202110243-TRIUMPH-STUDENT-WORKBOOK-PHYSICAL_SCIENCE-G09-PART1

(vi) A force acts on body of mass 1 kg and producing an acceleration of 1m/s2. This force will be (A) 1 N (B) 2 N (C)3 N (D)4 N (vii) The product of net force and interaction time is called as (A) impulse (B) momentum (C) acceleration (D) inertia AS2-Asking questions and making hypothesis 27. Choose the correct answer. (i) Which of the following object displays Newton’s third law of motion on operation? (A) Helicopter (B) Bicycle (C) Rocket (D) Car (ii) A body is moving with mass 5 kg and velocity 10 m/s. Another body with mass 25 kg and velocity 2 m/s is moving along the same line. Which of the following statements hold true? (A) Both bodies experience different forces (B) Both bodies experience same forces (C)Both bodies have different momentum (D)Both bodies have same momentum AS4-Information skills and projects 28. Choose the correct answer. (i) Linear momentum is given by (B) f x m (A) m x a (C)f x a (D)m x v CHAPTER 3. LAWS OF MOTION 99

4. REFRACTION OF LIGHT AT PLANE SURFACES SESSION 1 INTRODUCTION TO REFRACTION AND RELATIVE REFRACTIVE INDEX 1.1 Mind Map SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 100

1.2 Terminology i. Refraction - When light travels from one medium to another, its direction changes at the interface. The phenomenon of changing direction at the interface of two media is known as refraction. ii. Incident Ray - An incident ray is a ray of light that strikes a surface. iii. Refracted Ray - The incident ray becomes the refracted ray as soon as it enters a different medium from which it is travelling. iv. Angle of Incidence - The angle that the incident ray makes with the normal at the point of incidence is called the angle of incidence. v. Angle of Refraction - The angle that the refracted ray makes with the normal at the point of refraction is called the angle of refraction. vi. Absolute Refractive Index - The absolute refractive index of any medium is the refractive index of the medium with respect to vacuum. vii. Relative Refractive Index - The ratio of the refractive index of the second medium to the refractive index of the first medium is called the relative refractive index. 1.3 Key Concepts i. The light ray always travels between two points in a path which needs the shortest possible time to cover. This is known as Fermat’s principle. ii. When light travels from one medium to another, its direction changes at the interface. The phenomenon of changing direction at the interface of two media is known as refraction. iii. The refractive index of a medium with respect to another medium is defined as the ratio of speed of light in the first medium to the speed of light in the second medium. iv. When a light ray travels from a rarer medium (For e.g. air) to a denser medium (For e.g. glass) the value of r is less than the value of ‘i’ and the refracted ray bends towards the normal. v. When the light ray travels from a denser medium to a rarer medium, it bends away from the normal and r >i. 1.4 Reflection on Concepts Q1. [AS1] The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. [Refer to TB page 64 Q1] A. The speed of light in a diamond (v)= 1,24,000 km/s The speed of light in air (c) =3,00,000 km/s SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 101

The refractive index of diamond with respect to air (n) =SpeSepdeeodf liogfhltiginhtdiniamairon(cd) (v) = 3,00,000 = 2.419 1,24,000 The refractive index of diamond(n) = 2.42 Q2. [AS1] Refractive index of glass relative to water is 9 . What is the refractive index of 8 water relative to glass. [Refer to TB page 65 Q2] A. Relative refractive index (n21) = Refractive index of second medium (2) Refractive index of first medium (1) n21 = n2 n1 Refractive index of glass relative to water – ngw = Refractive index of glass = 9 Refractive index of water 8 Refractive index of water relative to glass – nwg = Refractive index of water = 8 Refractive index of glass 9 Q3. [AS7] Why do stars appear twinkling? [Refer to TB page 65 Q7] A. Stars appear twinkling due to atmospheric refraction. The layers of atmosphere on earth have different optical densities due to which they offer different refractive index values to the light coming from the stars. The light travels through the many layers of the Earth’s atmosphere, the light of star is bent many times and in random directions. This random refraction results in the star twinkling out. 1.5 Application of Concepts Q1. [AS7] A light ray is incident on air –liquid interface at 45 degree and is refracted at 30 degree. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refracted ray be 90 degree. [Refer to TB page 65 Q1] A. Case (i): angle of incidence (i)= 45◦ Angle of refection (r) = 30◦ Refractive index (n) = sin i sin r √1 = sin 45 = 2 sin 30 1 √ 2 =2 = √22 SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 102

