

vertical axis bisects the symmetrical lens into halves. The imaginary centre of the lens is

termed as optical centre .Light rays incident towards either face of the lens and

traveling parallel to the principal axis will either converge or diverge at or from a point

on the principal axis. This point is known as the focal point of the lens. The focal point

is denoted by the letter F. Each lens has two focal points - one on each side of the lens.

Unlike mirrors, lenses can allow light to pass through either face, depending on where

the incident rays are coming from. Subsequently, every lens has two possible focal

points. The distance from the lens to the focal point is known as the focal length

(abbreviated by f). A lens have an imaginary point referred to as the 2F point. This is

the point on the principal axis that is twice as far from the vertical axis as the focal point

is.

Refraction Rules for a Converging Lens

Any incident ray travelling parallel to the principal axis of a converging lens will

refract through the lens and pass through the focal point on the opposite side of

the lens.

Any incident ray travelling through the focal point on the way to the lens will

refract through the lens and become parallel to the principal axis.

An incident ray that passes through the optical center of the lens will pass

through the lens without any deviation or negligible deviation.

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Refraction Rules for a Diverging Lens

Any incident ray travelling parallel to the principal axis of a diverging lens will

refract through the lens and travel in line with the focal point (i.e., in a direction

such that its extension will pass through the focal point). These rays after

refraction appear to be diverging from the focus.

Any incident ray travelling towards the focal point on the way to the lens will

refract through the lens and become parallel to the principal axis.

An incident ray that passes through the optical center of the lens will pass

through the lens without any deviation or negligible deviation.

The rules merely describe the behavior of three specific incident rays. While there is a

multitude of light rays being captured and refracted by a lens, only two rays are needed

in order to determine the image location.

Converging Lens Image Formation

Converging lenses can produce both real and virtual images. Images are formed at

locations where any observer is sighting as they view the image of the object through

the lens. So if the path of several light rays through a lens is traced, each of these light

rays will intersect at a point upon refraction through the lens. While different observers

will sight along different lines of sight, each line of sight intersects at the image location.

The diagram below shows several incident rays emanating from an object - a light bulb.

Three of these incident rays correspond to our three strategic and predictable light rays.

Each incident ray will refract through the lens and be detected by a different observer

(represented by the eyes). The point where the refracted rays are intersecting is the

location of the image.

http://www.physicsclassroom.com/class/refrn/u14l5b.cfm#3rules

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In this case, the image is a real image since the light rays are actually passing through

the image location. To each observer, it appears as though light is coming from this

location.

Diverging Lens Image Formation

Diverging lens create virtual images since the refracted rays do not actually converge to

a point. In the case of a diverging lens, the image location is located on the object's side

of the lens where the refracted rays would intersect if extended backwards. Every

observer would be sighting along a line in the direction of this image location in order

to see the image of the object. As the observer sights along this line of sight, a refracted

ray would come to the observer's eye. This refracted ray originates at the object, and

refracts through the lens. The diagram below shows several incident rays emanating

from an object - a light bulb. Three of these incident rays correspond to our three

strategic and predictable light rays. Each incident ray will refract through the lens and

be detected by a different observer (represented by the eyes). The location where the

refracted rays are intersecting is the image location. Since refracted light rays do not

actually exist at the image location, the image is said to be a virtual image. It would only

appear to an observer as though light were coming from this location to the observer's

eye.

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Images of Objects That Do Not Occupy a Single Point

The above case was the formation of an image by a "point object" - in this case, a small

light bulb. The same principles apply to objects that occupy more than one point in

space. For example, a person occupies a multitude of points in space. As you sight at a

person through a lens, light is being reflected from each individual point on that person

in all directions. Some of this light reaches the lens and refracts. All the light that

originates from one single point on the object will refract and intersect at one single

point on the image. This is true for all points on the object; light from each point

intersects to create an image of this point. The result is that a replica or image of the

object is created as we look at the object through the lens. This is depicted in the

diagram below.

