ELECTRICITY - III : LONG ANSWERS

GIVE ANSWER IN DETAIL

*INDEX-TOPIC SEARCH

1. EXPLAIN ELECTRIC CHARGE
   

   ANS:
   

   
   
   • A comb used for combing dry hair picks up small pieces of paper brought near it because it acquires electric charge.
     
   • Similarly, a glass rod rubbed with silk cloth acquires positive charge.
     
   • An electric charge is a fundamental property associated with protons and electrons.In addition to mass, protons and electrons possess electrical charge.
     
   • Electric charge is measured in coulomb(C).
     
   • The magnitude of charge on both protons and electrons is same, i.e. 1.6 x 10^(-19) coulomb.We use negative(-) sign for electron (as it has negative charge) and positive(+) sign for proton (as it has positive charge).
     
   • Coulomb's law describes the interaction between two charges and gives the magnitude of the force between them.
     

   
   

   
   
2. DEFINE ELECTRIC CURRENT AND EXPLAIN DIFFERENCE BETWEEN ELECTRON CURRENT AND CONVENTIONAL CURRENT
   

   ANS:
   

   
   
   • Electric current is the net amount of charge that passes through the cross-sectional area of the conductor per unit time.
     
   • The charge will flow if the electrons move. In fact, electric current is the flow of electrons.
     
   • Electrons in an atom revolve around nucleus(protons) in different orbits.
     
   • In metals the electrons in the outermost orbit are relatively free to move and are called FREE ELECTRONS. These 'free' electrons contribute to the flow of charge(electric current).
     
   • Metals like silver, copper, aluminium, etc. possess maximum free electrons and are good conductors of electricity.
     
   • If we provide energy to these free electrons, we get flow of electrons in one direction which we call current.
     
   • The flow of electrons is from cathode(negative terminal) to anode(positive terminal).
     
   • The conventional current ,however, is shown as flow from anode to cathode as initially it was believed that the positive charge is responsible for electric current.
     

   
   

   
   
3. EXPLAIN : ELECTRIC POTENTIAL AND ELECTRICAL POTENTIAL DIFFERENCE
   

   ANS:
   

   
   
   • The work done on a charge is stored in it as potential energy.This work done (potential energy) is represented by electric potential.
     
   • The electrical potential energy per unit charge at a point in an electric field is called the ELECTRICAL POTENTIAL at that point.
     
   • Electrical Potential = potential energy/charge = work done/charge
     
     V = W/Q
     
     The unit of electric potential is joule/coulomb and it is called volt in the memory of Volta.
     
   • To establish electric current in a conductor we need flow of charge(free electrons).
     
   • The electrical potential difference across the conductor causes the flow of charge(current).It is the driving force that causes current.
     
   • The charges flow from the end at a higher potential to the end at lower potential.
     
   • To create potential difference a cell(source of energy) is used.
     
   • Electrical potential difference between two points A and B on a conductor is the work done to move a unit charge from A to B.
     
   • Electrical potential difference is generally called voltage and its unit is volt(V).
     

   
   

   
   
4. EXPLAIN THE CONSTRUCTION AND WORKING OF A SIMPLE CELL (VOLTAIC CELL)
   

   ANS:
   
   

   
   
   • Volta's cell(Voltaic cell) is a simple cell which converts chemical energy into electrical energy.
     
   • As shown in figure, a simple voltaic cell can be made by taking some dilute sulphuric acid in a glass container.Two metallic plates, one of zinc and the other of copper are used as electrodes.
     
   • When we join two electrodes by a conducting wire and place a lamp in between, we find that the lamp glows as electric current flows.
     
   • When two plates are placed in dilute sulphuric acid, a chemical reaction takes place between the plates and the acid. Zinc plate gets negatively charged and the copper plate gets positively charged.Thus zinc behaves as cathode and copper as anode.
     
   • A potential difference is created between the two plates due to this and current flows.
     
   • Since a chemical reaction causes current, in Volta's cell, chemical energy is converted to electrical energy.
     

   
   

   
   
5. DESCRIBE OHM'S EXPERIMENT AND ITS CONCLUSION
   

   ANS:
   
   AIM:To study the relation between voltage(V) and current(I).
   
   APPARATUS:Conducting wire, ammeter, voltmeter, resistors, battery, key.
   
   
   
   PROCEDURE:Connect various apparatus in a circuit as shown in figure.The ammeter measures current and the voltmeter measures voltage(V) across the resistor(R).
   
   Press the key(K) and adjust the battery such that the voltmeter shows voltage of 1 volt.Record the value of current(I) as shown by the ammeter.
   
   Record different values of current obtained for voltage of 2V, 3V, 4V, 5V,etc.
   
   Record all the observations in the table as shown.
   
   Calculate the ratio of V and I for each observation and record it in the last column of the table. This gives the value of resistance(R).
   
   
   
   Plot the graph of V v/s I as shown.
   
   
   
   CONCLUSION:
   

   
   
   1. The current(I) increases linearly with voltage(V),i.e, I α V.
      
   2. The value of V/I (i.e. R) is constant for all observations. R is called resistance of the conductor.
      
   3. The graph of V v/s I is a straight line passing through the origin.
      

   
   
   Based on above experiment,Ohm formulated the following law:The current passing through a conductor is directly proportional to the potential difference(voltage) across the conductor.
   

   
   
6. EXPLAIN SERIES CONNECTION OF RESISTORS AND DERIVE THE FORMULA FOR EQUIVALENT RESISTANCE
   

   ANS:
   
   

   
   
   • When the resistors are connected end-to-end, we get series connection.
     
   • For series connection of resistors the current flowing through each resistor is the same as the overall current of the circuit.
     
   • The overall voltage of the circuit, however, gets divided among the resistors according to their magnitudes.
     
     
   • Consider that three resistors R₁, R₂ and R₃ are connected in series in a circuit having overall potential difference(voltage) V and current I.
     
   • If V₁, V₂ and V₃ are voltage drops across R₁, R₂ and R₃, respectively, then
     
     V₁ = IR₁, V₂ = IR₂, and V₃ = IR₃
     
     
   • Similarly, if R is the equivalent resistance, then
     
     V = IR.
     
     
   • Now,
     
     V = V₁ + V₂ + V₃
     
     ∴ IR = IR₁ + IR₂ + IR₃
     
     ∴ R = R₁ + R₂ + R₃
     
     Thus, equivalent resistance(R) of the resistors(R₁, R₂, R₃) connected in series is the algebraic sum of all the resistances.
     

   
   

   
   
7. EXPLAIN PARALLEL CONNECTION OF RESISTORS AND DERIVE THE FORMULA FOR EQUIVALENT RESISTANCE
   

   ANS:
   
   

   
   
   • When resistors are connected between two common points, all having same connecting point on higher potential side as well as on lower potential side, we get parallel connection.
     
   • For resistors connected in parallel, the voltage drop(V) across all the resistors remains the same as the potential difference(voltage) of the circuit.
     
   • The overall current, however, gets distributed among the resistors according to their magnitudes.
     
     
   • Consider that three resistors R₁, R₂ and R₃ are connected in parallel in a circuit having overall potential difference(voltage) V and current I.
     
     
   • If I₁, I₂ and I₃ are the values of current passing through R₁, R₂ and R₃, respectively, then
     
     I₁ = V/R₁
     
     I₂ = V/R₂ and
     
     I₃ = V/R₃
     
     
   • Similarly, if R is the equivalent resistance, then
     
     I = V/R.
     
     
   • Now,
     
     I = I₁ + I₂ + I₃
     
     ∴ V/R = V/R₁ + V/R₂ + V/R₃
     
     ∴ 1/R = 1/R₁ + 1/R₂ + 1/R₃
     
     Thus, the reciprocal of equivalent resistance is equal to the sum of the reciprocals of individual resistances.
     

   
   

   
   
8. DISTINGUISH BETWEEN SERIES CONNECTION AND PARALLEL CONNECTION.
   
   ANS:
   
9. EXPLAIN : ELECTRICAL ENERGY AND POWER
   

   ANS:
   

   
   
   • To maintain the flow of charge(current), a battery has to do work continuously on the charge.
     
   • This work done on charge is converted into electrical energy.
     
   • If
     
     W = work done by the battery
     
     V = voltage of battery
     
     Q = electric charge
     
     I = current
     
     R = resistance
     
     t = time
     
     then we have,
     
     W = V x Q
     
     ∴ W = V x I x t        [as Q = I x t]
     
     ∴ W = I² x R x t        [as V = I x R]
     
     Thus, W is the energy consumed if the current(I) flows through resistance(R) for time (t).The unit of electrical energy is joule(J).
     
   • The rate at which the electrical energy is consumed is known as electrical power.In other words electrical energy consumed per unit time is called power.
     
     
   • If P is electrical power, then
     
     P = W/t
     
     ∴ P = (I² x R x t)/t        [as W = I² x R x t]
     
     ∴ P = I² x R
     
     or P = V x I        [as I x R = V]
     
     The unit of power is watt.
     
   • When 1 ampere current passes through a circuit having voltage of 1 V, the power is called 1 watt.
     