= 1.414 Case (ii) Angle between reflected ray and refracted ray 90◦ As per the figure : i + 90◦ + r = 180◦ r = 90◦ − i n= sin i = sin i = sin i = tan i sin r sin(90−i) cos i tan i = n tan i = 1.414 tan i = tan 54.7◦ i = 54.7◦ Q2. [AS7] In what cases does a light ray not deviate at the interface of two media? [Refer to TB page 65 Q2] A. In the following cases the light ray does not deviate at the interface. Case(i): If the refractive indices of two media are equal. Case(ii): When the incident ray strikes normally at the point of incidence. 1.6 Higher Order Thinking Skills Q1. [AS1] Why is it difficult to shoot a fish swimming in water? [Refer to TB page 65 Q1] SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 103

A. Due to refraction at water and air interface, the fish appears to be raised and seems to be close to the surface. The is called apparent depth. When we shoot the fish in water, we can not judge the actual position and aim the gun to apparent position of fish instead of real position. Hence it is very difficult to shoot a fish swimming in water. Q2. [AS7] When we sit at a camp fire, objects beyond the fire are seen swaying. Give the reason for it. [Refer to TB page 65 Q3] A. From the campfire, heat is carried into surrounding air by the process of convection. Dur- ing this process the density of surrounding air changes continuously, thus changes its re- fractive index slightly. This continuous change in refractive index gives rise to continuous change in angle of refraction. As a result the objects beyond the fire are seen swaying. 1.7 Suggested Experiments Q1. [AS3] Conduct an experiment to find the relation between angle of incidence and angle of refraction, when light rays travel from denser to rarer medium. [Refer to TB page 66 Q3] A. Take a metal disk. Use a protractor and mark angles along its edge as shown in the figure (a). Arrange two straws at the centre of the disk in such a way that they can be rotated freely about the centre of the disc. SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 104

Adjust one of the straws to make an angle 10o. Immerse half of the disc vertically into the water, filled in a transparent vessel. While dipping, verify that the straw at 10o is inside the water. From the top of the vessel try to view the straw which is inside the water as shown in figure (b). Then adjust the other straw which is outside the water until both straws appear to be in a single straight line. Then take the disc out of the water and observe the two straws on it. We will find that they are not in a single straight line. Measure the angle between the normal and second straw. Draw a table as shown below and note the value. Do the same for various angles. Find the corresponding angles of refraction and note them in the table drawn. We will observe that in the above activity, ‘r’ is greater than‘i’ in all cases. S.No. (i) (r) 1 10° 2 15° 3 20° 4 25° 5 30° 6 35° 7 40° SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 105

We will observe that the angle of refraction(r) is always greater than the angle of inci- dence (i). From this activity we can generalize that when the light ray travels from denser to rarer, it bends away from the normal and r>i. The relation between angle of incidence and angle of refraction can be given by n1sin i = n2sin r. This is called snell’s law. Q2. [AS3] Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear and why? (Make hypothesis and do the above experiment). [Refer to TB page 66 Q4] A. Take a metal ball and make it black with soot in a candle flame. Immerse the ball in a water beaker. A thin air/empty layer is formed between water and soot. The light ray travels from denser medium (water) to rarer medium(air/empty layer). If the angle of incidence is greater than the critical angle, then total internal reflection takes place. Hence metal ball appears in shining. 1.8 Suggested Projects Q1. [AS4] Collect the refractive indices of the following media. Compare them with those are given in the given table. SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 106