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IMAGE FORMATION BY A LENS FOR OBJECTS PLACED AT DIFFERENT

POSITIONS IN FRONT OF IT

CONVERGING/CONVEX LENS

Step-by-Step Method for Drawing Ray Diagrams

Case 1. An object located beyond the 2F point of a double convex lens.

1. Pick a point on the top of the object and draw three

incident rays traveling towards the lens. Using a

scale, accurately draw one ray so that it passes

exactly through the focal point on the way to the

lens. Draw the second ray such that it travels

exactly parallel to the principal axis. Draw the third

incident ray such that it travels directly to the exact center of the lens. Place

arrowheads upon the rays to indicate their direction of travel.

2. Once these incident rays strike the lens, refract them according to the three rules of

refraction for converging lenses. The ray that passes through the focal point on the

way to the lens will refract and become parallel to the principal axis. Use a scale to

accurately draw its path. The ray that traveled parallel to the principal axis on the

way to the lens will refract and pass through the focal point. And the ray that

traveled to the exact center of the lens will continue in the same direction. Place

arrowheads upon the rays to indicate their direction of travel. Extend the rays past

their point of intersection.

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3. Mark the image of the top of the object.

The image point of the top of the object is the point where the three refracted rays

intersect. All three rays should intersect at exactly the same point. This point is

merely the point where all light from the top of the object would intersect upon

refracting through the lens. Of course, the rest of the object has an image as well

and it can be found by applying the same three steps to another chosen point.

4. After completing the first three steps, only the image location of the top extreme of

the object has been found. Thus, the process must be repeated for the point on the

bottom of the object. If the bottom of the object lies upon the principal axis (as it

does in this example), then the image of this point will also lie upon the principal

axis and be the same distance from the mirror as the image of the top of the object.

At this point the entire image can be filled in.

The ray diagram above illustrates that when the object is located at a position

beyond the 2F point, the image will be located at a position between the 2F point

and the focal point on the opposite side of the lens. The image will be inverted,

diminished (smaller than the object), and real.

Similarly ray diagrams for the image formation by a convex lens for other position

of the object can be drawn. It should be noted that the process of constructing a ray

diagram is the same regardless of where the object is located. While the result of

the ray diagram (image location, size, orientation, and type) is different, the same

three rays are always drawn. The three rules of refraction are applied in order to

determine the location where all refracted rays appear to diverge from (which for

real images, is also the location where the refracted rays intersect).

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An object is located at the 2F point (fig 1)

An object is located between the 2F and the focal point. (fig 2)

Fig 1 Fig 2

Ray Diagram for Object Located in Front of the Focal Point

In the three cases described above - the case of the object being located beyond 2F, the

case of the object being located at 2F, and the case of the object being located between 2F

and F - light rays are converging to a point after refracting through the lens. In such

cases, a real image is formed. As shown above, real images are formed when the object

is located a distance greater than the focal length from the convex lens. A virtual image

is formed if the object is placed between the optical centre and the focal point of the

convex lens.

http://www.physicsclassroom.com/class/refrn/U14l5da.cfm#method#methodhttp://www.physicsclassroom.com/class/refrn/U14l5da.cfm#atC#atChttp://www.physicsclassroom.com/class/refrn/U14l5da.cfm#atC#atChttp://www.physicsclassroom.com/class/refrn/U14l5da.cfm#atC#atChttp://www.physicsclassroom.com/class/refrn/U14l5da.cfm#atC#atC

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A ray diagram for the case in which the object is located the focal point and the optical

centre is shown in the diagram at the left. Observe that in this case the light rays diverge

after refracting through the lens. When refracted rays diverge, a virtual image is

formed. The image location can be found by tracing all light rays backwards until they

intersect. For every observer, the refracted rays would seem to be diverging from this

point; thus, the point of intersection of the extended refracted rays is the image point.

Since light does not actually pass through this point, the image is referred to as a virtual

image. Observe that when the object in located in front of the focal point of the

converging lens, its image is an upright/erect and enlarged image that is located on the

object's side of the lens. In fact, one generalization that can be made about all virtual

images produced by lenses (both converging and diverging) is that they are always

upright/erect and always located on the object's side of the lens.

