Physics (CBSE 2009) - · PDF fileLenses Power (P) Aperture (A) L 1 3 D 8 cm L ... Physics...

1. What is sky wave propagation? [1]

Solution

In the frequency range from a few MHz up to 30–40 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back toward the earth. This mode of propagation of electromagnetic waves is called sky wave propagation, and it is used in shortwave radio communication.

2. Write the following radiations in ascending order in respect of their frequencies :

X-rays, microwaves, UV rays and radio waves. [1]

Solution

Radio waves, microwaves, UV rays and X-rays.

3. Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why? [1]

Solution

If field lines were entirely confined between two ends of a straight solenoid, the flux through the cross-section at each end would be non-zero. But the flux of field

B

through any closed surface must always be zero. For a toroid, this difficulty is absent because it has no ‘ends’. [1]

4. You are given following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope? [1]

Solution

Objective: L1, because the lens L1 has larger aperture compared to other lenses and the resolving power and light gathering directly depend on the aperture diameter.

Eyepiece: The magnification power of an astronomical telescope is directly proportional to the power of eyepiece, therefore the lens L3 is chosen as this eyepiece.

Lenses Power (P) Aperture (A)

L1 3 D 8 cm

L2 6 D 1 cm

L3 10 D 1 cm

5. If the angle between the pass axis of polarizer and the analyzer is 45°, write the ratio of the intensities of original light and the transmitted light after passing through the analyzer. [1]

Solution

The intensity of the light after passing through the polarizer is

I I= 02cos ,θi

where angle qi is the angle between the initial polarization and the axis of polarizer.

Physics (CBSE 2009)Time: 3 hours Max. Marks: 70

General Instructions

1. All questions are compulsory. 2. There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry two marks each, questions 19 to 27

carry three marks each and questions 28 to 30 carry five marks each. 3. There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three

marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions. 4. Use of calculators is not permitted. 5. You may use the following values of physical constant wherever necessary:

c = 3 × 108 ms–1

h = 6.626 × 10–34 Jse = 1.602 × 10–19 Cm0 = 4p × 10–7 T m A–1

1

49 109 2

πε0

2Nm C= × −

Mass of electron me = 9.1 × 10–31 kgMass of neutron mn = 1.675 × 10–27 kgBoltzmann’s constant k = 1.381 × 10–23 JK–1

Avogadro’s number NA = 6.022 × 1023mol–1

Radius of earth = 6400 km

Physics Special market Book 2_2009 CBSE.indd 1 12/21/2011 12:27:13 PM

Physics (CBSE 2009) 2

Let I0 be the intensity of the incident light, which after passing through the polarizer becomes

I Ip 0= °cos .2 45

After passing through the analyzer, it becomes

I II I

a p= ° = × =cos .2 0 0452

1

2 4

Therefore, the intensity of the light becomes one quarter of the initial intensity after passing the analyzer.

6. The figure shows a plot of three curves a, b, c showing the variation of photocurrent vs. collector plate potential for three different intensities I1, I2 and I3 having frequencies u1, u2 and u3, respectively, incident on a photosensitive surface.

Point out the two curves for which the incident radiations have same frequency but different intensities. [1]

Photoelectriccurrent

I1

I2

I3

Collector plate potential

c

b

a

Solution

The stopping potential of material depends only on the frequency and work function of the material. It is independent of intensity. Since both curves a and b have the same stopping potential, the frequency of radiation which cause curves a and b are the same.

7. Two nuclei have mass numbers in the ratio 1:2. What is the ratio of their nuclear densities? [1]

Solution

Radius of the nucleus is given by

R R A A= 031/ where is the atomic mass

ρπ π π

= =( )

=mA

R

mA

R A

mR4

343

343

03 1 3 3

03

/.

So, the nuclear density is a constant.

8. What type of wavefront will emerge from a: (a) point source and (b) distant light source?

Solution

A point source produces a spherical wavefront and a distant light source produces a plane wavefront.

