TOPPER SAMPLE PAPER 5

Class XII- Physics

Time: Three Hours Max. Marks: 70 General Instructions

(a) All questions are compulsory.

(b) 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.

(c) 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.

(d) Use of calculations is not permitted.

(e) You may use the following physical constants wherever necessary.

19

8 1

34

7 1

23 1

23

27

1.6 10

3 10

6.6 10

4 10

1.38 10

6.023 10 /

1.6 10

µ λ

−

−

−

− −

−

−

= ×

= ×

= ×

= ×

= ×

= ×

= ×

o

B

A

n

k

e C

c ms

h Js

TmA

JK

N mole

m kg

Q. No

1. A uniform electric field E

exists between two charged plates as shown in the figure. What would be the work done in moving a charge q along the

closed rectangular path ABCDA?

(1)

2. How does the (i) pole strength and (ii) magnetic moment of each part of a bar magnet change if it is cut into two equal pieces transverse to its length? (1)

3. Figure given below shows two positions of a loop PQR in a perpendicular uniform magnetic field. In which position of the coil is there an induced emf?

(1)

4. Why are microwaves used in RADAR? (1)

5. The image of a candle is formed by a convex lens on a screen. The lower half of the lens is painted black to make it completely opaque. Draw the ray diagram to show the image formation. How will this image be different from the one obtained when the lens is not painted black? (1)

6. In an experiment on photoelectric effect, the following graphs were obtained between the photoelectric current (I) and the anode potential (V). Name the characteristic of the incident radiation that was kept constant in this experiment.

(1)

7. Name two factors on which electrical conductivity of a pure semiconductor at a given temperature depends. (1)

8. Give the conditions for inputs A and B for the following Boolean expressions.

( ) 1 ( ) 0i A B ii A B+ = + = (1)

9. Four point charges are placed at four corners of a square in the two

ways (i) and (ii) as shown below. Will the (a) Electric field (b) Electric potential, at the center of the square be the same or different in the two configurations and why?

(2)

10. The I – V characteristics of a resistor are observed to deviate from the straight line for higher values of current as shown below. Why is this so?

(2)

11. A charged particle moving with uniform velocity v

enters a region where

uniform electric and magnetic fields E

and B

are present. It passes through the region without any change in its velocity. What can we conclude about the

(i) Relative directions of v

, E

and B

(ii) Magnitude of E

and B

(2)

12. Figure shows two long coaxial solenoids, each of length L . The outer

solenoid has an area of cross - section 1A and number of turns be 1n .

The corresponding values for the inner solenoid are 2 2andA n . Write the

expression for self – inductance 1 2,L L of the two coils and their mutual

inductance M . Hence, show that 1 2 .M L L<

(2)

13. Two identical metallic surfaces A and B are kept parallel to each other in air, separated by a distance of 1. 0 cm as shown in figure.

Surface A is given a positive potential of 10V and the other surface B is earthed.

(i) What is the magnitude and direction of the uniform electric field between points y and z?

(ii) What is the work done in moving a charge of 20 Cµ from point x

to point y? (2)

14. In the circuit shown below R represents a resistance. If the frequency ν of the supply is doubled, how the value of C and L should be changed so that the glow in the bulb remains unchanged?

OR An air cored coil L and a bulb B are connected in series to AC mains as shown in the given figure:

The bulb glows with some brightness. How would the glow of the bulb change if an iron rod were inserted in the coil? Give reason in support of your answer. (2)

15. Experimental observations have shown that X – rays

(i) travel in vacuum with a speed of 83 10 / .m s×

(ii) exhibit the phenomenon of diffraction and can be polarized. What conclusions can be drawn about the nature of X - rays from each of these observations? (2)

16. Write the relation between the angle of incidence ( i ), the angle of emergence) ( e ), the angle of prism ( A ) and the angle of deviation (δ ) for rays undergoing refraction through a prism. What is the relation between refractive index of material of a prism in terms of Aandδ∠ .(2)

17. A radioactive material is reduced to 1

16 of its original amount in 4 days.

