l l RA lA rB 2 RA 1 1 2 1
r r2 RB lB rA RB 2 2 8
Solution : (d) Resistance R= ρ π 2 ⇒ R∝ ⇒ = × ⇒ = × =
Example: 24
Solution : (b) By using H = V 2t HA = RB = 8
R⇒ HB RA 1
Example: 25
Solution : (a) A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the
same source. The energy now liberated per second is [CBSE PMT 1995]
(a) 200 J (b) 400 J (c) 25 J (d) 50 J
Let resistance of the heating coil be R, when coil cut in two equal parts, resistance of each part will be R .
2
When these two parts are corrected in parallel, Req = R i.e. resistance becomes, so according to
4
P ∝ 1 ; Power becomes 4 times i.e. P′ = 4P = 400 J/sec
R
Two identical electric lamps marked 500 W, 220 V are connected in series and then joined to a 110 V line.
The power consumed by each lamp is
(a) 125 W (b) 25 W (c) 225 W (d) 125 W
4 4 4
Both bulbs are identical so voltage across each bulb will be 55V. 500 W 500 W
220 V 220 V
VA 2 55 2 125 55 V
VR 220 4 55 V
Hence power consumed by each bulb is × PR = × 500 = W
110 V
Tricky Example: 2
Electric bulb 50 W – 100 V glowing at full power are to be used in parallel with battery 120 V, 10 Ω.
Maximum number of bulbs that can be connected so that they glow in full power is [CPMT 2002]
(a) 2 (b) 8 (c) 4 (d) 6
Solution : (c) When bulb glowing at full power, current flows through it i = P = 50 = 1 amp ⇒ i = n and voltage
n V 100 2 2
across the bulb is 100 V. If suppose n bulbs are connected in parallel with cell as shown in figure then
according to the cell equation E = V + ir ⇒ 120 = 100 + n × 10 ⇒ n = 4.
2
i/n 1
i/n 2
i i/n n
120 V, 10 Ω
Thermo Electric Effect of Current
If two wires of different metals are joined at their ends so as to form two junctions, then the resulting
arrangement is called a “Thermo couple”.
Seebeck Effect.
(1) Definition : When the two junctions of a thermo couple are maintained at different temperatures, then a
current starts flowing through the loop known as thermo electric current. The potential difference between the
junctions is called thermo electric emf which is of the order of a few micro-volts per degree temperature difference
(µV/oC). i Fe
Hot Cold
Cu G Cu
(2) Origin of thermo emf : The density of free electrons in a metal is generally different from the density of
free electrons in another metal. When a metal is brought into intimate contact (say by soldering) with other metal,
the electrons tend to diffuse from one metal to another, so as to equalise the electron densities. As an illustration,
when copper is brought into intimate contact with iron, the electrons diffuse from iron to copper. But this diffusion
cannot go on continuously because due to diffusion, the potential of copper decreases and the potential of iron
increases. In other words, iron becomes positive with respect to copper. This is what stops further diffusion. In the
case of thermocouple whose junctions are at the same temperature, the emf’s at the junctions will be equal in
magnitude but opposite in direction. So, the net emf for the whole of thermocouple will be zero.
Let us now consider the case when the temperature of one junction of the thermocouple is raised. Raising the
temperature of one junction will affect the electron density in the two metals differently. Moreover, the transfer of
electrons at the junction will be easier than the transfer of electrons at the cold junction. Due to both these reasons,
the emf’s at the two junctions will be different. This produces a net emf in the thermocouple. This emf is known as
Seebeck emf.
(3) Seebeck series : The magnitude and direction of thermo emf in a thermocouple depends not only on
the temperature difference between the hot and cold junctions but also on the nature of metals constituting the
thermo couple.
Seebeck arranged different metals in the decreasing order of their electron density. Some of the metals forming
the series are as below.
Sb, Fe, Ag, Au, Sn, Pb, Cu, Pt, Ni, Bi
(i) About magnitude thermo emf : Thermo electric emf is directly proportional to the distance between the
two metals in series. Farther the metals in the series forming the thermo couple greater is the thermo emf. Thus
maximum thermo emf is obtained for Sb-Bi thermo couple.
(ii) Direction of thermo electric current : If a metal occurring earlier in the series is termed as A and the
metal occurring later in the series is termed as B, then the rule for the direction of conventional current in
thermocouple made of elements A and B is ABC. That is, at the cold junction current will flow from A to B. e.g. in
Fe-Cu thermocouple, at the cold junction current flows from A to B that is from Fe to Cu. At the hot junction, the
current flows from Cu to Fe. This may be remembered easily by the hot coffee.
(4) Law of thermoelectricity
(i) Law of successive temperature : If initially temperature limits of the cold and the hot junction are t1 and
t2, say the thermo emf is E t2 . When the temperature limits are t2 and t3, then say the thermo emf is E t3 then
t1 t2
E t2 + E t3 = E t3 where E t3 is the thermo emf when the temperature limits are E t3
t1 t2 t1 t1 t1
(ii) Law of intermediate metals : Let A, B and C be the three metals of Seebeck series, where B lies
between A and C. According to this law, E B + E C = E C
A B A
When tin is used as a soldering metal in Fe-Cu thermocouple then at the junction, two different thermo couples
are being formed. One is between iron and tin and the other is between tin and copper, as shown in figure (i)
Now iron is thermoelectrically more positive as compared to tin and tin is more positive with respect to copper
(the element which occurs earlier in the seebeck series gets positively charged on losing the electrons at the
junction), so as clear from the figure below, the thermo emf’s of both the thermocouples shown in the figure (ii) are
additive
Cu E Cu
Sn Sn
Fe E Sn
Fe
(i) (ii)
∴ If soldering metal in a thermocouple is an intermediate metal in the series then thermo emf will not be affected.
It is also clear from the above discussions that if the soldering metal does not lie between two metals (in
Seebeck series) of thermocouple then the resultant emf will be subtractive.
(5) Effect of temperature on thermo emf : In a thermocouple as the temperature of the hot junction
increases keeping the cold junction at constant temperature (say 0oC). The thermo emf increases till it becomes
maximum at a certain temperature.
(i) Thermo electric emf is given by the equation E =αt+ 1 β t2
2
where α and β are thermo electric constant having units are volt/oC and volt/oC2 respectively (t = temperature
of hot junction).
(ii) The temperature of hot junction at which thermo emf becomes maximum is called neutral temperature (tn).
Neutral temperature is constant for a thermo couple (e.g. for Cu-Fe, tn = 270oC)
(iii) Neutral temperature is independent of the temperature of cold junction.
(iv) If temperature of hot junction increases beyond neutral temperature, thermo emf start decreasing and at a
particular temperature it becomes zero, on heating slightly further, the direction of emf is reversed. This temperature
of hot junction is called temperature of inversion (ti).
(v) Graphical representation of thermo emf Thermo
emf
(a) tn ti + tc
= 2
(b) Graph is parabolic O tn Temperatutire of hot
junction
(c) For E to be maximum (at t = tn)
dE = 0 i.e. α + β tn = 0 ⇒ tη = −α
dt β
(6) Thermo electric power : The rate of change of thermo emf with the change in the temperature of the
hot junction is called thermoelectric power.
It is also given by the slope of parabolic curve representing the variation of thermo emf with temperature of the
hot junction, as discussed in previous section.
It is observed from the above graph that as temperature of hot junction increases from that of the cold junction
to the neutral junction, though the thermo emf is increasing but the slope of the graph, that is the rate of change of
thermo emf with temperature of hot junction is decreasing. Note that, at the neutral temperature, the thermo emf is
maximum but the slope i.e. the thermoelectric power is zero.
The thermo electric power dE is also called Seebeck coefficient. Differentiating both sides of the
dt
equation of thermo emf with respect to t, we have P = dE = d (α t + 1 β t2) ⇒ P =α + β t
dt dt 2
The equation of the thermo electric power is of the type y = mx + c, so the graph of thermo electric power is
as shown below.
P
Slope β
α
t
Peltier Effect.
(1) If a current is passed through a junction of two different metals, the heat is either evolved or absorbed at
the junction. This effect is known as Peltier effect. It is the reverse of Seebeck effect. Before going into the detailed
explanation, we will first revise an important concept about absorption and evolution of energy when electric charge
is made to pass through two points having some potential difference.
When a positive charge flows from high potential to low potential, it releases energy and when positive charge
flows from low potential to high potential it absorbs energy.
(2) Explanation of Peltier effect : In the light of above statement it can be seen that if current is made to
flow in Fe-Cu thermocouple by connecting it to a battery then the junction at which current goes from Fe to Cu
becomes hot because here positive charge is flowing from high potential to low potential, so energy is released.
Remember that, in iron-copper thermocouple, the polarity of the contact potential at each junction is such iron is at
higher potential. Similarly the junction where current flows from Cu to Fe becomes colder because at this junction
current is flowing from negative to positive potential, so energy is absorbed. Thus it is observed that on application
of potential difference in a thermocouple temperature difference is automatically created. The amount of heat
absorbed at cold junction is equal to the heat released at hot junction.
Cu
––
++
Fe
(3) Peltier co-efficient (π) : Heat absorbed or liberated at the junction is directly proportional to the charge
passing through the junction i.e. H ∝ Q ⇒ H = πQ ; where π is called Peltier co-efficient. It’s unit is J/C or volt.
(i) If Q = 1 then H = π i.e. Peltier co-efficient of a junction is defined as heat absorbed or liberated at the
junction when a unit quantity of electric charge flows across the junction (H is also known as Peltier emf).
(ii) Relation between π and absolute temperature : Suppose the temperature of the cold junction is T
and that of the hot junction is T + dT and let dE be the thermo emf produced, then it is found that
π = T dE = T × S ; where T is in Kelvin and dE = P = Seebeck coefficient S
dT dT
(iii) π-depends on : (a) Temperature of junction (b) Difference in electron density of the two metal used in
thermocouple.
(iv) Comparison between Joule and Peltier effect
Joules effect Peltier’s Effect
(a) In joule’s effect energy is only released. (a) In peltier’s effect energy is released at one junction
and absorbed at the other junction.
(b) Heat produced depends upon i2, so, heat is always (b) Heat produced depends upon i1, ∴ junction at which
released, whether i is positive or negative. the heat is released or absorbed changes when the
direction of current changes.
(c) It is identically produced by ac or dc (c) In Peltier’s effect if ac is passed, at the same junction
heat is released when current flows in one direction and
absorbed when the direction of current reverses. The net
amount of heat released or absorbed at a junction is
therefore zero. Thus, Peltier’s effect cannot be observed
with ac.
(d) Joules effect is irreversible. (d) Peltier effect is reversible, its complimentary is
Seebeck effect.
(e) In Joule’s effect heat is released throughout the length (e) In this effect heat is released or absorbed only at the
of wire. junctions.
Thomson’s Effect.
(1) Definition : In Thomson’s effect we deal with only metallic rod and not with thermocouple as in Peltiers
effect and Seebeck’s effect. (That’s why sometimes it is known as homogeneous thermo electric effect. When a
current flows thorough an unequally heated metal, there is an absorption or evolution of heat in the body of the
metal. This is Thomson’s effect.
(2) Types of Thomson’s effect
(i) Positive Thomson’s effect
In positive Thomson’s effect it is found that hot end is at high potential and cold end is at low potential.
e.g. Cu, Sn, Ag, Sb
Element’s occurring before lead in Seebeck series are called thermoelectrically negative but this does not
mean that their Thompson effect is negative.
(ii) Negative Thomson’s effect
In the elements which show negative Thomson’s effect, it is found that the hot end is at low potential and
the cold end is at higher potential e.g. Fe, Co, Bi
(3) Thomson’s co-efficient : In Thomson’s effect it is found that heat released or absorbed is proportional
to Q∆θ i.e. H ∝ Q∆θ ⇒ H = σQΔθ where σ = Thomson’s coefficient. It’s unit is Joule/coulomboC or volt/oC and
∆θ = temperature difference.