Material Refractive Material Refractive medium index medium index Air 1.0003 Canada 1.53 balsam Ice 1.31 1.54 Water 1.33 Rock salt 1.63 Kerosene 1.44 Carbon 1.65 Disulphide Fused quartz 1.46 1.71 Dense flint 1.77 Turpentine oil 1.47 glass 2.42 Crown glass 1.52 Ruby Sapphire Diamond Benzene 1.50 Find out the pairs of media in which light travels almost with same speed. Coconut oil, Cooking oil, Hydrogen gas, Petrol, Diesel, Glycerine, Vinegar, Hydrochloric acid, Transparent plastic. [Refer to TB page 66 Q1] A. Students’ activity SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 107

Objective Questions (1) Refractive indices of Ice, Benzene, Ruby and Kerosene are 1.31, 1.50, 1.71 and 1.44 respectively. In which of the above media, light travels slowly ? (Pg 66; TB Q 6) (A) Ice (B) Benzene (C) Ruby (D) Kerosene Correct Answer: C (2) The relative refractive index of water with respect to air is 4 . Then relative refrac- 3 tive index of air with respect to water is (Pg 66; TB Q 7) (A) 4 (B) 3 (C) 4 (D) 3 3 4 Correct Answer: D SESSION 1. INTRODUCTION TO REFRACTION AND RELATIVE REFRACT... 108

SESSION 2 SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND ITS APPLICATION 2.1 Mind Map 2.2 Terminology i. Snell’s law – The relation between angle of incidence and angle of refraction can be given by n1 sin i = n2 sin r. ii. Total internal reflection – When the angle of the incidence is greater than critical angle, the light ray gets reflected into the denser medium at the interface i.e., light never enters the rarer medium. This phenomenon is called total internal reflection. iii. Mirage – Mirage is an optical illusion where it appears that water has collected on the road at a distant place but when we get there, we don’t find any water. SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 109

2.3 Key Concepts i. The relation between angle of incidence and angle of refraction can be given by n1 sin i = n2 sin r. This is called snell’s law. ii. The angle at which the refracted ray just grazes the interface separating the two media is called as the critical angle. iii. When the angle of the incidence is greater than the critical angle, the light ray gets reflected into the denser medium at the interface i.e., light never enters the rarer medium.This phenomenon is called as total internal reflection. iv. Mirage is an optical phenomenon which occurs naturally in which the light rays get bent to produce a displaced image. An example of mirage would be the appearance of water on roads from a far distance but is not when you get to that place. 2.4 Reflection on Concepts Q1. [AS1] The absolute refractive index of water is 43. What is the critical angle ? [Refer to TB page 65 Q3] A. The absolute refractive index of water = 4 3 Critical angle (c) =? We know that n = 1 c sin sin c = 1 n sin c = 1 = 3 = 0.75 4 4 3 c = sin−1(0.75) Critical angle (c) = 48.59° Q2. [AS1] Determine the refractive index of benzene if the critical angle is 42 degree. [Refer to TB page 65 Q4] A. Refractive index of Benzene (n) = ? SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 110

Critical angle of the benzene = sinc = sin 42o = 0.6691 We know that n = 1 sin c n = 1 sin 42o n = 1 0.6691 n = 1.49 n =1.5 Q3. [AS1] Explain the formation of mirage? [Refer to TB page 65 Q5] A. Mirage is an optical illusion where it appears that water has collected on the road at a distant place but when we get there, we don’t find any water. The formation of a mirage is the best example where refractive index of a medium varies throughout the medium. During a hot summer day, air just above the road surface is very hot, acts as rarer medium. And the air at higher altitudes is cool, acts as denser medium. Thus the refractive index of the cooler air at the top is greater than the refractive index of hotter air just above the road. When the light from a tall object such as tree passes through a medium just above the road, whose refractive index decreases towards ground, it suffers refraction and takes a curved path because of total internal reflection. Hence we feel the illusion of water being present on road which is the virtual image (mirage) and an inverted image of tree on the road. SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 111