How much material should one begin with so that 34 10 kg−× of the

material is itself after 6 days? (2)

18. Distinguish between ‘point to point’ and ‘broadcast’ communication modes. Give an example of each. (2)

19. When two known resistances, R and S , are connected in the left and

right gaps of a meter bridge, the balance point is found at a distance ‘ l ’ from the zero end of the meter bridge wire. A unknown resistance X is now connected in parallel to the resistance s and the balance point is

now found at a distance 2l from the zero end of the meter bridge wire.

Obtain a formula of X in terms of 1 2, and .l l S

(3)

20. An a. c. circuit consists of a series combination of circuit elements X

andY . The current is ahead of the voltage in phase by4

π . If element

X is a pure resistor of 100Ω

(i) Name the circuit element Y and

(ii) Calculate the r. m. s. value of current when r. m. s. value of voltage is 141 V. (3)

21. (a) Draw a labeled ray diagram to show the formation of an image by a compound microscope. Write expression for its magnifying power.

b) How does the resolving power of a compound microscope change, when (i) refractive index of the medium between the object and the objective lens is increased and (ii) wavelength of the radiation used is increased?

OR

A concave lens has the same radius of curvature for both sides and has a refractive index 1.5 in air. In the second case, it is immersed in a liquid of refractive index 1. 4. Calculate the ratio of the focal length of the lens in the two cases. (3)

22. Why is diffraction of sound waves more easily observed than diffraction of light waves? Light of wavelength 600 nm is incident normally on a single slit of width 0.5 mm. Calculate the separation between two dark bands on the sides of the central maximum. The diffraction pattern observed on a screen placed at 2m from the slit. (3)

23. Obtain Einstein’s photoelectric equation. Explain how it enables us to understand the:

(i) linear dependence of the maximum kinetic energy of the emitted electrons on the frequency of the incident radiation

(ii) existence of threshold frequency for a given photo emitter. (3)

24. The potential energy (V ), for a pair of nucleus varies with the separation ( r ) between them in the manner shown below. Use this graph to explain why the force between the nucleon must be regarded as

(i) strongly repulsive for separation values less than or

(ii) attractive nuclear force ( )or r>

(b) Write the two characteristic features of nuclear forces.

(3)

25. Define the term half life period and decay constant of a radioactive substance. Write their SI units and establish the relationship between the two. (3)

26. State the principle of working of p - n diode as a rectifier. Explain with the help of a circuit diagram the use of p - n diode as a full wave rectifier. Draw a sketch of the input output waveforms. (3)

27. In a diode AM demodulator, the output circuit consists of

1 and 10 .R k C pF= Ω = A carrier signal of 100 kHz is to be

demodulated. Is the given set up good for this purpose? If not, suggest a value of C that would make the diode circuit good for demodulating this carrier signal. (3)

28. An electric dipole is held in a uniform electric field.

(i) Show the no translatory force acts on it. (ii) Derive an expression for the torque acting on it.

(iii) The dipole is aligned parallel to the field. Calculate the work

done in rotating it througho180 . (5)

OR

(a) Two extremely small charged copper spheres have their centers separated by a distance of 50 cm in vacuum. What is the mutual force of

electrostatics repulsion if the charge on each is 76.5 10 ?C

−×

(b) What will be the force of repulsion if:

(i) The charge on each sphere is doubled and their separation is halved?

(ii) The two spheres are placed in water. (5)

29. (a) With the help of a labeled diagram, explain the principle and working of a moving coil galvanometer. (b) Two parallel coaxial circular coils of equal radius R and equal number of turns N carry equal currents I in the same direction and are separated by a distance 2 R . Find the magnitude and direction of the net magnetic field produced at the mid – point of the line joining their centers. (5)

OR How are materials classified according to their behaviour in a magnetic field? How will you judge as to which of the two given similar magnets is stronger without using a third magnet? (5)

30. Using the data given below, state which of the given lenses you will use

as an eyepiece and as an objective to construct an astronomical telescope.

Lenses Power (P) Aperture

(A) L1

L2

L3

3 D

6 D

10 D

8 cm

1 cm

1 cm

Draw a ray diagram to show the formation of the image of a distant object in the normal adjustment position for the astronomical telescope so formed. Write the expression for its (i) magnifying power and (ii) length of the telescope. (5)

OR

Draw a ray diagram to show the formation of the image of a small object due to compound microscope. Derive an expression for its magnifying power. You are given two convex lenses of short aperture having lengths 4 cm and 8 cm respectively. Which one of these will you use as an objective and which one as an eyepiece for constructing a compound microscope? (5)

MOCK PAPER 5

PHYSICS - XII

Q. No. Solutions Marks

1. The net work done in moving a charge q along the closed rectangular

path ABCDA is zero. 1

2. (i) Pole strength does not change. 1

2

(ii) Magnetic moment reduces to half. 1

2

3. In second position there will be an induced emf. 1 4. Because microwaves possess greater energy and least angular speed.