(i) If Q = 1 and ∆θ = 1 then σ = H so the amount of heat energy absorbed or evolved per second between
two points of a conductor having a unit temperature difference, when a unit current is passed is known as
Thomson’s co-efficient for the material of a conductor.
(ii) It can be proved that Thomson co-efficient of the material of conductor σ = −T d2E also Seebeck co-
dT 2
efficient S = dE so dS d 2 E hence σ = −T dS = T × β ; where β = Thermo electric constant = dS
dT dT = dT 2 dT dt
Application of Thermo Electric Effect.
(1) To measure temperature : A thermocouple is used to measure very high (2000oC) as well as very low
(– 200oC) temperature in industries and laboratories. The thermocouple used to measure very high temperature
is called pyrometer.
(2) To detect heat radiation : A thermopile is a sensitive instrument used for detection of heat radiation
and measurement of their intensity. It is based upon Seebeck effect.
A thermopile consists of a number of thermocouples of Sb-Bi, all connected in series.
Sb T1
Heat radiations G
Bi T2
This instrument is so sensitive that it can detect heat radiations from a match stick lighted at a distance of 50
metres from the thermopile.
(3) Thermoelectric refrigerator : The working of thermo-electric refrigerator is based on Peltier effect.
According to Peltier effect, if current is passed through a thermocouple, heat is absorbed at one junction and is
evolved at the other junction of the thermocouple. If on the whole, the heat is absorbed, then the thermocouple acts
as thermoelectric refrigerator. It’s efficiency is small in comparison to conventional refrigerator.
(4) Thermoelectric generator : Thermocouple can be used to generate electric power using Seebeck effect
in remote areas. It can be achieved by heating one junction in a flame of kerosene oil lamp and keeping the other
junction at room or atmospheric temperature. The thermo emf so developed is used to operate radio receivers or
even radio transmitters.
Concepts
The emf developed in a thermo couple is rather small i.e. of the order of a few µV/oC.
A current is passed in a thermocouple formed with dissimilar metals whose one junction is heated and other is cooled. If π1 and
π2 are the Peltier co-efficient of cold and hot junction respectively then the net emf across the junction is proportional to (π2 –
π1)
Example
Example: 26 The smallest temperature difference that can be measured with a combination of a thermocouple of thermo
Solution : (a) e.m.f. 30 µV per degree and a galvanometer of 50 ohm resistance capable of measuring a minimum current
Example: 27
Solution : (d) of 3 × 10–7 ampere is [MP PET 2000]
Example: 28
(a) 0.5 degree (b) 1.0 degree (c) 1.5 degree (d) 2.0 degree
By using E = aθ ⇒ i R = aθ ⇒ 3 × 10–7 × 50 = 30 × 10–6 × θ ⇒ θ = 0.5 degree
The expression for thermo e.m.f. in a thermocouple is given by the relation E = 40 θ − θ2 , where θ is the
20
temperature difference of two junctions. For this, the neutral temperature will be [AMU (Engg.) 2000]
(a) 100o C (b) 200o C (c) 300o C (d) 400o C
Comparing the given equation of thermo e.m.f. with E = α t + 1 β t 2 we get α = 40 and β = 1 . By
2 − 10
using tn = −α ⇒ tn = 400o C .
β
One junction of a certain thermoelectric couple is at a fixed temperature Tr and the other junction is at
temperature T. The thermo electromotive force for this is expressed by E = K(T − Tr ) T0 − 1 (T + Tr ) at
2
temperature T = 1 T0 , the thermo electric power will be [MP PMT 1994]
2
(a) 1 KT0 (b) KT0 (c) 1 KT02 (d) 1 K(T0 − Tr )2
2 2 2
Solution : (a) As we know thermo electric power S = dE . Hence by differentiating the given equation and putting
Example: 29 dT
T = 1 T0 we get S = 1 KT0 .
2 2
The cold junction of a thermocouple is maintained at 10o C. No thermo e.m.f. is developed when the hot
junction is maintained at 530o C. The neutral temperature is [MP PMT 1994]
(a) 260o C (b) 270o C (c) 265o C (d) 520o C
Solution : (b) Given tc = 10o C and ti = 530o C hence by using tn = ti + tc ⇒ tn = 270o C
Example: 30 2
The thermo emf develops in a Cu-Fe thermocouple is 8.6 µV/oC. It temperature of cold junction is 0oC and
temperature of hot junction is 40oC then the emf obtained shall be
(a) 0.344 mV (b) 3.44 µV (c) 3.44 V (d) 3.44 mV
Solution : (a) By using thermo emf e = aθ where a = 8.6 µV and θ = temperature difference = 40oC
oC
Example: 31 So e = 8.6 × 10–6 × 40 = 344 µV = 0.344 mV.
A thermo couple develops 200 µV between 0oC and 100oC. If it develops 64 µV and 76 µV respectively
between (0oC – 32oC) and (32oC – 70oC) then what will be the thermo emf it develops between 70oC and
100oC
(a) 65 µV (b) 60 µV (c) 55 µV (d) 50 µV
Solution : (b) By using e 100 = e 32 + e 70 + e 100 ⇒ 200 = 64 + 76 + e71000 ⇒ e 100 = 60µV
Example: 32 0 0 32 70 70
Solution : (d) A thermo couple is formed by two metals X and Y metal X comes earlier to Y in Seebeck series. If temperature
Example: 33 of hot junction increases beyond the temperature of inversion. Then direction of current in thermocouple will
so
(a) X to Y through cold junction (b) X to Y through hot junction
(c) Y to X through cold junction (d) Both (b) and (c)
In the normal condition current flows from X to Y through cold. While after increasing the temperature of hot
junction beyond temperature of inversion. The current is reversed i.e. X to Y through hot junction or Y to X
through cold junction.
Peltier co-efficient of a thermo couple is 2 nano volts. How much heat is developed at a junction if 2.5 amp
current flows for 2 minute
(a) 6 ergs (b) 6 × 10–7 ergs (c) 16 ergs (d) 6 × 10–3 erg
Solution : (a) H = π i t = (2 × 10 −9 ) × 2.5 × (2 × 60) = 6 × 10 −7 J = 6 erg
Example: 34 A thermo couple develops 40 µV/kelvin. If hot and cold junctions be at 40oC and 20oC respectively then the
emf develops by a thermopile using such 150 thermo couples in series shall be
(a) 150 mV (b) 80 mV (c) 144 mV (d) 120 mV
Solution : (d) The temperature difference is 20oC = 20 K. So that thermo emf developed E = aθ = 40 µV × 20K = 800µV.
K
Hence total emf = 150 × 800 = 12 × 104 µV = 120 mV
Chemical Effect of Current
Current can produce or speed up chemical change, this ability of current is called chemical effect (shown by dc
not by ac).
When current is passed through an electrolyte, it dissociates into positive and negative ions. This is called
chemical effect of current.
Important Terms Related to Chemical Effect.
(1) Electrolytes : The liquids which allows the current to pass through them and also dissociates into ions on
passing current through them are called electrolytes e.g. solutions of salts, acids and bases in water, etc.
Note : ≅ These liquids which do not allow current to pass through them are called insulators (e.g.
vegetable oils, distilled water etc.) while the liquids which allows the current to pass through them
but do not dissociates into ions are called good conductors (e.g. Hg etc.)
≅ Solutions of cane sugar, glycerin, alcohol etc. are examples of non-electrolytes.
(2) Electrolysis : The process of decomposition of electrolyte solution into ions on passing the current
through it is called electrolysis. Positive ions (called cations)
Electric current Electrolyte Decomposed
into
Negative ions (called Anions)
Note : ≅Practical applications of electrolysis are Electrotyping, extraction of metals from the ores, Purification of
metals, Manufacture of chemicals, Production of O2 and H2, Medical applications and electroplating.
≅ Electroplating : It is a process of depositing a thin layer of one metal over another metal
by the method of electrolysis. The articles of cheap metals are coated with precious metals like silver
and gold to make their look more attractive. The article to be electroplated is made the cathode and
the metal to be deposited in made the anode. A soluble salt of the precious metal is taken as the
electrolyte. (If gold is to be coated then auric chloride is used as electrolyte).
(3) Electrodes : Two metal plates which are partially dipped in the electrolyte for passing the current through
the electrolyte.
Anode : Connected to positive terminal of battery Anode Cathode
Cathode : Connected to negative terminal of battery + –
(4) Voltameter : The vessel in which the electrolysis is
carried out is called a voltameter. It contains two electrodes Electrolyte
and electrolyte. It is also known as electrolytic cell.
(5) Equivalent weight : The ratio of the atomic weight of an element to its valency is defined as it’s
equivalent weight.
(6) Types of voltameter : Voltameter is divided mainly in following types
Cu-voltameter Ag voltameter Water voltameter
In copper voltameter, electrolyte is In silver voltameter electrolyte is a In water voltameter the electrolyte
solution of copper e.g. CuSO4, solution of silver, e.g. AgNO3. used is acidic water, because it is
CuCl2, Cu(NO3)2 etc. Cathode may Cathode may be of any metal but much more conducting than that of
be of any material, but anode must anode must be of silver. The pure water. So acid CH2SO4
be of copper. dissociation reaction is as follows increases the concentration of free
ions in the solution. The electrodes
CuSO4 in water dissociates as follows AgNO3 → Ag+ + NO3– are made of platinum, because it
does not dissolve into electrolyte and
CuSO4 → Cu++ + SO4– – The silver dissolves from the anode does not react with the products of
gets deposited on the cathode. electrolysis. When current flows
Cu++ moves towards cathode and During this process, the through the electrolyte, hydrogen gas
takes 2 electron to become neutral concentration of the electrolyte is collected in the tube placed over
and deposited on cathode remains unchanged. In this process the cathode (– ve electrode) and
one electron per reaction is active oxygen is collected in the tube placed
Cu++ + 2e → Cu and valence of Ag atom is also one. over the anode (+ve electrode).
SO4– – moves towards anode and • – + Anode Hydrogen and oxygen are liberated
looses 2 electrons their. Copper is Rh in the proportional in which they are
deposited on the cathode and an AgNO3 solution found in water i.e. the volume ratio
equivalent amount of copper is lost Ag of hydrogen and oxygen is 2 : 1.
by the anode, but the concentration
of copper sulphate solution remains O2 H2
the same. In this process, two
electrons per reaction are active and
valence of copper atom is also two.
Rh
•
A Cu C Cathode
plates Cu solution
Cu lost Cu deposited +–
– A+
Rh
Faraday’s Law of Electrolysis.
(1) First law : It states that the mass of substance deposited at the cathode during electrolysis is directly
proportional to the quantity of electricity (total charge) passed through the electrolyte.
Let m be the mass of the substance liberated, when a charge q is passed through the electrolyte. Then,
according to the Faraday’s first law of electrolysis m ∝ q or m = zq, where the constant of proportionality z is called
electrochemical equivalent (E.C.E.) of the substance. If a constant current i is passed through the electrolyte for time
t, then the total charge passing through the electrolyte is given by q = i t
Therefore we have m = zit . If q = 1 coulomb, then we have m = z × 1 or z = m
Hence, the electrochemical equivalent of substance may be defined as the mass of its substance deposited at
the cathode, when one coulomb of charge passes through the electrolyte.
S.I. unit of electrochemical equivalent of a substance is kilogram coulomb–1 (kg-C–1).
(2) Second law : If same quantity of electricity is passed through different electrolytes, masses of the
substance deposited at the respective cathodes are directly proportional to their chemical equivalents.
Let m be the mass of the ions of a substance liberated, whose chemical equivalent is E. Then, according to
Faraday’s of electrolysis, m∝ E or m = constant × E or m = constant
E
Note : ≅ Chemical equivalent E also known as equivalent weight in gm i.e. E = Atomic mass (A)
Valance (V)
(3) Relation between chemical equivalent and electrochemical equivalent : Suppose that on passing
same amount of electricity q through two different electrolytes, masses of the two substances liberated are m1 and
m2. If E1 and E2 are their chemical equivalents, then from Faraday’s second law, we have m1 = E1
m2 E2
Further, if z1 and z2 are the respective electrochemical equivalents of the two substances, then from Faraday’s
first law, we have m1 = z1q and m2 = z2q ⇒ m1 = z1
m2 z2
So from above equation z1 = E1 ⇒ z∝E ⇒ z2 = z1 × E2
z2 E2 E1
(4) Faraday constant : As we discussed above E ∝ z ⇒ E = Fz ⇒ z = E = A
F VF
‘F’ is proportionality constant called Faraday’s constant.