2.5 Application of Concepts Q1. [AS7] Why does a diamond shine more than a glass piece cut to the same shape? [Refer to TB page 65 Q4] A. Refractive index of diamond is very high (2.42). It is more than normal glass. Due to high refractive index, critical angle for diamond is very less. So most of the rays enter into the diamond surface, gets total internal reflection. So it shines more. 2.6 Higher Order Thinking Skills Q1. [AS7] Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface for a certain viewing position. [Refer to TB page 65 Q2] A. Critical angle for water is 42°. When the rays of light travelling through water, strike glass—air interface of test tube at angle of more than 42° , they get totally internally reflected. When these reflected rays reach the eye, they appear to come from the surface of the tube itself. Thus, the test tube appears silvery. 2.7 Suggested Experiments Q1. [AS3] Verify experimentally that S in i is a constant. [Refer to TB page 66 Q1] S in r A. Procedure: Take a wooden plank. Cover with white chart. Draw two perpendicular lines, passing through the middle of the paper. Let the point of intersection be 'O'. Mark one line as NN which is normal to the another line marked as MM. Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at ‘O’ SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 112

Take a protractor and place it along NN (its centre coincides with O). Then mark the angles from 0° to 90° on both sides of the line NN. Repeat the same on the other side of the line NN. The angle should be indicated on the curved line. Now place a semi–circular glass disc so that its diameter coincides with the interface line (MM) and its center coincides with the point O. Point a laser light along NN in such a way that the light propagates from air to glass through the interface at point O and observe the path of laser light coming from other side of disc. Now send Laser light along a line which makes 15° (i) with NN and see that it passes through point O. Measure its corresponding angle of refraction (r). Note these values in table. Find sin i, sin r and also the ratio sin i /sin r. Do the same experiment for the angles of incidence such as 20°, 30°, 40° and 50° .In each and every case, we get the ratio sin i/ sin r as a constant. S. No. (i) (r) sin i sin r sin i sin r 1 15 ˚ 2 20 ˚ 3 30 ˚ 4 40 ˚ 5 50 ˚ Q2. [AS3] Organize some activities to understand the phenomenon of total internal reflection. [Refer to TB page 66 Q2] A. Take a cylindrical transparent vessel of 1 lit. Place a coin at the bottom of the vessel. Now pour water until you get the image of the coin on the water surface (look at the surface of water from a side). This is the phenomenon of total internal reflection. One of that is a mirage which we witness while driving or while walking on a road during a hot summer day. If light ray passes from denser medium to rarer medium then the refractive angle is more than the incident angle. The incident angle for which the angle of refraction is 90°, is called critical angle. If the angle of incidence is more than critical angle, then total internal reflection occurs. SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 113

Q3. [AS3] Conduct the activity–7 again. How can you find critical angle of water? Explain your steps briefly. [Refer to TB page 66 Q6] A. Take a cylindrical transparent vessel of 1 lit. Place a coin at the bottom of the vessel. Now pour water until we get the image of the coin on the water surface (look at the surface of water from a side). This is the phenomenon of total internal reflection. When light ray passes from denser medium to rarer medium and If the angle of incidence is more than critical angle, then total internal reflection occurs. SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 114

Critical angle of water : Applying Snell’s law n1. sin i = n2. sin r nwater. sin C = nair. sin 90◦ 1.33 × sin C = 1.0003 × 1 sin C = 1.0003 = 0.7521 1.33 sin C = sin 48◦46 C = 48◦46 = 48.75◦(approx) 2.8 Suggested Projects Q1. [AS4] Collect information on working of optical fibres. [Refer to TB page 67 Q2] A. Students’ activity. A reference answer is given - 1. Total internal refraction is the basic principle behind working of optical fibre. 2. An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer. 3. A bunch of such thin fibres forms a light pipe. 4. The above figure shows the principle of light transmission by an optical fibre. 5. The following figure sketches an optical fibre cable. SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 115