1 5.

1

2

Image will be less bright. 1

2

6. Frequency of the incident radiation was kept constant. 1

7. (i) Band gap 1

2

(ii) Biasing 1

2

8. (i) A = 0, B = 0 1

2

(ii) A = 1. B = 1 1

2

9. The length of edge of cube is a , so

2

aOA OB OC OD= = = =

o

1 2and and 90E E E AOB= = ∠ =

The net electric field at o in fig. (i)

2 2

o 1 2 2 ( )E E E E i= + =

oE is directed at o45 from OD.

1

2

In figure (ii), the magnitude of E

remains same but direction changes.

1

2

For potential in (i)

Potential at ( ) ( ) ( ) ( )

o

1

4

Q Q Q QO

OA OB OC ODπε

+ + − − = + + +

= 0 1

2

and in case (ii)

Potential at ( ) ( ) ( ) ( )

o

1

4

Q Q Q QO

OA OB OC ODπε

− − + + = + + +

= 0

Hence, the potential remains same. 1

2

10. At higher value of voltage, the resistor shows the non- ohmic character.

This is the reason for the deviation of I – V graph from straight line. 2

11. (i) When ,v EandB

are mutually perpendicular to each other then the

particle will pass through the field without any change in its velocity. 1

(ii) In this situation, EandB

are such in magnitude that

0e m

F F+ =

1

12. The self inductance of outer solenoid

2

1 o 1 1 (1)L n Alµ= 1

2

The self – inductance of inner solenoid

2

2 o 2 2 (2)L n A lµ= 1

2

The mutual inductance of given solenoids

o 1 2 2oi ioM M n n A lµ= = (3)

1

2

From equation (1) and (2)

1 2 o 1 2 1 2L L n n l A Aµ=

Now, as 1 2 1 2 2, SoA A A A A> >

So 1 2oi ioM M L L= <

1

2

13. (i) 10V volts= x =1 cm = 0.01m

310

10 /0.01

VE volt meter

x= = =

1

2

Direction of E

is from surface A towards surface B. 1

2

(ii) 620 20 10q C Cµ −= = ×

Work done 6 420 10 10 2 10W qV J− −= = × × = × 1

14. For some glow of bulb, the current in the resistance R (i.e., bulb) should

remain same, i.e., the value of impedance 1

2

( )22

L CZ R X X= + − will remain same. i.e.,

the value of ( )L CX X− should remain same.

We have,

2L

X Lπν= 1

2

So, on doubling the value of ν , the L should be halved for some value of

.L

X

Also, we have,

1

2C

Xcπν

= 1

2

So, on doubling the value of ν , C should be halved for some value

ofC

X .

1

2

OR

On inserting the magnetic core (rod) inside the solenoid, the value of

2

o rN A

Ll

µ µ =

will increase, hence induced emf also increases which

opposes the flow of current in circuit, so bulb will glow with less brightness. 2

15. As X-rays travel with speed of light in vacuum, it shows that these are electromagnetic waves. 1

Phenomena of diffraction supports wave like characteristics of X- rays.

Polarization depicts that X-rays are transverse in nature. 1

2

16. Here, i is angle of incidence,

e is angle of emergence.

The required expression is

( )i e Aδ = + − 1

2

1

2

The refractive index µ of material of prism is related with A∠ and m

δ as

( )

sin2

sin 2

mA

A

δ

µ

+ = 1

17. Let the half life period of material is T, we have

1

2

nN

No

=

1

2

Here, 1

and 4days16

Nt

No= =

1 1

416 2

n

n

= ⇒ =

t n= × half life time

half life period 4

1day.4

t

n= = =

1

2

Now, we have

0N =?, 34 10 , 6days, T = 1 dayN kg t

−= × =

So no. of half lives in 6 days 1

66

1n= = =

1

2

( )

1

o

o

63

0

1

1

2

2

4 10 2

0.256

n

n

N

N

N N

N kg

−

=

= ×

= × ×

=

∵

1

2

18. Point to point communication mode is that communication medium which

is used in signal communication for point to point contract between the transmitter and receiver.