Ray Diagram for Object Located at the Focal Point

A convex lens produces a real image is when an object

is placed beyond the focus and a virtual image when

an object is placed within the focal point of the lens

(i.e., in front of F). But what happens when the object is

located at F? That is, what type of image is formed

when the object is located exactly at the focus of a

converging lens? The diagram below shows two

incident rays and their corresponding refracted rays.

For the case of the object located at the focal point (F), the light rays neither converge

nor diverge after refracting through the lens. As shown in the diagram above, the

refracted rays are traveling parallel to each other. Subsequently, the light rays may

converge beyond the range of the lens hence forming an enlarged, real, inverted image

at infinity.

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Converging Lenses - Object-Image Relations

Hence to Summarise

1. When the object is located at a location beyond the 2F point, the image will always

be real, inverted and smaller in size than the object and located somewhere in

between the 2F point and the focal point (F) on the other side of the lens.

2. When the object is located at the 2F point, the image will also be located at the 2F

point on the other side of the lens. In this case, the image will be inverted and of

the same size as the object.

3. When the object is located between F and 2F point, the image will be formed

beyond the 2F point on the other side of the lens. The image will be inverted and

magnified/enlarged.

4. When the object is located at the focal point, highly magnified real and inverted

image is formed at infinity.

5. When the object is located between the focal point and the optical centre, the image

will always be located somewhere on the same side of the lens as the object. It will

be virtual, erect and highly magnified.

It might be noted from the above descriptions that there is a relationship between the

object distance and object size and the image distance and image size. Starting from a

large value, as the object distance decreases (i.e., the object is moved closer to the lens),

the image distance increases; meanwhile, the image height increases. At the 2F point,

the object distance equals the image distance and the object height equals the image

height. Eight different object locations are drawn in red and labeled with a number; the

corresponding image locations are drawn in blue and labeled with the identical

number.

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REVISION EXERCISE

1. For which positions of an object in front of a converging lens a real image is

formed?

2. With the help of a ray diagram show the formation of a real image, of the same

size as the object, by a convex lens.

3. A converging lens is sometimes used as a magnifying glass. Depict with the help

of a ray diagram the position of the object in front of the lens to produce the

magnified effect.

Diverging /Concave Lens

Step-by-Step Method for Drawing Ray Diagrams

The method of drawing ray diagrams for a double concave lens is described below.

1. Pick a point on the top of the object and draw three

incident rays traveling towards the lens.

Using a scale, accurately draw one ray so that it

travels towards the focal point on the opposite side

of the lens; this ray will strike the lens before

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reaching the focal point; stop the ray at the point of incidence with the lens. Draw

the second ray such that it travels exactly parallel to the principal axis. Draw the

third ray to the exact center of the lens. Place arrowheads upon the rays to indicate

their direction of travel.

2. Once these incident rays strike the lens, refract them according to the three rules of

refraction for double concave lenses.

The ray that travels towards the focal point

will refract through the lens and travel parallel

to the principal axis. Use a straight edge to

accurately draw its path. The ray that traveled

parallel to the principal axis on the way to the

lens will refract and travel in a direction such

that its extension passes through the focal

point on the object's side of the lens. Align a straight edge with the point of

incidence and the focal point, and draw the second refracted ray. The ray that

traveled to the exact center of the lens will continue to travel in the same direction.

Place arrowheads upon the rays to indicate their direction of travel. The three rays

should be diverging upon refraction.

3. Locate and mark the image of the top of the object.

The image point of the top of the object is the

point where the three refracted rays intersect.

Since the three refracted rays are diverging,

they must be extended behind the lens in

order to intersect. Using a scale, extend each of

the rays using dashed lines. Draw the

extensions until they intersect. All three

http://www.physicsclassroom.com/class/refrn/U14l5ea.cfm#rules#ruleshttp://www.physicsclassroom.com/class/refrn/U14l5ea.cfm#rules#rules

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extensions should intersect at the same location. The point of intersection is the

image point of the top of the object. The three refracted rays would appear to

diverge from this point. This is merely the point where all light from the top of the

object would appear to diverge from after refracting through the double concave

lens. Of course, the rest of the object has an image as well and it can be found by

applying the same three steps to another chosen point.