6. Because of the small radius of the fibre, light going into it makes a nearly glancing incidence on the wall. 7. All organs of human body are not accessible to the naked eye of the doctor, for example intestines. 8. The doctor inserts optical fibre into the stomach through the mouth. Q2. [AS4] Prepare a report about various uses of optical fibres in our daily life. [Refer to TB page 67 Q3] A. Students’ activity. Q3. [AS4] Take a thin thermocol sheet. Cut it in circular discs of different radii like 2 cm, 3 cm, 4 cm, 4.5 cm, 5 cm etc and mark centers with sketch pen. Now take needles of length nearly 6 cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2 cm radius in such a way that the needle is inside the water as shown in fig–P4. Now try to view the free end (head) of the needle from surface of the water. • Are you able to see the head of the needle? Now do the same with other discs of different radii. Try to see the head of the needle, each time. Note: The position of your eye and the position of the disc on water surface should not be changed while repeating the activity with other discs. SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 116

a. At what maximum radius of disc, were you not able to see the free end of the needle? b. Why were you not able to view the head of the nail for certain radii of the discs? c. Does this activity help you to find the critical angle of the medium (water)? d. Draw a diagram to show the passage of light ray from the head of the nail in different situations. [Refer to TB page 67 Q4] A. First we have to calculate the maximum radius of disc for which we can able to see the free end of the needle. Here c: critical angle of water – air interface r = radius of circular disc h = height of the needle Apply Snell’s law to water – air interface nwater × sin c = nair × sin r nwater × sin c = nair × sin 90◦ nwater × sin c = 1 × 1 sin c = 1 = 1 (∵ nwater = 4 ) nwater 4 3 3 SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 117

From right angle triangle – tan c = r = 1 h ( 4 )2−1 3 r= h 4 )2−1 ( 3 r = 6 = √6×3 = √18 16−9 16−9 7 9 r = 18 = 6.8 ≈ 7 2.645 a. Yes b. 7 cm c. As the angle of incidence on water surface is greater than critical angle, total internal reflection takes place; no light ray incident on eye. Hence needle head cannot be seen. d. Yes, applying Snell’s law to water – air interface. sin c = 1 n water sin c = 1 = 3 4 4 3 c = sin−1( 3 ) 4 c = 49◦ e. In all the cases SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 118

Objective Questions (1) Which of the following is Snell’s law? (Pg 65; TB Q 1) sin r (A) n1sin i = n2 (B) n1 = sin r n2 sin i (C) n2 = sin r n1 sin i (D)n2sin i = constant Correct Answer: B (2) The refractive index of glass with respect to air is 2. Then the critical angle of glass–air interface is (Pg 65; TB Q 2) (A) 0° (B) 45° (C) 30° (D) 60° Correct Answer: C (3) Total internal reflection takes place when the light ray travels from (Pg 65;TB Q 3) (A) Rarer to denser medium (B) Rarer to rarer medium (C)Denser to rarer medium (D)Denser to denser medium Correct Answer: C (4) If the angle of incidence is equal to critical angle, then the angle of refraction is (Pg 66;TB Q4) (A) 0° (B) 20° (C) 90° (D) 180° Correct Answer: C SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 119

(5) Mirage is a best example for the phenomenon of (Pg 66;TB Q 5) (A) Reflection (B) Refraction (C)Total internal reflection (D) Shift Correct Answer: C SESSION 2. SNELL’S LAW AND TOTAL INTERNAL REFLECTION AND I... 120

SESSION 3 REFRACTION THROUGH A GLASS SLAB 3.1 Mind Map 3.2 Terminology i. Incident ray – The ray which originally falls on a surface of a medium is called as incident ray. ii. Emergent ray – The ray which finally emerges out of the other medium back to the original one is called emergent ray. iii. Lateral Shift – The distance between the incident ray and the emergent ray when a light ray enters a glass slab is called as the lateral shift. SESSION 3. REFRACTION THROUGH A GLASS SLAB 121