As z = E and z = m (from I law) so E = m hence if Q =1 Faraday then E =m i.e. If electricity supplied to
F Q F Q
a voltameter is 1 Faraday then amount of substance liberated or deposited is (in gm) equal to the chemical
equivalent. e.g. to deposit 16 gm O2; 2 Faraday electricity is required.
Note : ≅Remember Number of gm equivalent = given mass × valency
atomic mass
≅ 1 Faraday = 96500 C
≅ Also F = Ne {where N = Avogrado number)
Electro Chemical Cell.
It is an arrangement in which the chemical energy is converted into electrical energy due to chemical action
taking place in it. The total amount of energy that can be provided by this cell is limited and depends upon the
amount of reactants. Electro chemical cells are of two types.
(1) Primary cell : Is that cell in which electrical energy is produced due to chemical energy. In the primary
cell, chemical reaction is irreversible. This cell can not be recharged but the chemicals have to be replaced after a
long use examples of primary cells; Voltaic cell, Daniel cell, Leclanche cell and Dry cell etc.
(i) Voltaic cell
Cu Zn Positive electrode – Cu rod
+ –
Negative electrode – Zn rod
Polarisation Cu
Zn Electrolyte Electrolyte – dil. H2SO4
Dil. H2SO4 Emf – 1.08 V
Local action
Main chemical reactions
H2SO4 → 2H+ + SO4– –
Zn → Zn++ + 2e–
(ii) Daniel cell C
Cu – Zn A+ Cu Pot Positive electrode – Cu rod
Crystals Rod Negative electrode – Zn rod
CuSO4 Solution Electrolyte – dil. H2SO4
(Depolariser) Emf – 1.1 V
Porous pot
Main chemical reactions
Dil. H2SO4
(electrolyte) Zn + H2SO4 → ZnSO4 + 2H+ + 2e–
2H+ + CuSO4 → H2SO4 + Cu++
(iii) Lechlanche cell
Positive electrode – Carbon rod
Zn rod C A Graphite rod Negative electrode – Zn rod
– +
Porous pot
Glass pot MnO2 + charcoal dust Electrolyte – NH4Cl solution
(depolariser) Emf – 1.45 V
NH4Cl Solution
Main chemical reactions
(Electrolyte)
Zn + 2NH4Cl → 2NH3 + ZnCl2 + 2H+ + 2e–
2H+ + 2MnO2 → Mn2O3 + H2O + 2 unit of positive
charge
(iv) Dry cell
C +A Positive electrode – Carbon rod with
– brass cap
Graphite rod
Zn pot Negative electrode – Zn vessel
Porous pot Electrolyte – Paste of NH4Cl
MnO2 + charcoal dust and saw dust
(depolariser)
NH4Cl Solution Emf – 1.5 V
(Electrolyte) Main chemical reactions – Similar to
Leclanche cell
(2) Secondary cell : A secondary cell is that cell in which the electrical energy is first stored up as a chemical
energy and when the current is taken from the cell, the chemical energy is reconverted into electrical energy. In the
secondary cell chemical reaction are reversible. The secondary cells are also called storage cell or accumulator. The
commonly used secondary cells are
In charged Lead accumulator Alkali accumulator
+– + – Ni(OH)2
Glass vessel Fe(OH)2
PbO2 Perforated
steel grid
Pb
dil. H2SO4 KOH 20%
+ Li(OH), 1%
Positive electrode Perforated lead plates coated with PbO2 Perforated steel plate coated with Ni(OH)4
Negative electrode Perforated lead plates coated with pure lead Perforated steel plate coated with Fe
During charging Chemical reaction Chemical reaction
At cathode : PbSO4 + 2H+ + 2e– → Pb + H2SO4 At cathode :
During At anode : Fe(OH)2 + 2OH+ – 2e– → Ni(OH)4
discharging PbSO4 + SO4– – + 2H2O – 2e– → PbO2 + 2H2SO4 At anode :
Specific gravity of H2SO4 increases and when specific Fe(OH)2 + 2K+ + 2e– → Fe + 2KOH
Efficiency gravity becomes 1.25 the cell is fully charged. Emf of cell : When cell is fully charge then E
Emf of cell : When cell is full charged them E = 2.2 volt = 1.36 volt
Chemical reaction Chemical reaction
At cathode : Pb + SO4– – – 2e– → PbSO4 At cathode :
At anode : Fe + 2OH– – 2e– → Fe(OH)2
PbO4 + 2H+ – 2e– + H2SO4 → PbSO2 + 2H2O At anode :
Specific gravity of H2SO4 decreases and when specific Ni(OH)4 + 2K+ + 2e– → Ni(OH)2 + 2KOH
gravity falls below 1.18 the cell requires recharging. Emf of cell : When emf of cell falls below 1.1 V
Emf of cell : When emf of cell falls below 1.9 volt it requires charging.
the cell requires recharging.
80% 60%
(3) Defects In a primary cell : In voltaic cell there are two main defects arises.
Local action : It arises due to the presence of impurities of iron, carbon etc. on the surface of commercial Zn rod
used as an electrode. The particles of these impurities and Zn in contact with sulphuric acid form minute voltaic cell in which
small local electric currents are set up resulting in the wastage of Zn even when the cell is not sending the external current.
Removal : By amalgamating Zn rod with mercury (i.e. the surface of Zn is coated with Hg).
Polarisation : It arises when the positive H2 ions which are formed by the action of Zn on sulphuric acid,
travel towards the Cu rod and after transferring, the positive charge converted into H2 gas atoms and get deposited
in the form of neutral layer of a gas on the surface of Cu rod. This weakens the action of cell in two ways.
Removal : Either by brushing the anode the remove the layer or by using a depolariser (i.e. some oxidising
agent MnO2, CuSO4 etc which may oxidise H2 into water).
Note : ≅The end point voltage of dry cell is 0.8 V.
Concepts
Electrolysis takes place for dc and low frequency ac, as at high frequency, due to inertia (i.e. mass) ions cannot follow the
frequency of ac.
Electrolytes are less conducting then the metallic conductors because ions are heavier than electrons.
If ρ is the density of the material deposited and A is the area of deposition then the thickness (d) of the layer of the material deposited in
electroplating process is d = m Zi t ; where m = deposited mass, Z = electro chemical equivalent, i = electric current.
ρA = ρA
Example
Example: 35 In an electroplating experiment, m gm of silver is deposited when 4 ampere of current flows for 2 minute. The
Solution : (b) amount (in gm) of silver deposited by 6 ampere of current for 40 second will be
Example: 36 [MNR 1991; MP PET 2002]
Solution : (c)
Example: 37 (a) 4 m (b) m/2 (c) m/4 (d) 2m
Solution : (b)
Example: 38 By using m= zit ⇒ m1 = i1t1 ⇒ m = 4 × 2 × 60 ⇒ m2 =m/2
m2 i2t2 m2 6 × 40
Solution : (b)
Example: 39 A current of 16 ampere flows through molten NaCl for 10 minute. The amount of metallic sodium that
appears at the negative electrode would be
Solution : (b) [EAMCET 1984]
Example: 40
(a) 0.23 gm (b) 1.15 gm (c) 2.3 gm (d) 11.5 gm
Solution : (a)
By using m = zit = A it ⇒ m = 1× 23 × 16 × 10 × 60 = 2.3 gm
Example: 41 VF 96500
For depositing of 1 gm of Cu in copper voltameter on passing 2 amperes of current, the time required will be
(For copper Z = 0.00033 gm/C)
(a) Approx. 20 minutes (b) Approx. 25 minutes (c) Approx. 30 minutes (d) Approx. 35 minutes
By using m = zit ⇒ 1 = 0.00033 × 2 × t ⇒ t = 1515.15 sec ≈ 25 min.
Two electrolytic cells containing CuSO4 and AgNO3 respectively are connected in series and a current is
passed through them until 1 mg of copper is deposited in the first cell. The amount of silver deposited in the
second cell during this time is approximately (Atomic weights of copper and Silver are respectively 63.57 and
107.88) [MP PMT 1996]
(a) 1.7 mg (b) 3.4 mg (c) 5.1 mg (d) 6.8 mg
By using m1 = E1 ⇒ 1 = 63.57 / 2 = 31.7 ⇒ m2 = 3.4 mg
m2 E2 m2 107.88 / 1 107.88
When a copper voltameter is connected with a battery of emf 12 volts, 2 gms of copper is deposited in 30
minutes. If the same voltameter is connected across a 6 volt battery, then the mass of copper deposited in 45
minutes would be
[SCRA 1994]
(a) 1 gm (b) 1.5 gm (c) 2 gm (d) 2.5 gm
By using m = zi t = zV t ⇒ m1 = V1 t 1 2 = 12 × 30 ⇒ m2 = 1.5 gm
R m2 V2 t 2 ⇒ m2 6 × 45
Silver and copper voltameter are connected in parallel with a battery of e.m.f. 12 V. In 30 minutes, 1 gm of
silver and 1.8 gm of copper are liberated. The energy supplied by the battery is (ZCu = 6.6 × 10–4 gm/C and
ZAg = 11.2 × 10–4 gm/C)
[IIT 1975]
(a) 24.13 J (b) 2.413 J (c) 0.2413 J (d) 2413 J
By using m = z i t, for Ag voltameter 1 = 11.2 × 10–4 × i1 × 30 × 60 ⇒ i1 = 0.5 amp. i1 Ag volta
For Cu voltameter 1.8 = 6.6 × 10–4 × i2 × 30 × 60 ⇒ i2 = 1.5 amp i Cu volta
Main current i = i1 + i2 = 1.5 + 0.5 = 2A. i2
So energy supplied = Vi = 12 × 2 = 24 J
12 V
Amount of electricity required to pass through the H2O voltameter so as to liberate 11.2 litre of hydrogen will be
(a) 1 Faraday (b) 1 Faraday (c) 2 Faraday (d) 3 Faraday
2
Solution : (a) Mass of hydrogen in 11.2 litres of hydrogen = 11.2 × M = 11.2 × 2 = 1gm
Example: 42 22.4 22.4
Solution : (b)
Example: 43 We know that 1 gm of hydrogen is equal to 1 gm equivalent wt. of hydrogen. It means that 11.2 litre of
Solution : (c)
hydrogen at NTP represents 1 gm equivalent of hydrogen, so for liberation it requires 1 Faraday electricity.
Example: 44
Solution : (b) Amount of electricity required to liberate 16 gm of oxygen is
(a) 1 Faraday (b) 2 Faraday (c) 1 Faraday (d) 3 Faraday
2
Number of gm equivalent = Given mass = 16 2 = 2. Hence 2 Faraday electricity is needed.
gm equivalent weight 16 /
Total surface area of a cathode is 0.05 m2 and 1 A current passes through it for 1 hour. Thickness of nickel
deposited on the cathode is (Given that density of nickel = 9 gm/cc and it’s ECE = 3.04 × 10–4 gm/C)
(a) 2.4 m (b) 2.4 cm (c) 2.4 µm (d) None of these
Mass deposited = density × volume of the metal …… (i)
……(ii)
m=ρ×A×x
Hence from Faraday first law m = Zit
So from equation (i) and (ii) Zit = ρ × Ax ⇒ x = Zit = 3.04 × 10 −4 × 10 −3 × 1 × 36 × ...... = 2.4 × 10 −6 m = 2.4 µm
ρA 9000 × 0.05
Resistance of a voltameter is 2 Ω, it is connected in series to a battery of 10 V through a resistance of 3 Ω. In a
certain time mass deposited on cathode is 1 gm. Now the voltameter and the 3Ω resistance are connected in
parallel with the battery. Increase in the deposited mass on cathode in the same time will be
(a) 0 (b) 1.5 gm (c) 2.5 gm (d) 2 gm
Remember mass of the metal deposited on cathode depends on the current through the voltameter and not on the
current supplied by the battery. Hence by using m = Zit, we can say m Parallel = i Parallel ⇒ mParallel = 5 ×1 = 2.5gm .
mSeries iSeries 2
Hence increase in mass = 2,5 – 1 = 1.5 gm
3Ω Volta i2 2Ω
i1 2Ω Volta
10 10
i1 = 5 = 2A 3Ω i2 = 2 = 5A
10V 10V
Tricky Example: 3
In a copper voltameter, the mass deposited in 30 s is m gram. If the current time graph is as shown in
the figure, the e.c.e. of copper, in gm/coulomb, will be [Pantnagar 1987]
(a) m
(b) m i (mA)
2 100 mA
(c) 0.6 m
(d) 0.1 m 10 20 30 t (sec)
Solution : (b) Area of the given curve on x-axis = it = 1 (10 + 30) × 100 × 10 −3 = 2 Coulomb
2
From Faraday's first law m = zit ⇒ z = m = m .
it 2
Electromagnetic waves
Maxwell’s Contribution.