   • Using the definition of power,
     
     W = P x t
     
     ∴ joule = watt x second
     
     Thus, watt-second is a unit of energy.
     
   • For practical purpose watt-second is a small unit of energy.Therefore, practical unit of energy is kilowatthour(kWh).
     
     1 kWh = 3.6 x 10⁶ joule
     
   • The practical unit of energy(kWh) is called 'unit'.
     

   
   

   
   
10. EXPLAIN : CHEMICAL EFFECT OF ELECTRIC CURRENT GIVING EXAMPLE.
    

    ANS:
    
    

    
    
    • The phenomenon of a chemical reaction taking place due to electric current is called chemical effect of electric current.
      
    • The chemical effect of electric current can be utilised to deposit a thin layer of metal on another metal.
      
    • The process of depositing a desired metal on another metallic object by using electricity is called ELECTROPLATING.
      
    • As shown in figure, take aqueous solution of copper sulphate in a container.
      
    • Dip a copper plate and a metal spoon (not of copper) in the solution and connect the circuit such that copper plate works as anode and the spoon as cathode.
      
    • When we allow electric current to pass through the circuit for some time, the electrolyte (CuSO₄) gets ionised and Cu⁺² ions formed during ionization travel to cathode (spoon) and accumulate there. Similarly, SO₄^-2 ions travel to anode and react with copper to form copper sulphate.
      
    • Thus, copper from anode is removed and it gets deposited on cathode through the electrolyte.
      
    • Ultimately, the spoon gets electroplated, i.e. the spoon gets coated with a thin layer of copper.
      

    
    

    
    
11. COMPARE THE WORK OF A BATTERY WITH A MECHANICAL PUMP.
    
    ANS:
    
    As shown in the figure, a pump supplies the work to maintain a difference of height(h) and thus provides energy to the flow of water. This flow of water can utilise the energy to do work to move some object. For example, it can turn a paddle as shown.
    
    In electric circuit, we can compare electric current(flow of electrons) with the flow of water, the lamp with the paddle, the wire with the pipe and the battery(source of energy) with the pump.
    
    

ELECTRICITY -II :BRIEF ANSWERS

ANSWER IN BRIEF

*INDEX-TOPIC SEARCH

1. HOW DO WE GET FLOW OF ELECTRONS IN A CONDUCTOR ?
   

   ANS:
   

   
   
   • In an atom the electrons revolving around the nucleus in different orbits are tightly bound to the protons in the nucleus.
     
   • In metals the electrons in the outermost orbit are relatively free to move and are called free electrons.
     
   • These free electrons move randomly in the metal and their net movement in a particular direction is zero.Therefore we do not get current.
     
   • If we provide energy to these free electrons, they move in one direction and we get flow of electrons(current).
     
   • Thus we get current(flow of electrons) by connecting the conductor to a source of energy which creates potential difference across the conductor.
     

   
   

   
   
2. STATE OHM'S LAW AND WRITE ITS FORMULA(EQUATION).
   

   ANS:
   Ohm's Law: The current passing through a conductor is directly proportional to the potential difference(voltage) across the conductor.
   
   Its formula is:
   
         V = I R
   

   Where V = potential difference
   
   I = current in the conductor
   
   R = resistance of the conductor.

   
   

   
   
3. WRITE THE DISADVANTAGES OF SERIES CONNECTION.
   

   ANS:
   The disadvantages of series connection are :
   

   
   
   1. The total resistance in this case is the sum of all resistances.This causes more power loss.
      
   2. The source voltage gets divided among all resistances in series.The equipment either do not work or they work less efficiency.For example if two bulbs of 230V rating are connected in series in a line of 230V then each gets only 115V and gives dim light.
      
   3. If a bulb connected in series with other equipment blows off then it breaks the whole circuit and remaining equipment(like TV, fridge, computer etc.) also stop working.
      

   
   

   
   
4. WRITE ADVANTAGES OF PARALLEL CONNECTION.
   

   ANS:
   The advantages of parallel connection are:
   

   
   
   1. For parallel connection, the voltage drop across each resistor is same.Since the voltage drop is uniform and is equal to that of the source,an equipment like bulb does not glow dim even if there are more appliances .
      
   2. If an equipment or bulb fails to work, the remaining equipment continue to work as the current is not disrupted in remaining circuit.
      
   3. As the equivalent resistance of the circuit is smaller than the smallest resistance of the circuit, there is minimum power loss.
      

   
   

   
   
5. EXPLAIN THE CAUSE OF RESISTANCE IN A CONDUCTOR AND DIFFERENTIATE BETWEEN CONDUCTORS AND INSULATORS
   

   ANS:
   

   
   
   • Electrical resistance is a property of the conducting wire.
     
   • When electrons try to move within the conducting wire, their movement is opposed by the atoms and molecules of the conductor.
     
   • The collision of electrons with atoms(molecules) gives rise to resistance to the movement of electrons.
     
   • Metallic elements are good conductors because of free electrons in them.These free electrons overcome the resistance and produce current.
     
   • Materials like plastic, wood, rubber,etc. are insulators(bad conductors) as they do not have free electrons to produce current.
     

   
   

   
   
6. EXPLAIN : ELECTRICAL ENERGY
   

   ANS:
   

   
   
   • To maintain the flow of charge(current), a battery has to do work continuously on the charge.
     
   • This work done on charge is converted into electrical energy.
     
   • If
     
     W = work done by the battery
     
     V = voltage of battery
     
     Q = electric charge
     
     I = current
     
     R = resistance
     
     t = time
     
     then we have,
     
     W = V x Q
     
     ∴ W = V x I x t        [as Q = I x t]
     
     ∴ W = I² x R x t        [as V = I x R]
     
     Thus, W is the energy consumed if the current(I) flows through resistance(R) for time (t).The unit of electrical energy is joule(J).
     

   
   

   
   
7. EXPLAIN : ELECTRICAL POWER
   

   ANS:
   

   
   
   • The rate at which the electrical energy is consumed is known as electrical power.In other words electrical energy consumed per unit time is called power.
     
     
   • If P is electrical power, then
     
     P = W/t
     
     ∴ P = (I² x R x t)/t        [as W = I² x R x t]
     
     ∴ P = I² x R
     
     or P = V x I        [as I x R = V]
     
     The unit of power is watt.
     
     
   • When 1 ampere current passes through a circuit having voltage of 1 V, the power is called 1 watt.
     

   
   

   
   
8. WHAT IS ELECTROLYSIS ? WRITE FARADAY'S LAWS OF ELECTROLYSIS
   

   ANS:
   
   Electrolysis is the process of separation of ions of a subsatnce with the help of electric current.
   
   Faraday's Laws of Electrolysis
   

   
   
   1. The mass of the substance deposited on cathode is proportional to the charge passing through the electrolyte.
      
   2. For a given amount of charge passed, the masses of different elements deposited on the cathode are in proportion to their chemical equivalents.
      

   
   

   
   
9. GIVE REASON : ONE SHOULD NOT TOUCH ELECTRICAL SWITCH WITH WET HANDS.
   

   ANS:Pure water is bad conductor of electricity but the tap water used by us contains dissolved impurities. Due to this water becomes good conductor of electricity. If we touch a switch with wet hands, there is a chance that current will flow through our body due to water on our hands and this may give us an electric shock and harm us.
   



10. STATE THE ADVANTAGES OF HEATING EFFECT OF CURRENT IN DAILY LIFE.
    

    ANS:

    
    
    • Domestic appliances such as electric heater, electric iron, electric lamp, toaster, etc. use heating effect of electric current.
    • Fuse is a safety device consisting of a low melting point wire. When current increases abruptly due to some reason, fuse melts and breaks the circuit. It thus protects us and the costly appliances from being damaged.
      

    
    



11. DISTINGUISH BETWEEN AMMETER AND VOLTMETER.
    

    ANS:
    

ELECTRICITY - I :MCQs AND SHORT ANSWERS

MULTIPLE-CHOICE QUESTIONS

*INDEX-TOPIC SEARCH

SELECT THE CORRECT ALTERNATIVE:

1. Who had patented more than 1000 inventions during his life time ?
   
   
   
   1. Edison
   2. Volta
   3. Ampere
   4. Faraday
      
   
   
2. The unit of electric current is _______.
   
   
   
   1. coulomb
   2. volt
   3. ampere
   4. ohm
      
   
   


3. The unit of electric charge is _______.
   
   
   
   1. coulomb
   2. volt
   3. ampere
   4. ohm
      
   
   
4. The unit of electric current is _______.
   
   
   
   1. coulomb
   2. coulomb/sec
   3. coulomb-sec
   4. coulomb/volt
      
   
   
5. 1 ampere = _______.
   
   
   
   1. coulomb sec
   2. coulomb/sec
   3. volt/sec
   4. ohm/sec
      
   
   
6. Work done on an electric charge is stored in it as _______.
   
   
   
   1. potential energy
   2. kinetic energy
   3. thermal energy
   4. nuclear energy
      
   
   
7. The unit of electrical potential is _______.
   
   
   
   1. coulomb/joule
   2. joule/coulomb
   3. watt/coulomb
   4. joule/ampere
      
   
   
8. The unit of electrical potential is _______.
   
   
   
   1. ampere
   2. coulomb
   3. volt
   4. watt
      
   
   
9. _______ is not essential to obtain electric current.
   