For example, parallel wire lines, twisted pair and co – axial cables. It is used in line communication. 1 Broadcast communication is that where there is no point to point contact between transmitter and receiver. It is used in space communication and satellite communication. 1

19. In first case,

1

1

(1)100

R l

S l=

−

1

2

In second case,

2

2

(2)100

R l

X S l=

+ −

1

2

Dividing eqn. (2) by (1), we get

( )( )

2 1

1 2

100

100

l lS

X S l l

−=

+ −

1

2

( )( )

1 2

2 1

100

100

l lX S

S l l

−+=

−

1

2

( )( )

1 2

2 1

1001

100

l lX

S l l

−+ =

−

1

2

( )( )

1 2

2 1

1001

100

l lX S

l l

−= − −

1

2

20. (i) The circuit element Y is capacitor. 1

(ii) The impedence 100

141.41

2

Z = = Ω 1

2

Irms rmsV

Z=

1

2

141

141.4=

1

2

1rmsI A≅ 1

2

21. (a)

1 Here, O = objective lens

E = Eye piece

AB = object, A” B” final image Magnifying power ( m ) of a compound microscope is given by

1o

L dm

f fe

= +

1

2

where L = distance between 1 2andC C

o,ef f are focal length of eye-piece and objective respectively

d = distance of final image formed from eye-piece. (b) Resolving power of a compound microscope is given by

2 sin

.R Pµ θ

λ=

1

2

where µ is the refractive index of the medium between object and

objective lens, θ is half the angle of cone of light from the point object and λ is the wavelength of radiations used.

(i) on increasing µ , resolving power will increase. 1

2

(ii) on increasing λ , the resolving power will decrease. 1

2

OR

Given that

1R r=

and 2R r= −

refractive index of lens with respect to air 1.5aµ = refractive index of

lens with respect to liquid 1.5

1.4lµ =

1

2

Then, ( )1 1 2

1 1 11a

f R Rµ

= − −

1

2

( )1

1 1 11.5 1 .

f r r

= − +

1

1

1 20.5

f r

f r

= ×

=

1

2

The focal length of lens for air medium is equal to the radius of curvature of its surfaces. Again, for liquid medium, we have

( )2 1 2

1 1 11l

f R Rµ

= − −

1

2

( )1.5 1 1 1

1.4 r r

− = +

2

2

2

1 0.1 2

1.4

1 1

7

7

f r

f r

f r

= ×

=

=

1

2

The focal length for liquid medium is 7 times the radius of curvature of its surfaces. Hence, the ratio

1

2

1

7

f

f=

1

2

22. For diffraction of a wave, an obstacle of the size of wavelength of the

wave is needed. As wavelength of light is of the order of 610 ,m−

and

obstacles of this size are rare, so diffraction of light is not common while diffraction of sound is common because of its large wavelength. 1 Separation between two dark bands on either side of central maximum = width of central maximum 1

10

o 3

2 2 2 6000 10

0.5 10

D

d

λβ

−

−

× × ×= =

× 1

o 4.8 .mmβ =

where d = slit width

D = separation between screen and slit.

λ = Wavelength of light used.

23. Einstein’s photoelectric equation is in accordance with the energy conservation law as applied to the photons absorbed and to the emitted electron.

Energy of incident photon = max. K. E. + work function of metal

1

2

2

max o

2

max max o

1

2

1

2

hv mv W

K mv hv W

= +

= = −

1

2

At threshold frequency ( )oν , no K.E. is given to the electron, so

o ohv w= 1

2

Therefore, max oK hv hv= − 1

2

This is the Einstein’s photoelectric equation.

(i) Above the threshold frequency ov , maxK of emitted electrons depends

linearly on the frequency of incident radiation. 1

2

(ii) When frequency o max,v v K< is negative this has no physical meaning.

Hence, there is no photoelectric emission below the threshold frequency.

1

2

24.