9. A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor ‘R’. Plot a graph showing the variation of terminal potential ‘V’ with resistance R. [2]

Predict for the graph the condition under which ‘V’ becomes equal to ‘E’.

Solution

The terminal potentialV E Ir= −

where V is the terminal potential, E is the emf of the cell and r is internal resistance of the cell.

I R r E( )+ =

or V EEr

R rER

R r= −

+=

+.

R

V

The terminal potential approaches the emf when the external resistances approach infinity. Or equivalently, the terminal potential becomes equal to the emf if the circuit is open.

10. (a) Can two equipotential surfaces intersect each other? Give reasons. [2]

(b) Two charges –q and +q are located at points A (0, 0, –a) and B (0, 0, +a), respectively. How much work is done in moving a test charge from point P (7, 0, 0) to Q (–3, 0, 0)?

Solution

(i) If two equipotential surfaces intersect, at the point of intersection, the potential will have two distinct values.

(ii) W Q V V= −’( ( ) ( ))Q P

V

kqr

kqr

kq

a

kq

a( )P = − + = −

++

+=

AP BP

volt.7 7

02 2 2 2

Vkqr

kqr

( )QAQ BQ

= − + = 0 volt.

ThereforeW Q= × =′ 0 0 J.

Zero work is needed to move a charge from P (7, 0, 0) to Q (–3, 0, 0).

11. By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by 21%? [2]

Solution

The transmission range of a TV tower is given by

d Rh= 2 where d is the range, R is radius of earth and h, is height of

the tower.The new height

h h h h′ = + =0 21 1 21. ( . ).

Therefore, new transmission range

d Rh d′ = =2 1 21 1 21( . ) . .



Or the transmission range will increase by

d dd

′ − × =100 45.82%.

12. Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time. [2]

Solution

When two ends of the conductor are connected to the battery, an electric field

E is generated inside the

conductor. Therefore, force acting on the electron is

F am eE

= = −e

where –e is the charge of the electron, and me is mass of electron. Therefore acceleration is

aeEm

= −e

.

Therefore the average velocity of the electron which is called drift velocity is

v aeEmd = = −τ τ

e

.

where t is the average time between two successive collisions and is called drift velocity.

13. How does a charge q oscillating at certain frequency produce electromagnetic waves ? [2]

Sketch a schematic diagram depicting electric and magnetic fields for an electromagnetic wave propagating along the z-direction.

Solution

A charge q which oscillates produces a time-varying magnetic field in the neighborhood which in turn produces a time-varying electric field in the neighbor-hood. This process continues since both time-varying electric and magnetic fields act as sources to each other; thus the electromagnetic wave is generated.

B

E

y

x

z

14. A charge ‘q’ moving along the x-axis with a velocity v

is subjected to a uniform magnetic field

B acting along the

z-axis as it crosses the origin O. [2]

B

x

z - axis

O y

q

(a) Trace its trajectory.(b) Does the charge gain kinetic energy as it enters the

magnetic field? Justify your answer.

Solution

(a)

Electron

Magnetic fieldNo field

x

y

Direction of magnetic field is in tothe page

(b) Magnetic force does not change kinetic energy of particle, because this force is always perpendicular to velocity.

15. The following figure shows the input waveforms (A, B) and the output waveform (Y) of a gate. Identify the gate, write its truth table and draw its logic symbol. [2]

A

B

Y

0 1 2 3 4 5 6 7

Solution

The truth table of the given diagram is

A B Y

0 0 1

1 0 1

0 1 1

1 1 0

Therefore the gate used in the circuit is NAND gate.

AB

Y

16. State Biot–Savart law. A current I flows in a conductor placed perpendicular

to the plane of the paper. Indicate the direction of the magnetic field due to small element dl

at point P situated

at a distance r from the element as shown in the figure.

Iz

r PO

d�

x

y

ασ



Solution

(I) According to the Biot–Savart’s law, the magnitude of the magnetic field dB at any point P located at position vector

r with respect to current element dl carrying a

current I is(a) Proportional to the current I,(b) Proportional to the length element dl of the

current carrier,(c) Proportional to sin q, where q is the angle between

r and dli.e.

dBI d r

r

l

∝ ×( )3

(II) From Biot – Savart′s law the magnetic field at P is along the –x direction.