   
   
   1. Free electrons
   2. Potential difference
   3. Electric circuit
   4. potential
      
   
   
10. _______ invented a simple electrochemical cell.
    
    
    
    1. Archimedes
    2. Edison
    3. Volta
    4. Coulomb
       
    
    
11. _______ energy is converted into electrical energy in Voltaic cell.
    
    
    
    1. Thermal
    2. Mechanical
    3. Kinetic
    4. Chemical
       
    
    
12. The unit of resistance is _______.
    
    
    
    1. volt/ampere
    2. coulomb/second
    3. volt/coulomb
    4. ampere/volt
       
    
    
13. The unit of resistance is _______.
    
    
    
    1. coulomb
    2. volt
    3. ohm
    4. ampere
       
    
    
14. When resistors are connected in series _______.
    
    
    
    1. voltage drop is uniform
    2. current is uniform
    3. both voltage and current are uniform
    4. neither of two is uniform
       
    
    
15. When resistors are connected in parallel _______.
    
    
    
    1. voltage drop is uniform
    2. current is uniform
    3. both voltage and current are uniform
    4. neither of two is uniform
       
    
    
16. Three resistors of 6 Ω, 12 Ω and 12 Ω are connected in parallel. Their equivalent resistance is _______.
    
    
    
    1. 6 Ω
    2. 12 Ω
    3. 3 Ω
    4. 1/3 Ω
       
    
    
17. Which of the following is not correct for electrical work ?
    
    
    
    1. W = VQ
    2. W = VIt
    3. W = I²Rt
    4. W = I²RQ
       
    
    
18. _______ is not a unit of energy.
    
    
    
    1. joule
    2. watt
    3. watt second
    4. kWh
       
    
    
19. 1 joule = _______.
    
    
    
    1. 1 watt-second
    2. 1 watt/second
    3. 1 coulomb/second
    4. 1 unit
       
    
    
20. Which of the following is not correct ?
    
    
    
    1. P = W/t
    2. P = I²R
    3. P = WI
    4. P = VI
       
    
    
21. Silicon is a/an _______.
    
    
    
    1. conductor
    2. insulator
    3. semiconductor
    4. superconductor
       
    
    
22. Who discovered electron ?
    
    
    
    1. Coulomb
    2. Volta
    3. Ampere
    4. Thomson
       
    
    
23. 1 A = _______ mA
    
    
    
    1. 100
    2. 10³
    3. 10^(-3)
    4. 10^(-6)
       
    
    
24. Ohm's law states that
    
    
    
    1. resistance increases as current increases
    2. resistance increases as voltage increases
    3. current increases as voltage increases
    4. current increases as resistamce increases
       
    
    
25. Equivalent resistance of resistances in parallel is _______.
    
    
    
    1. smaller than the smallest resistance
    2. larger than the largest resistance
    3. an average of all resistances
    4. algebraic sum of all resistances
       
    
    
26. 1 unit of domestic energy is _______.
    
    
    
    1. 1 joule
    2. 1 watt second
    3. 3.6 x 10⁶ j
    4. 3.6 x 10⁶ kwh
       
    
    
27. Three resistors of equal value are connected in parallel.The circuit current is 3 A.Then the current passing through each resistor is _______.
    
    
    
    1. 3 A
    2. 1 A
    3. 9 A
    4. 1/3 A
       
    
    
28. How much electric charge is present on 100 protons ?
    
    
    
    1. 6 x 10^-19 C
    2. 6 x 10^-17 C
    3. 1.6 x 10^-19 C
    4. 1.6 x 10^-17 C
       
    
    
29. Pure water is electrically...
    
    
    
    1. a good conductor
    2. a bad conductor
    3. a semiconductor
    4. a super conductor
       
    
    
30. The frequency of direct current in India is______ Hz.
    
    
    
    1. 0
    2. 50
    3. 60
    4. 220
       
    
    

ANSWERS


(1) A (2) C (3) A (4) B (5) B (6) A (7) B (8) C (9) D (10) C (11) D (12) A (13) C (14) B (15) A (16) C (17) D (18) B (19) A (20) C (21) C (22) D (23) B (24) C (25) A (26) C (27) B (28) D (29) B (30) A

SHORT QUESTIONS

ANSWER IN SHORT:

1. What type of charge does a glass rod acquire when it is rubbed with silk cloth ?
   
   ANS:A glass rod acquires positive charge when it is rubbed with silk cloth
   
   
2. what is an electric charge ?
   
   ANS:An electric charge is a fundamental property associated with protons and electrons.
   
   
3. What is the unit of electric charge ?
   
   ANS:The unit of electric charge is coulomb.
   
   
4. What is the magnitude of charge on proton ?
   
   ANS:The magnitude of charge on proton is 1.6 x 10^(-19).
   
   
5. What is the magnitude of charge on electron ?
   
   ANS:The magnitude of charge on electron is 1.6 x 10^(-19).
   
   
6. Mention some metals which contain maximum free electrons.
   
   ANS:Copper, aluminium and silver are the metals containing maximum free electrons.
   
   
7. What are conductors ?
   
   ANS:Conductors are materials which conduct electricity readily.
   
   
8. What are insulators ?
   
   ANS:Insulators are materials which do not conduct electricity.
   
   
9. Give some examples of insulators.
   
   ANS:Rubber, leather, plastic, glass, etc. are examples of insulators.
   
   
10. Define : Electric current.
    
    ANS:Electric current is the net amount of charge that passes through an area(cross-sectional area) of the conductor per unit time.
    
    
11. Who gave the concept of electron ?
    
    ANS:Sir J.J.Thomson gave the concept of electron.
    
    
12. How many electrons should flow in one second to contribute an electric current of 1 ampere ?
    
    ANS:6.25 x 10¹⁸ electrons should flow in one second to contribute an electric current of 1 ampere.
    
    
13. What should be done to get electric current ?
    
    ANS:Energy should be provided o the free electrons in a conductor to obtain electric current.
    
    
14. Define : Electric potential.
    
    ANS:At a point in an electric field, the electrical potential energy per unit charge is called the electrical potential at that point.
    
    
15. What is the direction of actual current ?
    
    ANS:The direction of actual current is from cathode to anode.
    
    
16. What is the direction of conventional current ?
    
    ANS:The direction of conventional current is from anode to cathode.
    
    
17. Define : Electrical potential difference.
    
    ANS:The work done to move a unit electric charge from one point of a conductor to another point of that conductor is called the electrical potential difference between these two points.
    
    
18. State Ohm's law.
    
    ANS:The current passing through a conductor is directly proportional to the potential difference (voltage drop) across the conductor.
    
    
19. Define : 1 ohm resistance.
    
    ANS:If a potential difference of 1 volt between two terminals of a conductor causes a current of 1 ampere in it, then the resistance of the conductor is 1 ohm.
    
    
20. On which factors does the heat produced in a conductor depend ?
    
    ANS:The heat produced in a conductor depends on the electric current and the resistance in it.
    
    
21. A fuse wire has a low melting point.True or false ?
    
    ANS:Yes, this is a true statement.
    
    
22. Define : Electric Power.
    
    ANS:Electrical energy consumed per unit time is called electric power.
    
    
23. What is the unit of power ?
    
    ANS:The unit of power is watt (or, joule/sec).
    
    
24. How many kj equal 1 kwh ?
    
    ANS:3.6 x 10³ kj equal 1 kwh.
    
    
25. How many joule equal 1 kwh ?
    
    ANS:3.6 x 10⁶ kj equal 1 kwh.
    
    
26. Pure water does not conduct electricity. True or false ?
    
    ANS:True, pure water does not conduct electricity.
    
    
27. The unit of electric charge is faraday. True or false ?
    
    ANS:False, the unit of electric charge is coulomb.
    
    
28. What are electrolytes ?
    
    ANS:The solutions that conduct electricity are known as electrolytes.
    
    
29. To which terminal is the object connected during electroplating ?
    
    ANS:The object is connected to the negative terminal(cathode) during electroplating.
    
    
30. What is electroplating ?
    
    ANS:The process of depositing a desired metal on another mettalic object by using electricity is called electroplating.
    
    
31. A fuse works on the principle of chemical effect of electric current. True ?
    
    ANS:No, a fuse works on the principle of thermal effect of electric current.
    
    
32. How do we get flow of electrons in a conductor ?
    
    ANS:We get flow of electrons in a conductor by connecting the conductor to a source of energy(a cell or battery) which drives free electrons in a definite direction.
    
    
33. Current flows in a wire.Can we call the wire charged ? Why ?
    
    ANS:No. Because the flow of current in a wire does not add or remove any charged particles in it.
    
    
34. What causes resistance in a conductor ?
    
    ANS:The collision of free electrons moving in a conductor with the atoms or molecules of the conductor causes resistance in a conductor.
    
