Tuesday, 4 March 2014

Chapter 12. Electricity | CBSE Class 10th | Science | Solved Exercises

Question 1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is –
(a) 1/25
(b) 1/5
(c) 5
(d) 25

Question 2. Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R

Question 3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

Explanation :
 V2 Electric Power P = R V2 R = P 220 ×220 = 100 R = 484 Ω
Now, When this bulb with Resistance 484 Ω is operated at 110 V
 V2 Electric Power P = R 110 × 110 = 484 Power consumed by bulb = 25 W
Question 4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

Explanation :
Suppose the conducting wire has resistance R Ω
When ther wires are connected in series :
 Equivalent resistance R(series) = R + R = 2R
When ther wires are connected in parallel :
 R ×R R2 R Equivalent resistance R(parallel) = = = R + R 2R 2
When same potential difference (V) is applied across series and parallel combination of resistors, let the the Power produced in the form of heat be P(series) and P(parallel)
 V2 P(series) R(series) V2 R(parallel) R / 2 1 = = × = = P(parallel) V2 R(series) V2 2R 4 R(parallel)

Question 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer. Voltmeter is connected in parallel in the circuit to measure the potential difference between two points.

Question 6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

 l R = ρ A
Where;
 R = Resistance of conductor ρ = Electrical resistivity of copper l = Length of conductor A = Area of cross-section of conductor R = 10 Ω Diameter of wire = 0.5 mm = 5/(10×1000) m Radius of wire r = (5 × 10−4)/2 = 2.5 × 10−4 m = 25 × 10−5 m
 A = Πr2 = (22/7)(25 × 10−5 m )2 = 3.14 ×625 × 10−10 m2 = 1962.50 × 10−10 m2 = 19.625 × 10−8 m2
 RA l = ρ
 10 × 19.625 × 10−8 l = m 1.6 × 10− 8 196.25 l = m 1.6 l = 122.65625 m
Hence; the required length of wire l is 122.65625 m
When the diameter 'D' is doubled, the radius 'r' will become 2r and its area A' will increase by 4 times and will become 4A.
 A' = Π(2r)2 ∴ A' = 4× Πr2 = 4 × A
Since, as we know :
 l R ∝ A
∴ Resistance of wire with double the diameter
 l R ∝ 4A
Hence the resistance of the wire will be reduced to 1/4 th of the earlier, if diameter of the wire is doubled

Question 7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

 I (amperes) 0.5 1 2 3 4 V (volts) 1.6 3.4 6.7 10.2 13.2

Plot a graph between V and I and calculate the resistance of that resistor.

Answer. Let us plot given values of Potential difference 'V' along Y-axis and that of current 'I' along X-axis.

 V (volts) I (amperes) R(resistance) =V/I 1.6 0.5 3.2 3.4 1.0 3.4 6.7 2.0 3.35 10.2 3.0 3.4 13.2 4.0 3.3
While plotting a graph with given corresponding values of V and I, we have found that approximately the same value for V/I = 3.33 (approx.) is obtained in each case. Thus the V–I graph is a straight line that passes through the origin of the graph, as shown in Fig., V/I is a constant ratio.
A straight line plot shows that as the current through a wire increases, the potential difference across the wire increases linearly – this is Ohm’s law.

Question 8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer. As we know, according to Ohm's law
 V R = I
 V = 12 Volts I = 2.5 mA = 2.5×10−3 A

 12 R = Ω 2.5×10−3

 12×10−3 R = Ω 2.5 R = 4.8 × 103 Ω

Question 9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer. When resistors are connected in series, the strength of current 'I' passing through each resistor is same.
REq. = R1+R2+R3+R4+R5

REq. = 0.2+0.3+0.4+0.5+12 Ω = 13.4 Ω

 V 9V 90 ∴ I = = = REq. 13.4 Ω 134
I = 0.67 Ampere
∴ value of current passing through 12 Ω resistor as well others = 0.67 Ampere.

Question 10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Let 'n' be the number of resistances, connected in parallel to 220 V to draw a maximum of 5 A current
 1 1 1 = + + ... n times REq. 176 176
 176 REq. = Ω n

Now, according to Ohm's law :
 V R = I

 V Or   I = R

 220 × n 5 A = 176

 5 ×176 880 n = = = 4 220 220
Hence, Number of Resistors, which can be connected in parallel across 220 V line to carry 5 A of current is 4

Question 11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer. (i) To get a combination resistance of 9 Ω, we can connect the given three resistance of 6 Ω each as given below in the diagram :

As shown, a parallel combination of two resistance is connected in series with the third resistance.
 R1×R2 REq = + R3 R1 + R2

 6×6 REq = + 6 6 + 6

 36 REq = + 6 12

 REq = 3 + 6 = 9 Ω
(ii) To get a combination resistance of 4 Ω, we can connect the given three resistance of 6 Ω each as given below in the diagram :

As shown, a series combination of two resistance is connected in parallel with the third resistance.
 1 1 1 = + REq R1 + R2 R3

 (R1+ R2) × R3 REq = (R1 + R2)+R3

 (6+ 6) × 6 12×6 REq = = = 4 Ω (6 + 6 ) + 6 18

Question 12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Voltage applied across the bulbs = 220 V
Maximum allowable current = 5 A
Power rating of each bulb W = 10 W
Let R be resistance of each bulb
 V R = I P I = V V2 ∴ R = P 220 × 220 ∴ R = 10 ∴ R = 4840 Ω

Let 'n' be the number of bulbs, connected in parallel to 220 V to draw a maximum of 5 A current
 1 1 1 = + + ... n times REq. 4840 4840