17. Why are high-frequency carrier waves used for transmission? [2]

OR What is meant by the term ‘modulation’? Draw a block

diagram of a simple modulator for obtaining an AM signal.

Solution

High-frequency carrier waves are used in communication due to the following reasons :(a) Size of the antenna. For efficient transmission and

reception, the transmitting and receiving antennas must have a length equal to one by fourth of the wavelength of the audio signal. It is not practical to use a huge antenna for transmission and reception.

(b) Effective power radiated by the antenna. The power radiated by the linear antenna of length l is

given by power radiated ∝

lλ

2

. Therefore for higher

wavelengths the power radiated by the antenna decreases.

(c) Mixing up of the signals. If many transmitters are transmitting the base band information signals simultaneously, these signals will get mixed up and there is no way to distinguish between them. This can be avoided by using high frequency transmission and then allotting a band of frequencies.

OR

Solution

The process of superimposing the information signal on to a high-frequency carrier wave is called modulation.The block diagram of the AM modulator is given below.

m(t) x(t) y(t)

Bx(t)+Cx(t)2c(t)

Ac sinwct(carrier)

Am sinwmt(Modulating

signal)

Squarelaw device

Bandpass filtercentred

at wc

AM wave+

18. A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme: [2]

A A A A A1 2 3 4→ → → →a aβ γ

The mass number and atomic number of A are 180 and 72, respectively. What are these numbers for A4?

Solution

Emission of one a-radiation decreases the mass number of the atom by 4 and atomic number by two.

Therefore,

A A72180

70176→

a.

Again

A A70176

68172→

a.

Emission of gamma radiation reduces the energy of the nucleus. Therefore,

A A4 68172= .

Therefore, atomic number of A4 is 68 and mass number is 172.

19. A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.

Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞. [3]

Solution

According to the Gauss’s law,

E da

Qi =∫ enc

ε0

Therefore for a charged spherical shell, the electric field outside the shell

E rQ

i4 2

0

πε

=

or EQr

r = 1

4 02π ε

.

Electric field for a point inside the shell

E ri4 02π = .

Therefore E

= 0.

1

4πε0

QEr = r 2

rR

Er= 0

Er

O

20. Three identical capacitors C1, C2 and C3 of capacitance 6 mF each are connected to a 12 V battery as shown.



C1

C2

C312 V

+

–

Find(a) charge on each capacitor(b) equivalent capacitance of the network(c) energy stored in the network of capacitors.

Solution

(a) Charge on capacitor C1 = C2 = 36 µC; Charge on capacitor C3 = = 72 µC.

(b) Ceq = 3 µF + 6 µF = 9 µF.(c) W = 648 µW.

21. (a) The energy levels of an atom are as shown below. Which of them will result in the transition of a photon of wavelength 275 nm? [3]

0 eV

–2 eVDC

BA

–4.5 eV

–10 eV

(b) Which transition corresponds to emission of radiation of maximum wavelength?

Solution

(a) The difference between the energy levels = energy of the photon emitted.

hcE E

λ= −1 2 .

But it is given that the wavelength is 275 nm. Therefore

E EV

1 2

1240

2754 5− = =e nm

nmeV.

Therefore the transition B emits radiation of wavelength 275nm.

(b) The radiation of maximum wavelength is emitted when the change in energy is minimum.

Transition A has the minimum change of energy therefore transition A emits the maximum wavelength.

22. A proton and an alpha particle are accelerated through the same potential. Which one of the two has (a) greater value of de Broglie wavelength associated with it and (b) less kinetic energy? Justify your answers. [3]

Solution

The energy of a particle accelerating in potential difference V is

E QV= . Since the charge on the alpha particle is twice that of the

proton

KE KE .pa = 2

Therefore, the kinetic energy of the proton is less than that of the alpha particle.

We have

KE = Pm

2

2.