    
35. Write the equations for equivalent resistance R of two resistors R₁ and R₂ connected in (i)series (ii)parallel.
    
    ANS:(i)Series: R = R₁ + R₂(ii)Parallel : (1/R) = (1/R₁) + (1/R₂)
    
    
36. Define the unit of electrical energy.
    
    ANS:The unit of electrical energy is joule and it is defined as : the electrical energy consumed when one coulomb of charge passes through a point in a conductor having 1 volt potential difference.
    
    
37. What type of wire should be used in a fuse ?
    
    ANS:A wire made from a low melting point metal/alloy having high resistance should be used in a fuse.
    
    
38. How much electric energy is spent if 200 units are consumed ?
    
    ANS:200 kwh or 200 x (3.6 x 10⁶) = 7.2 x 10⁸ j energy is spent if 200 units are consumed.
    
    
39. What is the principle of electroplating ?
    
    ANS:The principle of electroplating is that a metal can be deposited on the surface of another metal with the help of electrolysis.
    
    
40. What is electrolysis ?
    
    ANS:The process of separation of ions of a substance with the help of electric current is called electrolysis.
    
    
41. How much electric charge is present on 100 neutrons ?
    
    ANS:Zero.
    
    
42. Which effect of electric current is used in fuse ?
    
    ANS:The heating effect of electric current is used in fuse.
    
    
43. Write an expression for the amount of heat produced in a wire of resistance R and carrying current I for time t.
    
    ANS:Heat produced = I²Rt
    
    
44. What is meant by saying that the potential difference between two points is 1 volt ?
    
    ANS:The meaning of the statement is that 1 joule work needs to be done to move 1 coulomb of charge from one point to the another.
    

DISPERSION OF LIGHT AND OPTICAL INSTRUMENTS - III :LONG ANSWERS

GIVE ANSWER IN DETAIL

*INDEX-TOPIC SEARCH

1. EXPLAIN THE DISPERSION OF WHITE LIGHT BY PRISM.
   

   ANS:
   
   When a narrow beam of white light is incident on the transparent side of a prism, the prism decomposes it into its constituent colours and we obtain a group of bands of different colours called SPECTRUM OF WHITE LIGHT. This separation of white light into its constituent colours is called DISPERSION OF WHITE LIGHT.
   
   
   White light consists of different colours having different wavelength. When white light passes through vacuum the velocity of all colours is same. When the light passes through a refractive medium (like water) the velocities of all components differ because the waves of different colours are refracted in different proportions. This results in dispersion of white light.
   
   The colours obtained on the screen starting from the bottom are : violet, indigo, blue, green, yellow, orange, red.
   
   This shows that in a transparent medium, the velocity of violet light is the least and that of the red light is the highest. Hence the violet colour at the bottom and the red at the top of the spectrum.
   

   
   
2. DESCRIBE IN SHORT HOW THE SEVEN COLOURS OF LIGHT CAN BE COMBINED TO GET WHITE LIGHT.
   

   ANS:
   
   
   Take two prisms P₁ and P₂ with same prism angles and arrange them as shown in the figure. When you allow a narrow beam of white light to be incident on prism P₁, it decomposes into its seven constituent colours (which can be obtained on a screen).
   
   When these waves are allowed to be incident on the second prism P₂, they recombine and emerge from P₂ as a beam of white light. Thus, recombination of colours of white light is a reverse process of dispersion.
   

   
   
3. EXPLAIN : PRIMARY COLOURS OF LIGHT AND THEIR SUPERIMPOSITION.
   

   ANS:
   

   
   
   • Red, blue and green are three primary colours of light.
     
   • White light can be obtained with proper combination of these three colours of light and we need not combine all seven constituent colours.
     
   • Additive mixture of these primary colours in different proportions produce a wide range of different colours(shades). This method is called ADDITIVE MIXTURE METHOD.
     
   • Colours obtained by such method are called COMPOSITE COLOURS.Magenta, cyan and yellow are examples of composite colours.
     
     
   • SUPERIMPOSITION:
     
     
     
     • Take three torches fitted with transparent glass plates of red, blue and green colour.
       
       
       
     • If the torches are arranged as shown in the figure, and their lights after passing through the coloured plates are projected on a screen (or wall), we find different regions with different colours.
       
       
     • The portion where all three colours superimpose appears white.
       
       
     • The portion where red and blue colours superimpose appears magenta.
       
       
     • The portion where red and green colours superimpose appears yellow.
       
       
     • The portion where blue and green colours superimpose appears cyan.
       
     
     

   
   

   
   
4. EXPLAIN : SUBTRACTIVE METHOD OF MIXING OF PIGMENTS.
   

   ANS:
   

   
   
   • The mixing of primary colours of light(i.e/ red, blue, green) by additive method gives white light.
     
   • The mixing of red, blue and green pigments, however, does not give white pigment as this mixing is done by subtractive method.
     
     
   • As shown in the figure when white light is incident on blue pigment; violet, green and blue colours of light are reflected and remaining colours are absorbed.
     
   • When white light is incident on yellow pigment; only yellow, orange and green colours are reflected and remaining colours are absorbed.
     
   • Thus, both yellow and blue pigments do not absorb green colour. If blue and yellow pigments are mixed, then that mixture will reflect only green colour of light.
     
   • Cyan, magenta and yellow are primary pigments and pigments of any colour can be produced by appropriately mixing these pigments.
     

   
   

   
   
5. EXPLAIN : DEFECTS OF VISION
   

   ANS:
   

   
   
   • We can see an object clearly when its image is formed exactly on retina.
     
   • The thickness of the lens of the eye changes according to the distance of the object.
     
   • Ciliary muscles help the lens to change its thickness.
     
   • When the lens can focus the object properly its image is formed exactly on retina and we can see it clearly.
     
   • When the lens cannot change its thickness properly, the defects of vision arise.
     
   • There are two types of defects of vision :
            (1)Near-sightedness (Myopia)
            (2)Far-sightedness
     

   
   

   NEAR-SIGHTEDNESS

   
   

   
   
   • If the lens of the eye does not become sufficiently thin to focus a distant object, the rays coming from the object after being refracted are focussed before reaching the retina.
     
   • As the lens remains thick, it can focus nearby objects perfectly on the retina. Therefore the defect is called nearsightedness.
     
   • This defect arises due to excessive convergence of the light rays.
     
   • A concave lens of appropriate focal length corrects this defect.
     

   
   

   FAR-SIGHTEDNESS

   
   

   
   
   • If the lens of the eye does not become sufficiently thick to focus a nearby object, the rays coming from the object, after being refracted are focussed behind the retina.
     
   • As the lens remains thin, it can focus distant objects perfectly on the retina. Therefore the defect is called far-sightedness.
     
     
   • This defect arises due to less convergence of the light rays.
   
   
   
   • A convex lens of appropriate focal length corrects this defect.
   
   

   
   
   

   
   
6. EXPLAIN : COMPOUND MICROSCOPE
   

   ANS:
   

   
   
   • An instrument using two convex lenses to obtain the magnified image of a small object is called COMPOUND MICROSCOPE.
     
   • The lens towards the object is called objective lens whereas the lens towards the eye is called eyepiece.
     
   • The focal length of the objective is smaller than that of the eyepiece.
     

   
   
   

   WORKING

   
   

   
   
   • The object to be observed(AB) is placed at a distance slightly greater than f_o (focal length of objective lens).
     
   • A magnified, real and inverted image(A'B') is formed on the other side of the objective lens.
     
   • A'B'(image of AB) works as an object for eyepiece.
     
   • The position of the eyepiece is so adjusted that A'B' lies within its focal length(f_e).
     
   • The eyepiece now acts as simple microscope and forms a highly magnified, virtual and erect image A"B".
     

   
   

   
   
7. EXPLAIN : ASTRONOMICAL TELESCOPE
   

   ANS:
   

   
   
   • Astronomical telescope is used to observe far away heavenly objects like planets and stars which appear very small and closer to one another because of their great distances.
     
   • An astronomical telescope consists of two convex lenses arranged coaxially.
     
   • The lens towards the far away object is called objective lens and that near the eye is called eyepiece.
     
   • The focal length of the objective lens is larger than that of the eyepiece.
     

   
   
   

   WORKING

   
   

   
   
   • When the telescope is focussed on a distant object, parallel rays coming from the object form a real, inverted and small image A₁B₁.
     
   • This image (A₁B₁) acts as an object for the eyepiece.
     
   • The eyepiece can be moved to and fro in such a way that virtual and magnified image A₂B₂ of A₁B₁ is formed.
     
   • The final image A₂B₂ formed by the telescope is inverted (compared to object).
     
   • Thus, far away objects can be seen as magnified image.
     

   
   

   
   
8. DISTINGUISH BETWEEN NEAR-SIGHTEDNESS AND FAR-SIGHTEDNESS.
   

   ANS:
   

   
   

DISPERSION OF LIGHT AND OPTICAL INSTRUMENTS - BRIEF ANSWERS

*INDEX-TOPIC SEARCH

ANSWER IN BRIEF

1. WHY DOES DISPERSION OF LIGHT OCCUR ? EXPLAIN.
   

   ANS:
   
   White light consists of different colours having different wavelength.When white light passes through vacuum the velocity of all colours is same.When light passes through a refractive medium( like water) the velocities of all components differ because waves of different colours are refracted in different proportions.This results in dispersion of white light.
   