4. Repeat the process for the bottom of the object.

If the bottom of the object lies upon the principal axis (as

it does in this example), then the image of this point will

also lie upon the principal axis and be the same distance

from the lens as the image of the top of the object. At this

point the complete image can be filled in.

Diverging Lenses - Object-Image Relations

The diagrams above show that for any position of an object in front of a concave

lens, the image formed is

located on the object' side of the lens

a virtual image

an erect image

reduced in size (i.e., smaller than the object)

Another characteristic of the images of objects formed by diverging lenses pertains

to how a variation in object distance affects the image distance and size. The

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diagram below shows five different object locations (drawn and labeled in red)

and their corresponding image locations (drawn and labeled in blue).

The diagram shows that as the object distance is decreased, the image distance is

decreased and the image size is increased. So as an object approaches the lens, its

virtual image on the same side of the lens approaches the lens as well; and at the

same time, the image becomes larger.

REVISION EXERCISE

1. Which of the following, a converging lens or a diverging lens can be used to produce

(a) a real image that has the same size as the object

(b) a virtual and diminished image?

Support your answer with ray diagrams.

2. The image of an object is found to be

(a) real and reduced in size.

(b) real and magnified

(c) Virtual and magnified

(d) virtual and diminished.

What type of lens is used to produce each image?

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The Lens Formula

We have seen above how to find the details of the image of an object, formed by the lens

by drawing appropriate ray diagrams. This, however, is not the only method available

for this purpose. It is possible to find all the relevant details of the image by doing

calculations based on a simple formula called the lens formula.

The Lens Formula

1/v- 1/u = 1/f

Here v and u denotes the positions of the image and object and f denotes the focal

lengths of the lens.

All the terms in this formula are algebraic quantities i.e. we have to attach a plus or

minus sign to them, on the basis of standard accepted sign convention. The sign

convention is based on the following rules.

1. The optical centre of the lens is the reference point or origin for measuring all the

distances.

2. As per the standard (co-ordinate geometry ) sign convention :

I. We draw all the ray diagrams with the light rays propagating from left to

right.

II. All distances (measured from the optical centre of points, situated to the left

/right of the optical centre), are attached a minus/plus sign.

III. The focal length of the convex lens if taken with a positive sign while that of

concave lens is taken with negative sign.

IV. The size of an object/image ,situated above the principal axis is taken with a

plus(minus)

It implies that size of a (real) inverted image would be taken with a minus

sign while that of a (virtual) erect image would be taken with a plus sign.

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Magnification

The ray diagrams, drawn above have shown that the size of the image is in general

different from that of the object. We can call the ratio of the size of image (formed by a

lens ), to the size of the object ,as the magnification due to lens.

Thus Magnification, m= size of the image (I)/ /size of the object (O)

We can calculate this through a formula. The magnification formula, for a lens is

m=v/u.

Here v and u denote, as before, the positions of the image and the object respectively.

Magnification, m as this formula indicates would again be an algebraic quantity i.e. it

would have a plus or minus sign associated with it As per the sign convention, stated

above we can say :

For a real image inverted and enlarged image of an object, the magnification m

would be a negative quantity whose magnitude is greater than one.

For a real inverted and diminished image of an object, the magnification m

would be a negative quantity whose magnitude is less than 1 i.e is a fraction

For a virtual erect and enlarged of an object, the magnification m would be a

positive quantity whose magnitude is greater than 1

For a virtual, erect and diminished image of an object the magnification m,

would be a positive quantity whose magnitude is less than 1,i.e,a fraction

Power of a Lens

A convex lens, as we know, usually converges an incident parallel beam of light to a

point on its principal axis. The distance of this point (called the focus) from the optical

centre of the lens, is known as its focal length.