Therefore,

Pm

P

ma

a

2 2

2= p

p

or

PP

mm

a a

p p

= 2.

The de-Broglie wavelength is inversely proportional to the momentum of the particle. Therefore proton has the longer wavelength.

21. In a single slit diffraction experiment, when a tiny circular obstacle is placed in the path of light form a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why? [3]

State two points of difference between the interference patterns obtained in Young’s double slit experiment and the diffraction pattern due to a single slit.

Solution

The bright spot appears in the middle of the shadow because of the diffraction of light at the edges of the tiny object.Difference between interference and diffraction :(a) The interference pattern has a number of bright and

dark fringes of equal intensity. The diffraction pattern has a central bright fringe of maximum intensity and intensity falls as we go away from the centre.

(b) Interference is due to the superposition of two waves coming from two coherent sources, whereas diffraction is the superposition of secondary wavelets coming from different parts of same wavefront.

24. (a) Define self inductance. Write its S.I. units. (b) Derive an expression for self inductance of a long

solenoid of length l, cross-sectional area A having N number of turns. [3]

Solution

(a) The effect in which a changing current in a circuit induces an emf in the same circuit is called self induction. According to Lenz’s law, self-induction always opposes any change in the current in the circuit and is generally termed as back emf.

The total flux linked to the circuit due to current I in the circuit is given by

Φ = LI where L is the proportionality constant called self

inductance.

or LI

= Φ.

(b) Let length of the solenoid be l, cross-sectional area be A and number of turns per unit length be n. If I is the current in the winding, the magnetic field inside the solenoid has the value

B nI= µ0



Then the self inductance is

LI

= Φ

Ln BA

In n A

I= =l l l( )( )( )µ0

That is, Self inductance of the solenoid is

L n A= µ02 l .

25.

X

B

G

A D C

Y

The figure shows experimental setup of a meter bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of 10 Ω is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 10 Ω resistance is instead connected in series with resistance ‘Y’. Determine the values of the resistances X and Y. [3]

Solution

The balancing condition

ll

XY1−

=

It is given that l = 0.4 mTherefore,

2

3= X

Y

or Y X= 3

2. (1)

If the resistance 10 Ω connected in series to X, then the new balancing condition is

XY+ =10

1

Y X= +10. (2)

From (1) and (2)

X Y= Ω = Ω20 30and .

Therefore,

XY +

=10

1

2.

Therefore the new null point is 33.33 cm away from the point A.

26. Derive the expression for force per unit length between two long straight parallel current-carrying conductors. Hence define one ampere. [3]

OR Explain the principle and working of a cyclotron with the

help of a schematic diagram. Write the expression for cyclotron frequency. [3]

Solution

a

b

d

Fba

BaIb

L

Ia

Let two straight parallel conductors a and b be separated by a distance d.

According to Ampere’s circuital law the magnetic field experienced by the conductor b due to the current in a is

BIdaa= µ

π0

2

along the direction perpendicular to wire b. Therefore, the magnetic force acting on the wire at a

wire segment of length L is

F I Bba dL= ×→ →

So, F I LBI I L

dba b aa b= = µ

π0

2.

Similarly the force on ‘a’ due to ‘b’ can be found. From considerations similar to above we can find the force

Fab ,

on a segment of Length L of ‘a’ due to the current in ‘b’. It is equal in magnitude to

Fba , and directed towards ‘b’. Thus, F Fba ab= −

Therefore, two current carrying conductors attract each other if the current is flowing in the same direction.The force per unit length

FI I

daba b= µ

π0

2

The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one meter apart in vacuum, would produce on each of these conductors a force equal to 2 × 10−7 newtons per metre of length.



Solution

Top view Side view

Uniformmagneticfieldregion.

Electricacceleratingfield betweenthe magneticfield regions.

N

Output beam of highvelocity ion

Injection ofion

De

S

N S

Cyclotron is a machine invented by E.O Lawrence which is used for accelerating charged particles, such as protons and neutrons to high energies. The basic working principle is that a charged particle undergoes circular motion with frequency independent of its speed if the velocity of the particle and the direction of magnetic field are directly perpendicular to each other, whereas an electric field will accelerate it.