3.3 Key Concepts i. The distance between the incident ray and the emergent ray when a light ray enters a glass slab is called as the lateral shift. ii. The angle between emergent ray and normal is called angle of emergence. 3.4 Reflection on Concepts Q1. [AS5] Explain the refraction of light through a glass slab with a neat ray diagram. [Refer to TB page 65 Q6] A. When light travels from one medium to another medium, its direction changes at the in- terface. This phenomenon is called refraction. If light travels from rarer medium to denser medium, it bends towards the normal and if light travels from denser medium to rarer medium, it bends away to the normal.The refracting surfaces of glass slab are parallel to each other. When light ray incident on one surface of the glass slab, it refracted twice and finally emerges from the second surface. At first refraction it travels from rarer medium to denser medium. And at the second refraction it travels from denser medium to rarer medium. The perpendicular distance between the incident ray and emergence ray is called as lateral shift, if the slab is placed horizontally on the plane. The perpendicular distance between the incident ray and emergent ray is called as vertical shift, if the slab is placed vertically on the plane. The angle between the actual path of ray and refracted ray is called angle of deviation (s). SESSION 3. REFRACTION THROUGH A GLASS SLAB 122

3.5 Application of Concepts Q1. [AS5] Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation. [Refer to TB page 65 Q3] A. Place of the object is O Place of the image is O´ Arrow marks shows the path of the light ray. 3.6 Suggested Experiments Q1. [AS3] Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel. • What changes do you notice? • What could be the reasons for these changes? [Refer to TB page 66 Q5] A. The part of glass rod in the glycerine disappears. As the refractive index of glass rod and glycerine are same, we can not identify the interface of these two mediums. Here no refraction takes place. But the refractive indices of water and glass rod are different. So the glass rod appears with big size due to refraction. SESSION 3. REFRACTION THROUGH A GLASS SLAB 123

Q2. [AS3] Find the critical angle of glass and water with respect to air using activity - 5. [Refer to TB page 66 Q7] A. a) Critical angle of glass with respect to air : We know from Snell’s law n1 sin i = n2 sin r ⇒ 1/n = sin c / sin 90o ⇒ 1/n = sin c /1 ⇒ n = 1 / sin c ⇒ c = sin-1 (1/n) b) Critical angle of water with respect to air We know from Snell’s law n1 sin i = n2 sin r ⇒ 1/n = sin c / sin 90o ⇒ 1/n = sin c /1 ⇒ n = 1 / sin c ⇒ sin c = 1/ n12 SESSION 3. REFRACTION THROUGH A GLASS SLAB 124

Objective Questions (1) In an experiment to trace the path of ray through a glass slab, Shiva traced as shown in the figure. The teacher asked to identify the emergent ray. Which of the following would Shiva indentify. (Pg 66; TB Q 8) (A) AB (B) BC (C) CD (D)N1, N2 Correct Answer: C —— CCE Based Practice Questions —— AS1-Conceptual Understanding Very Short Answer Type Questions 1. State true or false. [Refer to Session 4.1 ] (i) Absolute refractive index = Speed of light in vacuum/ Speed of light in medium → n=c/v. [] 2. Answer the following questions in one sentence. [Refer to Session 4.1 ] CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 125

(i) Define refractive index. 3. Fill in the blanks. [Refer to Session 4.1 ] (i) Speed of light in vacuum is . (ii) Refractive index of kerosene is . (iii) The refractive index of ice is . (iv) The process of refraction involves bending of light ray except when it is incident . 4. State true or false. [ ] [Refer to Session 4.3 ] [ ] (i) R.I = Thickness of the slab / (Thickness of slab − Vertical shift) 5. State true or false. [Refer to Session 4.2 ] (i) Refraction is a result of change in speed of light at the interface. CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 126

6. Fill in the blanks. [Refer to Session 4.2 ] (i) Mirage is an example of _____________________________________________ . (ii) The angle of incidence at which the light ray, travelling from denser to rarer medium, grazes the interface is called __________________________________________. (iii) At critical angle of incidence, the angle of refraction is _________________ . Short Answer Type Questions 7. Answer the following questions in 3-4 sentences. (i) [(Session 4.1)] The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (ii) [(Session 4.1)] The absolute refractive index of water is 4/3. What is the critical angle of it ? CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 127

(iii) [(Session 4.1)] Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (iv) [(Session 4.1)] Write the laws of refraction. CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 128