 4840 REq. = Ω n

Now, according to Ohm's law :
 V R = I

 V Or   I = R

 220 × n 5 A = 4840

 5 ×4840 24200 n = = = 110 220 220
Hence, Number of bulbs which can be connected in parallel with each other across the two wires of 220 V line when the maximum allowable current is 5 A = 110

Alternate Solution: !!
Vlotage applied across the bulbs = 220 V
Maximum allowable current = 5 A
Power rating of each bulb W = 10 W
Max. Power capacity of the line , P = VI
P= 220 × 5 = 1100 W
let 'n' be the Numbers of bulbs of 10 Watt each which can be connected in parallel
Therefor n= Total Power Capacity of line÷Power Rating of each bulb
n= 1100÷10 = 110
Hence 110 bulbs can be connected in parallel across the line

Question 13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Case 1. When electric oven is connected with just one resistance coil
 V 220 I1 = = 9.2 A R 24

Case 2. When electric oven is connected with two resistance coils in series
∴ R = 24 + 24 = 48 Ω

 V 220 I2 = = = 4.6 A approx. R 48

Case 3. When electric oven is connected with two resistance coils in parallel
 24 × 24 24 × 24 R = = = 12 Ω approx. 24 + 24 48 V 220 I3 = = = 18.3 A approx. R 12

Question 14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Case 1. : When resistor is connected in series :

 V = 6 Volt R = 1 + 2 = 3 Ω
 V 6 I = = R 3 I = 2 A
 Let P1 be power used in 2 Ω resistor P1 = VI = 6×2 = 12 W
Case 2. : When resistor is connected in parallel :

 V = 4 Volt R1×R2 R = R1+R2 12×2 R = 12+2 R = 1.71 Ω
 V 4 I = = R 1.7 I = 2.34 A
 Let P2 be power used in 2 Ω resistor P2 = VI = 4×2.34 = 9.36 W Comparison of power used : Ratio P1 : P1 :: 12 : 9.36 P1 : P1 :: 1 : 0.78

Question 15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

 Power rating of first lamp = 100 W Power rating of second lamp = 60 W Voltage applied across lamps = 220 V Let, Resistance of first lamp = R1 Resistance of second lamp = R2 V2 As we know, R = P 220 ×220 R1 = = 480 Ω 100 220 ×220 R2 = = 806.66 Ω 60
When R1 and R2 are connected in parallel, Let be REq equivalent resistance and I be current drawn through R1 and R2 connected in parallel.
 1 1 1 = + REq R1 R2 R1×R2 REq = R1+R2 484×806.66 REq = Ω 484+806.66 390423.44 REq = = 302.5 Ω 1290.66 V 220 I = = R 302.5 I = 0.727 A

Question 16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Energy = Power × Time
 For TV set : E1 = 250 W × 1 × 3600 sec = 900000 J = 9 × 105 J For Toaster : E2 = 1200 W × 10 × 60 sec = 720000 J = 7.2 × 105 J => E1 >E2
∴ TV Set uses more power.

Question 17. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Heat energy developed in the heater= I2Rt
Rate of Heat Energy= Power
 P = I2Rt / t P = I2R P = 152 × 8 P = 1800 Watt
∴ Heat is developed at the rate of 1800 W or 1.8 KW or 1800 joule per sec.

Question 18. Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminum wires usually employed for electricity transmission?

(a) Tungsten offers high resistance against electric current and it has high melting point. Mainly for these two characteristics, Tungsten is used almost exclusively for filament of electric lamps. When the electric current flows through Tungsten filament, due to high resistance of Tungsten, it gets heated up to a very high temperature and starts glowing with light. High melting point of Tungsten prevents it from melting away at higher temperature.

(b) The conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal because an alloy has certain favorable characteristics, which are as follows :
1. It offers high resistance against an electric current
2. It can withstand high temperature due to high melting poin.
3. It has high density
As per Joule’s law of heating,
H = I2Rt
The that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance
(ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor.
Higher resistance, high melting point and higher density of an alloy in conductors of electric heating devices, all to gather, help in producing a large amount of heat at higher temperature, at the same time offsetting any deformation arising out of temperature variations.

(c) When appliances are connected in series, the value of equivalent resistance will be very large, as shown below :-
Req = R1 + R2 + R3 ......
Due to resulting higher value of Equivalent Resistance, there will be a considerable drop in line voltage and at the same time a large amount of heat will be produced in the domestic circuit As per Joule’s law of heating,
H = I2Rt
This may result in either loss of electric energy due to heating or melting away of electric wiring in domestic circuit due to over heating. Any fault in a series arrangement results in complete shutdown of other devices in circuit. When appliances are connected in parallel, the value of equivalent resistance will be very small hence damage or energy losses due to heating will be minimum. Also break down of a certain device does not affect others in the parallel circuit
This is the main reason, why the series arrangement is not used for domestic circuits

(d) As we know :
 l R = ρ A
Where;
 R = Resistance of conductor ρ = Electrical resistivity of copper l = Length of conductor A = Area of cross-section of conductor
 l R ∝ A
 l =>    R ∝ Πr2
 l =>    R ∝ r2
From the above, it is clear that as the radius of cross sectional area of a conductor increases, the resistance of conductor decreases and vice-versa

(e) Certain elements like Silver, Copper and Aluminum are the best conductors of electricity due to theirs low electric resistivity. Hence, voltage drop and heating loss due to resistance against electric current is minimum. Silver being too costly, Copper and Aluminum wires are usually employed for electricity transmission

Intext Questions|Page 200 |Chapter 12. Electricity | CBSE Class 10th Science

Question 1.