(1) Ampere’s Circuital law
According to this law the line integral of magnetic field along any closed path or circuit is µ0 times the total
∫current threading the closed circuit i.e., →B.d→l = µ0i
(2) Inconsistency of Ampere’s law
Maxwell explained that Ampere's law is valid only for steady current or when the electric field does not change
with time. To see this inconsistency consider a parallel plate capacitor being
charged by a battery. During the charging time varying current flows through
connecting wires.
∫Applying Ampere's law for loop l1 and l2 l1 →B.d→l = µ0i
∫But →B.d→l = 0 (Since no current flows through the region between the
l2
plates). But practically it is observed that there is a magnetic field between the plates. Hence Ampere's law fails
∫i.e. →B.d→l ≠ µ0i .
l1
(3) Modified Ampere’s Circuital law or Ampere- Maxwell’s Circuital law
Maxwell assumed that some sort of current must be flowing between the capacitor plates during charging
process. He named it displacement current. Hence modified law is as follows
∫ ∫→ → or →B. d→l = µ0 (ic + ε0 dφ E )
B. dl = µ0 (ic + id ) dt
where ic = conduction current = current due to flow of charges in a conductor and
id = Displacement current = ε0 dφ E = current due to the changing electric field between the plates of
dt
the capacitor
Note : ≅ Displacement current ( id ) = conduction current (ic ).
≅ ic and id in a circuit, may not be continuous but their sum is always continuous.
(4) Maxwell’s equations
E→.d→s = q (Gauss’s law in electrostatics) (ii) B→. → (Gauss's law in magnetism)
∫ ∫(i) = 0
s ε0 ds
s
∫ ∫(iii)→→ dφ B (iv) → → dφ E (Maxwell- Ampere's Circuital law)
. = − dt (Faraday’s law of EMI) = µo (ic + εo dt
B dl B dl
EM Waves.
(1) Definition
A changing electric field produces a changing magnetic field and vice versa which gives rise to a transverse
wave known as electromagnetic wave. The time varying electric and magnetic field are mutually perpendicular to
each other and also perpendicular to the direction of propagation of this wave.
The electric vector is responsible for the optical effects
of an EM wave and is called the light vector.
(2) History of EM waves :
(i) Maxwell : Was the first to predict the EM wave.
(ii) Hertz : Produced and detected electromagnetic waves experimentally at wavelengths of 6 m.
Experimental setup
Hertz experiment based on the fact that a oscillating charge is accelerating continuously, it will radiate
electromagnetic waves continuously. In the following figure
• The metallic plates (P1 and P2) acts as a capacitor.
• The wires connecting spheres S1 and S2 to the plates provide a low inductance.
Input Gap P1 S'1
Induction coil S'2
S1
Receiver
S2
P2
When a high voltage is applied across metallic plates these plates get discharged by sparking across the narrow
gap. The spark will give rise to oscillations which in turn send out electromagnetic waves. Frequency of these wave
is given by ν = 2π 1
LC
The succession of sparks send out a train of such waves which are received by the receiver.
(iii) J.C. Bose : Produced EM waves of wavelength ranging from 5mm to 25 mm.
(iv) Marconi : Successfully transmitted the EM waves up to a few kilometer. Marconi discovered that if one
of the spark gap terminals is connected to an antenna and the other terminal is Earthed, the electromagnetic waves
radiated could go upto several kilometers.
(3) Source of EM waves
A charge oscillating harmonically is a source of EM waves of same frequency.
(4) Production of EM waves
A simple LC oscillator and energy source can produce waves of desired frequency.
Frequency of oscillating discharge in LC
circuit = Frequency of EM waves = 1
2π LC
Note : ≅ In an atom an electron circulating around the nucleus in a stable orbit, although accelerating
does not emit electromagnetic waves; it does so only when it jumps from a higher energy orbit to a
lower energy orbit.
≅ Electromagnetic waves (X-rays) are also produced when fast moving electrons are suddenly stopped
by a metal target of high atomic number.
≅ Most efficient antennas are those which have a size comparable to the wavelength of the of
electromagnetic wave they emit or receive.
(5) Nature of EM waves
The EM Waves are transverse in nature. They do not require any material medium for their propagation.
(6) Properties of EM waves
(i) Speed : In free space it's speed c = 1 = E0 = 3 × 108 m / s.
µ0ε 0 B0
In medium v = 1 ; where µ0 = Absolute permeability, ε 0 = Absolute permittivity
µε
E0 and B0 = Amplitudes of electric field and magnetic field vectors.
(ii) Energy : The energy in an EM waves is divided equally between the electric and magnetic fields.
Energy density of electric field ue = 1 ε 0 E 2 , Energy density of magnetic field uB = 1 B2
2 2 µ0
It is found that ue = uB . Also uav = ue + uB = 2ue = 2uB = ε0E2 = B2
µ0
(iii) Intensity (I) : The energy crossing per unit area per unit time, perpendicular to the direction of propagation
of EM wave is called intensity. I = uav × c = 1 ε 0 E 2c = 1 B2 .c
2 2 µ0
(iv) Momentum : EM waves also carries momentum, if a portion of EM wave of energy u propagating with
speed c, then linear momentum = Energy (u)
Speed (c)
Note : ≅ When the incident EM wave is completely absorbed by a surface, it delivers energy u and
momentum u / c to the surface.
≅ When a wave of energy u is totally reflected from the surface, the momentum delivered to surface is 2u / c.
(v) Poynting vector(S). : In EM waves, the rate of flow of energy crossing a unit area is described by the
poynting vector. It's unit is watt / m2 and S = 1 (E × B) = c 2ε0(E × B) . Because in EM waves E and B are
µo
perpendicular to each other, the magnitude of S is | S |= 1 E B sin 90o = EB = E2 .
µ0 µ0 µC
Note : ≅ The direction of the poynting vector S at any point gives the wave's direction of travel and
direction of energy transport the point.
(vi) Radiation pressure : Is the momentum imparted per second pre unit area. On which the light falls.
For a perfectly reflecting surface Pr = 2S ; S = Poynting vector; c = Speed of light
c
For a perfectly absorbing surface Pa = S .
c
Note : ≅ The radiation pressure is real that's why tails of comet point away from the sun.
EM Spectrum.
The whole orderly range of frequencies/wavelengths of the EM waves is known as the EM spectrum.
Uses of EM spectrum Uses
Gives informations on nuclear structure, medical treatment etc.
Radiation
γ-rays Medical diagnosis and treatment study of crystal structure, industrial radiograph.
X-rays Preserve food, sterilizing the surgical instruments, detecting the invisible writings,
UV- rays finger prints etc.
To see objects
Visible light To treat, muscular strain for taking photography during the fog, haze etc.
Infrared rays In radar and telecommunication.
Micro wave and radio wave
Earth’s Atmosphere.
The gaseous envelope surrounding the earth is called it's atmosphere. The atmosphere contains 78% N 2,
21% O2 , and traces of other gases (like helium, krypton, CO2 etc.)
(1) Division of earth's atmosphere
Earth atmosphere has been divided into regions as shown.
(i) Troposphere : In this region, the temperature decreases with height from 290 K to 220 K.
(ii) Stratosphere : The temperature of stratosphere varies from 220 K to 200 K.
(iii) Mesosphere : In this region, the temperature falls to 180 K.
(iv) Ionosphere : Ionosphere is partly composed of charged particles, ions and electrons, while the rest of the
atmosphere contains neutral molecules.
(v) Ozone layer absorbs most of the ultraviolet rays emitted by the sun.
(vi) Kennelly heaviside layer lies at about 110km from the earth's surface. In this layer concentration of
electron is very high.
(vii) The ionosphere plays a vital role in the radio communication.
Ionosphere Appleton layer ≃480km
Earth ≃80km
Kennelly Heaviside ≃50km
Thermosphere
Mesosphere ≃12km
Ozone layer
Stratosphere
Troposphere
(2) Green house effect
The warming of earth's atmosphere due to the infrared radiations reflected by low lying clouds and carbon
dioxide in the atmosphere of earth is called green house effect.
(3) Modulation and demodulation
The audio waves can be heard only over short distances. To overcomes this difficulty, an audio wave (low
frequency) to be transmitted is superimposed on the carrier wave (high frequency). This process of superimposing is
called modulation.
The process of separating the audio frequency wave from the carrier wave is called demodulation.
Amplitude modulation (Amplitude of
carrier wave modifies in accordance
with the amplitude of modulating
wave)
Frequency modulation (frequency of
carrier wave changes in accordance
with the amplitude of modulating
wave)
(4) Role of earth's atmosphere in propagation of radio waves
(i) Radio waves classification :
(a) Very low frequency (VLF) → 10 KHz to 30 KHz
(b) Low frequency (LF) → 30 KHz to 300 KHz
(c) Medium frequency (MF) or medium wave (MW) → 300 KHz to 3000 KHz
(d) High frequency (HF) or short wave (SW) → 3 MHz to 30 MHz
(e) Very high frequency (VHF) → 30 MHz to 300 MHz
(f) Ultra high frequency (UHF) → 300 MHz to 3000 MHz
(g) Super high frequency or micro waves → 3000 MHz to 300, 000 MHz
(ii) Amplitude modulated transmission : Radio waves having frequency less than or equal to 30 MHz
form an amplitude modulation band (or AM band). The signals can be transmitted from one place to another place
on earth's surface in two ways
(a) Ground wave propagation : The radio waves following the surface of the earth are called ground waves.
(b) Sky wave propagation : The amplitude modulated radio waves which are reflected back by the ionosphere
are called sky waves.
(iii) Frequency modulated (FM) transmission : Radio waves having frequencies between 80 MHz and
200 MHz form a frequency modulated bond. T.V. signals are normally frequency modulated.
Note : ≅ Ionosphere cannot reflect back the waves of frequencies greater than 40 MHz as these waves
easily penetrate through the ionosphere.
(5) T.V. Signals :
(i) T.V. signals are normally frequency modulated. So
T.V. signals can be transmitted by using tall antennas.
(ii) Distance covered by the T.V. signals d = 2hR
(h = Height of the antenna, R = Radius of earth)
(iii) Area covered A = πd 2 = 2πhR
(iv) Population covered = area × population density.