2. WHAT ARE PRIMARY AND COMPOSITE COLOURS ? EXPLAIN GIVING EXAMPLES.
   

   ANS:
   
   

   
   
   • Red, blue and green are three primary colours of light.
   • White light can be obtained with proper combination of these three colours.
   • Additive mixture of these primary colours in different proportions produces a wide range of different colours.
   • Colours obtained by such mixing are called composite colours.
   • Mixing of BLUE and RED gives MAGENTA. Mixing of BLUE and GREEN gives CYAN and mixing of RED and GREEN gives YELLOW. Thus, magenta, cyan and yellow are composite colours of light.
     

   
   
   
3. WHY DO DEFECTS OF VISION ARISE ? EXPLAIN IN BRIEF.
   

   ANS:
   
   

   
   
   • We can see an object clearly when its image is formed exactly on retina.
   • The thickness of the eye lens changes according to the distance of the object.
   • Ciliary muscles help the lens to change its thickness.
   • When the lens cannot change its thickness properly, it cannot focus the object exactly on retina and the defects of vision arise.
   • If the lens of the eye does not become sufficiently thin to focus a distant object, the defect is called nearsightedness or myopia.
   • If the lens cannot become sufficiently thick to focus a nearby object, the defect is called farsightedness or hypermetropia.
     

   
   
   
4. MENTION DEFECTS OF VISION AND THEIR REMEDY.
   

   ANS:
   
   There are two types of defects of vision:

   • Nearsightedness or myopia
   • Farsightedness or hypermetropia.
     

   
   
   A concave lens of appropriate focal length corrects the defect of nearsightedness whereas a convex lens of appropriate focal length corrects the defect of farsightedness.
   

   
   
5. EXPLAIN : COMPLIMENTARY COLOURS.
   

   ANS:
   
   Any two colours which on mixing produce white light are called complimentary colours. For example :

   • Blue and yellow are complimentary colours of each other.
   • Green and magenta are complimentary colours of each other.
   • Red and cyan are complimentary colours of each other.
     

   
   

   
   
6. MENTION TWO POINTS OF DIFFERENCE BETWEEN COMPOUND MICROSCOPE AND ASTRONOMICAL TELESCOPE.
   

   ANS:
   
   

   
   

DISPERSION OF LIGHT AND OPTICAL INSTRUMENTS-I : MCQs & SHORT ANSWERS

MULTIPLE-CHOICE QUESTIONS

*INDEX-TOPIC SEARCH

SELECT THE CORRECT ALTERNATIVE:

1. ________ decomposes white light into its constituent colours.
   
   
   
   1. Flat glass slab
   2. Convex mirror
   3. Convex lens
   4. Prism
      
   
   
   
2. Arranging in the ascending order of wavelength, which one is true ?
   
   
   
   1. Blue,Green,Red
   2. Orange,Green,Red
   3. Blue,Yellow,Green
   4. Orange, Yellow, Green
      
   
   
   
3. The velocity of waves of all colours is same in _______
   
   
   
   1. water
   2. air
   3. vacuum
   4. glass
      
   
   
   
4. In a transparent medium, the velocity of ______ light is the least.
   
   
   
   1. red
   2. green
   3. yellow
   4. violet
      
   
   
   
5. _______ is not a primary colour of white light.
   
   
   
   1. Red
   2. Blue
   3. Violet
   4. Green
      
   
   
   
6. Blue + red = _______
   
   
   
   1. magenta
   2. cyan
   3. yellow
   4. violet
      
   
   
   
7. Blue + green = _______
   
   
   
   1. magenta
   2. cyan
   3. yellow
   4. violet
      
   
   
   
8. Red + green = _______
   
   
   
   1. magenta
   2. cyan
   3. yellow
   4. violet
      
   
   
   
9. Red + green + blue = _______
   
   
   
   1. white
   2. cyan
   3. magenta
   4. black
      
   
   
   
10. Blue + yellow = _______
    
    
    
    1. magenta
    2. red
    3. cyan
    4. white
       
    
    
    
11. Green + magenta = _______
    
    
    
    1. cyan
    2. yellow
    3. red
    4. white
       
    
    
    
12. Red + cyan = _______
    
    
    
    1. white
    2. green
    3. violet
    4. magenta
       
    
    
    
13. Pigments are mixed according to _______ method.
    
    
    
    1. additive
    2. subtractive
    3. multiplicative
    4. fractional
       
    
    
    
14. ________ can vary thickness and hence the focal length of eye-lens.
    
    
    
    1. Retina
    2. Vitreous humour
    3. Cornea
    4. Ciliary muscles
       
    
    
    
15. In the spectrum of white light which colour is deviated maximum /
    
    
    
    1. Red
    2. Yellow
    3. Violet
    4. Blue
       
    
    
    
16. The splitting of white light into seven colours is known as ______
    
    
    
    1. reflection
    2. refraction
    3. interference
    4. dispersion
       
    
    
    
17. In spectrum obtained with prism which colour is deviated maximum ?
    
    
    
    1. Violet
    2. Red
    3. Green
    4. Orange
       
    
    
    
18. Primary colours are :
    
    
    
    1. Red, blue, yellow
    2. Red, orange, yellow
    3. Red, blue, green
    4. Red, green, cyan
       
    
    
    
19. Which of the following is not a primary colour ?
    
    
    
    1. Red
    2. Blue
    3. Green
    4. Yellow
       
    
    
    
20. Primary pigments are :
    
    
    
    1. Yellow, green, magenta
    2. Magenta, yellow, cyan
    3. Blue, green, yellow
    4. Blue, green, violet
       
    
    
    
21. Hypermetropia is caused by...
    
    
    
    1. low converging power of eye lens
    2. low diverging power of eye lens
    3. high converging power of eye lens
    4. retinal displacement
       
    
    

ANSWERS TO MCQs:
(1) D (2) A (3) C (4) D (5) C (6) A (7) B (8) C (9) A (10) D (11) D (12) A (13) B (14) D (15) A (16) D (17) A (18) C (19) D (20) B (21) A

SHORT ANSWERS

1. Rainbow is the evidence of which fact ?
   
   

   ANS:Rainbow is the evidence of the fact that white light is composed of seven colours.

   
   
2. What is spectrum of white light ?
   

   ANS:The colours obtained on screen after the dispersion of white light form a group of bands called 'spectrum of white light'.

   
   
3. Define : Dispersion of light.
   

   ANS:The separation of white light into its constituent colours is called dispersion of light.

   
   
4. Why does dispersion of light occur ?
   

   ANS:Dispersion of light occurs because different colours are refracted in different proportion by the prism.

   
   
5. Which colour of light has maximum velocity while passing through a tarnsparent medium ?
   

   ANS:Red colour of light has maximum velocity while passing through a transparent medium.

   
   
6. What determines the colour of the object ?
   

   ANS:The colour of the light which is reflected from the object determines the colour of the object.

   
   
7. What happens to the light falling on an opaque object ?
   

   ANS:When light is incident on an opaque object, some of its colours are absorbed and some(remaining) colours are reflected from the object.

   
   
8. Why does a green leaf appear green ?
   

   ANS:A green leaf appears green because when white light falls on it, other colours of light are absorbed by the leaf and only the green light is reflected.

   
   
9. Why does a red rose appear black when seen in green light ?
   

   ANS:When green light falls on a red rose, it absorbs all the green light. So no light is reflected from the surface of red rose and it appears black.

   
   
10. Mention the primary colours of white light.
    

    ANS:Red, blue and green are the three primary colours of white light.

    
    
11. Why are red, blue and green called primary colours of white light ?
    

    ANS:Red, blue and green are called primary colours of white light because combination of these three colours produces white light.

    
    
12. What do we mean by composite colours ?
    

    ANS:Colours obtained by the mixing of primary colours are aclled composite colours.

    
    
13. Give examples of composite colours.
    

    ANS:Magenta, cyan and yellow are composite colours.

    
    
14. What are complimentary colours ?
    

    ANS:Any two colours which produce white light on mixing are called complimentary colours.

    
    
15. Give two examples of complimentary colours.
    

    ANS:Blue and yellow; green and magenta are examples of complimentary colours.

    
    
16. How are coloured pictures are produced on T.V. screen ?
    

    ANS:Coloured pictures are produced on T.V. screen by mixing of primary colours of light in various proportions.

    
    
17. If red, blue and green pigments are mixed then white pigment is not obtained.Why ?
    

    ANS:If red, blue and green pigments are mixed then white pigment is not obtained because pigments are mixed according to subtractive method whereas the colours of light are mixed by additive method.

    
    
18. Mention three primary pigments.
    

    ANS:Cyan, magenta and yellow are three primary pigments.

    
    
19. Mention two defects of sight.
    

    ANS:Nearsightedness (myopia) and farsightedness (hypermetropia) are the two defects of sight.

    
    
20. Which defect is caused due to excessive convergence of the light rays ?
    

    ANS:Nearsightedness is caused due to excessive convergence of the light rays.

    
    
21. Which defect is caused due to excessive divergence of the light rays ?
    

    ANS:Farsightedness is caused due to excessive divergence of the light rays.