Question 1.What does an electric circuit mean?

Answer : An electrical circuit essentially consists of load devices, connected together by conducting material such as wire and electric power source. The naure of load devices, depending upon the application, may be either Resistive or Inductive or Capacitive .

Question 2. Define the unit of current.

Answer : The rate of flow of electric charges, flowing through a particular area of a conductor is called Electric current. S.I. unit of Electric current is ampere (A). One ampere of current represents a flow of one coulomb of charge, passing through a particular area of a conductor, per second. Coulomb is the SI unit of electric charge and one coulomb (C) of charge is equivalent to the charge contained in nearly 6 × 1018 electrons

Question 3.Calculate the number of electrons constituting one coulomb of charge.

We know that an electron possesses a negative charge of 1.6 × 10–19 C
Let 'n' be the number of electrons present in one coulomb of charge

Number of electrons in Coulomb × Charge per Electron = One coulomb of charge

∴ n ×1.6 × 10–19 C = 1 C

 1 1019 100×1018 n = = = 1.6 × 10–19 1.6 16

n = 6.25 × 1018 electrons

Intext Questions|Page 202 |Chapter 12. Electricity | CBSE Class 10th Science

Question 1. Name a device that helps to maintain a potential difference across a conductor.

Answer : Electric cell is the device that helps to maintain a potential difference across a conductor. The chemical action within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When the cell is connected to a conducting circuit element, the potential difference sets the charges in motion in the conductor and produces an electric current

Question 2. What is meant by saying that the potential difference between two points is 1 V?

Answer : One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other. Volt (V) is the SI unit of electric potential difference

Question 3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer : We define the electric potential difference (V) between two points in an electric circuit carrying some current as the work (W) done or Energy in joule to move a unit charge(Q) from one point to the other –

Potential difference (V) between two points = Work done (W)/Charge (Q)
V=
 W Q

W = V×Q

W = 6×1

W = 6 joules

Intext Questions|Page 209 |Chapter 12. Electricity | CBSE Class 10th Science

Question 1. On what factors does the resistance of a conductor depend?

Answer : Resistance of the conductor depends on following factors :

(i) Length of conductor (l)
: Resistance of a uniform metallic conductor is directly proportional to its length

R ∝ l

(ii) The Cross-section area of the conductor (A) :
Resistance of a uniform metallic conductor is inversely proportional to the area of cross-section (A).

R
 1 A
(iii) Nature of conducting material or its electrical resistivity (ρ):
Different materials exhibit different amount of electrical resistivity. It is denoted by ρ (rho) and also know as constant of proportionality.
R =ρ
 l A

Question 2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer : As we know :
Current (I)
 1 R

Also
R
 1 A

 ∴ Current (I) ∝ A (Area of cross-section)
The current is inversely proportional to Resistance and further Resistance of a uniform metallic conductor is inversely proportional to the area of cross-section (A). Which means more the area of cross section, lesser will be the resistance and therefor more will be the current flow. Hence, current Will flow more easily through a thick wire than a thin wire of the same material, when connected to the same source

Question 3. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer : As we know :
V (Potential difference) = Current (I) × Resistance (R)

As the Resistance (R) is constant.

When V (Potential difference) decreases to half of its former value

ILater=
 VLater R
ILater=
 VFormer 2R
=
 IFormer 2
Clearly, Resistance being constant, as potential difference decreases to half of its former value, the value of current (I) will also be reduced to half.

Question 4.Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer : The resistivity of an alloy is generally higher than that of its constituent pure metals. Higher resistance results in increased electrical heating. Alloys do not oxidise (burn) easily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc

Question 5. Use the data in Table 12.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

Answer : (a) Iron is a better conductor than mercury as its Electrical resistivity is less (10.0 × 10–8Ω m), whereas that of mercury is very high.( 94.0 × 10–8 Ω m)

(b) Silver is the best conductor of electric current because of its Electrical resistivity is 1.60 × 10–8 Ω m, which is least of all other metals.

Friday, 21 February 2014

CBSE Class IX (9th) Science | Chapters Exercises Solutions

Chapter 1. Matter In Our Surroundings | CBSE Class IX (9th) Science | Solved Exercises

Chapter 2. Is Matter Around Us Pure ? | CBSE Class IX (9th) Science | Solved Exercises

Chapter 3. Atoms And Molecules | CBSE Class IX (9th) Science | Solved Exercises

Chapter 4. Structure Of The Atom | CBSE Class IX (9th) Science | Solved Exercises

Chapter 5. The Fundamental Unit Of Life | CBSE Class IX (9th) Science | Solved Exercises

Chapter 6. Tissues | CBSE Class IX (9th) Science | Solved Exercises

Chapter 7. Diversity in Living Organisms | CBSE Class IX (9th) Science | Solved Exercises

Chapter 8. Motion | CBSE Class IX (9th) Science | Solved Exercises

Chapter 9. Force Aand Laws Of Motion | CBSE Class IX (9th) Science | Solved Exercises

Chapter 10. Gravitation | CBSE Class IX (9th) Science | Solved Exercises

Chapter 11. Work And Energy | CBSE Class IX (9th) Science | Solved Exercises

Chapter 12. Sound | CBSE Class IX (9th) Science | Solved Exercises

Chapter 13. Why Do We Fall Ill | CBSE Class IX (9th) Science | Solved Exercises

Chapter 14. Natural Resources | CBSE Class IX (9th) Science | Solved Exercises

Chapter 15. Improvement In Food Resources | CBSE Class IX (9th) Science | Solved Exercises

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Science : Sample Paper CBSE Class IX-Term I (First Term September)

Science : Sample Paper CBSE Class IX  -Term II (Second Term March)
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Solved Exercises | Chapter 11. The Human Eye and the Colourful World |Science | CBSE Class 10th

Question 1. The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia.
(b) accommodation.
(c) near-sightedness.
(d) far-sightedness.