Example
Example: 1 A flash light is covered with a filter that transmits red light. The electric field of the emerging beam is
represented by a sinusoidal plane wave Ex = 36 sin(1.20 × 107 z − 3.6 × 1015 t) V / m . The average intensity of
Solution : (b) the beam will be
Example: 2
(a) 0.86 W / m2 (b) 1.72 W / m2 (c) 3.44 W / m2 (d) 6.88 W / m2
I av = cε 0 E 2 = 3 × 108 × 8.85 × 10 −15 × 36 2 = 1.72 W / m2
0 2
2
What should be the height of transmitting antenna if the T.V. telecast is to cover a radius of 128 Km
(a) 1560 m (b) 1280 m (c) 1050 m (d) 79 m
Solution : (b) Height of transmitting antenna h= d2 = (128 × 10 3 )2 = 1280 m
Example: 3 2 Re 2 × 6.4 × 106
Solution : (a) A T.V. tower has a height of 100 m. How much population is covered by T.V. broadcast, if the average
population density around the tower is 1000 / Km2
Example: 4
Solution : (c) (a) 39.5 × 105 (b) 19.5 × 106 (c) 29.5 × 107 (d) 9 × 10 4
Example: 5
Solution : (b) Radius of the area covered by T.V. telecast d = 2hRe
Example: 6 Total population covered = πd 2 × population density = 2πhRe × Population density
Solution : (c)
Example: 7 = 2 × 3.14 × 100 × 6.4 × 106 1000 = 39.503 × 105
× 106
Solution : (d)
Example: 8 An electromagnetic radiation has an energy 14.4 KeV. To which region of electromagnetic spectrum does it belong
Solution : (c) (a) Infra red region (b) Visible region (c) X-rays region (d) γ -ray region
Example: 9
λ = hc = 6.6 × 10 −34 × 3 × 108 = 0.8 × 10 −10 m = 0.8 Å . This wavelength belongs to X-ray region.
E 14.4 × 10 3 × 1.6 × 10 −19
A point source of electromagnetic radiation has an average power output of 800W. The maximum value of
electric field at a distance 3.5 m form the source will be
(a) 56.7 V/m (b) 62.6 V/m (c) 39.3 V/m (d) 47.5 V/m
Intensity of electromagnetic wave given is by I = Pav = Em2
4πr 2 2µ0c
Em = µ0cPav = (4π × 10 −7 ) × (3 × 108 ) × 800 = 62.6 V /m
2πr 2 2π × 3.5 2
In the above problem, the maximum value of magnetic field will be
(a) 2.09 × 10 −5 T (b) 2.09 × 10 −6 T (c) 2.09 × 10 −7 T (d) 2.09 × 10 −8 T
The maximum value of magnetic field is given by Bm = Em = 62.6 = 2.09 × 10 −7 T
c 3 × 108
A plane electromagnetic wave of wave intensity 6W / m2 strikes a small mirror area 40cm2 , held
perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second
will be
(a) 6.4 × 10−7 kg − m / s 2 (b) 4.8 × 10−8 kg − m / s 2 (c) 3.2 × 10−9 kg − m / s 2 (d) 1.6 × 10−10 kg − m / s 2
In one second p = 2U = 2Sav A = 2 × 6 × 40 × 10−4 = 1.6 × 10−10 kg − m / s 2
c c 3 × 108
The charge on a parallel plate capacitor is varying as q = q0 sin 2πnt . The plates are very large and close
together. Neglecting the edge effects, the displacement current through the capacitor is
(a) q (b) q0 sin 2πnt (c) 2πnq0 cos 2πnt (d) 2πnq0 cos 2πnt
ε0A ε0 ε0
ID = dq = d q0 sin 2πnt = 2πnq0 cos 2πnt
dt dt
The value of magnetic field between plates of capacitor, at distance of 1m from centre where electric field
varies by 1010 V / m / s will be
(a) 5.56T (b) 5.56µT (c) 5.56mT (d) 5.56nT
Solution : (d) B = µ0ε 0r dE = 1 × 1010 = 5.56 × 10−8 T e = 1
2 dt 2 × 9 × 1016
µ0ε 0
Reflection of light
When a ray of light after incidenting on a boundary separating two media comes back into the same media,
then this phenomenon, is called reflection of light.
Incident ray Normal ⇒ ∠i = ∠r
⇒ After reflection, velocity, wave length and frequency
Reflected ray
of light remains same but intensity decreases
ir ⇒ There is a phase change of π if reflection takes place
Boundary
from denser medium
Note : ≅ After reflection velocity, wavelength and frequency of light remains same but intensity
If light ray incident normally on a surface, after reflection it retraces the path.
decreases.
≅
Real and virtual images
If light rays, after reflection or refraction, actually meets at a point then real image is formed and if they
appears to meet virtual image is formed.
I O Real image (Real image)
(Real image) (Virtual object) O
I
(Virtual object)
O I (Virtual image) (Real object)
(Real object) (Virtual image) (Virtual image)
Plane Mirror.
The image formed by a plane mirror is virtual, erect, laterally inverted, equal in size that of the object and at a
distance equal to the distance of the object in front of the mirror.
xx
(1) Deviation : Deviation produced by a plane mirror and by two inclined plane mirrors.
ir δ 2θ Final path
δ θ Original path δ
δ = (180 – 2i) δ = (360 – 2θ)
Note : ≅ If two plane mirrors are inclined to each other at 90o, the emergent ray is anti-parallel to
incident ray, if it suffers one reflection from each. Whatever be the angle to incidence.
(2) Rotation : If a plane mirror is rotated in the plane of incidence through angle θ, by keeping the incident
ray fixed, the reflected ray turned through an angle 2θ.
IR RR IR θ
θ
θ
RR
(3) Images by two inclined plane mirrors : When two plane mirrors are inclined to each other at an angle
θ, then number of images (n) formed of an object which is kept between them.
(i) n = 360 − 1 ; If 360 = even integer
θ θ
(ii) If 360 = odd integer then there are two possibilities
θ
(a) Object is placed symmetrically (b) Object is placed asymmetrically
Object n = 360 − 1 Object n = 360
θ α θ
θ/2
θ/2 β
Note : ≅ If θ = 0o i.e. mirrors are parallel to each other so n = ∞ i.e. infinite images will be formed.
≅ If θ = 90o, n = 360 −1 = 3
90
≅If θ = 72o, n= 360 −1 = 4 (If nothing is said object is supposed to be symmetrically placed).
72
(4) Other important informations
(i) When the object moves with speed u towards (or away) from the plane mirror then image also moves
toward (or away) with speed u. But relative speed of image w.r.t. object is 2u.
(ii) When mirror moves towards the stationary object with speed u, the image will move with speed 2u.
O I OI
u u
Rest 2u
Mirror at rest u
Mirror is moving
(iii) A man of height h requires a mirror of length at least equal to h/2, to see his own complete image.
(iv) To see complete wall behind himself a person requires a plane mirror of at least one third the height of
wall. It should be noted that person is standing in the middle of the room.
H M' H M'
E E h
h h 3
2 hE
L M' M'
B d
d
Concepts
The reflection from a denser medium causes an additional phase change of π or path change of λ/2 while reflection from rarer
medium doesn't cause any phase change.
We observe number of images in a thick plane mirror, out of them only second is brightest.
Incident light (100%)
10% Brightest image
80%
I 9%
0.9%O
M I
M Wrong
OM = MI
O Correct
OM = MI
To find the location of an object from an inclined plane mirror, you have to see the perpendicular distance of the object from the mirror.
Example
Example: 1 A plane mirror makes an angle of 30o with horizontal. If a vertical ray strikes the mirror, find the angle between
Solution : (c)
mirror and reflected ray [RPET 1997]
(a) 30o (b) 45o (c) 60o (d) 90o
Since angle between mirror and normal is 90o and reflected ray (RR) makes an IR
angle of 30o with the normal so required angle will be θ = 60o . 30o
RR 30o
θ = 60o
30o
Example: 2 Two vertical plane mirrors are inclined at an angle of 60o with each other. A ray of light travelling horizontally
Solution : (d) is reflected first from one mirror and then from the other. The resultant deviation is
Example: 3
Solution : (c) (a) 60o (b) 120o (c) 180o (d) 240o
Example: 4 By using δ = (360 − 2θ ) ⇒ δ = 360 − 2 × 60 = 240o
Solution : (a)
Example: 5 A person is in a room whose ceiling and two adjacent walls are mirrors. How many images are formed
Solution : (b)
Example: 6 [AFMC 2002]
(a) 5 (b) 6 (c) 7 (d) 8
The walls will act as two mirrors inclined to each other at 90o and so sill form 360 −1 = 3 images of the person.
90
Now these images with object (Person) will act as objects for the ceiling mirror and so ceiling will form 4
images as shown. Therefore total number of images formed = 3 + 4 = 7
I1 O I′1 O′
I′2 I′3
I2 I3 Four images by ceiling
Three images by walls
Note : ≅ The person will see only six images of himself (I1, I 2 , I3, I ' , I ' , I ' )
1 2 3
A ray of light makes an angle of 10o with the horizontal above it and strikes a plane mirror which is inclined at
an angle θ to the horizontal. The angle θ for which the reflected ray becomes vertical is
(a) 40o (b) 50o (c) 80o (d) 100o
From figure Vertical RR
θ + θ + 10 = 90
⇒ θ = 40o θ IR
θ 10o Horizontal line
θ Plane mirror
A ray of light incident on the first mirror parallel to the second and is reflected from the second mirror parallel
to first mirror. The angle between two mirrors is
(a) 30o (b) 60o (c) 75o (d) 90o
From geometry of figure
θ + θ + θ = 180o θ
⇒ θ = 60o θ
θθ
θ
A point object is placed mid-way between two plane mirrors distance 'a' apart. The plane mirror forms an
infinite number of images due to multiple reflection. The distance between the nth order image formed in the
two mirrors is
(a) na (b) 2na (c) na/2 (d) n2 a
Solution : (b) III order II order I order M' M I order II order III order
image image image O image image image
I3' I2' I1' a/2 a/2 a/2 a/2 I1 I2 I3
3a/2 a 3a/2
5a/2 5a/2
Example: 7 From above figure it can be proved that seperation between nth order image formed in the two mirrors = 2na
Two plane mirrors P and Q are aligned parallel to each other, as shown in the figure. A light ray is incident at
an angle of θ at a point just inside one end of A. The plane of incidence coincides with the plane of the figure.
The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is
(a) l l
d tanθ
(b) d dθ
l tanθ
(c) ld tanθ
Solution : (a) (d) None of these
Suppose n = Total number of reflection light ray undergoes before exist out.
x = Horizontal distance travelled by light ray in one reflection.
So nx = l also tanθ = x l
d x
θθ
n l d
tanθ
⇒ = d
Example: 8 A plane mirror and a person are moving towards each other with same velocity v. Then the velocity of the
Solution : (c) image is
Example: 9
(a) v (b) 2v (c) 3v (d) 4v
Solution : (b)
If mirror would be at rest, then velocity of image should be 2v. but due to the motion of mirror, velocity of
image will be 2v + v = 3v.
A ray reflected successively from two plane mirrors inclined at a certain angle undergoes a deviation of 300o.
The number of images observable are
(a) 10 (b) 11 (c) 12 (d) 13
By using δ = (360 − 2θ ) ⇒ 300 = 360 − 2θ
⇒ θ = 30o . Hence number of images = 360 − 1 = 11
30
Tricky example: 1
A small plane mirror placed at the centre of a spherical screen of radius R. A beam of light is falling on
the mirror. If the mirror makes n revolution. per second, the speed of light on the screen after reflection
from the mirror will be
(a) 4πnR (b) 2πnR (c) nR (d) nR
2π 4π
Solution : (a) When plane mirror rotates through an angle θ, the reflected ray rotates through an angle 2θ. So spot on
the screen will make 2n revolution per second
∴ Speed of light on screen v = ωR = 2π (2n)R = 4πnR
Tricky example: 2
A watch shows time as 3 : 25 when seen through a mirror, time appeared will be
(a) 8 : 35 (b) 9 : 35 (c) 7 : 35 [RPMT 1997; JIPMER 2001, 2002]
(d) 8 : 25
Solution : (a) For solving this type of problems remember
Actual time = 11 : 60 – given time
So here Actual time = 11 : 60 – 3 : 25 = 8 : 35
Tricky example: 3
When a plane mirror is placed horizontally on a level ground at a distance of 60 m from the foot of a
tower, the top of the tower and its image in the mirror subtend an angle of 90o at the eye. The height of
the tower will be [CPMT 1984]
(a) 30 m (b) 60 m (c) 90 m (d) 120 m
Solution : (b) Form the figure it is clear that h = tan 45o
60
⇒ h = 60 m Tower
h
45o
45o 60 m
Image
Curved Mirror.