    
    
22. What type of lens is used to correct the defect of nearsightedness ?
    

    ANS:A concave lens of appropriate focal length is used to correct the defect of nearsightedness(myopia).

    
    
23. What type of lens is used to correct the defect of farsightedness ?
    

    ANS:A convex lens of appropriate focal length is used to correct the defect of farsightedness (hypermetropia).

    
    
24. Mention the principle of simple microscope.
    

    ANS:The principle of simple microscope is that when an object is placed between the optical center and focus of a convex lens; a magnified, erect and virtual image of the object is obtained.

    
    
25. What is a simple microscope ?
    

    ANS:A simple microscope is a convex lens used to get the magnified image of a small object.

    
    
26. What is a compound microscope ?
    

    ANS:An instrument using two convex lenses to obtain the magnified image of a small object is called a compound microscope.

    
    
27. In compound microscope, the focal length of objective lens is smaller than that of the eyepiece. True or false ?
    

    ANS:This is a true statement.

    
    
28. If an object reflects all colours of incident light, what is its colour ?
    

    ANS:If an object reflects all colours of incident light, its colour is white.

    
    
29. If an object absorbs all colours of incident light, what is its colour ?
    

    ANS:If an object absorbs all colours of incident light, its colour is black.

    
    
30. What is the function of ciliary muscles ?
    

    ANS:The function of ciliary muscles is to vary the focal length of the eyelens by changing the thickness of the lens.

    
    
31. When white light is incident on blue pigment which colours of light are reflected from it ?
    

    ANS:When white light is incident on blue pigment; violet, blue and green colours of light are reflected from it.

    
    
32. When white light is incident on yellow pigment which colours of light are reflected from it ?
    

    ANS:When white light is incident on yellow pigment; yellow, orange and green colours of light are reflected from it.

    
    
33. Green colour is absorbed by yellow and blue pigment. True or false ?
    

    ANS:This is a false statement.

    
    
34. If blue and yellow pigments are mixed, then which colour of light will be reflected ?
    

    ANS:If blue and yellow pigments are mixed, then only green colour of light will be reflected.

    
    
35. What type of image with respect to original object is obtained by telescope ?
    

    ANS: Real and inverted image is obtained by telescope with respect to the original object.

    
    

LIGHT:REFLECTION AND REFRACTION-IV : NUMERICALS

SOLVED NUMERICALS OF TEXT BOOK (GSEB)

*INDEX-TOPIC SEARCH

1. If an object of height 4 cm is placed at distance of 12 cm from a concave mirror having focal length 24 cm, find the position, nature and the height of the image.
   
   SOLUTION
   
   Here,
   
   object height h = 4 cm
   
   object dis tance u = -12 cm
   
   focal length f = -24 cm
   
   Putting these values in mirror formula,
   
   (1/u) + (1/v) = (1/f)
   
   ∴ (-1/12) + (1/v) = (-1/24)
   
   ∴ (1/v) = (1/12) - (1/24)
   
   ∴ (1/v) = (1/24)
   
   ∴ v = 24
   
   Now, magnification m = (-u/v) = (-24)/(-12) = 2
   
     ∴ (h'/h) = 2
   
     ∴ h' = 2h = 2 x 4 = 8 cm
   
   Since v is positive, the image is formed behind the mirror at the distance of 24 cm from the mirror.It is virtual and has height of 8 cm.
   


2. An object of height 6 cm is placed at a distance of 10 cm from a convex mirror with radius of curvature 30 cm.Find the position, nature and the height of its image.
   
   SOLUTION
   
   Here,
   

   Object distance u = -10 cm
   Radius of curvature R = 30 cm
   Object height h = 6 cm

   
   
   Now,
   
   focal length f = (R/2) = 30/2 = +15
   
   Using mirror formula,
   
    (1/u) + (1/v) = (1/f)
   
     ∴ (1/v) = (1/f) - (1/u)
   
     ∴ (1/v) = (1/15) - (1/-10)
   
     ∴ (1/v) = (1/15) + (1/10) =(1/6)
   
     ∴ v = 6 cm
   
   Also, magnification m = -(u/v) = -(6/-10) = 0.6
   
     ∴ (h'/h) = 0.6
   
     ∴ h' = mh =0.6 x 6 = 3.6 cm
   
   Thus, the image is formed behind the mirror(as v is positive), it is virtual and its height is 3.6 cm.
   


3. The refractive index of water with respect to air is 1.33. The velocity of light in vacuum is 3 x 10⁸ m/s. Calculate the velocity of light in water.
   
   SOLUTION
   
   Here, the refractive index of water with respect to air is 1.33.
   
     ∴ absolute refractive index of water = 1.33
   
   Now, Absolute refractive index of water =
   

   Velocity of light in vacuum
   Velocity of light in water
   

   
   
     ∴ Velocity of light in water =
   

   Velocity of light in vacuum
   Absolute refractive index of water
   
   = 3 x 10⁸
   1.33
   
   = 2.25 x 10⁸ m/s.
   

   
   


4. A ray of light enters from water to glass.Refractive index of glass with respect to water is 1.12. Find absolute refractive index of water if absolute refractive index of glass is 1.5.
   
   SOLUTION
   
   Taking water as "medium1" and glass as "medium2",
   
   Absolute refractive index of water, η₁ = ?
   
   Absolute refractive index of glass, η₂ = 1.5
   
     η₂₁ = 1.12_________(given)
   
   ∴ η₂/η₁ = 1.12
   
   ∴ η₁ =η₂/1.12
   
   ∴ η₁ = (1.5)/(1.12) = 1.34
   
   Hence,absolute refractive index of water = 1.34
   


5. A swimmer lights a torch under sea water.Light from the torch is incident on water surface in such a way that incident light makes an angle of 37° with water surface.Find the angle of refraction if absolute refractive indices of water and air are 1.33 and 1.0 respectively.
   
   SOLUTION
   
   Taking water as "medium1" and air as "medium2",we have
   
     η₁ = 1.33
   
     η₂ = 1.00
   
   Angle of incidence, θ₁ = 37°
   
   Angle of refraction, θ₂ = ?
   
   Now, (sin θ₁/sin θ₂) = (η₂/η₁)
   
   ∴ (sin 37°/sin θ₂) = (1.00/1.33)
   
   ∴ sin θ₂ = [ (0.6015 x 1.33)/1.00]
   
   ∴ sin θ₂ = 0.8000
   
   ∴ θ₂ = 53°_____(from the table of sines)
   
   Hence, angle of refraction = 53°
   


6. An object is placed at 30 cm in front of a convex lens of focal length 20 cm. Find the position of the image.
   
   SOLUTION
   
   Here,
   
   Focal length f = +20 cm
   
   Object distance u = -30 cm
   
   Image distance v = ?
   
   Using lens formula,
   
     (1/f) = (1/v) - (1/u)
   
   ∴ (1/v) = (1/u) + (1/f)
   
   ∴ (1/v) = (1/-30) + (1/20)
   
   ∴ (1/v) = (-2 + 3)/60
   
   ∴ (1/v) = (1/60)
   
   ∴ v = 60 cm.
   
   The positive value of v indicates that the image is formed at 60 cm on the right side of lens from the center of the lens.
   


7. At what distance the object should be placed so that the image will be formed at a distance 10 cm from a concave lens ? Focal length of the lens is 20 cm.
   
   SOLUTION
   
   Since the lens is concave, the image will be formed on the same side as that of the object.
   
   Here,
   
   Image distance v = -10 cm
   
   Focal length f = -20 cm
   
   Object distance u = ?
   
   Using lens formula,
   
     (1/f) = (1/v) - (1/u)
   
   ∴ (1/u) = (1/v) - (1/f)
   
   ∴ (1/u) = (1/-10) - (1/-20)
   
   ∴ (1/u) = (-2 + 1)/20
   
   ∴ (1/u) = -(1/20)
   
   ∴ u = -20 cm
   
   Thus the object should be placed 20 cm from the concave lens on the left side of the lens.
   

UNSOLVED NUMERICALS OF TEXT BOOK (GSEB)

1. An object of 5 cm height is placed at a distance of 15 cm from a concave mirror.Find the position, height and nature of its image.The focal length of the mirror is 10 cm.
   
   SOLUTION
   
   Here
   
   height of object, h = 5 cm
   
   Distance of object, u = -15 cm
   
   Focal length of concave mirror, f = -10 cm
   
   Distance of image, v = ?
   
   Using mirror formula,
   
     (1/v) + (1/u) = (1/f)
   
   ∴ (1/v) = (1/f) - (1/u)
   
   ∴ (1/v) = (1/-10) - (1/-15)
   
   ∴ (1/v) = -(1/10) + (1/15)
   
   ∴ (1/v) = (-3 + 2)/30 = -(1/30)
   
   ∴ v = -30 cm
   
   Also,(image height)/(object height) = (-v/u)
   
   ∴ (h'/h) = (-v/u)
   
   ∴ (h'/5) = (-30/-15)
   
   ∴ h' = 10 cm
   
   Hence the image is formed at a distance of 30 cm ; its height is 10 cm and it is real and inverted.
   