The cyclotron consists of a chamber in which there are two semicircular hollow metal D-shaped boxes, D1, D2 called dees insulated from each other. The chamber is placed between the poles of a very strong electromagnet so that the magnetic field passes across the dees perpendicular to the path of the ion. An electric oscillator establishes an accelerating potential difference across the gap between the dees.The force acting on the circulating charge is

qvBmv

r=

2

.

Therefore the frequency of the particle is

ω = qBm

fqB

m= =ω

π π2 2.

27. Three light rays red (R), green (G) and blue (B) are incident on a right angled prism ‘abc’ at face ‘ab’. The refractive indices of the material of the prism for red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. Out of those which color ray will emerge out of face ‘ac’? Justify your answer. Trace the path of these rays after passing through face ‘ac’. [3]

b

45°

a

BGR

c

Solution

In the first face AC of the prism all the rays are incident normally. Therefore, the rays go undeviated.

In the face AC of the prism all three rays are incident at angle of 45°.The angle of refraction of the blue ray is

n rB Bsin sin45° =

sin Br = × =1 471

21 03944. . .

Therefore the blue ray undergoes total internal reflection.The angle of refraction of the green ray is

n rG Gsin sin45° =

sin Gr = × =1 441

21 0188. . .

Therefore, the green ray also undergoes total internal reflection.The angle of refraction of the red rays is

n rR Rsin sin45° =

sin Rr = × =1 391

20 9828. .

rR = °79 3822.

B

45°

79.38°

45°

G

R

28. (a) Derive an expression for the average power consumed in a series LCR circuit connected to a.c. source in which the phase difference between the voltage and the current in the circuit is f. [5]

(b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value in receiving circuits? Name the factors on which it depends.

OR (a) Derive the relationship between the peak and the rms

values of current in an a.c. circuit.

(b) Describe briefly, with the help of a labeled diagram, working of a step-up transformer.

A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain.



Solution

Let the voltage in the circuit be V V t= 0 sin( )ω and the current I I t= +0 sin( )ω ϕ where ϕ is phase difference between the voltage and current

IVZ

V

R X XI C

00 0

2 2= =

+ −( )

Therefore the power in the circuit is

P VIVZ

t t= = +02

sin sinω ω ϕ( ).

The average power is

PV

Z= rms cos

2

ϕ.

(| (sin ( )| | ( ) | )2 2ω ω ϕ ϕ ω ϕ ϕt t tsin cos cos cos+ = − + =

The quality factor Q of the circuit is defined by how sharp the resonance is. If the circuit is more sharp, it becomes more selective.Mathematically

QL

R= ω0

where ω0 is the natural frequency of the circuit, L is the inductance, and R the resistance in the circuit.

OR

Solution

(a) The rms value of the current is defined as

I

I t

TI

T

rms =

=

∫0

02 2

0

1 2

2

sin

/

ω

where I0 is the peak current.

(b) Transformer is a device which works on the principle of mutual induction. A transformer consists of two sets of coils, insulated from each other. One of the coils called the primary coil has Np turns. The other coil is called the secondary coil; it has Ns turns.

A/C source More turns, higher voltage

Current passing through the primary coil induces a magnetic field around them since both the primary and the secondary are wound on same iron core. The flux due to the current in the primary passes through the secondary coil. So if an alternating current is passing through the primary, it produces an alternating magnetic flux which links the secondary coil and induces an emf in it.

The induced emf in the secondary coil

εs s= −NddtΦ

The same magnetic flux Φ induces a back emf in the primary also

εp p= −NddtΦ

If the resistance of the transformer is very low, εp p= v . And if only a little current taken from the secondary or

it is open circuited, then

εp p= v

Then

v Nddts s= − Φ

and v Nddtp p= − Φ

or vv

NN

s

p

s

p

= .