It is a part of a transparent hollow sphere whose one surface is polished.
C P PC
F Principle axis F
Converges the light rays Diverges the light rays
(1) Some definitions : : Mid point of the mirror
(i) Pole (P) : Centre of the sphere of which the mirror is a part.
(ii) Centre of curvature (C)
(iii) Radius of curvature (R) : Distance between pole and centre of curvature.
(iv) Principle axis (Rconcave = –ve , Rconvex = +ve , Rplane = ∞)
: A line passing through P and C.
(v) Focus (F) : An image point on principle axis for which object is at ∞
(vi) Focal length (f) : Distance between P and F.
(vii) Relation between f and R : f = R (fconcare = –ve , fconvex = + ve , fplane = ∞ )
2
(viii) Power : The converging or diverging ability of mirror
(ix) Aperture : Effective diameter of light reflecting area. Intensity of image ∝ Area ∝ (Aperture)2
(x) Focal plane : A plane passing from focus and perpendicular to principle axis.
(2) Rules of image formation and sign convention :
Rule (i) Rule (ii) Rule (iii)
F F F FC C
(3) Sign conventions : Incident ray +
(i) All distances are measured from the pole. – +
(ii) Distances measured in the direction of incident rays are taken as Mirror or Lens – Principle
positive while in the direction opposite of incident rays are taken negative. axis
(iii) Distances above the principle axis are taken positive and below the
principle axis are taken negative.
Note : ≅ Same sign convention are also valid for lenses.
Use following sign while solving the problem :
Concave mirror Convex mirror
Real image (u ≥ f) Virtual image (u< f) u →–
v →+
Distance of object u →– u→– f →+
Distance of image v →– v→+ O→+
f →– I →+
Focal length f →– O→+ R →+
Height of object O →+ I →+ m→+
Height of image I →– R→ –
m→+
Radius of curvature R→–
Magnification m→ –
(4) Position, size and nature of image formed by the spherical mirror
Mirror Location of the Location of the Magnification, Nature
object image Size of the image
(a) Concave At infinity At focus i.e. v = f m << 1, diminished Real Erect
virtual inverted
inverted
Real
i.e. u = ∞
Away from centre of Between f and 2f i.e. m < 1, diminished Real inverted
curvature (u > 2f) f < v < 2f
At centre of At centre of m = 1, same size as Real inverted
curvature u = 2f curvature i.e. v = 2f that of the object
Between centre of Away from the centre m > 1, magnified Real inverted
curvature and focus : of curvature
∞ C F P F < u < 2f
v > 2f
At focus i.e. u = f At infinity i.e. v = ∞ m = ∞, magnified Real inverted
Virtual erect
Between pole and v > u m > 1 magnified
focus u < f Virtual erect
(b) Convex At infinity i.e. u = ∞ At focus i.e., v = f m < 1, diminished Virtual erect
P F C Anywhere between Between pole and m < 1, diminished
∞
infinity and pole focus
Note : ≅In case of convex mirrors, as the object moves away from the mirror, the image becomes smaller
and moves closer to the focus.
≅ Images formed by mirrors do not show chromatic aberration.
≅For convex mirror maximum image distance is it’s focal length.
≅In concave mirror, minimum distance between a real object and it's real image is zero.
(i.e. when u = v = 2f)
Mirror formula and magnification.
For a spherical mirror if u = Distance of object from pole, v = distance of image from pole, f = Focal length,
R = Radius of curvature, O = Size of object, I = size of image, m = magnification (or linear magnification ), ms =
Areal magnification, Ao = Area of object, Ai = Area of image
(1) Mirror formula : 1 = 1 + 1 ; (use sign convention while solving the problems).
f v u
Note : ≅ Newton’s formula : If object distance (x1) and image distance (x2) are measured from focus
instead of pole then f 2 = x1 x2
(2) Magnification : m = Size of object
Size of image
Linear magnification Areal magnification
Transverse Longitudinal
When a object is placed When object lies along the principle
perpendicular to the principle axis then its longitudinal magnification
axis, then linear magnification is
called lateral or transverse m = I = −(v2 − v1) b mb
magnification. O (u2 − u1) a ma
It is given by dv v 2
du u
If object is small; m = − =
m= I = − v = f = f −v Also Length of If a 2D-object is placed with it's plane
O u f −u f perpendicular to principle axis
(* Always use sign convention v 2 It's Areal magnification
while solving the problems) u
image = × Length of object (L0 ) Area of image (Ai ) ma × mb
Area of object (Ao ) ab
Ms = = = m2
(Li ) = u f f 2 .Lo Ai
− Ao
⇒ ms = m2 =
Note : ≅ Don't put the sign of quantity which is to be determined.
≅ If a spherical mirror produces an image ‘m’ times the size of the object (m = magnification) then
u, v and f are given by the followings
u = m−1 f, v = −(m − 1) f and f = m u (use sign convention)
m m − 1)
(3) Uses of mirrors
(i) Concave mirror : Used as a shaving mirror, In search light, in cinema projector, in telescope, by E.N.T.
specialists etc.
(ii) Convex mirror : In road lamps, side mirror in vehicles etc.
Note : ≅ Field of view of convex mirror is more than that of concave mirror.
Different graphs Graph between 1 and 1
v u
(a) Real image formed by
concave mirror 1 (b) Virtual image formed by (c) Virtual image formed by
convex mirror 1
v concave mirror 1
v
v
1 11
u uu
Graph between u and v for Graph between u and m for Graph between u and m for
real image of concave mirror virtual image by concave mirror virtual image by convex mirror.
Hyperbola m m
2f 1
1
f fu u
f 2f
Concepts
Focal length of a mirror is independent of material of mirror, medium in which it is placed, wavelength of incident light
Divergence or Convergence power of a mirror does not change with the change in medium.
If an object is moving at a speed vo towards a spherical mirror along it’s axis then speed of image away from mirror is
vi = − u f f 2.vo (use sign convention)
−
When object is moved from focus to infinity at constant speed, the image will move faster in the beginning and slower later on,
towards the mirror.
As every part of mirror forms a complete image, if a part of the mirror is obstructed, full image will be formed but intensity will be
reduced.
O
C P
∞ F
I I O
FC
Can a convex mirror form real images? Real
yes if (distance of virtual object) u < f (focal length) image Virtual object
Example
Example: 10 A convex mirror of focal length f forms an image which is 1/n times the object. The distance of the object from
Solution : (a) the mirror is
Example: 11
(a) (n – 1) f (b) n − 1 f (c) n + 1 f (d) (n + 1) f
n n
By using m= f
f −u
Here m = + 1 , f → +f So, + 1 = + +f ⇒ u = −(n − 1)f
n n f −u
An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm.
The size of the image is [MP PET 1993]
(a) 0.11 cm (b) 0.50 cm (c) 0.55 cm (d) 0.60 cm
Solution : (c) By using I = f f
O −u
Example: 12
Here O + 5 cm, f = − R = −10 cm , u = −1m = −100 cm
Solution : (d) 2
Example: 13
Solution : (c) So, I = −10 ⇒ I = – 0.55 cm.
Example: 14 +5 − 10 − (−100)
Solution : (b) An object of length 2.5 cm is placed at a distance of 1.5 f from a concave mirror where f is the magnitude of
Example: 15 the focal length of the mirror. The length of the object is perpendicular to the principle axis. The length of the
image is
Solution : (a)
Example: 16 (a) 5 cm, erect (b) 10 cm, erect (c) 15 cm, erect (d) 5 cm, inverted
Solution : (a)
By using I = f f u ; where I =?, O = + 2.5 cm. f → − f , u = – 1.5 f
O −
∴ + I = − f −f ⇒ I = −5 cm. (Negative sign indicates that image is inverted.)
2.5 − (−1.5 f)
A convex mirror has a focal length f. A real object is placed at a distance f in front of it from the pole produces
an image at
(a) Infinity (b) f (c) f / 2 (d) 2f
By using 1 11 ⇒ 1 = 1 + 1 ⇒ v = f
f =v+u +f v 2
(− f )
Two objects A and B when placed one after another infront of a concave mirror of focal length 10 cm from
images of same size. Size of object A is four times that of B. If object A is placed at a distance of 50 cm from
the mirror, what should be the distance of B from the mirror
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
By using I = f ⇒ IA × OB = f − uB ⇒ 1 × 1 = −10 − uB ⇒ uB = −20cm .
O f −u IB OA f −uA 1 4
− 10 − (− 50)
A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length 10 cm. The centre
of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of
the wire is
(a) 4 cm2 (b) 6 cm2 (c) 16 cm2 (d) 36 cm2
By using m2 = Ai ; where m= f
Ao f −u
−10 −2 Ao = 9 cm2 −2 2 4 cm 2
3 3
Hence from given values m= − 10 − (− 25) = and ∴ Ai = ×9 =
A convex mirror of focal length 10 cm is placed in water. The refractive index of water is 4/3. What will be the
focal length of the mirror in water
(a) 10 cm (b) 40/3 cm (c) 30/4 cm (d) None of these
No change in focal length, because f depends only upon radius of curvature R.
Example: 17 A candle flame 3 cm is placed at distance of 3 m from a wall. How far from wall must a concave mirror be
Solution : (c) placed in order that it may form an image of flame 9 cm high on the wall
Example: 18
Solution : (b) (a) 225 cm (b) 300 cm (c) 450 cm (d) 650 cm
Example: 19 Let the mirror be placed at a distance x from wall 3cm
Solution : (b) By using
I = −v ⇒ − 9 = − (− x) ⇒ x = −4.5m = − 450cm. 3m (x–3)m
O u + 3 − (x − 3) x
A concave mirror of focal length 100 cm is used to obtain the image of the sun which subtends an angle of 30'.
The diameter of the image of the sun will be
(a) 1.74 cm (b) 0.87 cm (c) 0.435 cm (d) 100 cm
Diameter of image of sun d = fθ Image of θ = 30' = 30 o
sun 60
⇒ d = 100 × 30 × π
60 180 d
θ
F
⇒ d = 0.87 cm .
A thin rod of length f / 3 lies along the axis of a concave mirror of focal length f. One end of its magnified
image touches an end of the rod. The length of the image is [MP PET 1995]
(a) f (b) 1 f (c) 2 f (d) 1 f
2 4
If end A of rod acts an object for mirror then it's image will be A' and if u = 2f − f = 5f
3 3
So by using 1 = 1 + 1 ⇒ 11 1 ⇒ v = − 5 f 2f
f v u − f = v + −5f 2
f / 3 u = 2f – (f/3)
3 A
5 f A' C F
2 2
∴ Length of image = f − 2f =
v
Example: 20 A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let O be the pole of
Solution : (d)
the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C. If
the mirror is now filled with water, the image will be [IIT-JEE 1998]
(a) Real, and will remain at C (b) Real, and located at a point between C and ∞
(c) Virtual and located at a point between C and O (d) Real, and located at a point between C and O
C Object C Object
image
Image
O O
Initially Finally
Tricky example: 4
An object is placed infront of a convex mirror at a distance of 50 cm. A plane mirror is introduced
covering the lower half of the convex mirror. If the distance between the object and plane mirror is 30
cm, it is found that there is no parallel between the images formed by two mirrors. Radius of curvature
of mirror will be
(a) 12.5 cm (b) 25 cm (c) 50 cm (d) 18 cm
3
Solution : (b) Since there is no parallel, it means that both images (By plane mirror and convex mirror) coinciding
each other.
According to property of plane mirror it will form image at a distance of 30 cm behind it. Hence for
convex mirror u = – 50 cm, v = + 10 cm Object
A
By using 1 = 1 − 1 ⇒ 1 = 1 + 1 = 4
f v u f + 10 − 50 50 30 cm
20 cm
⇒ f = 25 cm ⇒ R = 2 f = 25cm. 50 cm 10 cm
2
Tricky example: 5
A convergent beam of light is incident on a convex mirror so as to converge to a distance 12 cm from
the pole of the mirror. An inverted image of the same size is formed coincident with the virtual object.