2. An object of 10 cm height is placed at a distance of 10 cm from a convex mirror.The radius of curvature of the mirror is 30 cm.Find the position, height and nature of its image.
   
   SOLUTION
   
   Here
   
   Height of object, h = 10 cm
   
   Distance of object, u = -10 cm
   
   Radius of curvature of convex mirror, R = 30 cm
   
   Distance of image, v = ?
   
   Since f = (R/2),
   
     f = (30/2) = 15 cm
   
   Using mirror formula,
   
     (1/u) + (1/v) = (1/f)
   
   ∴ (1/v) = (1/f) - (1/u)
   
   ∴ (1/v) = (1/15) - (1/-10) = (1/15) + (1/10)
   
   ∴ (1/v) = (2 + 3)/30 = (1/6)
   
   ∴ v = 6 cm
   
   Also,(image height)/(object height) = (-v/u)
   
   ∴ (h'/h) = (-v/u)
   
   ∴ (h'/10) = (-6/-10)
   
   ∴ h' = 6 cm
   
   Hence the image is formed at a distance of 6 cm; its height is 6 cm and it is virtual and erect.
   


3. Rays of light are entering from glass to glycerin.If the absolute refractive index of glass is 1.5 and that of glycerin is 1.47 then find the refractive index of glycerin with respect to glass.
   
   SOLUTION
   
   Let's consider glass as medium1 and glycerin as medium2.
   
   Absolute refractive index of glass, η₁ = 1.5
   
   Absolute refractive index of glycerin, η₂ = 1.47
   
   Now,
   
   Refractive index of glycerin with respect to glass = η₂₁
   
                 =η₂/η₁ = 1.47/1.5 = 0.98
   
   Hence the refractive index of glycerin with respect to glass is 0.98.
   


4. Rays of light are entering from air to water. If the angle of incidence at the surface separating two mediums is 70°, find the angle of refraction of light in water.Absolute refractive index of water is 1.33.
   
   SOLUTION
   
   Let's consider air as medium1 and water as medium2.
   
   Absolute refractive index of air, η₁ = 1.00
   
   Absolute refractive index of water, η₂ = 1.33
   
   Angle of incidence θ₁ = 70°
   
   Angle of refraction θ₂ = ?
   
   Now, sin θ₁/sin θ₂ = η₂/η₁
   
   ∴ (sin 70°/sin θ₂) = (1.33/1.00)
   
   ∴ (0.9397/sin θ₂) = (1.33/1.00)
   
   ∴ sin θ₂ = 0.7065
   
   ∴ θ₂ = 44° 57'
   
   Hence the angle of refraction = 44° 57'
   


5. An object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm at a distance of 20 cm from the lens. Find the position of the image.
   
   SOLUTION
   
   Here
   
   Focal length of convex lens, f = 10 cm
   
   Distance of object, u = -20 cm
   
   Distance of image, v = ?
   
   Using lens formula,
   
      (1/f) = (1/v) - (1/u)
   
   ∴ (1/v) = (1/f) + (1/u)
   
   ∴ (1/v) = (1/10) + (1/-20) = (1/10) - (1/20)
   
   ∴ (1/v) = (2 - 1)/20 = (1/20)
   
   ∴   v = 20 cm
   
   Hence the image is formed at the distance of 20 cm from the lens.
   


6. An object is placed perpendicular to the principal axis of a concave lens of focal length 30 cm at a distance of 20 cm from the lens. Find the position of the image.
   
   SOLUTION
   
   Here
   
   Focal length of concave lens, f = -30 cm
   
   Distance of the object, u = -20 cm
   
   Distance of the image, v = ?
   
   Using lens formula,
   
   
      (1/f) = (1/v) - (1/u)
   
   ∴ (1/v) = (1/f) + (1/u)
   
   ∴ (1/v) = (1/-30) + (1/-20) = (-2 - 3)/60
   
   ∴ (1/v) =(-5/60) = (-1/12)
   
   ∴ v = (-12) cm
   
   Hence the image is formed at the distance of 12 cm from the lens on the same side of the object.
   


7. Find the power of a convex lens with focal length 0.4 m
   
   SOLUTION
   
   Power of lens, p = (1/f)
   
   ∴ p = (1/0.4) ______________(for convex lens f is +ve)
   
   ∴ p = 2.5 D
   
   Hence the power of lens is 2.5 D (dioptre).
   

LIGHT:REFLECTION AND REFRACTION-III : LONG ANSWERS

GIVE ANSWER IN DETAIL

*INDEX-TOPIC SEARCH

1. EXPLAIN WITH DIAGRAM : REFLECTION FROM PLANE MIRROR.
   
   
   

   ANS:
   
   Consider an object AO of height h placed in front of a plane mirror MM' at a distance u.
   
   Each small portion of this extended object AO acts like a point source.
   
   Consider two incident rays AN and AQ from the endpoint A of the object AO.
   
   According to the law of reflection, NA and QR are corresponding reflected rays.
   
   Since ray AN is normal to the mirror, the ray NA is reflected back on the same path.
   
   Similarly, OQ is normal to the mirror. Therefore QO is reflected back on the same path.
   
   Rays NA and QR are reflected rays which meet after extending backwards (behind the mirror) at A'. Thus A' is the virtual image of A. Similarly all the points of object AO form their virtual image behind the mirror. Thus the image A'I of all the points of AO is formed at a distance v behind the mirror. This image is virtual, erect and of the same size (height) as the object. The distances u and v are equal,i.e. the image is at the same distance behind the mirror as the object is in front of the mirror.
   The image is laterally inverted,i.e.,if a person raises the left hand the image appears to raise the right hand.
   



2. EXPLAIN WITH DIAGRAM : CARTESIAN SIGN CONVENTION FOR SPHERICAL MIRROR.
   

   ANS:
   

   
   
   • The pole P of the mirror is taken as the origin.
     
     
   • The principal axis of the mirror is taken along X-axis.
     
     
   • The object is considered as placed on the left of the mirror, i.e. the light is incident on the mirror from the left hand side of the mirror.
     
     
   • All the distances along the principal axis are measured from the pole (P).
     
     
     
   • The distances measured in the direction of the incident ray are taken as positive.
     
     
   • The distances measured in the direction opposite to the direction of incident ray are taken as negative.
     
     
   • The heights measured upwards and perpendicular to the principal axis of the mirror are taken as positive.
     
     
   • The heights measured downwards and perpendicular to the principal axis of the mirror are taken as negative.
     

   
   



3. DERIVE MIRROR FORMULA.
   

   ANS:
   
   The relation between object distance(u), image distance(v) and the focal length (f) of the mirror is known as MIRROR FORMULA.
   
   
   Consider a concave mirror MM' (of small aperture).
   
   An object AB of height h is placed on the left of the mirror beyond its center of curvature so that the image formed is real, inverted and diminished in size(height = h').
   
   Here :
   
   Object distance = PB = -u
   
   Image distance = PB'= -v
   
   Focal length = PF = -f
   
   Radius of curvature = PC = -R
   
   As shown in Figure, Δ ABP ∼ Δ A'B'P.
   
   ∴ A'B'/AB = PB'/PB
   
   ∴ A'B'/AB = (-v)/(-u)
   
   ∴ A'B'/AB = v/u.............(I)
   
   Similarly, as shown in the figure, Δ ABC ∼ Δ A'B'C.
   
   ∴ A'B'/AB = CB'/CB
   
   ∴ A'B'/AB = (PC-PB')/(PB-PC)........[as CB'=(PC-PB') and CB=(PB-PC)]
   
   ∴ A'B'/AB = [-R - (-v)]/[-u - (-R)]
   
   ∴ A'B'/AB = (-R + v)/(-u + R)........(II)
   
   By (I) & (II),
   
   v/u = (-R + v)/(-u + R)
   
   ∴ -uv + vR = -uR + uv
   
   ∴ vR + uR = 2uv
   
   ∴ (vR/uvR) + (uR/uvR) =2uv/uvR.......[Dividing each side by uvR]
   
   ∴ (1/u) + (1/v) = (2/R)..............(III)
   
   Now, for an object placed at infinite distance (u = ∞), the image is formed at the focus F (v = f).Thus, eq.(III) reduces to
   
   (1/∞) + (1/f) = (2/R)
   
   ∴ (2/R) = (1/f)....................(IV)
   
   comparing eq.(III) and (IV),
   
   (1/u) + (1/v) = (1/f)
   



4. EXPLAIN : MAGNIFICATION AND DERIVE ITS FORMULA.
   

   ANS:
   The ratio of the height of the image to the height of the object is called MAGNIFICATION.
   
   Magnification, m = image height/object height = h'/h
   
   From the figure(see the figure in derivation of MIRROR FORMULA), it is clear that
   
   Δ ABP ∼ Δ A'B'P
   
   ∴ A'B'/AB = PB'/PB

   A'B' = image height = h'
   AB = object height = h
   PB = object distance = u
   PB' = image distance = v

   
   
   ∴ h'/h = u/v
   
   ∴ m = u/v
   
   According to the sign convention, m = (-h')/(h).
   
   ∴ m = -(u/v).
   