(a) Potential energy of a pair of nucleon as function of their separation (r) (i) The graph shows the short range character of nuclear force of the

order of 2-3 fm. It is attractive but becomes strongly repulsive when the separation is less than about 1 fm (known as hard core region)

i.e., or r> . 1

(ii) From the graph, it evident that for or r> , the potential energy is

negative thereby implying the attractive nature of nuclear force

for or r> . 1

(b) Characteristics of nuclear forces:

(i) These are short range forces: 1

2

(ii) These are charge independent. 1

2

25. Half life period – This is the time taken by a radioactive substance to

become half of its original amount. It is represented by 1 2T .

Its unit is sec. 1

2

Decay Constant – Decay constant is the reciprocal of the half life period.

If decay constant is large. The substance will decay rapidly. SI of unit of it

is 1sec .−

1

2

Relation between Decay constant and half life –

Considering the number of atom initially to be 0N and after time t it

becomes N .

By decay law, dN

Ndt

λ= − 1

2

Where λ is decay constant

dN

dtN

λ= −

0

Nt

oN

dNdt

Nλ= −∫ ∫

0n nl N l N tλ− = −

o

Ne t

Nλ= −

o

tN N e

λ−= 1

2

after o1 2 , , Then

2

Nt T N= =

1/ 200

2

TNN e

λ−= 1

2

1/ 2

1/ 2

1 2

2

n 2

0.693

Te

T l

T

λ

λ

λ

=

=

=

1

2

26. Device which converts alternative current (a. c.) into direct current (d. c.) is

known as rectifier. 1

2

Principle – Junction diode conducts only when forward biased and it does not conduct when reverse biased. This makes the junction diode to

work as a rectifier. 1

2

Full wave rectifier –

1

2

In the figure, P is the primary coil and S is the secondary coil of the

transformer. 1 2andD D are the diodes which are connected to the

secondary coil of the transformer. The load R, across which output voltage is obtained, is connected to a mid point on secondary coil.

1

2

Working – When the half cycle of input ac signal flows through the primary coil, induced emf is set up in the secondary winding coil due to

mutual induction. Now the diode 1D becomes forward biased and diode

2D becomes reverse biased. Thus, output is there due to diode 1.D

1

2

During the negative half cycle of the ac input signal, diode 2D becomes

forward and conducts while diode 1D becomes reverse biased and does

not conduct. Thus, in this case, the output is due to diode 2.D

1

2

27. Given that,

3

11

1 10

10 10

R k

C pF F−

= Ω = Ω

= =

3 11 810 10 10RC s

− −∴ = × = 1

2

and 5100 10cf kHz Hz= =

51

10 secc

f

−∴ = 1

2

We find that1

cf

is not less than RC as is required for demodulation;

therefore, the arrangement is not for the desired purpose. 1

2

For a satisfactory arrangement, let us try

6

3 6 3

1 10

10 10 10

C F F

RC s

µ −

− −

= =

= × =

1

2

Now 1

c

RCf

<< 1

2

∴ The condition is satisfied. This is a good enough for demodulation.

1

2

28. (i) Considering an electric dipole of dipole moment p

, held at an angle θ

in a uniform electric field E

as shown in given figure.

1

2

The force experienced by the charges q and – q are qE

and - qE

respectively.

Hence, net force = qE

+ (- qE

) = 0 1

2

Therefore, net translatory force on the dipole is zero.

(ii) Toque = Force × perpendicular distance 1

2

.qE AC=

sinqEABτ θ= 1

2

2 sinqE l θ=

sinpEτ θ= 1

2

or p Eτ = ×

1

2

(iii) Work done in turning an electric dipole in an electric field by angle

θ ( )1 2from toθ θ

( )2cos , cosW pE θ θ= − 1

2

Here, o o

1 20 and 180θ θ= = 1

2

So, o ocos0 cos180W pE = −

1

2

( )1 1pE= − −

2W pE= 1

2

OR

Given: For two extremely small charged copper spheres in vacuum.