8. Answer the following questions in 3-4 sentences. (i) [(Session 4.2)] Define total internal reflection? Give two examples. (ii) [(Session 4.2)] What is the reason behind the brilliance of diamond? Long Answer Type Questions 9. Answer the following questions in 6-8 sentences. (i) [(Session 4.2)] Explain the phenomenon of total internal reflection with one or two activities. CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 129

(ii) [(Session 4.2)] Mirage is an optical illusion where it appears that water has collected on the road at a distant place but when we get there, we don’t find any water. What is the reason? Why does it appear so? CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 130

(iii) [(Session 4.2)] A rectangular glass wedge (prism) is immersed in water as shown in figure E–a. For what value of angle α , will the beam of light, which is normally incident on AB, reach AC entirely as shown in figure E–b? Take the refractive index of water as 4/3 and the refractive index of glass as 3/2. CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 131

(iv) [(Session 4.2)] Write the mathematical derivation of Snell’s Law. CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 132

CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 133

AS2-Asking questions and making hypothesis Short Answer Type Questions 10. Answer the following questions in 3-4 sentences. (i) [(Session 4.1)] On what factors does the refractive index of a medium depend? 11. Answer the following questions in 3-4 sentences. (i) [(Session 4.2)] What is the use of refractive index of a light? CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 134

AS3-Experimentation and field investigation Long Answer Type Questions 12. Answer the following questions in 6-8 sentences. (i) [(Session 4.1)] Conduct an experiment to obtain a relation between angle of incidence and angle of refraction. CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 135

CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 136

13. Answer the following questions in 6-8 sentences. (i) [(Session 4.3)] Conduct an experiment to determine the ray tracking by a glass slab and lateral shift of the glass slab. CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 137

AS4-Information skills and projects Long Answer Type Questions 14. Answer the following questions in 6-8 sentences. (i) [(Session 4.1)] Write the refractive index of the materials in the below table. Material Medium Refractive index Air Water Kerosene Rock salt Dense flint glass CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 138

AS5-Communication through drawing and model making Short Answer Type Questions 15. Answer the following questions in 3-4 sentences. (i) [(Session 4.3)] What is the angle of deviation produced by a glass slab? Explain with ray diagram. AS6-Appreciation and aesthetic sense, Values Short Answer Type Questions 16. Answer the following questions in 3-4 sentences. (i) [(Session 4.1)] How do you appreciate the role of Fermat principle in drawing ray diagram? CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 139

AS7-Application to daily life, concern to bio diversity Short Answer Type Questions 17. Answer the following questions in 3-4 sentences. (i) [(Session 4.2)] A ray of light travels from an optically denser to rarer medium. The critical angle of the two media is ‘c’. What is the maximum possible deviation of the ray? Long Answer Type Questions 18. Answer the following questions in 6-8 sentences. (i) [(Session 4.2)] Write a note on application of total internal reflection. CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 140

CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 141

Objective Questions AS1-Conceptual Understanding 19. Choose the correct answer. (i) The relation between refractive index and optical density (A) inversely proportional (B) directly proportional (C)no relation (D)none of these (ii) Change in direction of light when travelling from one medium to another is called (A) reflection (B) multiple reflection (C) refraction (D)total internal reflection (iii) If refractive index of a medium increases, the speed of light in that medium (A) increases (B) decreases (C)does not change (D)none of these (iv) The basic cause of refraction is (A) when light is incident on a boundary (B) when the refractive indices of two media are equal (C)the change in the speed of light in going from one medium to another (D) none AS2-Asking questions and making hypothesis 20. Choose the correct answer. (i) Which one of the following is Snell’s law? CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 142

(A) n1 sini=sinr/n 2 (B) n1 /n2 =sinr/sini (C)n2 /n1 =sinr/sini (D)n2 sin i=constant (ii) When a pencil kept in a glass tumbler filled with water seen from the side of the glass, it seems to be bend due to (A) reflection (B) dispersion (C) refraction (D) slattering AS3-Experimentation and field investigation 21. Choose the correct answer. (i) The critical angle of a diamond is (B) 23.4° (A) 24.4° (C) 25.4° (D) 26.4° AS4-Information skills and projects 22. Choose the correct answer. (i) The factor on which refractive index does not depend (A) nature of material (B) wavelength (C) frequency (D)none of these AS6-Appreciation and aesthetic sense, Values 23. Choose the correct answer. (i) A patient’s stomach can be viewed by inserting pipe. (A) optical fiber (B) vacuum (C) water (D)none of these CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 143