Question 2. The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.

Question 3. The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm.
(c) 25 cm.
(d) 2.5 m.

Question 4. The change in focal length of an eye lens is caused by the action of the
(a) pupil.
(b) retina.
(c) ciliary muscles.
(d) iris.

Question 5. A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Answer. (i) In case of distant vision
 1 P = f 1 f = P 1 f = − 5.5 10 f = − 55 2 f = − 11 2 f = − 11 f = − 0.181 m f = − 18.1 cm

(ii) In case of near vision.
 1 P = f 1 f = P 1 f = − 1.5 10 f = − 15 2 f = 3 f = 0.667 m f = 66.7 cm

Question 6. The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Answer. A concave lens is required to correct a myopic vision .
Also, given, far point of a myopic person, f = 80 cm = 0.8 m
 1 Power = − (in metre) Far point (f) 1 P = − D 0.8 10 P = − D 8 f = 1.25 D

Question 7. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

As we know that hypermetropia is corrected by using a convex lens.
Here, The near point of a hypermetropic eye is 1 m and the near point of the normal eye is 25 cm.
Therefor, v= − 1 metre = − 100 cm and u= −25 cm
As per lens formula :
 1 1 1 − v u f 1 1 1 − − 100 − 25 f − 1 + 4 1 − 100 f 100 f = cm 3 1 f = m 3 1 P = f 1 × 3 P = Dioptre 1 P = + 3.0 D

Question 8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer. The eye lens is composed of a fibrous, jelly-like material. To see closure and distant object, Its focal length can me changed to certain extent by contraction and relaxation of ciliary muscles. The ability of the eye lens to adjust its focal length by contracting and relaxing its ciliary muscles is called accommodation. To see closer object, ciliary muscles, must decrease the focal length of eye lens. However, the focal length of the eye lens cannot be decreased below a certain minimum limit. To see an object comfortably and distinctly, the distance of object should be at least 25 cm from the eyes. Due to this, a normal eye is not able to see clearly the objects placed closer than 25 cm. This distance at which objects can be seen is clearly, is called the least distance of distinct vision or the near point of the eye.

Question 9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer. There is no change to the image distance in the eye when we increase the distance of an object from the eye. To see closer or distant object clearly, the eyes, due to its ability of accommodation, can increase or decrease focal length of its lens, so that the image is always formed at retina. The eye lens is composed of a fibrous, jelly-like material. To see closer or distant objects, focal length of its lenses can be increased or decreased to some extent by the contraction and relaxation of the ciliary muscles.

Question 10. Why do stars twinkle?

Answer. The earth atmosphere consists of layers of different gases of varying density. Since the physical conditions of the earth’s atmosphere are not stationary. The refractive index of atmosphere goes on changing continuously. As the starlight, enters the earth’s atmosphere, it undergoes refraction multiple times before it reaches the eyes on earth. This makes the apparent position of the star changing slightly and continuously. Since the stars are very distant, they approximate point-sized sources of light. As the path of rays of light coming from the star goes on varying slightly, the apparent position of the star fluctuates and the amount of starlight entering the eye flickers – the star sometimes appears brighter, and at some other time, fainter, Hence the twinkling of stars.

Question 11. Explain why the planets do not twinkle.

Answer. The planets are much closer to the earth, and are thus seen as extended sources. If we consider a planet as a collection of a large number of point-sized sources of light, the total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying the twinkling effect. Hence, the planets do not twinkle

Question 12. Why does the Sun appear reddish early in the morning?

Answer. As we know, the Sun light consists of seven colours. Due to the presence of molecules of air and other fine particles in the atmosphere, scattering of visible light into different colours takes place. The colours with shorter wavelengths such as blue are scattered easily than colours with longer wavelengths such as Red. In early morning, as the Sun rises above horizon, light from the Sun has to cover a larger distance in the denser medium of earth’s atmosphere before reaching our eyes. The blue light, due to shorter wavelength get scattered in midway and not visible in the sky. Only Red light having larger wavelength is able to reach our eyes due to less scattering. Hence, the Sun appear reddish early in the morning

Question 13. Why does the sky appear dark instead of blue to an astronaut?

Answer. As an astronaut moves far away from the earth atmosphere and fly at very high altitude in the space,  the medium becomes rarer.   Due to the absence of air molecules or fine particles, there is hardly any scattering of light. That is why, the sky appears dark to an astronaut flying at very high altitudes, as scattering of light is not prominent at such heights.

Sunday, 1 December 2013

Chapter 10. Light – Reflection and Refraction | Science | CBSE Class 10th |Solved Exercises

Solved Exercises | Chapter 10. Light – Reflection and Refraction | Science | Class 10th |CBSE

Question 1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay

Question 2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.

Answer. (d) Between the pole of the mirror and its principal focus.

Question 3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.

Question 4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.

Question 5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane.
(b) concave.
(c) convex.
(d) either plane or convex.

Answer. (d) either plane or convex.

Question 6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.

Answer. (c) A convex lens of focal length 5 cm.

Question 7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer. (a). The range of distance of the object from the mirror should be less than 15 cm i.e. from 0 to 15 cm in the front of mirror from the pole.
(b). The nature of image so formed will be virtual and erect.
(c)  The size of image will be larger than object

Question 8. Name the type of mirror used in the following situations.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.

Answer. (a) In headlights of a car, type of mirror used is concave as the light of the lamp, under goes divergence from reflector surface and cover a large area in the front.
(b) Side/rear-view mirror of a vehicle is a convex mirror as it gives a diminished, Virtual and an erect image of the side or rear with wider field of view . A convex mirrors enable the driver to view much larger area than would be possible with a plane mirror
(c) Solar furnace is a concave mirror as Sun rays after reflection from its surface, get converged at focus with much intense heat.

Question 9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer. Yes, the half covered lens will still produce a complete image of the object but the image so formed may not be as intense as formed with uncovered lens. In fact, parts or broken pieces of lens behave like a complete lens and form complete image.
Verification : Take a convex lens. Light a candle. Now form an image of a burning candle on a white surface on the other side of lens by adjusting the distance between lens and candle. We can observe, a complete real and an inverted image of candle. Now cover half of lens with black paper, and try to form an image now. We can observe that again a complete, real and an inverted image of candle is formed. Image formed was less intense from earlier.

Question 10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer. Height of the object h = + 5 cm
Focal length f = + 10 cm
object-distance u = –25 cm
Image-distance v = ?
Height of the image h′ = ?

 1 1 1 − = v u f => 1 1 1 = − v 10 25 => 5 − 2 3 = = 50 50 => 50 v = = 16.66 cm 3
A real and inverted image will be formed on other side of lense at 16.66 cm from its optical centre
 => v m = u => h2 16.66 = h1 − 25 => h2 16.66 = 5 − 25 => −16.66 × 5 h2 = 25 => h2 = − 3.33 cm

An inverted, 3.33 cm high image will be formed.

Question 11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer. Focal length f = − 15 cm
Image-distance v = − 10 cm
object-distance u = ?

 1 1 1 − = v u f 1 1 1 − = − 10 u − 15 − 1 1 1 + = 10 15 u => 1 − 3 + 2 = u 30 => u = − 30 cm

Object is placed at a distance of 30 cm from concave lens

Question 12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Answer. Focal length f = 15 cm
object-distance u = –10 cm
Image-distance v = ?
 1 1 1 + = v u f 1 1 1 + = v − 10 15 1 1 1 = + v 15 10 1 2+3 5 = = v 30 30 v = 6 cm

A virtual and erect image is formed 6 cm behind the mirror.

Question 13. The magnification produced by a plane mirror is +1. What does this mean?

Answer. The magnification produced by a plane mirror is +1 implies that the image formed by a plane mirror is virtual , erect and of the same size, as that of object.

Question 14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Radius of curvature, R = + 30 cm
Focal length, f = R/2 = + 30/2 cm = + 15 cm
Object-distance, u = – 20 cm
Height of object
h1′= 5 cm
Image-distance, v = ? Height of image h2′= ?

 1 1 1 + = v u f 1 1 1 + = v − 20 + 15 1 1 1 = + v 15 20 1 4 + 3 7 = = v 60 60 60 v = = 8.57 cm 7
The image is formed behind the mirror at a distance of 8.6 cm
 h2 − v m = = h1 u h2 8.57 => = = 5 cm 20 8.57 × 5 cm = h2 = 20 Height (Size ) of Image = h2 = 2.175 cm

Thus, a 2.175 cm high, virtual and erect image is formed

Question 15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Answer. Focal length, f = - 18 cm
Object-distance, u = – 27 cm
Height of object, h1′= 5 cm
Image-distance, v = ?
Height of image h2′= ?
 1 1 1 + = v u f 1 1 1 + = v − 27 -18 15 − 1 1 1 = + v 18 27 1 − 3 + 2 1 = = v 54 54 60 v = = 54 cm 7
The screen should be kept at a distance of 54 cm in front of mirror
 h2 − v m = = h1 u h2 (− 54 ) => = = 7 cm (− 27 ) (− 54 ) × 7 cm => h2 = (− 27 ) Height (Size ) of Image = h2 = 2 × 7 cm = 14 cm cm

Thus, a 14 cm high, virtual and an inverted image is formed

Question 16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?

 1 P = f 1 - 2.0 = f −1 f = m 2 − 1 f = × 100 cm 2 f = − 50 cm = - 0.50 cm
The lens is a concave lens

Question 17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer. Power of lens, P = + 1.5 D
 1 P = f 1 1.5 = f 1 f = m 1.5 10 f = m 15 10 f = m 15 2 f = m = + 0.67 m 3

Focal length of the lens is + 0.67 m. The prescribed lens is converging type in nature.

Additional Questions | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science

Question 1. Write the mirror formula :

Answer. In mirror, the distance of the object from its pole is called the object distance (u)
The distance of the image from the pole of the mirror is called the image distance (v)
The distance of the principal focus from the pole is called the focal length (f).
There is a relationship between these three quantities given by the mirror formula which is expressed as :
 1 1 1 + = v u f

Question 2. Write one use of concave and convex mirrors each.

Answer. Concave mirror is used in some telescopes and also as magifying looking mirror while applying make-up or shaving. Convex mirror is used in vehicles as rear view mirror.

Question 3. Why do we use convex for side-view ?

Answer. Convex mirrors are used in vehicles for side-view because they give a virtual, upright, though diminished, image. Although images so formed are smaller but this results in showing a large area in the back drop with a wider field of view .

Question 4. How does a light ray bend when it travels from :
(i) A denser to rarer medium ?
(ii) Ararer to denser medium >

Answer.An optically denser medium has a larger refractive index, where as optically rarer medium has a lower refractive index. Due to refration, the speed of light is higher in a rarer medium than a denser medium also direction of propagation of light chages as it chages the medium. . (i) When light travels from a denser medium to a rarer medium, it speeds up and bends away from the normal.
(ii) When light travels from a rarer medium to a denser medium, it slows down and bends towards the normal.