7

1 2 6.5 10q q C−= = ×

Separation between the spheres, 50 0.5r cm m= =

1

2

Force of mutual repulsion 1 2

2

o

1

4

q qF

rπε=

1

2

( )( )

( )

9 7 7

2

9 10 6.5 10 6.5 10

0.5F

− −× × × ×=

1

2

21.52 10F N

−= × 1

2

(b) (i) As 1 2

2,

q qF

rα When charge on each sphere is doubled and their

separation is halved, the new force of repulsion will become 16 times,

i.e., 1

2

' 216 16 1.52 10F F

−= = × × 1

2

1' 2.43 10 0.243F N N

−= × = 1

2

(ii) In a liquid (water) of dielectric constant ( )80 ,k = new force of

repulsion, ''f will be ,F k i.e.,

1

2

2

'' 1.52 10

80F

−×=

1

2

'' 41.90 10F N

−= × 1

2

29. (a) Principle of moving coil galvanometer – Current carrying coil placed

in a magnetic field experiences a torque. 1

2

Suppose current I is passing through the coil. This coil experiences a torque

sinNIBAτ θ=

Where N is total number of turns in the coil, θ is the angle between

magnetic field vector and area vector of the coil. 1

2

Counter torque is developed in the spring. If k is the restoring torque per unit angular twist, for an angular twist of ' 'θ we have

Restoring torque ' kτ θ=

In equilibrium 'τ τ= 1

2

NIBASin kθ θ=

sin

kI

NBA

θ

θ∴ =

1

2

If the magnetic field is radial, i.e., plane of the coil is parallel to the direction of magnetic field.

Then, o90 ,θ =

k

I INBA

θ α θ∴ = ⇒

Thus, following current is proportional to deflection. 1

2

1

2

1

2

Magnetic field induction at an axial point at a distance x from the centre of a coil of radius R , number of turns N and carrying current I is given by

( )

2' o

3 22 22

NIRB

R x

µ=

+

1

2

Resultant field B due to both the coils at the mid point P is obtained by putting x R=

( ) ( )

2 2

o o

3 32 2 22 2

2

2 2

INR NIRB

R R R

µ µ= × =

+

1

2

2o /

2 2

NIB Wb m

R

µ= .

This field acts along the common axis of the coils. 1

2

OR

Materials are classified as ferromagnetic, paramagnetic and diamagnetic on the basis of their behaviour in a magnetic field. The magnetic susceptibility of a paramagnetic substance is inversely proportional to its absolute temperature. Hence on cooling, its magnetic susceptibility

increases and therefore, its magnetization increases. 1

2

Diamagnetic materials - Such materials get magnetized in a direction opposite to the direction of magnetic field. Hence, their relative permeability, susceptibility and intensity of magnetization are negative. 1 Paramagnetic materials - Such materials get magnetized in the direction of the magnetizing field. Hence, their relative permeability, susceptibility and intensity of magnetization is positive. 1

Ferromagnetic materials - Such materials get strongly magnetized in the direction of the magnetizing field. So their relative permeability, susceptibility and intensity of magnetization are positive and are much greater than one. 1

Time period of vibration is given by

2H

IT

MBπ=

1

2

1

orI

T TM m

α ≈ 1

2

where m is the pole strength. Thus, higher is the pole strength, lesser will be the time period. Thus, the

magnet vibrating faster is stronger. 1

2

30. Eye-piece: lens 3l 1

Objective: lens 1l 1

We know

Magnifying power o 1 e

e

f f

f r

= − +

1

For larger magnifying power objective lens must have large aperture and large focal length. Focal length of eye-piece must be as small as possible.

1

(i) Magnifying power, o

e

fm

f=

1

2

(ii) Length of the telescope o ef f= +

1

2

OR

Formation of image of a small object due to compound microscope

1

2

Magnifying power of compound microscope mβ

α=

1

2

Where L = length of microscope tube

D = least distance of distinct vision. 1

2

tan " " " " "

tan "' " " "' "

A B P B A Bm

A B P B A B

β β

α α= = = =∵

1

2

" " ' '.

"' ''o e

A B A BM M

A B AB=

1

2

But and 1o e

e

v Dm m

u f= = +

1

2

1e

v Dm

u f

∴ = +

As

o

u , , length of microscope

1= 1

o

e

f v L

Dm

f f

→ →

⇒ +

1

2

Magnifying power can be increased by taking objective and eye-piece of

small focal lengths. 1

2

However, .o e

f f<

Hence, if focal length 4 cm is to be used as objective and that of focal length 8 cm to be used as eye piece. 1