Section I [40 Marks]

1. A. Choose the correct alternative (a), (b), (c) or (d) for each of the questions givenbelow

(i) The electrostatic potential energy of two point charges, 1 µC each, placed 1metre apart in air is

(a) 9 103× J (b) 9 109× J

(c) 9 10 3× − J (d) 9 10 3× − eV

(ii) A wire of resistance R is cut into n equal parts. These parts are thenconnected in parallel with each other. The equivalent resistance of thecombination is

(a) nR (b) R n/

(c) n R/ 2 (d) R n/ 2

(iii) Magnetic susceptibility of platinum is 0.0001. Its relative permeability is

(a) 1.0000 (b) 0.9999

(c) 1.0001 (d) 0

(iv) When a light wave travels from air to glass

(a) its wavelength decreases.

(b) its wavelength increases.

(c) there is no change in wavelength

(d) its frequency decreases.

ISCSOLVED PAPER 2017

Issued by Council for the Indian Certificate of Secondary Education

Fully Solved (Questions-Answers)

2

General Instructions

1. Answers to this paper must be written on the paper provided separately.

2. You will not be allowed to write during the first 15 min.

3. This time is to be spent in reading the question paper.

4. The time given at the head of this paper is the time allowed for writing the answers.

5. Section I is compulsory. Attempt any four questions from Section II.

6. The intended marks for questions or parts of questions are given in brackets [ ].

(v) A radioactive substance decays to 1/16th of its initial mass in 40 days. Thehalf life of the substance, in days, is

(a) 20 (b) 10

(c) 5 (d) 2.5

B. Answer all questions given below briefly and to the point(i) Maximum torque acting on an electric dipole of moment 3 10 29× − cm in a

uniform electric field E is 6 10 25× − Nm. Find E .

(ii) What is meant by drift speed of free electrons?

(iii) On which conservation principle is Kirchhoff’s Second Law of electricalnetworks based?

(iv) Calculate magnetic flux density of the magnetic field at the centre of acircular coil of 50 turns, having radius of 0.5 m and carrying a current of 5 A.

(v) An AC generator generates an emf ε where ε π= 314 50sin( )t volt. Calculatethe frequency of the emf ε.

(vi) With what type of source of light are cylindrical wave fronts associated?

(vii) How is fringe width of an interference pattern in Young’s double slitexperiment affected if the two slits are brought closer to each other?

(viii) In a regular prism, what is the relation between angle of incidence andangle of emergence when it is in the minimum deviation position?

(ix) A converging lens of focal length 40 cm is kept in contact with a diverginglens of focal length 30 cm. Find the local length of the combination.

(x) How can the spherical aberration produced by a lens be minimised?

(xi) Calculate the momentum of a photon of energy 6 10 19× − J.

(xii) According to Bohr, ‘Angular momentum of an orbiting electron isquantised’. What is meant by this statement?

(xiii) Why nuclear fusion reaction is also called thermo-nuclear reaction?

(xiv) What is the minimum energy which a gamma ray photon must possess inorder to produce electron-positron pair?

(xv) Show the variation of voltage with time, for a digital signal.

Section II [40 Marks]

Answer ten questions in this part, choosing four questions from Section A, threequestions from Section B and three questions from Section C.

2. (a) Show that electric potential at a point P, at a distance r from a fixed pointcharge Q, is given by

VQ

r=

14 0πε

.

(b) Intensity of electric field at a perpendicular distance of 0.5 m from aninfinitely long line charge having linear charge density (λ) is 3 6 103. ×

Vm− 1. Find the value of λ.

3. (a) Three capacitors C m1 3= µ , C m2 6= µ and C F3 10= µ are connected to a 50 Vbattery as shown in the Figure 1 below

Calculate

(i) The equivalent capacitance of the circuit between points A and B.

(ii) The charge on C1.

(b) Two resistors R1 60= Ω and R2 90= Ω are connected in parallel. If electricpower consumed by the resistor R1 is 15 W, calculate the power consumedby the resistor R2.

4. (a) Figure 2 below shows two resistors R1 and R2 connected to a battery havingan emf of 40V and negligible internal resistance. A voltmeter having aresistance of 300 Ω is used to measure potential difference across R1. Findthe reading of the voltmeter.

(b) A moving coil galvanometer has a coil of resistance 59 Ω, It shows a fullscale deflection for a current of 50 mA. How will you convert it to anammeter having a range of 0 to 3A?

5. (a) In a meter bridge circuit, resistance in the left hand gap is 2 Ω and anunknown resistance X is in the right hand gap as shown in Figure 3 below.The null point is found to be 40 cm from the left end of the wire. Whatresistance should be connected to X so that the new null point is 50 cmfrom the left end of the wire?

(b) The horizontal component of earth’s magnetic field at a place is13

times

the vertical component. Determine the angle of dip at that place.

6. (a) Using Ampere’s circuital law, obtain an expression for the magnetic fluxdensity B at a point X at a perpendicular distance r from a long currentcarrying conductor. (Statement of the law is not required).

G

2Ω

( )R

X

0 cm 100 cm40 cm

R1

40 V

R2

200 Ω 880 Ω

V

C1

3µF

C2

6µF

C3

10µF

A B

50 V

(b) PQ is a long straight conductor carrying a current of 3A as shown in Figure 4below. An electron moves with a velocity of 2 107× ms− 1 parallel to it. Find

the force acting on the electron.

7. (a) (i) AB and CD are two parallel conductors kept 1 m apart and connected bya resistance R of 6 Ω, as shown in Figure 5 below. They are placed in amagnetic field B = × −3 10 2 T which is perpendicular to the plane of theconductors and directed into the paper. A wire MN is placed over ABand CD and then made to slide with a velocity 2 ms− 1. (Neglect theresistance of AB, CD and MN .)

Calculate the induced current flowing through the resistor R.

(ii) In an ideal transformer, an output of 66 kV is required when an inputvoltage of 220 V is available. If the primary has 300 turns, how manyturns should the secondary have?

(b) In a series L-C-R circuit, obtain an expression for the resonant frequency.

Section B [40 Marks]

Answer any three questions

8. (a) (i) State any one property which is common to all electromagneticwaves.

(ii) Arrange the following electromagnetic waves in increasing order oftheir frequencies (i.e. begin with the lowest frequency):

Visible light, γ-rays, X-rays, micro waves, radio waves, infraredradiations and ultraviolet radiations.

(b) (i) What is meant by diffraction of light?

(ii) In Fraunhoffer diffraction, what kind of source of light is used andwhere is it situated?

9. (a) In Young’s double slit experiment using monochromatic light ofwavelength 600 nm, 5th bright fringe is at a distance of 0.48 mm from thecentre of the pattern. If the screen is at a distance of 80 cm from the planeof the two slits, calculate

(i) Distance between the two slits.

1 m

B

D

A

C

R

M

N

0.6 m

P

3A

Q

electron

(ii) Fringe width, i.e. fringe separation.

(b) (i) State Brewster’s law.

(ii) Find Brewster’s angle for a transparent liquid having refractive index1.5.

10. (a) Find critical angle for glass and water pair, given refractive index of glass is1.62 and that of water is 1.33.

(b) Starting with an expression for refraction at a single spherical surface,obtain Lens Maker’s Formula.

11. (a) A compound microscope consists of two convex lenses of focal length 2 cmand 5 cm. When an object is kept at a distance of 2.1 cm from the objective,a virtual and magnified image is formed 25 cm from the eye piece. Calculatethe magnifying power of the microscope.

(b) (i) What is meant by resolving power of a telescope?

(ii) State any method of increasing the resolving power of anastronomical telescope.

Section C [40 Marks]

Answer any three questions

12. (a) (i) Plot a labelled graph of | |Vs where Vs is stopping potential versusfrequency f of the incident radiation.

(ii) State how will you use this graph to determine the value of Planck’sconstant.

(b) (i) Find the de-Broglie wavelength of electrons moving with a speed of7 106× ms−1.

(ii) Describe in brief what is observed when moving electrons are allowedto fall on a thin graphite film and the emergent beam falls on afluorescent screen.

13. (a) Draw energy level diagram for hydrogen atom, showing first four energylevels corresponding to n = 1, 2, 3 and 4. Show transitions responsible for

(i) Absorption spectrum of Lyman series.

(ii) Emission spectrum of Balmer series.

(b) (i) Find maximum frequency of X-rays produced by an X-ray tubeoperating at a tube potential of 66kV.

(ii) State any one difference between characteristic X-rays andcontinuous X-rays.

14. (a) Obtain a relation between half life of a radioactive substance and decayconstant ( )λ .

(b) Calculate mass defect and binding energy per nucleon of 1020Ne, given

Mass of 1020 19 992397Ne = . u

Mass of 11 1007825H = . u

Mass of 01 1008665n = . u

15. (a) With reference to a semi-conductor diode, what is meant by

(i) Forward bias (ii) Reverse bias

(iii) Depletion region

(b) Draw a diagram to show how NAND gates can be combined to obtain an ORgate. (Truth table is not required).

Useful Constants and Relations

1. Charge of a proton (e) = × −16 10 19. C

2. Planck’s constant (h) = × −66 10 34. Js

3. Mass of an electron (m) = × −9 1 10 31. kg

4. Permittivity of vacuum (ε0) = × −885 10 12. Fm− 1

5. 14 0πε

= ×9 109 mF− 1

6. Permeability of vacuum (µ 0) = × −4 10 7π Hm− 1

7. µπ0

4

= × −1 10 7 Hm− 1

8. Speed of light in vacuum (c) = ×3 108 ms− 1

9. Unified atomic mass unit (u) = 931 MeV

10. Electron volt (1eV) = × −16 10 19. J

1. A.(i) The electric potential energy of two pointcharges 1 µC each, placed, metre apart inair is

Uq q

r= 1

4 0

1 2

12πε.

= × × ×− −9 10

10 10

19

6 6

= × × −9 10 109 12

= × −9 10 3 J

(ii) The equivalent resistance of thecombination is

1 1 1R R

n

R

neq

= + +...

1R

n

R

n

Req

= + +...=

nn

R

RR

neq =

2

(iii) Magnetic susceptibility of platinum

χm =00001.

Relative Permeability µ χr m= +1

= +1 00001. = 10001.

(iv)(a) Its wavelength decreases becausethey are related by the formula

µ λλ

= vac

med

here, µ λλ

= air

glass

As radioactivity ∝ number of atomspresent

NN= 0

16…(i)

But as we know

N Nn

=

0

1

2

where, n is the number of half lives.

∴ NN

n0

016

1

2=

⇒ 1

2

1

2

4

=

n

⇒ n =4

Half-life period

= Time of disintegration

No. of half lives

Half life period = 40 days

4

= 10 days

B. (i) Torque, τ = × −3 10 29 cm

p = × −6 10 25 Nm

τ = ×p E = pE sin θFor maximum value θ = °90

τ = pE

3 10 29× − = × ×−6 10 25 E

⇒ E = ××

−

−3 10

6 10

29

25= × − +0 5 10 29 25.

= × −0 5 10 4. N/C

(ii) Refer Q. 17 (i) (a) → Page 51 (ch-5) CodeF263

Insert Average drift velocity, veE

md = τ

v d = =τ average relaxation time

E = electric field

e = charge on electron

Answer sAnswer s

(iii) Kirchoff’s second law of electricalnetworks is passed on the law ofconservation of energy. Because the netchange in energy of a charge, after thecharge complete a closed path must bezero. It means that in any closed path ofan electrical circuit, the algebric sum ofthe emfs is equal to the algebric sum ofthe products of resistances and currentsflowing through them.

(iv) Magnetic flux φ =∫ B dA.

for maximum value φ =BA

Given, I = 5A

n = 50 turns

r =0 5. m

BA

= φ

BIn

r= µ

ππ0

42

= × × × ×−4 10

4

2 3 14 5 50

0 5

7ππ

..

.

= × −3 14 10 4. T

(v) In AC generator ε π= 314 50sin( )t V

Frequency is 50π because

ε ε ω= 0 sin t,

where ε0 = peak value of AC

ω =angular frequency

ε = voltage at any time t

(vi) Refer Ch 13 (Page 161) Code F263

Cylindrical wave front → Source of lightis linear.

(vii) According to Young’s double slit

experiment, fringe width β λ= D

d

where, D = distance between slit andscreen

d = distance between two slits

If d decreases, i.e. distance between twoslits decreases, β will increase.

(viii) Refer Pg-199 ch-17 Topic-2

till δ = + −i i A1 2

where, i1 = angle of incidence

i2 = angle of emergence

A = angle of the prism.

(ix) focal length of convex lens,f1 40= cm

Focal length of the concave lens,f2 30= − cm.

(ix) Hence the focal length of thecombination will be

1 1 1

1 2f f f= +

= +−

140

130

= −140

130

= − = −3 4

1201

120

f = − 120cm

= − = −120100

1 2. m

(x) Refer Page-208 ch-17 (Refraction oflight) (topic-3) code F263

(xi) As we know,

momentum of photon = mc

and according to Planck’s quantum

theory, energy of the photon is Ehc=λ

⇒ λ = hc

E

According to the question,

E = × −6 10 19 J

λ =× −

hc

6 10 19…(i)

According to Einstan’s theory ofmass-energy equivalence,

E mc= 2 …(ii)

mch=λ

mc = momentum of photon

Now, substituting value of h and λ

mch

hc=

× −6 10 19

= × × −h

hc

6 10 19

= ××

−6 10

3 10

19

8

= × −2 10 27 kgms− 1

(xii) According to Bohr, ‘angular momentumof an orbiting electron is quantised’ - itmeans of all the possible circular orbits,the electrons are permitted to circulateonly in those orbits in which the angularmomentum of an electron is an integral

multiple ofh

2π, h being Planck’s

constant. Therefore, for any permittedorbit,

L mvrnh= =2π

, n = 1 2 3, , , ...

where L, m and v are the angularmomentum, mass and the speed of theelectrons, r is the radius of thepermitted orbit and n is positive integercalled principal quantum number.

Thus, the angular momentum of theorbiting electron is quantised, has

certain valuenh

2π.

(xiii) To carry nuclear fusion in a bulkmaterial, the temperature of the

material has to be raised to 106

K . Ahigh temperature is necessary for thelight nuclei to have sufficient kineticenergy so that they can overcome theirmutual coulombic repulsion and comecloser than the range of nuclear force.Colliding nuclei should have enoughenergy due to their thermal motion andcan penetrate the coulomb barrier.That’s why nuclear fusion reaction isalso known as thermo - nuclearreaction as it involves lot of heat.

(xiv) γ-rays are neutral photons ofelectromagnetic nature. After an α- orβ- decay, the daughter nuclei is usuallyleft in the excited state. It attains theground state by single or successivetransitions by emitting one or morephotons. As these energies of the orderof MeV.

(xv) Refer Pg-338 (Ch-30) Ist graph

Part-II2.(a) Let P be the point at a distance r from

the origin O at which the electricpotential due to charge +Q is required.

The electric potential at a point P is theamount of work done in carrying a unitpositive charge from ∞ to P.

As, work done is independent of thepath, we choose a convenient pathalong the radial direction from infinityto the point P without acceleration.

Let A be an intermediate point on thispath where OA x= .

The electrostatic force on a unit positivecharge A is given by

FQ

x= 1

4 02πε

, along OA …(i)

Small work done in moving the chargethrough a distance dx from A to B isgiven by

dW dx Fdx Fdx= = ° = −F. cos 180

[Qcos 180 1° = − ]

dW Fdx= − …(ii)

Total work done in moving a unitpositive charge from ∞ to the point P isgiven by

W FdxQ

xdx

r r= − = −

∞ ∞∫ ∫ 14 0

2πε

= − = − −

−

∞∞∫Q

x dxQ

x

rr

4 4

1

0

2

0πε πε

= −∞

Q

r41 1

0πεQ x dx

x−∫ = −

2 1

WQ

r=

4 0πε…(iii)

From the definition of electricpotential, this work is equal to thepotential at point P.

VQ

r

Q

r= =

41

4 0πε πε0…(iv)

(b) By applying Gauss’s theorem, electricfield due to an infinitely long straightcharged wire is,

Er

= 12 0πε

λ.

where1

22 9 10

0

9

πε= × × mF− 1

λ = linear charge density = ?

r =0 5. m

E = ×3 6 103. Vm − 1

3 6 102 9 10

0 53

9

..

× = × × × λ

λ = × ×× ×

3 6 10 0 5

2 9 10

3

9

. .

= −10 7 cm − 1

3. (a) (i)

From the above figure, as C 1 and C2

are connected in series,

hence1 1 1

1 2C C Ceq

= +

1 1

3

1

6C eq

= +

C eq F= 2µ

Now, C eq is in parallel with C 1 and C2 ,

hence equivalent capacitance of thecircuit between points A and B will be10 2+ = 12µF

(ii) The charge on Cq

V1 =

q C V= 1 = ×3 50 = 150 C

r

x

O P B A E ∞

dx

+Q + C1

C1

3µF

C2

6µF

C3

10µF

50 V

(b)

Equivalent resistance of the circuit = Req

1 1 1

1 2R R Req

= +

= + = + =160

190

3 2

1805

180

Req = 36ΩNow, given P1 15= W

R1 60= Ω

Hence applying PV

R1

2

1

=

V P R21 1= = ×15 60

V2 900=V = 900 = 30 V

Same V is applied for both R1 and R2 ,

Total Power PV

RTotal =

2

eq

= ( )3036

2

= ×30 30

36= 25 W

Hence, P2 (Power consumed byresistance R2 ) = −P PTotal 1

= − =25 15 10 W

4.(a)

Total current = I

According to Ohm’s law

V IR=

IV

R= = 40

1080

R R Rs = += +=

1 2

200 880

1080 Ω

= 127

A

Now, for R1, PdV IR= 1

= ×127

200 =7 40. V

(b) R = 59 Ω

According to the formula,

SI

I IGg

g

=−

G = 59 ΩIg = 50 mA = × −50 10 3 A

I = 3A

∴ Shunt resistance

SI G

I I

g

g

=−

= × ×− ×

−

−50 10 59

3 50 10

3

3( )

= ×

×

50 59

10005920

= × ××

50 59 20

1000 59= 1 Ω

Hence, to convert galvanometer intoammeter, a shunt of 1 Ω is connected inparallel with the galvanometer.

5. (a)

Applying the condition of balancedwheatstone bridge,

R

X

l

l=

−100

According to the question, R = 2ΩX = ?

l =40 cm2 40

100 40X=

−2 40

60X=

⇒ X = × =2 60

403 Ω

The equivalent resistance of 3 Ω and 5 Ω

in parallel is3

3S

S+Ω

Again applying the condition ofbalanced Wheatstone bridge,

R1

40 V

R2

200 Ω 880 Ω

V

300 Ω

50 V

R1

60 Ω

R2

90 Ω

G

2Ω

( )R

X

0 cm 100 cm40 cm

2Ω

S

3Ω

X

⇒ 23

3

5050

1S

S+

= =

⇒ 23

3=

+S

S

⇒ 2 3 3( )+ =S S

⇒ 6 2 3+ =S S

⇒ 3 2 6S S− =⇒ S = 6 Ω

(b) Let Bv =component of Earth’s magneticfield

δ =angle of dip

BH = Horizontal component of earth’smagnetic field.

According to the question,

B Bv H= 3

Using tan δ = =B

B

B

Bv

H

H

H

3 = = °3 75tan

δ = °75

6. (a) Refer Page-99 (Ch-8)

Ampere’s circuital law along withdiagram

(b)

From the given figure, if we have tocalculate force on a long straightconductor.

(b) Force acting on the electron

F qvB=Due to wire, the field created is

BI

R= µ

π0

2= ×

× ×µ

π0 3

2 0 6.

= × × ×× ×

−4 10 3 10

2 0 6

7ππ .

= −10 6 T

F Bqv=

Now, F = × × × ×− −1 6 10 2 10 1019 7 6.

= × × × ×− −1610

10 2 10 1019 7 6

= × × − + −16 2 10 19 6 6

F = × −32 10 19 N

7. (a) (i) Given, B = × −3 10 2 T

MN BD= = 1 m (according to thefigure)

v = −2 1ms

The emf induced in the coil, ed

dt= − φ

ed BA

dt= − ( )

A = area of the loop = ×l x

ed Bl x

dt= − ×( ) = − = −Bl

dx

dtBlv

= − × × ×−3 10 1 22 = − × −6 10 2 V

Current,Ie

R= = × −6 10

6

2

= =−10 0012 . A

(ii) According to the question,

N P = 300 turns (turns of prime of)

N S = ? (turns of secondary)

V VS P> because VS = 66 kV

VP = 220 V

Hence,V

V

N

NS

P

S

P

= ⇒ 66 1000

220 300

× = N S

⇒ N S = × ×66 1000 300

220

= ×9 104 turns

= 90000 turns

(b) Refer page-137, 138 (ch-11)

Alternating current Series L-C-R circuit(page-137)

Section-B

8. (a) (i) One property which is common to allelectromagnetic waves is as follows.

The energy associated with the EMwaves is equally divided betweenelectric and magnetic field.

Energy density of electric frield,

u Ee = 1

20

2ε

Energy density of magnetic field,

uB

B = 1

2

2

0µ

So, the total energy per unit volume isgiven as, u u ue B= +

= +1

2

1

20

22

0

εµ

EB

Also, the average energy density is,

u EB

av = =1

2 20 0

2 02

0

εµ

(ii) Electromagnetic waves in increasingorder of their frequencies are asfollows. (3 103× to 3 108× Hz)

Radio waves < Micro waves < Infrared

( )4 10 8 1014 14× → ×Hz Hz

radiations < Visible light < Ultraviolet

radiations < X-rays < Gamma rays

( )5 10 5 1018 22× − × Hz

0.6 m

P

3A

Q

electron

(b) (i) The phenomenon of bending of lightaround the edges or the sharp cornersand then spreading of light within thegeometrical shadow of opaqueobstacles is called diffraction of light.Diffraction is more prominent whenslit width a is smaller thanwavelength ( )λ of light ( )a < λ .

Refer (ch-15) Diffraction of light.

(ii) Fraunhofer diffraction

In this type of diffraction, the sourceof light is not necessary to bemonochromatic or white light but theexpression calculated for the centralmaxima and minima, we considered itto be monochromatic. And the sourceof light is being placed at largedistance from the screen i.e.effectively at infinity. Nature of thewave front is planar. A converging lensis used to observe diffraction pattern.

9. (a) Given,

wavelength, λ = 600 nm = × −600 10 9 m

5th bright fringe observed at distance,x =048. mm = × −048 10 3. m

Distance between screen and slits,D = 80 cm

= × −80 10 2 m.

(i) Distance between two slits,

n th bright fringe is observed at,

xnD

dn = λ

⇒ 048 10 3. × −

= × × × ×− −5 80 10 600 102 9

d

⇒ d = × × × ××

− −

−5 80 10 600 10

048 10

2 9

3.

= × −5 10 3 m

(ii) Fringe width i.e. fringe separation

It is given by, β λ= D

d

= × × ××

− −

−80 10 600 10

5 10

2 9

3= × −9 6 10 5. m

(b) (i) When an unpolarised light isincident on the boundary betweentwo transparent media, the reflectedlight is polarised with its electricfield vector perpendicular to theplane of incidence when thereflected and refracted rays make aright angle with each other.

Thus, when reflected wave isperpendicular to the refracted wave,the reflected wave is a totally polarisedwave. The angle of incidence in thiscase is called Brewster’s angle iB .

Since, i rB + = π2

From Snell’s law,

µπ

= =−

sin

sin

sin

sin

i

r

i

i

B B

B2

= =sin

costan

i

iiB

BB

∴ µ = tan iB

This is Brewster’ law.

(ii) Here, refractive index, µ = 1 5.

iB = ?

From Brewster’ law,

tan .iB = =µ 1 5

∴ iB = = °−tan ( . ) .1 1 5 56 3

10. (a) Given,

refractive index of glass is µg = 1 62.

refractive index of water is, µw = 1 33.

From Snell’s law of refraction,sinsin

i

rw

g

= µµ

⇒ For the critical for glass-water pair,r = °90

⇒ sin

sin..

iC

901 331 62°

=

⇒ sin .iC =0 820

⇒ iC = −sin ( . )1 0 820 = °5508.

(b) Lens Maker’s Formula

O

r1

i

AN1

n1 n2

B

C1 I1

C

R1

(b)

i2

A

n1n2

N2

I1

C

R2

C2

r2

ID

(c)

O

u v

B D

AN1 N2

C(a)

The above figure shows the imageformation by a convex lens.

Applying the equation for sphericalrefracting surface to the first interfaceABC, we obtain

n

OB

n

BI

n n

BC1 2

1

2 1

1

+ = −…(i)

A similar procedure applied to thesecond interface ADC gives,

− + = −n

DI

n

DI

n n

DC2

1

1 2 1

2

…(ii)

For a thin lens, BI DI1 1=Adding Eqs. (i) and (ii), we obtain

n

OB

n

DIn n

BC DC1 1

2 11 2

1 1+ = − +

( ) …(iii)

Suppose the object is at infinity i.e.

OB → ∞ and DI f→Eq. (iii) can be written as

n

fn n

BC DC1

2 11 2

1 1= − +

( ) …(iv)

The point where image of an objectplaced at infinity is formed is called thefocus (f) of the lens and the distance fgives its focal length. A lens has twofoci, F and ′F on either side of it by signconvention.

BC R1 1=or DC R2 2= −Therefore, Eq. (iv) can be written as

11

1 11 2

1 2fn

R R= − −

( ) Q 1 2

2

1

nn

n=

…(v)

Eq. (v) is known as the Lens Maker’formula.

11. (a) Here,

f0 2= cm, fe = 5 cm, u0 2 1= . cm,

v e = 25 cm, v 0 = ?

For the objective lens,1 1 1

0 0 0f v u= −

⇒ 1 1 1

0 0 0v f u= + = +

−

1

212 1.

⇒ 1 0 14 20v

= ..

⇒ v 0 42= cm

Now, magnification i.e. magnifyingpower of the microscope,

Mv

u

D

fe

= +

0

0

1 = +

422 1

1255.

= =20 6 120( )

(b) (i) Resolving Power of a Telescope

The resolving power of a telescope isdefined as the reciprocal of thesmallest angular separation betweentwo distant objects whose images areto be seen separately.

Resolving power = =1

1 22d

D

θ λ.

Resolving power depends on (i)wavelength λ,

(ii) diameter of the objective D.

(iii) to increase the resolving power of anastronomical telescope, we need toincrease the numerical aperture ofthe objective.

Section-C

12. (i) When moving electrons are allowed tofall on a thin graphite film, then theelectrons from the graphite film will beknocked out, after gaining sufficientenergy from the collision with themoving electrons and then this beam ofemergent electrons falls on thefluorescent screen, due to which visibleradiation will be produced from thescreen (i.e. fluorescence), helping inidentifying that emission of electronshad been started from the thin film.

(ii) Emission spectrum of Balmer series InBalmer series, the electrons jumps fromany outer orbit to the second orbit.

Emission spectrum for Balmer seriesfrom energy level diagram.

dθS2

S1

Objective

of telescope

n=4

n=3

n=2

n=1

n=0

Balmor series

13. (a) (i) Absorption spectrum of Lymanseries In Lyman series, the electronsjumps from outer orbit to first orbit.So, its absorption spectrum will be

Absorption spectrum for Lymanseries.

(ii)

(b) (i) Given,

potential, V = 66 kV.

So, the minimum wavelengthcorresponding to maximum frequencyis,

λmin = hc

N=

×1242

66 103= × −18 8 10 3. nm

= × −18 8 10 12. m

So, maximum frequency, vC

maxmin

=λ

= ×× −

3 10

18 8 10

8

12.= ×0 15 1020. Hz

= ×1 5 1019. Hz

(ii) Continuous X-rays are those rayswhich comes out when an innerelectron is being knocked out, by thecollision with an atom of targetmaterial. In continuous X-ray, theintensity varies gradually.

But, when an electron suppose from aK shell is being knocked out, then avacancy is being created. To fill thisvacancy, an electron jumps fromanother orbit. When such transitiontakes place the energy is convertedand released in the form of photonand hence, the rays released areknown as characteristic X -rays. Incharacteristic X-rays, at certainwavelength at which intensitybecomes very large.

14. (a) Refer page 295, 296 (ch-24) code F263

(b) In a nucleus of 1020Ne

No. of protons = 10

No. of neutrons = − =20 10 10

Total mass of 10 protons and 10neutrons = +10 10m mp n

= +10( )m mp n

= × +10 1007825 1008665( . . )

= ×2 2 01649. u = 20 1649. u

Mass Defect

∆m = −20 1649 19 992397. . =0 172503. u

Total binding energy = ×0 172503 931.

= 160 600293. MeV

E b per nucleon = 160 60029320

.

= 8 03001465. MeV/nucleon

15. (a) Refer page 316, 317 During formation ofp-n junction, two important processoccur namely diffusion and drift. Due tothe concentration gradient across p andn-sides, holes diffuse from p-side ton-side ( )p n→ and electrons diffusefrom n-side to p-side ( )n p→ . Thediffused charge carriers combine withtheir counterpart in the vicinity ofjunction and neutralise each other.

Thus, in the vicinity of the junction,negative charge (in the form of fixedacceptor ions) on the p-side andpositive charge (in the form of fixeddonor ions) on the n-side.

This sets up a potential differenceacross the junction, which stops furtherdiffusion when it become strong. Theregion on either side of junction whichbecomes depleted (free) of the mobilecharge carrier is called depletionlayer/depletion region.

Depletion region it is free from mobilecharge carriers at equilibrium. Thewidth of depletion region is of the orderof 10 6− m. The potential differencedeveloped across the depletion region iscalled the potential barrier. Due to thepositive space charge region on n-sideand negative space charge region onp-side of the junction, an electric fieldis setup from +ve charge to negativecharge. Motion of charge carriers due tothis electric field is called drift.

Forward Biasing

A junction is said to be forward biasedwhen the positive terminal of theexternal battery is connected to thep -side and negative terminal to the

n=4

n=3

n=2

n=1

n=0

p n

Electron driftElectron diffusion

Depletion regionHole diffusion

Hole drift

Depletion region in p-n junction

n-side of the diode. As, the direction ofapplied voltage is opposite to the builtin potential or barrier potential,therefore depletion layer decreases.

The current in this situation of forwardbiasing is called forward current and isconstituted by the motion of majoritycharge carriers across the junction. Inforward bias, the junction diode offerslow resistance.

Reverse Biasing

A junction diode is said to be reversebiased when the positive terminal ofthe external battery is connected to then-side and negative terminal to the

p-side of the diode. As, the direction ofapplied voltage is in same direction ofbuilt in potential or barrier potential of,therefore depletion layer decreases..

The current in this situation of reversebiasing is very small due to minoritycharge carriers and flows across thejunction. This current is called reversecurrent. It offers high resistance.

(b)

P is NAND, Q is OR gate.

Ei

+ –

E

Hole Electronp-region n-region

Junction

+ –

Battery (Forward Biased)

Ei

+–

E

Hole Electronp-region n-region

Junction

+ –

Battery (Reverse Biased)

A

B X