AS7-Application to daily life, concern to bio diversity 24. Choose the correct answer. (i) The number of telephone signals transmitted through optical fiber is (A) 1000(Approximately) (B) 3000(Approximately) (C) 2000(Approximately) (D) 4000(Approximately) CHAPTER 4. REFRACTION OF LIGHT AT PLANE SURFACES 144

5. GRAVITATION SESSION 1 INTRODUCTION AND UNIFORM CIRCULAR MOTION 1.1 Mind Map 1.2 Terminology i. Uniform circular motion – The motion of a body along a circular path with constant speed. ii. Centripetal acceleration – This acceleration, that changes only the direction of velocity of a body is called “centripetal acceleration”. iii. Centripetal force – The net force which change only the direction of the velocity of a body is called “centripetal force”. 1.3 Key Concepts i. The force of attraction between two objects is called gravitational force. ii. Motion of a body with constant speed in a circular path is called uniform circular motion. iii. The acceleration which causes changes in direction of velocity of a body is called centripetal acceleration and it is directed always towards the center of the circle during the uniform circular motion of a body. iv. The net force required to maintain a body in uniform circular motion is known as ‘centripetal force’ and its magnitude is given by F = mv2 . R SESSION 1. INTRODUCTION AND UNIFORM CIRCULAR MOTION 145

1.4 Reflection on Concepts Q1. [AS1] How do you explain that an object is in uniform circular motion? [Refer to TB page 81 Q1] A. Uniform Circular Motion: Uniform circular motion is a motion of the body with a constant speed in circular path. Materials required: Electric motor, old C.D., small wooden block, battery, connecting chords, stop clock. Procedure: 1) Take an electric motor and fix a disc to the shaft of the electric motor. 2) Place a small wooden block on the disc, as shown in figure. 3) Switch on the motor. 4) Find the time required to complete ten revolutions by the block. 5) Repeat the same two to three times. 6) We observe that the wooden block moves in a circular path with a constant speed. 7) So, this motion of wooden block is called uniform circular motion. SESSION 1. INTRODUCTION AND UNIFORM CIRCULAR MOTION 146

1.5 Application of Concepts Q1. [AS1] A car moves with constant speed of 10 m/s in a circular path of radius 10 m.The mass of the car is 1000 kg. Who or what is providing required centripetal force for the car? How much is it? [Refer to TB page 81 Q1] A. The centripetal force for the car is provided by the mass of the car. Centripetal force, F = mv2 R Mass of car (m) = 1000 kg Speed of car (v) = 10 m/s R, radius of circular path = 10 m F=? F = 1000×102 = 104N 10 Centripetal Force (F) = 104 N Objective Questions (1) The acceleration which can change only the direction of velocity of a body is called (Pg 82;Q 1) (A) Acceleration due to gravity (B) Uniform acceleration (C)Centripetal acceleration (D)Centrifugal acceleration Correct Answer: C (2) The distance between the Earth and the Moon is (Pg 82;Q 2) (A) 3,84,400 Km (B) 3,84,400 cm (C)84,000 Km (D)86,000 Km Correct Answer: A SESSION 1. INTRODUCTION AND UNIFORM CIRCULAR MOTION 147

SESSION 2 UNIVERSAL LAW OF GRAVITATION AND FREE FALL 2.1 Mind Map 2.2 Terminology i. Law of gravitation – Every object in this universe attracts every other body. The force of attraction between the two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. ii. Weight – Weight of a body is the force of attraction on the body due to earth. It is given by the product of mass of the body and the acceleration due to gravity; W = mg. iii. Weightlessness – The state of the freely falling body. iv. Free–fall –A body is said to be in free fall when only gravity acts on it. (its acceleration is ‘g’). SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 148