A step up transformer is used to increase the voltage. Therefore, it has more number of turns in the secondary and the output voltage is

v vNNs p

s

p

= .

A step up transformer converts a low voltage to high voltage without violating the conservation of energy. During the conversion, the current in the circuit goes down so that the power which is the product of current and voltage remains constant.

29. (a) Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor in its common emitter configuration. Draw the typical input and output characteristics. [5]

(b) Explain, with the help of a circuit diagram, the working of n-p-n transistor as a common emitter amplifier.

OR

(a) How is a zener diode fabricated so as to make it a special purpose diode? Draw I-V characteristics of zener diode and explain the significance of breakdown voltage.

(b) Explain briefly, with the help of a circuit diagram, how a p-n junction diode works as a half wave rectifier.

Solution

R2

mA

µA

+

+

−

−

+ −R1

VBB

VCE VCC

VBE

IB

IC

BC

Input characteristics

EIE



40

20

0.2 0.4VBE/V

VCE = 10.0 V

IB/µA

0.6 0.8 1.0

60

80

100

Output characteristics

4

2

6

8

10

42Collector to emitter voltage (VCE) in volts

Col

lect

or c

urre

nt (

I C)

in m

A

0 6 8 10 12 14 16

10 mA

20 mA

30 mA40 mA

50 mA

60 mA

Base current (IB)

(c) CE amplifier

V i

V 0

VBB

VCCE

IC

IE

IB

R BRC

BC

The collector voltage is

V V I RCC CE C L= +

and in the input circuit, V V I RBB BE B B= + .

when a small voltage vi is added to the input voltage

V v V I R I R rBB i BE B B B B+ = + + Δ +( )

orv I R ri B B= Δ +( ).

And the output current is

Δ =I IC Bβ .

OR

Solution

Zener diode is fabricated by heavily doping both p- and n-sides of the junction. Due to this, depletion region formed is very thin (<10–6 m) and the electric field of the junction is extremely high (~5 × 106 V/m) even for a small reverse bias voltage of about 5V. It is seen that when the applied reverse bias voltage (V) reaches the breakdown

voltage (Vz) of the Zener diode, there is a large change in the current. Note that after the breakdown voltage Vz, a large change in the current can be produced by an almost insignificant change in the reverse bias voltage. In other words, Zener voltage remains constant, even though current through the Zener diode varies over a wide range. This property of the Zener diode is used for regulating supply voltages so that they are constant.

(b)I (mA)

I (mA)

Reverse biasForward bias

V(V)

Vz

I-V characteristics When the reverse bias voltage V = Vz, then the electric field strength is high enough to pull valence electrons from the host atoms on the p-side which are accelerated to n-side. These electrons account for high current observed at the breakdown. The emission of electrons from the host atoms due to the high electric field is known as internal field emission or field ionization. The electric field required for field ionization is of the order of 106 V/m.

Diode as a rectifier

Transformer

SecondaryPrimary

B

Α X

R L

Y

A diode allows current to pass only when it is forward biased. So if an alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier. During the positive half cycle of the a.c., the diode conducts and current flows through the load resistance from X to Y. During the negative half cycle of the ac, the diode does not conduct. Therefore, the current flows through the diode in only one direction.

Volta

ge a

cros

s R

LVo

ltage

at A

Input ac

(a)

Output voltage

t

t



30. Trace the rays of light showing the formation of an image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices n1 and n2. Establish the relation between the distances of the object, the image and the radius of curvature from the central point of the spherical surface.

Hence derive the expression of the lens maker’s formula. [5]

OR

Draw the labeled ray diagram for the formation of image by a compound microscope.

Derive the expression for the total magnification of a compound microscope. Explain why both the objective and the eyepiece of a compound microscope must have short focal lengths.