As per our common perception of the term, a lens would be more powerful if it does its

converging(diverging) job more effectively or quickly. Thus, for more powerful lens,

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the focal length would have a smaller magnitude and vice versa. We, therefore, define

the power (p),of a lens as the reciprocal of its focal length(F).Thus

Power (p) =1/focal length (f)

i.e p=1/f

The focal length, in SI units is measured in meters. The corresponding SI unit of power

(metre1)has been given the name DIOPTER (D)

the power of a convex lens ,of focal length 25cm (=0.25 ) would therefore, be

(1/+0.25d)=+4d.For a concave lens of focal length 20cm(0.2M),the power would be(1/-

0.20d)=-5d.

It follows that a lens of power +10d is a convex lens of focal length(1/10)m i.e,10cm

and.a lens of power -2d,would be a concave lens of focal length(1/2)m,ie,50cm

Try guessing the object positions, for a convex lens, for which m is a negative quantity

with a magnitude greater / less than 1

Can we have m as a positive fraction for a convex lens?

Is it possible for m to be negative for a concave lens? Can m be a positive quantity

greater than 1 for this lens?

It is time now to do some actual calculations based on these formulae.

Solved Examples

Example 1: an object is kept at a distance of 25 cm (ii) 45 cm from a convex lens of focal

length 20 cm. find the position , magnification and nature of image formed in each case

Solution: As per the standard sign convention, the object has to be kept to the left of the

lens.hence,in case (i) u = position of the object = -25 cm.The focal length of a convex lens

is taken with a positive sign . hence

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F = +20 cm

Substituting these values in the lens formula

1/v 1/u = 1/f

We get

1/v 1/-25 = 1/+20

Therefore, 1/v = 1/20 1/25 = {5-4} / 100 = 1/100

Therefore, v = +100 cm

The image is thus formed to the right of the lens at a distance of 100 cm from it

Therefore, magnification, m = v/u = +100 / -25 = -4

The image size is, therefore, 4 times as large as the object. The negative sign, with m,

implies an Inverted Image.

The image is, the therefore, a REAL IMAGE.

In case (ii) , we have u=45 cm, f= + 20cm

1/V - 1/-45 = 1/20

1/v =1/20 - 1/45 = 9-4/180= 5/180=1/36

V= +36 cm

The image is thus formed to the right of the lens at a distance of 36 cm from it.

Magnification, m=V/U = +36/-45= -4/5 = -0.8

The image formed is, therefore, a diminished image and its size is 0.8 times that of the

object.

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The negative sign, with m, implies an inverted image; the image is, therefore, a real

image.

Example2: A convex lens, of focal length 30 cm, forms an image of an object at a

distance of 30 cm from it. What is/ are the possible values of the distance of the object

from it?

Solution: Here it is not specified as to whether the image formed is to the right (real

image) or to the left (virtual image) the lens. We would, therefore, need to look at both

the possibilities.

Case I- Let the (real) image be formed to the right of the lens. We then have

V= + 30 cm

Also f= + 30 cm

Now 1/V- 1/U= 1/f

1/30- 1/u= 0 or u D

The object, in this case, would, therefore, be to the left of the lens and at an infinite (very

large ) distance from it.

Case II Let the (Virtual) image be formed to the left of the lens. We would then have

V= -30 cm and f= + 30 cm

1/-30 1/u = 1/30

-1/u= 1/30 + 1/30 = 2/30 = 1/15

U= -15cm

The object, in this case, would, therefore, be on the left of the lens and at a distance of 15

cm from it. Since this distance is less than the magnitude of the focal length, a convex

lens, as we know, forms a virtual (erect and enlarged) image to the left of the lens.

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Example: What is the power of concave lens of focal length 40cm?

This lens forms an image of the object whose magnification m=+0.4. Where is the object

located?

Solution: We have f= -40cm = - 0.4m

(For a concave lens, the focal length is taken with a negative sign)

Power = 1/f = 1/-0.4D = -2.5D

The power, of the given concave lens, is, therefore, -2.5D

Now m=+0.4 = v/u

(the positive value of m implies an erect an therefore, a virtual image)

V= +0.4u.

Substituting values in the lens formula

1/v-1/u= 1/f,

We get

1/+0.4u -1/u= -1/40

Or 1.5/u = -1/40

U= -60 cm

The object is, therefore, at a distance of 60 cm from the lens and to the left of it.