What is the focal length of the mirror
(a) 24 cm (b) 12 cm (c) 6 cm (d) 3 cm
Solution : (c) Here object and image are at the same position so this position must be centre of curvature
∴ R = 12 cm ⇒ f = R
2
C
Refraction of light
The bending of the ray of light passing from one medium to the other medium is called refraction.
Incident ray Denser medium
i
i Rarer medium
rδ
Denser medium Refracted ray r δ
Rarer medium
Deviation δ = (i – r) Deviation δ = (r – i )
Snell’s law
The ratio of sine of the angle of incidence to the angle of refraction (r) is a constant called refractive index
i.e. sin i =µ (a constant). For two media, Snell's law can be written as 1µ2 = µ2 = sin i
sin r µ1 sin r
⇒ µ1 × sin i = µ 2 × sin r i.e. µ sinθ = constant
Also in vector form : ˆi × nˆ = μ(rˆ × nˆ)
Refractive Index.
Refractive index of a medium is that characteristic which decides speed of light in it. It is a scalar, unit less and
dimensionless quantity.
(1) Types : It is of following two types
Absolute refractive index Relative refractive index
(i) When light travels from air to any transparent medium (i) When light travels from medium (1) to medium (2) then
then R.I. of medium w.r.t. air is called it’s absolute R.I. i.e. R.I. of medium (2) w.r.t. medium (1) is called it’s relative
air µmedium = c R.I. i.e. 1µ2 = µ2 = v1 (where v1 and v2 are the speed of
v µ1 v2
light in medium 1 and 2 respectively).
(ii) Some absolute R.I. (ii) Some relative R.I.
3 = 1.5 , 4 = 1.33 (a) When light enters from water to glass :
2 3
a µglass = a µwater = w µg µg 3/2 9
µw 4/3 8
= = =
a µdiamond = 2.4, a µCs2 = 1.62 (b) When light enters from glass to diamond :
a µ crown = 1.52, µvacuum = 1 , µair = 1.0003 ≈ 1 g µD = µD = 2.4 = 8
µg 1.5 5
Note : ≅Cauchy’s equation : µ = A+ B + C + ...... (λRed > λviolet so µ Red < µ violet ) µ∝ 1
λ2 λ4 λ
≅ If a light ray travels from medium (1) to medium (2), then 1µ2 = µ2 = λ1 = v1 µ ∝ 1
µ1 λ2 v2 v
(2) Dependence of Refractive index v∝λ
(i) Nature of the media of incidence and refraction.
(ii) Colour of light or wavelength of light.
(iii) Temperature of the media : Refractive index decreases with the increase in temperature.
(3) Principle of reversibility of light and refraction through several media :
Principle of reversibility Refraction through several media
1 Incident ray 1
i2
r 3
2
1µ2 = 1 1
2 µ1
1 µ2 × 2µ3 × 3µ1 = 1
Refraction Through a Glass Slab and Optical Path
(1) Lateral shift
The refracting surfaces of a glass slab are parallel to each other. When a light ray passes through a glass slab it
is refracted twice at the two parallel faces and finally emerges out parallel to i
it's incident direction i.e. the ray undergoes no deviation δ = 0. The angle of
emergence (e) is equal to the angle of incidence (i) t rδ
µ N
The Lateral shift of the ray is the perpendicular distance between the
incident and the emergent ray, and it is given by MN = t sec r sin (i – r) M
Normal shift : OO' = x = 1 − 1 t
µ
O x O' Glass
slab
µ
t
Or the object appears to be shifted towards the slab by the distance x
(2) Optical path :
It is defined as distance travelled by light in vacuum in the same time in which it travels a given path length in
a medium.
µ Time taken by light ray to pass through the medium = µx ; where x =
c
Light
geometrical path and µx = optical path
x
Light µ1 µ2 For two medium in contact optical path = µ1x1 + µ2 x2
x1 x2
Note : ≅ Since for all media µ > 1, so optical path length (µx) is always greater than the geometrical
path length (x).
Real and Apparent Depth.
If object and observer are situated in different medium then due to refraction, object appears to be displaced
from it’s real position. There are two possible conditions.
(1) When object is in denser medium and observer is (1) Object is in rarer medium and observer is in denser
in rarer medium medium.
O′ d h′
O h
h′ µ
h O′
dO
(2) µ = Real depth = h (2) µ = h'
Apparent depth h' h
Real depth >Apparent depth that's why a coin at the Real depth < Apparent depth that's why high flying
bottom of bucket (full of water) appears to be raised) aeroplane appears to be higher than it's actual height.
(3) Shift d = h − h' = 1 − 1 h (3) d = (µ − 1)h
µ
(4) For water µ= 4 ⇒d= h (4) Shift for water dw = h
3 4 3
For glass µ = 3 ⇒ d = h Shift for glass dg = h
2 3 2
Note : ≅If a beaker contains various immisible liquids as shown then
Apparent depth of bottom = d1 + d2 + d3 + .... µ1 d1
µ1 µ 2 µ 3 µ2 d2
µ3 d3
µ =combination d AC = d1 + d2 + ..... (In case of two liquids if d1 = d2 than µ = 2µ1 µ 2 )
d App. d1 + d2 + .... µ1 + µ2
µ1 µ2
Total Internal Reflection.
When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of
incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle,
angle of refraction becomes 90o, this angle of incidence is called critical angle (C).
When Angle of incidence exceeds the critical angle than light ray comes back in to the same medium after
reflection from interface. This phenomenon is called Total internal reflection (TIR).
Rarer r 90o θθ
Denser i i=C θ>C
TIR
Important formula μ = 1 = cosec C ; where µ → µRerer Denser
sinC
Note : ≅When a light ray travels from denser to rarer medium, then deviation of the ray is
δ = π − 2θ ⇒ δ → max. whenθ → min. = C δ
i.e. δ max = (π − 2C); C → critical angle θθ
(1) Dependence of critical angle
(i) Colour of light (or wavelength of light) : Critical angle depends upon wavelength as λ ∝ 1 ∝ sin C
µ
(a) λR > λV ⇒ CR > CV
(b) Sin C = 1 = µR = λD = vD (for two media) (c) For TIR from boundary of two media i > sin−1 µ R
R µD µD λR vR µD
(ii) Nature of the pair of media : Greater the refractive index lesser will be the critical angle.
(a) For (glass- air) pair → Cglass = 42o (b) For (water-air) pair → Cwater = 49o
(c) For (diamond-air) pair → Cdiamond = 24 o
(iii) Temperature : With temperature rise refractive index of the material decreases therefore critical angle increases.
(2) Examples of total internal reflection (TIR)
(i) Denser Rarer
Sky
i>θC O I
Rarer i>θC O
I Desner
Earth Earth
Mirage : An optical illusion in deserts Looming : An optical illusion in cold countries
(ii) Brilliance of diamond : Due to repeated internal reflections diamond sparkles.
(iii) Optical fibre : Optical fibres consist of many long high quality composite glass/quartz fibres. Each fibre
consists of a core and cladding. The refractive index of the material of the core (µ1) is higher than that of the
cladding (µ2).
When the light is incident on one end of the fibre at a small angle, the light passes inside, undergoes repeated
total internal reflections along the fibre and finally comes out. The angle of incidence is always larger than the
critical angle of the core material with respect to its cladding. Even if the fibre is bent, the light can easily travel
through along the fibre
A bundle of optical fibres can be used as a 'light pipe' in µ2 Cladding
medical and optical examination. It can also be used for µ1
optical signal transmission. Optical fibres have also been used
for transmitting and receiving electrical signals which are Core
converted to light by suitable transducers.
(iv) Field of vision of fish (or swimmer) : A fish (diver) inside the water can see the whole world through a
cone with.
(a) Apex angle = 2C = 98 o r
(b) Radius of base r = h tan C = h h C θ >C
µ2 −1
CC
(c) Area of base A = πh2 1)
(µ 2 −
Note : ≅ For water µ = 4 so r = 3h and A = 9πh2 .
3 7 7
(v) Porro prism : A right angled isosceles prism, which is used in periscopes or binoculars. It is used to
deviate light rays through 90o and 180o and also to erect the image.
45o 90o B A′
45o A B′
45o 45o
90o 45o 90o
45o 45o 45o 45o
Concepts
In case of refraction of light frequency (and hence colour) and phase do not change (while wavelength and velocity will change).
In the refraction intensity of incident light decreases at it goes from one medium to another medium.
A transparent solid is invisible in a liquid of same refractive index (Because of No refraction).
When a glass slab is kept over various coloured letters and seen from the top, the violet colour letters appears closer (Because
λv < λR so µV > µR and from µ = λ if µ increases then h' decreases i.e. Letter appears to be closer)
λ'
Water drop in air and air bubble in water behaves as a lens.
Air Water Air Water Air Water
Like convex lens Like concave lens
Example
Example: 1 A beam of monochromatic blue light of wavelength 4200 Å in air travels in water (µ = 4 / 3) . Its wavelength in
Solution: (c)
Example: 2 water will be [MNR 1991]
Solution: (b) (a) 2800 Å (b) 5600 Å (c) 3150 Å (d) 4000 Å
µ ∝ 1 ⇒ µ1 = λ2 ⇒ 1 = λ2 ⇒ λ2 = 3150 Å
λ µ 2 λ1 43 4200
On a glass plate a light wave is incident at an angle of 60o. If the reflected and the refracted waves are
mutually perpendicular, the refractive index of material is [MP PMT 1994; Haryana CEE 1996]
(a) 3 (b) 3 (c) 3 (d) 1
2 2 3
From figure r = 30o 60° 60°
∴ µ = sin i = sin 60o = 3 90°
sin r sin 30o r
Example: 3 Velocity of light in glass whose refractive index with respect to air is 1.5 is 2 × 108 m / s and in certain liquid
the velocity of light found to be 2.50 × 108 m / s . The refractive index of the liquid with respect to air is
Solution: (c)
Example: 4 (a) 0.64 (b) 0.80 (c) 1.20 [CPMT 1978; MP PET/PMT 1988]
(d) 1.44
µ ∝ 1 ⇒ µ li = vg ⇒ µl = 2 × 108 ⇒ µl = 1.2
v µg vl 1.5 2.5 × 108
A ray of light passes through four transparent media with refractive indices µ1.µ 2 , µ 3 , and µ 4 as shown in the
figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we
must have [IIT-JEE (Screening) 2001]
(a) µ1 = µ2 µ1 µ2 µ3 D
(b) µ 2 = µ3 B C µ4
(c) µ3 = µ4
(d) µ4 = µ1 A
Solution: (d) For successive refraction through difference media µ sinθ = constant.
Example: 5
Here as θ is same in the two extreme media. Hence µ1 = µ4
Solution: (b)
Example: 6 A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel to the surface of
Solution: (b)
Example: 7 water, then the value of µg would be [IIT-JEE (Screening) 2003]
Solution: (b)
Example: 8 (a) (4/3) sin i r µw=4/3
Solution: (b) (b) 1/ sin i Water r
Example: 9 (c) 4/ 3 Glass
(d) 1
i
For glass water interface g µω = sin i ......(i) and For water-air interface ω µa = sin r .....(ii)
sin r sin 90
∴ g µω ×ω µa = sin i ⇒ µg = 1
sin i
The ratio of thickness of plates of two transparent mediums A and B is 6 : 4. If light takes equal time in
passing through them, then refractive index of B with respect to A will be [UPSEAT 1999]
(a) 1.4 (b) 1.5 (c) 1.75 (d) 1.33
By using t = µx
c
⇒ µB = xA = 6 ⇒ AµB = 3 = 1.5
µA xB 4 2
A ray of light passes from vacuum into a medium of refractive index µ, the angle of incidence is found to be
twice the angle of refraction. Then the angle of incidence is
(a) cos−1 (µ / 2) (b) 2cos−1 (µ / 2) (c) 2 sin−1 (µ ) (d) 2 sin−1 (µ / 2)
By using µ = sin i ⇒ µ = sin 2r = 2 sin r cos r ( sin 2θ = 2 sinθ cosθ )
sin r sin r sin r
⇒ r = cos −1 µ . So, i = 2r = 2 cos −1 µ .
2 2
A ray of light falls on the surface of a spherical glass paper weight making an angle α with the normal and is
refracted in the medium at an angle β . The angle of deviation of the emergent ray from the direction of the
incident ray is [NCERT 1982]
(a) (α − β ) (b) 2(α − β ) (c) (α − β )/ 2 (d) (α + β )
From figure it is clear that ∆OBC is an isosceles triangle, B δ C
Hence ∠OCB = β and emergent angle is α α α
α –β β –α
Also sum of two in terior angles = exterior angle ββ
∴ δ = (α − β) + (α − β ) = 2(α − β ) A OD
A rectangular slab of refractive index µ is placed over another slab of refractive index 3, both slabs being
identical in dimensions. If a coin is placed below the lower slab, for what value of µ will the coin appear to be
placed at the interface between the slabs when viewed from the top
(a) 1.8 (b) 2 (c) 1.5 (d) 2.5
Solution: (c) Apparent depth of coin as seen from top = x + x = x µ2 = µ x
Example: 10 µ1 µ2 µ1 = 3 x
Solution: (c)
Example: 11 ⇒ 1 + 1 =1 ⇒ 1 + 1 =1 ⇒ µ = 1.5
Solution: (b) µ1 µ2 3 µ
Example: 12
Solution: (c) A coin is kept at bottom of an empty beaker. A travelling microscope is focussed on the coin from top, now
Example: 13 water is poured in beaker up to a height of 10 cm. By what distance and in which direction should the
microscope be moved to bring the coin again in focus
Solution: (d)
(a) 10 cm up ward (b) 10 cm down ward (c) 2.5 cm up wards (d) 2.5 cm down wards
Example: 14
When water is poured in the beaker. Coin appears to shift by a distance d = h = 10 = 2.5cm
4 4
Hence to bring the coil again in focus, the microscope should be moved by 2.5 cm in upward direction.
Consider the situation shown in figure. Water µw = 4 is filled in a breaker upto a height of 10 cm. A plane
3
mirror fixed at a height of 5 cm from the surface of water. Distance of image from the mirror after reflection
from it of an object O at the bottom of the beaker is
(a) 15 cm (b) 12.5 cm (c) 7.5 cm (d) 10 cm
From figure it is clear that object appears to be raised by 10 cm (2.5 cm)
4
5 cm
Hence distance between mirror and O' = 5 + 7.5 = 12.5 cm 10 O' 10 cm
So final image will be formed at 12.5 cm behind the plane mirror. 4 O
cm
The wavelength of light in two liquids 'x' and 'y' is 3500 Å and 7000 Å, then the critical angle of x relative to y will be
(a) 60o (b) 45o (c) 30o (d) 15o
sin C = µ2 = λ1 = 3500 = 1 ⇒ C = 30o
µ1 λ2 7000 2
A light ray from air is incident (as shown in figure) at one end of a glass fiber (refractive index µ = 1.5) making
an incidence angle of 60o on the lateral surface, so that it undergoes a total internal reflection. How much
time would it take to traverse the straight fiber of length 1 km
[Orissa JEE 2002]
(a) 3.33 µ sec Air
(b) 6.67 µ sec
(c) 5.77 µ sec Air 60o
Glass
(d) 3.85 µ sec
When total internal reflection just takes place from lateral surface then i = C i.e. C = 60o
From µ = 1 ⇒ µ = 1 = 2
sin C sin 60 3
2 × 1 × 10 3
3
3 × 108
( )Hence µx
time taken by light traverse some distance in medium t = C ⇒ t= = 3.85 µ sec.
A glass prism of refractive index 1.5 is immersed in water (µ = 4 / 3) . A light beam incident normally on the
face AB is totally reflected to reach the face BC if [CPMT 1981; IIT-JEE 1981]
(a) sinθ > 8 / 9 B θA
(b) 2 / 3 < sinθ < 8 / 9
(c) sinθ ≤ 2 / 3
(d) cosθ ≥ 8 / 9 C
From figure it is clear that
Solution: (a)
Example: 15 Total internal reflection takes place at AC, only if θ > C
Solution: (b, c) ⇒ sinθ > sin C ⇒ sinθ > 1 B θA
Example: 16 ω µg θ
⇒ sin θ > 1 ⇒ sin θ > 8 C
9/8 9
When light is incident on a medium at angle i and refracted into a second medium at an angle r, the graph of
sin i vs sin r is as shown in the graph. From this, one can conclude that
(a) Velocity of light in the second medium is 1.73 times the velocity of light in the I medium
(b) Velocity of light in the I medium is 1.73 times the velocity in the II
medium
(c) The critical angle for the two media is given by sin ic = 1 sin r
3
30o
(d) sin ic = 1 sin i
2
From graph tan 30o = sin r = 1 ⇒ 1µ2 = 3⇒ µ2 = v1 = 1.73 ⇒ v1 = 1.75 v2
sin i 1µ2 µ1 v2
Also from µ = 1 ⇒ sin C = 1 ⇒ sin C = 1 = 1 .
sin C Rarer µDenser 1µ2 3
A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive
indices of the material of the prism for the above red, green and blue wavelength are 1.39, 1.44 and 1.47
respectively. The prism will [IIT-JEE 1989]
(a) Separate part of red colour from the green and the blue colours
(b) Separate part of the blue colour from the red and green colours
(c) Separate all the colours from one another 45°
(d) Not separate even partially any colour from the other two colours
Solution: (a) At face AB, i = 0 so r = 0, i.e., no refraction will take place. So light will be incident on face AC at an angle of
Example: 17
incidence of 45o. The face AC will not transmit the light for which i > θ C , i.e., sin i > sinθ C
or sin 45o > (1 / µ ) i.e., µ > 2 (= 1.41) A
Now as µ R < µ while µG and µ B > µ, so red will be transmitted through the 45° 45°
face AC while green and blue will be reflected. So the prism will separate red B C
colour from green and blue.
An air bubble in a glass slab (µ = 1.5) is 6 cm deep when viewed from one face and 4 cm deep when viewed
from the opposite face. The thickness of the glass plate is
(a) 10 cm (b) 6.67 cm (c) 15 cm (d) None of these
Solution: (c) Let thickness of slab be t and distance of air bubble from one side is x
When viewed from side (1) : 1.5 x ⇒ x = 9cm 6 cm 4 cm
6
=
When viewed from side (2) : 1.5 = (t − x) ⇒ 1.5 = (t − 9) ⇒ t = 15cm Side 1 x Air Side 2
4 bubble
4
t
Tricky example: 1
One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the first face,
forms an image 12 cm behind the silvered face. The refractive index of the glass is [CPMT 1999]
(a) 0.4 (b) 0.8 (c) 1.2 (d) 1.6
Solution : (c) From figure thickness of glass plate t = 6 cm.
Let x be the apparent position of the silvered surface. x Image
According to property of plane mirror Object
x + 8 = 12 + 6 – x ⇒ x = 5 cm
Also µ = t ⇒ µ = 6 = 1.2 8 cm 12 cm
x 5
12 +(6–x)
t
Tricky example: 2
A ray of light is incident on a glass sphere of refractive index 3/2. What should be the angle of incidence
so that the ray which enters the sphere doesn't come out of the sphere
(a) tan −1 2 (b) sin −1 2 (c) 90o (d) cos −1 1
3 3 3
Solution : (c) Ray doesn't come out from the sphere means TIR takes place. iA B
Hence from figure ∠ABO = ∠OAB = C CC
O
1 1 2
∴ µ = sin C ⇒ sin C = µ = 3
Applying Snell's Law at A sin i = 3 ⇒ sin i = 3 sin C = 3 × 2 = 1 ⇒ i = 90o
sin C 2 2 2 3
Tricky example: 3
The image of point P when viewed from top of the slabs will be
(a) 2.0 cm above P (b) 1.5 cm above P (c) 2.0 cm below P (d) 1 cm above P
Solution: (d) The two slabs will shift the image a distance 1.5 cm
1.5 cm
d = 2 1 − 1 t = 2 1 − 1 (1.5) = 1cm 1.5 cm
µ 1.5 2.0 cm
P
Therefore, final image will be 1 cm above point P.
Refraction From Curved Surface.
1 P 2 1 P 2
O I O I
µ1 = Refractive index of the medium from which light rays are coming (from object).
µ 2 = Refractive index of the medium in which light rays are entering.
u = Distance of object, v = Distance of image, R = Radius of curvature
Refraction formula : µ2 − µ1 = µ2 − µ1 (use sign convention while solving the problem)
R v u
Note : ≅Real image forms on the side of a refracting surface that is opposite to the object, and virtual image
forms on the same side as the object.
≅ Lateral (Transverse) magnification m = I = µ1v .
O µ 2u
Specific Example
In a thin spherical fish bowl of radius 10 cm filled with water of refractive index 4/3 there is a small fish at a distance of 4
cm from the centre C as shown in figure. Where will the image of fish appears, if seen from E
(a) 5.2 cm (b) 7.2 cm (c) 4.2 cm (d) 3.2 cm
Solution : (a) By using µ2 − µ1 = µ2 − µ1
v u R
4 C E
3 4 cm
where µ1 = , µ 2 = 1, u = −6 cm, v = ?
On putting values v = −5.2 cm
Lens.
Lens is a transparent medium bounded by two refracting surfaces, such that at least one surface is spherical.
(1) Type of lenses
Convex lens (Converges the light rays) Concave lens (Diverges the light rays)
Double convex Plano convex Concavo convex Double concave Plane concave Convexo concave
Thick at middle Thin at middle
It forms real and virtual images both It forms only virtual images
(2) Some definitions
Optical axis
C2 C1 C1 C2 C1, C2 – Centre of curvature,
O O R1, R2 – Radii of curvature
Principle
axis
– R2 +R1 – R1 +R2
(i) Optical centre (O) : A point for a given lens through which light ray passes undeviated (Light ray passes
undeviated through optical centre).
(ii) Principle focus
First principle focus Second principle focus
F1 F1 F2 F2
Note : ≅Second principle focus is the principle focus of the lens.
≅ When medium on two sides of lens is same then | F1 |=| F2 |.
≅ If medium on two sides of lens are not same then the ratio of two focal lengths f1 = µ1
f2 µ2
µ1 µ2
(iii) Focal length (f) : Distance of second principle focus from optical centre is called focal length
fconvex → positive, fconcave → negative, fplane → ∞
(iv) Aperture : Effective diameter of light transmitting area is called aperture. Intensity of image ∝ (Aperture)2
(v) Power of lens (P) : Means the ability of a lens to converge the light rays. Unit of power is Diopter (D).
P = 1 = 100 ; Pconvex → positive, Pconcave → negative, Pplane → zero .
f (m) f (cm)
Note : ≅ Thick lens Thin lens
P↑f↓R↓ P↓f ↑R↑
(3) Image formation by lens
Lens Location of Location of the Nature of image
Convex the object image
Magnification Real Erect
2f f At infinity At focus i.e. v = f virtual inverted
i.e. u = ∞ m<1 Inverted
Concave Away from 2f Between f and 2f diminished Real
i.e. (u > 2 f) i.e. f < v < 2 f
At 2f or (u = 2 f) At 2f i.e. (v = 2 f) m<1 Real Inverted
diminished
Between f and 2f Away from 2f i.e. m=1 Real Inverted
same size Real Inverted
f 2if.e. f < u < 2 f (v > 2 f) Real Inverted
m>1 Virtual
At focus i.e. u = f At infinity i.e. v = ∞ magnified Erect
Between optical At a distance m=∞
centre and focus, greater than that of magnified
u< f object v > u
At focus i.e. v = f m>1
At infinity magnified
i.e. u = ∞ Between optical
centre and focus m<1 Virtual Erect
Anywhere diminished Virtual Erect
between infinity
and optical m<1
centre diminished
Note : ≅Minimum distance between an object and it’s real image formed by a convex lens is 4f.
≅ Maximum image distance for concave lens is it’s focal length.
(4) Lens maker’s formula
The relation between f, µ, R1 and R2 is known as lens maker’s formula and it is 1 = (µ − 1) 1 − 1
f R1 R2
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