Question 5. When a convex lens is focussed on a distant object, where will the image be formed ? Show it with a ray diagram.

Answer. When a convex lens is focussed on a distant object, the image will be formed at the focus of the lens.

Question 6. What is the meaning of
(i) Optical Centre
(ii) Principal Axis

Answer. (i) Optical Centre : The central point of a lens is its optical centre. It is usually represented by the letter O. A ray of light through the optical centre of a lens passes without suffering any deviation.
(ii) Principal Axis : An imaginary straight line passing through the two centres of curvature of a lens is called its principal axis. Optical centre and focus of lens lies on the Principal Axis.

Question 7. When does a convex lens form
(i) A virtual, erect, enlarged image ?
(ii) A real enlarged image ?

Answer. (i) A convex lens forms a virtual, erect and enlarged image, when the object is placed between focus and optical centre of a lens on its other side.
(ii) A convex lens forms a real and enlarged image when the object is placed between focus (f) and centre of curvature (2f) of a lens on its other side.

Question 8. What is the relationship between the focal length of a spherical mirror and radius of curvature ?

Answer. The focal length of a spherical mirror (f) is equal to half its radius of curvature (R) i.e.
Focal length, f = R/2

Question 9. Explain the term Magnification produced by a spherical mirror ?

Answer. Magnification produced by a spherical mirror gives the relative extent to which the image of an object is magnified with respect to the object size. It is expressed as the ratio of the height of the image to the height of the object. It is usually represented by the letter m.
If h1 is the height of the object and h2 is the height of the image, then the magnification m produced by a spherical mirror is given by
 Height of the image ( h1) m = Height of the object (h2 ) h1 m = h2
The magnification m is also related to the object distance (u) and image distance (v). It can be expressed as:
 h1 v Magnification (m) = = h2 u

Question 10. Name and explain the sign Convention for Reflection by Spherical Mirrors

Answer. While dealing with the reflection of light by spherical mirrors, we follow a set of sign conventions called the New Cartesian Sign Convention. In this convention, the pole (P) of the mirror is taken as the origin. The principal axis of the mirror is taken as the x-axis (X’X) of the coordinate system. The conventions are as follows –
(i) The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.
(ii) All distances parallel to the principal axis are measured from the pole of the mirror.
(iii) All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along – x-axis) are taken as negative.
(iv) Distances measured perpendicular to and above the principal axis (along + y-axis) are taken as positive.
(v) Distances measured perpendicular to and below the principal axis (along –y-axis) are taken as negative.

Question 11. Explain the following given terms
(i) Refraction of light
(ii) Laws of refraction of light
(ii) The Refractive Index

(i) Refraction of light : The direction of propagation of light, When traveling obliquely from one medium to another is subject to change. When light travels from a denser medium to a rarer medium, it speeds up and bends away from the normal. When light travels from a rarer medium to a denser medium, it slows down and bends towards the normal. This phenomenon of bending of light ray is known as refraction of light. Refraction is caused due to change in the speed of light as it enters from one transparent medium to another. The speed of light increases in rarer medium and decreases in denser medium.

(ii) Laws of refraction of light : The reflecting surfaces, of all types, obey the laws of reflection. The refracting surfaces obey the laws of refraction.
The following are the laws of refraction of light.
(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
(ii) The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is also known as Snell’s law of refraction.
If i is the angle of incidence and r is the angle of refraction, then,
 sin i = constant sin r

(ii) The Refractive Index : The refractive index of a transparent medium is the ratio of the speed of light in vacuum to that in the medium. It is also termed as the absolute refractive index of a given transparent medium .
If c is the speed of light in air and v is the speed of light in the medium, then, the refractive index of the medium nm is given by
 Speed of light in air c nm = = Speed of light in the medium v

Intext Questions | Page 168 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science

Question 1. Define the principal focus of a concave mirror.

Answer. The principal focus of the concave mirror is an point on the principal axis of the mirror, where incident rays of light, parallel to the principal axis, after reflection from mirror surface, intersect each other .

Question 2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer. For spherical mirrors of small apertures, the radius of curvature is twice the focal length i.e. R = 2f . Here radius of curvature R = 20 cm, focal length f = ?
 R 20 f = = = 10 cm 2 2

Focal length of mirror is 10 cm

Question 3. Name a mirror that can give an erect and enlarged image of an object.

Answer. Concave mirror gives an erect and enlarged image of an object, when the object is between pole (P) and principal focus of mirror (C).

Question 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer. Convex mirrors are commonly used as rear-view (wing) mirrors in vehicles because they give an erect, virtual, full size diminished image of distant objects with a wider field of view. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror.

Intext Questions | Page 171 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science

Question 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer.We know,for spherical mirrors of small apertures, the radius of curvature is twice the focal length i.e. R = 2f .
Given here, radius of curvature R is 32 cm. and focal length of a convex mirror f = ?
 R 32 f = = = 16 cm 2 2

Focal length of mirror is 16 cm

Question 2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Answer. Here given Magnification m = 3, Object-distance u = 10 cm
 v Magnification m = − Real image u v − 3 = − u 3 u = v v = 3u = 3 × − 10 cm = − 30 cm

The image will be formed at a distance of 30 cm in front of convex mirror from its the pole

Intext Questions | Page 176 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science

Question 1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Answer. The light ray bends towards the normal as it travels from a rarer medium of air to a denser medium of water, under goes refraction. Refraction is due to change in the speed of light as it enters from one transparent medium to another. The speed of light increases in rarer medium and decreases in denser medium

Question 2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m s–1.

Answer. Given, speed of light in vacuum C = 3 × 108 m s–1
Refractive index of glass ng = 1.50
Speed of light in the glass vg = ?
 C ng = Vg 3 × 10 8 ng = Vg 3 × 10 8 1.5 = Vg 3 × 10 8 Vg = 1.5 Vg = 2 × 108 ms−1

Speed of light in the glass vg = 2 × 108 ms−1

Question 3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer.The medium having highest optical density : Diamond ( Refractive Index 2.42 )
The medium having lowest optical density : Air ( Refractive Index 1.0003 )

Question 4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.

Answer. Using the information given in table, the refrative index of kerosene is 1.44, that of turpentine is 1.47 and that of water is 1.33. Clearly, water having lower refractive index 1.33, is optically rarer than kerosene and turpentine. Therefor the light travels fastest in water because of its lower optical density

Question 5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer. The diamond with refractive index of 2.42, is the most ‘optically denser medium' and there for the speed of light in diamond will be less i.e. 1.23 × 108 ms−1 (= 3 × 108 ms−1/2.42 ) compare to speed of light in vacuum C = 3 × 108 m s- 1

Intext Questions | Page 184 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science

Question 1. Define 1 dioptre of power of a lens.

Answer. 1 dioptre is SI unit of the power of a lens whose focal length is 1 metre. It is denoted by the letter D. Thus 1D = 1 m-1. Simply, when focal length ' f ' is expressed in metres, then, power is expressed in dioptres. The power of a convex lens is positive and that of a concave lens is negative.

Question 2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

 v = 50 m = 1

 1 P = = ? f

As the image formed by lens is real

 v m = u v 1 = u v = u u = − 50 cm

∴     Object distance is = 50 cm
 1 1 1 − = v u f 1 1 1 − = 50 − 50 f
 1 + 1 1 = 50 f 2 f = 50 f = 25 cm = .25 m

 1 100 P = + = + = + 4 dioptre .25 25

The power of convex lens is + 4 dioptre

Question 3. Find the power of a concave lens of focal length 2 m.

 Focal length of lens f = − 2 m Power of concave lens P = ?

 1 P = − f 1 P = − 2

 P = − 0.5 dioptre

The power of concave lens is − 0.5 dioptre

Activity 10.1 |Page 161 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Take a large shining spoon. Try to view your face in its curved surface.
• Q.1. Do you get the image? Is it smaller or larger?
• Answer : Yes, the image of the face formed on outer curved surface is smaller in size.
• Q.2. Move the spoon slowly away from your face. Observe the image. How does it change?
• Answer : The size of image gradually decreases with a increase in field of view.
• Q.3. Reverse the spoon and repeat the Activity. How does the image look like now?
• Answer : Earlier, when the spoon was close the image formed on the inner curved surface was erect and magnified and as we moved the spoon slowly away from our face,the image transitioned to a inverted image with gradual decrease in its size.
• Q.4. Compare the characteristics of the image on the two surfaces.
•  Outer Surface Inner Surface (i) Image is always erect (ii) Image size is gradually decreases as we move away the spoon (i) The image is erect when spoon is close and inverted when spoon is away (ii) Image size is larger when spoon is close and it is smaller when spoon is moved away
• Q. 5. Why do we see our image in the shining spoon?
• Answer : The surface of a shining spoon acts like a mirror. Due to reflection of light from its surfaces, we can see our image
• Q. 6. What types of mirrors are formed by the inner and outer curved surfaces of a spoon ?
• Answer : The inner curved surfaces of a spoon forms a concave mirror and the outer curved surfaces of a spoon forms a convex mirror

Activity 10.2 | Page 162 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science

CAUTION: Do not look at the Sun directly or even into a mirror reflecting sunlight. It may damage your eyes.
• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
• Q.1. Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?
• Answer : The light rays from the Sun, are concentrated at focus, and form a sharp spot of light on the sheet. Paper starts burning after some time due to increase intensity of reflected sun light from the mirror
• Q.2. Why, we should not look at the Sun directly or even into a mirror reflecting sunlight
• Answer : Because the intense heat resulting from the concentrated sun light through eye lens may burn the retina wall with dark spots. This may result in partial or complete vision impairment.

Activity 10.3 | Page 163 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science

You have already learnt a way of determining the focal length of a concave mirror. In Activity 10.2, you have seen that the sharp bright spot of light you got on the paper is, in fact, the image of the Sun. It was a tiny, real, inverted image. You got the approximate focal length of the concave mirror by measuring the distance of the image from the mirror.
• Take a concave mirror. Find out its approximate focal length in the way described above. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
• Mark a line on a Table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
• Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror. These lines will now correspond to the positions of the points P, F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus F lies mid-way between the pole P and the centre of curvature C.
• Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
• Observe the image carefully. Note down its nature, position and relative size with respect to the object size.
• Repeat the activity by placing the candle – (a) just beyond C, (b) at C, (c) between F and C, (d) at F, and (e) between P and F.
• In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Then, look for its virtual image in the mirror itself.
• Q. 1. Note down and tabulate your observations.
• Answer : Image formation by a concave mirror for different positions of the object
 Position of the object Position of the image Size of the image Nature of the image At infinity At the focus F Highly diminished,point-sized Real and inverted Beyond C Between F and C Diminished Real and inverted At C At C Same size Real and inverted Between C and F Beyond C Enlarged Real and inverted At F At infinity Highly enlarged Real and inverted Between P and F Behind the mirror Enlarged Virtual and erect
• Q. 2. What is the relation between focal length and radius of curvature ?
• Answer : Radius of curvature is equal to twice the focal length. We put this as R = 2f
• Q. 3. What is the nature of image formed when object is placed far beyond C ?
• Answer : A is real, inverted and highly dimnished image is formed at focus.
• Q. 4. What is the nature and position of image formed when object is placed just beyond C ?
• Answer : A is real, inverted and dimnished image is formed between Focus (F) and centre of curvature (C).
• Q. 5. Draw the ray diagram of formation of image when object is kept at center of curvature in front of a concave mirror.
• Q. 6. In which case the image formed by concave mirror is not obtained on the screen ?
• Answer : The image formed by concave mirror is not obtained on the screen, when the object is kept between focus (f) and pole (P) of a concave mirror.
• Q. 7. In which case the image of an object by concave mirror can be obtained on the screen ?
• Answer : The image of an object by concave mirror can be obtained on the screen, when the object is kept beyond focal length of the mirror.

Activity 10.4 | Page 166 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Q. 1. Draw neat ray diagrams for each position of the object shown in Table 10.1.
 Position of the object Ray diagram At infinity Beyond C At C Between C and F At F Between P and F
• You may take any two of the rays mentioned in the previous section for locating the image.
• Compare your diagram with those given in Fig. 10.7 of textbook.
• Answer : They were identical and matching
• Describe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.
 Position of the object Position of the image Size of the image Nature of the image At infinity At the focus F Highly diminished,point-sized Real and inverted Beyond C Between F and C Diminished Real and inverted At C At C Same size Real and inverted Between C and F Beyond C Enlarged Real and inverted At F At infinity Highly enlarged Real and inverted Between P and F Behind the mirror Enlarged Virtual and erect

Activity 10.5 | Page 167 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Take a convex mirror. Hold it in one hand.
• Hold a pencil in the upright position in the other hand.
• Q. 1. Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
• Answer : The image is erect and diminished
• Q. 2. Move the pencil away from the mirror slowly. Does the image become smaller or larger?
• Answer : The image becomes smaller.
• Repeat this Activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the mirror?
• Answer : The image will move closer to the focus

Activity 10.6 | Page 167-68 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Observe the image of a distant object, say a distant tree, in a plane mirror.
• Q. 1. Could you see a full-length image?
• Answer : No, We can not see a full-length image of a distant object in a plain mirror.
• Q. 1. Try with plane mirrors of different sizes. Did you see the entire object in the image?
• Answer : No, the result was same as before.
• Q. 4. Repeat this Activity with a concave mirror. Did the mirror show full length image of the object?
• Q. 4. Now try using a convex mirror. Did you succeed? Explain your observations with reason.
• Answer : Yes, now we could see full length image of distant object with wider field of view. The image formed was diminished, erect and virtual. Due to this reason, they are used as rear or side view mirrors in vehicles as distant objects in the backdrop can be seen clearly with much larger field of view .

Activity 10.7 | Page 172 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Place a coin at the bottom of a bucket filled with water.
• Q. 1. With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
• Q. 2. Repeat the Activity. Why did you not succeed in doing it in one go?
• Answer : Because on seeing, the coin appeared to be closer than its actual distance, so we are likely to miss the coin. Reflected light coming from the submerged coin in denser medium of water, on entering air which is a rarer medium, bend away from the normal due to refraction of light and image size becomes larger than its actual size, thus submerged object apparently seem closer.

Activity 10.8 | Page 172 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Place a large shallow bowl on a Table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.
• Q. Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?
• Answer : Yes, on pouring water into the bowl, the coin becomes visible again because due to refraction of light, for our eyes, the submerged coin apparently seems raised above its actual level and thus becomes visible on seeing from the same side and distance

Activity 10.9 | Page 172 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Draw a thick straight line in ink, over a sheet of white paper placed on a Table.
• Place a glass slab over the line in such a way that one of its edges makes an angle with the line.
• Q. 1. Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges?
• Answer :Yes, due to the refration of light, the line under the glass slab appear to be bent at the edges
• Q. 2. Next, place the glass slab such that it is normal to the line. What do you observe now? Does the part of the line under the glass slab appear bent?
• Answer : No, Now the part of the line under the glass slab appear in a straight line. Because a ray of light, which is perpendicular to the plain of a refracting medium, does not change its angle due to refraction.
• Q. 3. Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?
• Answer : Yes, the part of the line, beneath the slab, appear to be raised. Due to refraction of light, apparent position of image of object seems nearer than its actual position.

Activity 10.10 | Page 173 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O′.
• Join O and O′. Also produce EF up to P, as shown by a dotted line in Fig. 10.10.

Question 1. What happens to incident ray as it enters the glass slab ?

Answer : The incident ray as it enters from a rarer medium of air to a denser medium of glass, bends towards the normal, due to refraction of light . The refraction is cuased by change in the speed of light as it enters from one transparent medium to another.

Question 2. What happens to emergent ray as it leaves the glass slab ?

Answer : The emergent ray as it leaves the denser medium of glass and enters into a rarer medium of air, bends away from the normal, due to refraction of light. The refraction is caused by change in the speed of light as it enters from one transparent medium to another.

Question 3. What is the perpendicular distance between directions of incident and emergent rays ?

Answer : Lateral displacement. This gives a measure of path deviation of refracted rays due to refraction.

Question 4. As given in the activity above, the medium of incidence and emergent ray is same (air), what could be the possible observations for angle of incidence and angle of emergence ?

Answer : When the medium of incidence and emergent ray is same (air), angle of incidence is equal to angle of emergence.