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Total Internal Reflection

We have already studied about refraction, the observed effect of light waves changing

direction when entering a new medium, due to a change in the speed of the wave. We

found that the change in direction of the wave can be quantified using the refractive

indexes of the two materials.

Now, imagine a ray of light entering an optically less dense material, from an optically

denser one. What happens? The light ray bends away from the normal. As the

following diagram shows, the farther the incident ray is from the normal, the farther the

refracted ray will be from it as well.

However, with a small angular change in the angle of incidence comes a bigger change

in angle of refraction (due to the refractive indexes of the two materials).

Lets move on to an extreme case of this situation: when the ray exiting the optically

denser material is refracted to such an extent that it is bent to 900 from the normal.

The angle of incidence in this special case is called the critical angle because beyond

this point, there is a difference in the behavior of the light. When the critical angle for

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the two substances is exceeded, a phenomenon known as Total Internal Reflection

or T.I.R. occurs. This means that instead of exiting the optically denser material and

being refracted, the incident ray is reflected inside the material (i.e. internally).

After this point, normal laws of reflection are followed, by the ray, off of the surface

between the two materials.

The critical angle of a boundary can be found quite simply, using Snells Law, which

states:

where 1 and 2 correspond to the first and second media entered respectively, and

therefore where corresponds to the angle of incidence, and corresponds to the

angle of refraction. In the position of the critical angle, we know that the angle of

refraction, , is 90 . Therefore, sin is equal to 1. The angle of incidence is of course the

critical angle, so we now have:

The critical angle, c, can therefore be found simply by knowing the refractive indexes of

the two materials. It is also important to note that T.I.R. takes place only at the interface

of an optically denser material with one that is optically less dense, and not vice versa.

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Applications of Total Internal Reflection

Total Internal Reflection can be used as a way to make diamonds sparkle and

information travel rapidly on light waves through a fiber optic.

The cut of the diamond favors total internal reflection. Most rays

entering the top of the diamond will internally reflect until they reach

the top face of the diamond where they exit. This gives diamonds their

bright sparkle.

A fiber optic is a glass "hair" which is so thin that once light enters one end, it can never

strike the inside walls at less than the critical angle. The light undergoes total internal

reflection each time it strikes the wall. Only when it reaches the other end is it allowed

to exit the fiber.

Actually an optical fiber has two layers: a core made of a material of with a high

refractive index, and a second, outer layer with lower refractive index. The light waves

transmitted by an optical fiber are reflected off of the boundary between these two

substances, as shown in the diagram of a cross-section of a fiber below.

The smaller the refractive index of the cladding is compared to the refractive index of

the core, the smaller the critical angle is, allowing T.I.R. to take place in more conditions

(as it can be more often exceeded).

Optical fibers are used in a growing number of fields. In communication they are used

for carrying signals precisely, and at the speed of light. This is faster than the speed of

energy transmission by electrons, and therefore faster than electric signals. In medicine,

optical fibers are used by operating doctors to view previously inaccessible places, such

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as the inside of a lung. Optical fibers are helpful in that they allow the transmission of

light to or from places not usually possible. Because they are fibers, they can be bent,

allowing light to be bent easily and precisely around many corners, without the use of

more clumsy devices such as mirrors.

History of Total Internal Reflection

In 1854, British physicist named John Tyndall discovered the principle of optical fiber

by watching a stream of water flowing out of a barrel. His observation was that water

was carrying light, this illusion, of course, was due to total internal reflection, where the

light was bouncing off the sides of the water stream because the angle at which the light

was hitting the sides of the stream were larger than the critical angle between the two

media (air and water). This illusion is actually quite a popular act in magic shows where

the magician "pours light" using a physics principle.

Activity- 13

Light in a Test Tube

Materials

long test tube (the longer the better)

laser pointer

powdered milk or a few drops of liquid milk

water in container so water can be poured into test tube

Procedure

1. Take a clean test tube and put a small amount of powdered milk in it (only a

pinch). Fill the test tube with water and shake