   The object height(h) is always positive. The image height(h') is positive when the image is virtual and erect whereas h' is negative when the image is real and inverted.
   
   For a plain mirror,
   
   image height(h') = object height(h)
   
   ∴ m = (+1).
   
   If m > 1, the image is enlarged(actual magnification)
   
   If m = 1, the object and image are of same size
   
   If m < 1, the image is smaller than the object.
   
   Magnification is negative for real images and positive for virtual images. Thus, m is always positive for convex mirrors and it is negative for the real images formed by a concave mirror.
   



5. EXPLAIN : POWER OF LENS.
   

   ANS:
   

   
   
   • The power of lens is the reciprocal of its focal length (f).
     
     ∴ Power p = 1/f
     
     
   • The S. I. unit of power of a lens is dioptre (symbol D)
     
     Thus, 1 D = 1 m^-1
     
     
   • The power of a convex lens is positive(+) and that of a concave lens is negative(-).
     
     
   • If the focal length of a convex lens is 0.5 m, then its power is
     
     1/0.5 or 2.0 D.
     
     
   • If the focal length of a concave lens is (-0.2 m), then its power is
     
     -(1/0.2) or -5.0 D.
     
     
   • An ophthalmologist (or opticion), after examining a patient, prescribes corrective lenses for eyeglasses or spectacles in terms of dioptre (D).
     
     
   • Dioptremeter is used to measure the power of a lens.
     

   
   

   
   
6. State Laws of refraction and derive generalised form of Snell's law.
   

   ANS:
   The laws of refraction are:
   
   

   • The ratio of the sine of the angle of incidence(θ₁) to the sine of the angle of refraction(θ₂) is constant.
     
   • The incident ray, the normal to the surface separating two mediums and the refracted ray, all lie in the same plane.

   
   
   
   As shown in the figure,surface AB separates medium 1 and medium 2. PQ is the ray incident on surface AB at point Q, QR is the refracted ray and NQN' is the normal to the surface AB at Q.
   
   ∠ PQN = incident angle = θ₁
   
   ∠ RQN'= refraction angle = θ₂
   
   According to the law of refraction,
   
   sin θ₁ / sin θ₂ = constant
   
   The constant in the above equation is called the refractive index of medium 2 with respect to medium 1 and is denoted by η₂₁
   
   ∴ η₂₁ = sin θ₁ / sin θ₂
   
   The above equation is called Snell's law.The value of η₂₁ depends upon the nature of the two mediums.
   
   Now,
   
   η₂₁ = vel. of light in medium 1/vel. of light in medium 2
   
   ∴ η₂₁ = v₁/v₂
   

   [here,v₁ = velocity of light in medium 1
   v₂ = velocity of light in medium 2]

   
   
   Also, by the definition of absolute refractive index,
   
   Absolute refractive index of medium 1 =
   

   velocity of light in vaccum
   velocity of light in medium 1
   

   
   
   ∴ η₁ = c/v₁
   
   Similarly,
   
   the absolute refractive index of medium 2, η₂ = c/v₂.
   
   ∴ (η₂/η₁) = (v₁/v₂)
   
   ∴ (η₂/η₁) = η₂₁
   
   ∴ (η₂/η₁) = sin θ₁ / sin θ₂
   
   ∴η₁ sin θ₁ = η₂ sin θ₂
   
   This is called generalised form of SNELL'S LAW.
   



7. Derive LENS FORMULA.
   

   ANS:
   
   The relation between object distance(u), image distance(v) and the focal length(f) of a lens is known as LENS FORMULA.
   
   
   Consider a (thin spherical) convex lens MN (of small aperture).
   
   An object AB of height h is placed in front of the lens MN beyond 2F on its left side.
   
   The image A'B' formed by the lens is real, inverted and diminished in size (h').
   
   Here,
   

   Object Distance = OB = -u
   
   Image Distance = OB' = +v
   
   Focal Length = OF = +f

   
   
   As shown in figure, Δ ABO ∼ Δ A'B'O
   
   ∴ (AB/A'B') = (OB/OB')
   
   ∴ (AB/A'B') = (-u/v) _______________(I)
   
   Similarly, Δ OCF ∼ Δ A'B'F
   
   ∴ (OC/A'B') = (OF/FB')
   
   ∴ (AB/A'B') = (OF/FB')........(as AB = OC)
   
   ∴ (AB/A'B') = (OF)/(OB' - OF)
   
   ∴ (AB/A'B') = (f)/(v - f) __________(II)
   
   From Eq. (I) and (II), we have
   
     (-u/v) = (f)/(v - f)
   
   ∴ -u(v - f) = vf
   
   ∴ -uv + uf = vf
   
   ∴ (-uv/uvf) + (uf/uvf) = (vf/uvf)...(dividing by uvf)
   
   ∴ (-1/f) + (1/v) = (1/u)
   
   ∴ (1/f) = (1/v) - (1/u)
   
   This is known as LENS FORMULA.
   



8. EXPLAIN : TWINKLING OF STARS
   

   ANS:
   

   
   
   
   • The density of the atmosphere is not uniform everywhere as the atmosphere consists of layers of different densities.
     
     
   • The layer at the lower altitude (near earth's surface) is colder and denser than the layer at higher altitude.
     
     
   • Due to this , the refractive index of atmosphere decreases continuously as we move up from the surface.
     
     
   • The star light keeps on bending towards the normal at every point of the atmosphere till it reaches our eyes.
     
     
   • Thus, as shown in the figure(refer to the figure given in the answer for TOTAL INTERNAL REFLECTION) the position of the star appears different from its actual position.
     
     
   • Moreover, the atmosphere is not stationary, i.e. the physical condition of the refractive medium keeps on changing every moment.
     
     
   • This results in continuous shifting of the apparent position of the star as well as variation in the intensity of its light.
     
     
   • This leads to TWINKLING OF STARS.
     
     
   • Unlike stars the planets do not twinkle because they are far closer than stars. Therefore, we can consider stars as point-sized source of light and planets as a collection of large number of point-sized sources (extended objects) of light such that they nullify twinkling effect.
     

   
   
9. EXPLAIN : TOTAL INTERNAL REFLECTION (MIRAGE).
   

   ANS:
   

   
   
   
   • When a ray of light travels from optically denser medium to optically rarer medium it bends away from the normal at the surface separating two mediums due to refraction.
     
     
   • As the angle of incidence(θ_i) increases, the angle of refraction(θ_r) also increases and the ray of light moves further away from the normal.
     
     
   • For some value of angle of incidence, the angle of refraction becomes 90°.The angle of incidence for which the angle of refraction is 90°, is called CRITICAL ANGLE.
     
     
   • If the angle of incidence is greater than the critical angle, the angle of refraction becomes greater than 90° and the refracted ray is confined to the same medium only.This is calledTOTAL INTERNAL REFLECTION.
     
     
     
   • MIRAGE:
     
     
     
     
     • During summer, the air near the surface of the earth is hotter than the air above it. Therefore, the air near the surface of the earth is rarer than the air above it. Thus, as we move up from the surface of the earth, the refractive index of air increases.
       
       
     • As shown in figure when the light rays coming from the top point (P) of an object (like tree) pass thruogh different layers of air up to the ground, they move more and more away from the normal.
       
       
     • Thus the angle of refraction will go on increasing and the ray will undergo total internal reflection when it reaches the ground. This totally reflected ray enters our eye and the image of P is seen as P' as if reflection has occurred due to water.
       This is referred to as mirage
       
     
     

   
   
   
10. EXPLAIN : REFRACTION THROUGH A RECTANGULAR SLAB.
    

    ANS:
    
    
    As shown in figure, let []PQRS be a rectangular slab.
    
    A ray of light AB is incident at point B on surface PQ at an angle θ₁.
    
    At point B this ray is refracted making an angle θ₂ and travels as ray BC in glass.
    
    Ray BC is incident on surface RS at point C making an angle θ₃.
    
    From point C, the ray CD emerges from rectangular slab making an angle θ₄.
    
    If η₁,η₂ and η₃ are taken refractive indices of air, glass and air, we have
    
    η₁ = η₃ = 1
    
    According to the laws of refraction, at surface PQ,
    
      η₂ sin θ₂ = η₁ sin θ₁
    
    ∴ η₂ sin θ₂ = sin θ₁......(as η₁ = 1).......(I)
    
    At the surface RS parallel to PQ,
    
      η₂ sin θ₃ = η₃ sin θ₄
    
    ∴ η₂ sin θ₂ = η₃ sin θ₄......(as θ₂ = θ₃)
    
    ∴
    η₂ sin θ₂ = sin θ₄......(as η₃ = 1).......(II)
    
    By (I) and (II),
    
      sin θ₁ = sin θ₄
    
    ∴ θ₁ = θ₄
    
    Thus, the direction of the emergent ray CD is parallel to incident ray AB(EF) but because of the refraction, emergent ray is displaced by perpendicular distance CE. This kind of displacement (CE) is known as LATERAL SHIFT. The lateral shift is proportional to the perpendicular distance between two refracting surfaces.
    

    
    
11. DISTINGUISH BETWEEN CONVERGING AND DIVERGING LENS.
    

    ANS: