PHYSICS Complete Syllabus of Class XII...Model Test Paper-3 Subjective Test for Class-XII (Physics)...

Model Test Paper-3(for School / Board Exams.)

PHYSICS

(Complete Syllabus of Class XII)

MM : 70 Time : 3 Hrs.

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

CODE

A

GENERAL INSTRUCTIONS :

(i) All questions are compulsory.

(ii) Questions number 1 to 5 are very short answer type questions and carry 1 mark each.

(iii) Questions number 6 to 10 are short answer type questions and carry 2 marks each.

(iv) Questions number 11 to 22 are also short answer type questions and carry 3 marks each.

(v) Questions number 23 is a value based questions and carry 4 marks.

(vi) Questions number 24 to 26 are long answer type questions and carry 5 marks each.

(vii) There is no overall choice. However, an internal choice has been provided in one question of 2 marks,

one question of 3 marks and all three questions of 5 marks each. You have to attempt only one of the

given choices in such questions.

(viii) Use log tables if necessary, use of calculator is not allowed.

SECTION-A

1. Two identical sphere are placed on the insulated stand. Now one is given (+Q) while other is given (–Q) charge.

Which one have greater mass? [1]

2. What is the phase difference between voltage across capacitor and voltage across inductor in series LCR

circuit? [1]

3. The intensity of central maximum in Young's double interference is I0. If one of the slit is covered then what

will the intensity at central point? [1]

4. Define the polarization of dielectric material. [1]

5. What do you understand the breakdown of p-n junction diode? [1]

SECTION-B

6. What are the advantage of digital data over the analog data? [2]

7. Derive the total energy of electron in ground state of hydrogen atom in terms of proper constant. [2]

8. State the KVL and KCL briefly. [2]

(2)

Model Test Paper-3 Subjective Test for Class-XII (Physics)

9. Write down the various Bohr postulate. [2]

10. Prove that average intensity remains constant in interference phenomena. [2]

SECTION-C

11. A photosensitive surface when exposed by photon of wavelength 1 the maximum kinetic energy of photo-

electron emitted is K. If the same surface is exposed by photon of wavelength 2, the speed of photoelectron

get doubled. Find the work function of surface. [3]

12. In the adjacent circuit, the values of the following elements are given bias resistance. [3]

R1 = 15.8 k R

2 = 2.2 k

RL = 500

IB = 55 A = 200

VCC

= 9 V

(i) Find VQ

(ii) Find input resistance

C

E

B

R1

R2

Q

RL

VCC

= 9 V

13. Define AM signal. Draw the block diagram of amplitude modulator. [3]

14. Draw the phasor diagram for L-R, C-R, and L-C-R circuit. [3]

15. Find the equivalent resistance of given network between point P and Q. If resistance of each arm is R. [3]

Q

R

16. A conductor of length L rotate with constant angular speed inside the magnetic field as shown in the figure.

Find the potential difference induced between end O and point A at the other end. [3]

O

B

A

17. Draw the circuit diagram of full wave bridge rectifier. Show the conducting and non-conducting diode in both

half cycle. [3]

18. What is polarisation of light? Discuss two method to obtain plane polarised light. [3]

OR

Prove that intensity of light emerging from a polarised get reduce to 1

2 of its original value.

(3)

Subjective Test for Class-XII (Physics) Model Test Paper-3

19. Prove that radiation pressure of a electromagnetic wave at perfectly reflecting surface is 2l

C where l is intensity

of incident radiation and wave incident normally on the surface. [3]

20. A parallel plate capacitor is partially field with a dielectric having dielectric constant k and thickness t, prove

that equivalent capacitance of the system is [3]

0A

Ct

d tk

(All symbols have their usual meaning)

21. A radioactive material get decayed if decay constant is and initial number of nuclie is N0. Derive the number

of nuclie at an arbitrary time t0. [3]

22. A mark on the surface of a glass sphere (1.5) is viewed from a diametrically opposite position. It appearsto be at a distance of 10 cm from its actual position. Find the radius of sphere. [3]

23. Tushar was using a galvanometer in the practical class unfortunately it fell from his hand and broke. He was

upset, some of his friends advised him not tell the teachers but Tushar decided to tell his teacher. Teacher

listened to him patiently and on knowing that the act was not intentional but just on accident, did not scold

him and used the opportunity to show the internal structure of galvanometer to the whole class. [4]

(i) What are the values displayed by Tushar?

(ii) Explain the principle, construction and working of moving coil glavanometer.

SECTION-D

24. (i) State the Gauss’ theorem and with the help of the result of Gauss theorem, find the electric field intensity

due to

(a) A point charge q at a distance r from it.

(b) Infinite line charge having linear charge density C/m.(ii) Devide the charge Q into two parts such that force between them will be maximum [5]

OR

(i) Find the force of interaction between a line charge having linear charge density and point charge Q atdistance of a from its centre.

C/m

2l a

PO

(ii) Find the electric field at the centre of equilateral triangle having charge configuration shown in figure.

d

dd p

(+)Q

(+)Q

–Q

25. (i) Find the magnetic field intensity at point P and Q due to a solid conductor having uniformly distributed

current. [5]

r/2

r/2Q

PI I

(4)


(ii) A battery is connected between two point A and B on the circumference of a uniform conducting ring of radius

r and resistance R as shown one of arc AB of the ring subtends an angle at the centre. What is the value ofthe magnetic field at the centre due to current is the ring.

AI

I1

IB

I2

OR

(i) Find the magnetic field intensity at axial point of a solenoid if ends of solenoid formed angle and at a pointB on axis of solenoid.

II

S N B

Solenoid

(ii) A very long then strip of metal of width ‘b’ carries a current I along its length as shown in figure. Find the

magnitude of magnetic field in the plane of strip at a distance ‘a’ from the edge nearest to the point

I

b

a P

26. (i) Draw a neat and clean diagram for Telescope working outline the working principle and discuss the cases

when final image form at and at near point. [5](ii) In case of a thin lens of focal length f, if an object is placed at a distance x

1 from first focus and its image is

formed at a distance x2 from the second focus shown that x

1x

2 = f 2

OR

(i) Draw neat and clean diagram of compound microscope outline the working principle. Discuss the case of final

image formation at (i) infinity (ii) near point

(ii) In a compound microscope the object is 1 cm from the objective lens. The lenses are 30 m apart and the

intermediate image is 5 cm from eye-piece. What is magnification produced ?

� � �

(5)


PHYSICS

SECTION-A

1. The sphere having (–Q) charge is heavier than other sphere having charge (+Q). Because 1st sphere gain

electron while 2nd lose electron.

2. radian

3.2 2 0

0 1 2( ) (2 ) 4

4

II I I I I I ⇒

4. Induced dipole moment per unit volume is called polarization of dielectric.

5. When the reverse bias voltage is too large. Then very high abrupt current following through it in reverse direction

of that of normal operation. This is called breakdown of PN junction, which finally result of burning diode.

SECTION-B

6. Following are the advantages of digital data over the analog data.

(a) Digital data are immune from imperfection of electronic system.

(b) By coding the digital data we achieve secure transmission of data.

(c) Digital data have better noise immunity.

(d) Digital data can be processed by digital circuit component which are cheap and easily

produce.

7. Electrostatic attraction between electron and nucleus gas required centripetal force

2 2

20 00

1

4

e mv

aa

⇒ ...(i)

Again quantization of angular momentum givese+

e–

a0

02

nmva⇒

0

2

hv

ma⇒

putting a0 in equation (i) we get,

2

2

2 2 20 0 0

1

4 4

e hm

a m a

Model Test Paper-3(for School / Board Exams.)

MM : 70 Time : 3 Hrs.

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

CODE

A

SOLUTIONS

(6)


2

0

2

ha

m

⎛ ⎞⇒ ⎜ ⎟⎜ ⎟⎝ ⎠

Now

2

0 0

1P.E.

4

e

a

, 2

2

0 0

1 1K.E.1

2 2 4

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

emv

a

2 2

0 0 0 0

1 1P.E. K.E.

4 8

e e

a a

2

0 08

e

a

8. The sum of voltage drop across the all elements in a loop must be zero, it is called K-V-L.

Net current at any junction must be zero, it is called KCL.

9. (a) An atom has a number of stable orbits in which an electron can reside without the emission of radiant

energy. Each orbit corresponds, to a certain energy level.

(b) An electron may jump spontaneously from one orbit (energy level E1) to the other orbit (energy level E

2)

(E2 > E

1), than the energy charge E in the electron jump is given by E = E

2 – E

1 = hv.

(c) The motion of an electron in a circular orbit is restricted in such a manner that its angular momentum is

an integral multiple of h/2, thus

2

nhmvr

10. A special surface around nucleus which contained orbits of equal energy and radius was called shell. These

are numbered from inside to outside as 1, 2, 3, 4 ... and K, L, M, N etc. respectively.

OR

We know in interference

1 2 1 22 cos

RI I I I I

21 2 2

avg

0

( 2 4 cos )( )

2R

I I I dI

∫

2

2

1 2 0 1 2

0

( )[ ] 2 cos

2

I I I I d

∫

2 2( )2 0

2 2

I I

1 2I I sum of individual intensity

OR

When an object is placed between focus and optical centre of a convex lens its large and errected virtual image is

formed. This principle is used in simple microscope.

SECTION-C

11. We know Kmax

= hc

1

⇒ hc

K ...(i)

and,

2

4 hc

K ...(ii) ( velocity get twice i.e., get 4 times) dividing (i) by (ii), we get

(7)


1

2

4

hc

K

hcK

1 2

44

hc hc ,

1 2 1 2

4 4 13

hc hchc

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

1 2

4 1

3

hc ⎛ ⎞⇒ ⎜ ⎟ ⎝ ⎠

12. (i) 2

1 2

9 2.21 1volt

15.8 2.2

CCQ

V RV

R R

(ii) Since correct IB = 55 A

Input resistance 6

1.1200 k

55 10

Q

B

V

I

B

R1

R2

Q

RL

9 V

13. When the amplitude of high frequency carrier wave is changed in accordance with the intensity of the modulating

signal, it is called amplitude modulation. The frequency of the modulated wave is equal to carrier frequency.

Message Signal

AudioAmplifier

AmplitudeModulator

RF PowerAmplifier

Antena

14.

VL

90°

VR

(a)

(b)

VC

90°

VR

(c)

VL

V

VR

VC

VL

V

VR

VC

if V > VL C

if V < VL C

(8)


15. Current division at different junction is shown in the figure. We use the symmetry of circuit.

Now, 3 6 3 3

P R

I I I IV R R R R V

5

3 6 3 6P R

R R R RV V I I

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

R T

I/6I/6

I/6S

I/3

P

I

I/3

I/3

I/6

I/6

I/3

I/3

I/3

I/3

5 5

6 6

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

eq eq

R RI R I R

16. Let an element of length d induced potential difference across it.

dV = BVdx

= B xdxO

L

P

2

0

1( )

2 ∫

e

P

O

V V B xdx B l

17. During positive half cycle, the end P of the secondary winding positive and Q is negative. The diodes D1 and D

3 are

forward bised they can conduct current where diode D2 and D

4 are reverse biased offering infinite resistance.

During negative half cycle end P becomes negative and end Q positive. The diode D2 and D

4 are forward biased

whereas D1 and D

3 are now reversed biased.

+ +

–

t

Vin

Vin

Secondary voltage

t

Vout

D1

D3

D4

D2 D3D1

RL V

out

A.C.Supply

D1

D2

D3

D4

During positivecycle

Duringnegativecycle

P

Q

18. The phenomenon of confining the vibration of a light wave in a specific direction perpendicular to the direction of

wave motion is called polarisation of light. The plane containing the direction of vibration and wave motion is called

plane of polarisation.

Method of obtaining plane polarised light

(a) By reflection: Brewster discovered that when light is incident at a particular angle on a transparent substance

the reflected light is completely plane polarised with vibration in plane perpendicular to the plane of incidence.

The specific angle of incidence is called polarising angle QP

and is related to the refractive index is of the

material through relation tan PQ

(b) By dichronism: Some crystals such as tourmaline and sheets of idosulphate of quinone have the property

of strongly absorbing the light with vibration perpendicular to a specific direction (called transmission axis)

transmitting the light with vibration parallel to this selective absorption of light is called dichronism so if un

polarised light will be plane polarised with vibration parallel to transmission axis polaroids work on this principle.

(9)


OR

If plane polarised light of intensity I0 = kA2 is incident on a polarised and its vibration of amplitude A makes an

angle with transmission axis then the component of vibration parallel to transmission axis, will be Acos. Whileperpendicular to it A sin. Now as polarised light will pass only those vibration which are parallel to its transmissionaxis i.e.,

Acos, so the intensity of emergent light will be

2 2 2 2

0( cos ) cos cosI k A kA I

This is called Malus law

Now

2 22 2

0 0

0 0 0avg 2

0

cos cos

2 2

I d I d

II

d

∫ ∫

∫

19. An electromagnetic wave carries linear momentum, angular momentum as well as energy. Electromagnetic wave

give rise to pressure when they are reflected or absorbed at the surface of body.

We know energy of electromagnetic wave

E = PC (Where P is momentum)

The pressure exerted on the surface is force per unit area F

A

Pressure = 1 1 1⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

F dP d E dE I dE II

A A dt A dt C CA dt C dt A

This is pressure exerted on perfectily absorbed surface.

If surface is perpectly reflecting change in minimum = 2P

Hence pressure = 2I

C

20. We know

0

0

QE

A

, 0

QE

kA

EE

0E

0+Q –Q

tA Bd

After

0

B

A

V d

V

dvE dv E dx

dx ⇒ ∫ ∫

0

0

0 0

d t tE

E dx dxk

∫ ∫ 00 ( )E

E d t tk

⎧ ⎫ ⎨ ⎬⎩ ⎭

0( )

A B

tV V E d t

k

⎛ ⎞⇒ ⎜ ⎟

⎝ ⎠

Now

0

( )

Q QC

tVQ d t

k

A

⎧ ⎫ ⎨ ⎬

⎩ ⎭

0A

Ct

d tk

(Proved)

(10)


21. We know rate of decay

dNN

dt

dNdt

N⇒

0

0 0

tN

N

dNdt

N⇒ ∫ ∫ 0 0[ln In ] N N t 0

0

In⎛ ⎞

⎜ ⎟⎝ ⎠

Nt

N

0

0

Ne t

N

⇒ 0

0

tN N e

⇒

22. At P, 2 1 2 1

v u R

1 1.5 1 1.5

(2 10) 2R R R

⇒

I O

P

1 1.5 1

2 10 2 2R R R⇒

1 1 3

2 10 2 4R R R⇒

1 1

2 10 4R R⇒

4 2 10 10 5 cmR R R⇒ ⇒ ⇒

23. (i) Courage to tell the truth, gratitude to his teacher for his patience and tolerance.

(ii) In case of moving coil galvanometer, the deflecting torque due to current in the coil BINS is balanced by the

restoring couple due to elasticity of spring supporting the coil so if C is the restoring couple per unit tiwist

and is the deflection (i.e., rotation) of the coil

BINS C⇒

G

CI k

BNS⇒ (where G

Ck

BNS is galvanometer constant)

So in the case of moving coil galvanometer, the current passing through the coil is directly proportional to its

deflection.

SECTION-D

24. It relates the total flux of an electric field through a closed surface to the net charge enclosed by that surface

and according to it, the total flux linked with a closed surface is 0

1⎛ ⎞⎜ ⎟⎝ ⎠

times the charge enclosed by the closed

surface i.e.,

0s

qE ds

∫��

� ...(i)

Now let us find the electric field due to point charge and infinite line charge.

(11)


(i) (a) For point charge Gaussian surface will be spherical with charge at its centre. Angle between andE ds

��

is

O, because both are in radial direction

0 0s

q qE ds E ds⇒ ⇒

∫ ∫��

� �

20

4q

E r

rds

r

for spherical

surface

ds

2

04

qE

r⇒

(b) Gaussian surface for line charge will be cylindrical

Again E ds Eds ��

Now 0s

LE ds

∫

��

�

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

L dsE

0

2L

E rL

⇒ 02

Er

⇒

(ii) Let charge are q and Q – q

q Q – q

20

1 ( )

4

q Q qFe

r

For man( ) , 0dFe

Fedq

2

0

1( 2 ) 0

4Q q

r⇒

2 2Q

Q q q⇒ ⇒ , 2

2 2

0

20

4

d Fe

dq r

Hence Fe will be maximum

for q = 2

Q, Q – q =

2

Q

OR

(i) Let an elementry point of lengh dx

dq = dx

20 0

1

4 4 ( )

Q dxdF dq Q

a x

x

dx

a – x

net 2

04 ( )

l

l

Q dxF dF

a x

⇒

∫ ∫

0

1

4

l

l

Q

a x

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

0

1 1

4

Q

a l a l

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

(12)


(ii) From symmetry of problem, we must have

1 2 32

0

1| | | | | |

4

�� QE E E

d

Resultant of E1 and

1

2 2

22 cos120

rE E E E EE

d d

–Q

E3

E1

E2

+Q +Q2 2 2 1

22

E E E E

1 3 2 20 0

2 12

4 2R R

Q QE E E E

d d

25. (i) Applying ampere circuital law

0 enclosedPB dl I ∫�

Pr/2

I I

2

0

0 2

2 2

2 4P

rI

IrB

r

⎛ ⎞ ⎜ ⎟ ⎝ ⎠⇒

0

4p

Br

⇒

Similarly, 0 enclosedQb dl I ∫

3 /2r

0

32

2Q

rB I

⎛ ⎞⇒ ⎜ ⎟

⎝ ⎠

0

3Q

IB

r

⇒

(ii) Magntic field due to straight conductor is zero at centre.

1 2

2 1

(2 )I R r

I R r

1

1 2

2

2(2 )

II I

I

⎛ ⎞⇒ ⇒ ⎜ ⎟⎝ ⎠

I

O

I1

I

r

I2

0 1 2

0

(2 )

4

I IB

r r

⎛ ⎞ ⎜ ⎟ ⎝ ⎠�

0

1 2( (2 )) 0I I

g r

Hence net magnetic field at centre is 0.

OR

(i) If many time of insulated wire are wound around a cyllinder the resulting coil is called a solenoid. The field at

a point on the axis of solenoid can be obtained by superposition of field due to a large number of identical coils

all having their centre on the axis of solenoid.

(13)


So if we consider a coil of width dx at distance x from the point p on the axis solenoid as shown in figure.

I

x

dx

2

0

2 2 3/2

2( )

4 ( )

NIRdB N ndx

R x

∵ ( n is number of things per unit length)

20

2 2 3/22

4 ( )

dxdB nIR

R x

⇒

wh have x = R tan, dx = R sec2 d

2 2

0

3 3

sec

2 sec

nIR R ddB

R

⇒

0cos

2

nid

∫ 0 (sin sin )

2

ni

(ii) Assuming the strip to be made up of a large number of element parallel to its length consider an element of

length dx at a distance x from the point P

0 2

4

IdB

x

0 02

[ln ]4 4

a b

a

II dxB dB x

b x b

⇒

∫ ∫ xP

dx

0ln

2

a b

b a

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

26. (i) Telescope: It is an optical instrument used to increase the visual angle of distant large objects such as a

star, a planet or a cliff etc.

Astronomical telescope consists of two converging lenses, the one facing the object is called objective or

field lens and has large focal length and aperture while other facing the eye called eye-piece or ocular has

small focal length and aperture.

The distance between the two lenses is adjustable as a telescope is used to see distant object in it object

is between and 2F of objective and hence image formed by objective is real inverted and diminished andbetween F and 2F on the other side of it. This image act as an object for eye-piece and by shifting the position

of eye-piece, it is brought within its focus. So final image i with respect to intermediate image is errect, virtual

enlarge and at a distance D to from the eye. This in turn implies the final image with respect to object isinverted enlarge and at a distance D to from the eye.Magnifying power (MP)

0

Visual angle with instrumentMP

Visual angle for unaided eye

(14)


But from figure, 0

0

,

e

y y

f u

0

MP⎡ ⎤

⇒ ⎢ ⎥ ⎣ ⎦o

e

f

u

whith length of tube 0 e

L f ...(i)

Case (i)

When the final image is at infinity (for point) this situation is called normal adjustment as in this situation eye

is least strained or relaxed in this situation as for eye-piece v

1 1 1 1 11

e

e e e e

f

D u f u f D

⎛ ⎞⇒ ⇒ ⎜ ⎟ ⎝ ⎠

Substituting this value of ue in equation (i)

MP and oo e

e

fL f f

f

Case (ii)

If the final image is at D(near point) in this situation as for eye-piece v = D

1 1 1 1 11

e

e e e e

f

D u f u f D

⎛ ⎞⇒ ⇒ ⎜ ⎟ ⎝ ⎠

Substituding this value of ue in equation (i)

MP 1⎛ ⎞ ⎜ ⎟⎝ ⎠

o e

e

f f

f D

e

o

e

f DL f

f D

In this situation ue is minimum so for a given telescope MP is maximum while length of tube is minimum and

eye is most strained.

L

v = fo

FieldLens

ue

fo f

e

Eye Lens

(15)


(ii) As in case of this lens the distance of either foci from the optical centre is f.

|u| = (f + x1)

|u| = (f + x2)

2 1

1 1 1 1 1 1

( )v u f f x f x f⇒ ⇒

vu

x1

ff

x2

F2

IF1

2 21 2

1 2 1 2 1 2

1 2

2 12

( )( )

x x f

fx fx f f fx fx x x

f x f x f

⇒ ⇒

2

1 2x x f⇒

OR

(i) Compound microscope:

It consists of two convergent lenses of short focal lengths and aperture arranged co-axially. Lens of (focal

length f0) facing the object is called objective or field lens while the lens of focal length f

e facing the eye, eye

piece or ocular the objective has a smaller aperture and smaller focal length than eye-piece. The separation

between objective and eye-piece can be varied.

Working: The object is placed between F and 2F of objective, so the image is formed by objective is inverted,

real, enlarged and at a distance geneater than 2f0 on the other side of the lens. This image I

M acts as object

for eye-piece and is within its focus. So eye piece forms final image I whcih is errect, enlarge, and at a

distance D to from eye on the same side of eye-piece as IM.

Magnifing power

0

visual angle with instrumentMP

maximum visual angle for unaided eye

If the size of object is h and least distance of distinct vision is D

0

',

e

h h

D u⇒

MP⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⇒ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦e e

h D h D

u u h u

But for objective,

⇒ I V h vmo u h u

MP⎛ ⎞

⇒ ⎜ ⎟⎝ ⎠e

v D

u u

Length of tube L = v + ue

Case-I

When final image is at infinity this situation is called normal adjustment as in this situation eye is least

strained or relaxed. In this situation as for eye-piece v

1 1 1maximum

e e

e e

u f

u f

⇒ ⇒

MP with⎛ ⎞

⇒ ⎜ ⎟⎝ ⎠

e

e

v DL v f

u f

(16)


Case-II

When final image is at D (near point) in this case v = D

1 1 1 1 11

e e e e

D

D u f u D f

⎡ ⎤⇒ ⇒ ⎢ ⎥ ⎣ ⎦

MP 1⎛ ⎞

⇒ ⎜ ⎟⎝ ⎠e

v D

u f

e

e

fL v

f D⇒

(ii) As lenses are 30 m apart and intermediate image is formed 5 cm in front of eye-piece

ue = 5 cm, v = L – u

e = 30 – 5 = 25 cm

Now MP ⎛ ⎞

⎜ ⎟⎝ ⎠e

v Dm m

u u

Here u = 1 cm, D = 25 cm

25 25MP 125

1 5

⎡ ⎤⇒ ⎢ ⎥

⎣ ⎦

Negative sign implies that final image is inverted.

v ue

L fe

fo

h

Im

I

Compound Microscope

fe

fo

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PHYSICS Complete Syllabus of Class XII...Model Test Paper-3 Subjective Test for Class-XII (Physics)...

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