Solution

O

n1 n2

N

C IM R

u v

r

i

The given figure shows the geometry of formation of image I of an object O on the principal axis of a spherical surface with centre of curvature C, and radius of curvature R. The rays are incident from a medium of refractive index n1, to another of refractive index n2. As before, we take the aperture (or the lateral size) of the surface to be small compared to other distances involved, so that small angle approximation can be made. In particular, NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.

tan NOM =MN

OM∠

tan NCM =MN

MC∠

tan NIM =MN

MI∠

From ΔNOC, i = ∠ NOM + ∠ NCM

i =

MNOM

+MNMC

(1)

Similarlyr = ∠ − ∠NCM NIM

r =MNMC

MNMI

− (2)

But from Snell’s law

n i n r1 2=

By substituting (1) and (2) in this equation,

nOM

nMI

n nMC

1 2 2 1− = −

But OM = –u, MI = v and MC = rTherefore,

nv

nu

n nR

2 1 2 1− = −.

Lens maker’s formula

C2 P1 C P2 C1

P1I1

m1m2m1A

B

X

OP1 uY

v

v1

I I1O

Consider a convex lens (or concave lens) of absolute refractive index m2 to be placed in a rarer medium of absolute refractive index m1.

Considering the refraction of a point object on the surface XP1Y, the image is formed at I1 at a distance of v1.

CI1 = P1I1 = v1 (as the lens is thin)

CC1 = P1C1 = R1

CO = P1O = u

It follows from the refraction due to convex spherical surface XP1Y that

µ µ µ µ1 2

1

2 1

1−+ = −

u v R (1)

The refracted ray from A suffers a second refraction on the surface XP2Y and emerges along BI. Therefore I is the final real image of O.Here the object distance is

u Cl P l v= ≈ =1 2 1 1 (Note P1P2 is very small)

Let Cl P l v≈ =2 (Final image distance)

Let R2 be radius of curvature of second surface of the lens. It follows from refraction due to concave spherical

surface from denser to rarer medium that

− + = − = −−

µ µ µ µ µ µ2

1

1 1 2

2

2 1

2v v R R. (2)

Adding (1) & (2)

−−

+ = − −

µ µ µ µ1 12 1

1 2

1 1

u v R R( )

µ µ µ1 2 11 2

1 1 1 1

v u R R−

= − −

( )



But

1 1 1

v u f− =

and

µµ

µ µ2

1

12= =

∴ = − −

11

1 1

1 2f R R( ) .µ

OR

The angular magnification or magnifying power of a compound microscope is defined as the ratio of the angle b subtended by the final image at the eye to the angle a subtended by the object seen directly, when both are placed at the least distance of distinct vision. Therefore, angular magnification, M = β/a.

DL

A�

A�

B� B�

A�

ba

BFo

FeFo

Eye

E

Vo

fouo

fo

A

ye

Since the angles are small, a ≈ tan a and b ≈ tan b. Therefore,

M = tan

tan.

ba

Now,

tanb = ′′ ′′A B

D

and

tan .a = =A"B"

D

AB

D

Substituting the values of tan a and tan b in the expression for M, we get

M = ′′ ′′ × = ′′ ′′A B

D

D

AB

A B

AB

or MA B

= ′′ ′′′ ′

× ′ ′A B A B

AB.

Thus, the magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and the objective; that is,

M M Mo= ×e , (1)

Where Me and Mo are the magnifying powers of the eyepiece and the objective respectively. The linear magnification of the real, inverted image produced by the eyepiece is A”B” / A’B’. Linear magnification is given by

M

Dfe

e = +1 , (2)

where fe is the focal length of the eyepiece and A’B’ / AB is the linear magnification of the object produced by the objective. Also,

M

Vu

o

oo = . (3)

From Eqs. (1), (2) and (3), we have

Mvu

Df

o

o e

= +

1 .

We know that1 1 1

v u fo o o

− = .

Multiplying both sides by vo we get

vv

vu

vf

o

o

o

o

o

o

− =

− = − +vu

vf

o

o

o

o

1

vu

vf

o

o

o

o

= −1 .

Substituting this value in the expression for M, we have

Mvf

Df

o

o e

= −

+

1 1 .

From the above expression it can be concluded that for high magnification M the focal length of objective and eyepiece should be small as both the focal lengths are inversely proportional to magnification M.


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