cbse cbse class 12 physics solved model paper set 1

CBSE Class 12 Physics Solved Model Paper 2015

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Model Paper 2015Class: XII

Subject: Physics(Set – 1: Solved)

SECTION A

1. Depict the magnetic lines of force around a straight current carrying conductor.

Sol. The lines of force are in the form of concentric circles.

2. The refractive index of glass is 1.5. What is the velocity of light in glass?

Sol.

Speedof light in vacuumRefractive index of glass 1.5

Speedof light in glass

Since, Speed of light in vacuum is 3 × 108 ms–1.

8 18 13 10

Speedof light in glass 2 101.5

msms

3. What is the condition for maximum potential in Van de graff generator?

Sol. In a Van de graff generator inorder to attain maximum potential, the hollow conducting shell should be large and leakage of charge should be minimised by housing the generator in a steel chamer filled with nitrogen or methane at high pressure.



4. Does incident photon essentially eject a photoelectron?

Sol. No, incident photon does not essentially eject a photoelectron. If the frequency of the incident photon is less than the threshold frequency, there will be no emission of photoelectrons at all.

5. Calculate the bitrate for a signal which has a sampling rate of 10kHz and 8 quantisation levels have been used.

Sol.

Let n be the number of bits per sample then, the number of quantisation level = 2n

Since the number of quantisation level is 8,

2 8or 3

n

n

We know,

Bit rate Sampling rate Number of bits per sample10000 330,000bits/second

SECTION B

6. Why is it found experimentally difficult to detect neutrinos in nuclear β-decay?

Sol. Neutrinos are uncharged particles with almost no mass also, they interact very weakly with matter and are hard to detect. This is why; it is experimentally difficult to detect neutrinos in nuclear β-decay.

7. In a single slit diffraction experiment, the width of the slit of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Draw a plot of the intensity distribution?

Sol. The width of the central maximum in the single slit diffraction experiment is:



2 22

D fx

a a

When the width of the slit (a) is doubled, the size of the central diffraction band will become half. An area of central diffraction band will become (1/4) Thea and hence its intensity would become 4 times. The plot of intensity distribution is shown in the figure below. The dotted line in the curve is when the width of the slit is doubled.

8. A circular brass loop of radius a and resistance R is placed with its plane perpendicular to amagnetic field, which varies with time as o sinB B t . Obtain the expression for the induced current in the loop.

Sol. Induced current,

o

oo

o

Induced e.m.f.Resistance

1cos0

sin cos

cos

eI

Rd dt d

BAR R dt

ABA dB t t

R dt RA B

I tR

9. Using Kirchhoff's rules, determine:

(i) The voltage drop across the unknown resistance R.

(ii) The current flowing through the arm EF in the circuit as shown.



Sol.

The current distribution in the circuit is as shown.

By loop rule in AEFDA, we have 6 3 1 4 0 3 3 1A

x

x

x

1 2AApplying loop rule in EBCFE, we get

1+x 3 6 0

or 2 3 6 0 2 3

x

R x

R

R

3

2 Voltage drop across 1

3 2 3 volt.

2

R

R x R

OR

With the help of a circuit diagram explain how the internal resistance of a cell can be determined by using a potentiometer.

Sol.



First K1 is inserted, K2 is not inserted let the balancing length be L1

E L1

Where E is the EMF of cell,

Now ‘K2’ is closed, let the balance point be L2.

V L2

From (i) & (ii) equation

1

2

LE=

V L

Internal resistance of cell is given by

1

2

E1

V

L1

L

r R

r R

10. You are given a 2 µF parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?

Sol.

Here,

3

6

1mA 10 A

2μF 2 10 F

DI

C

sayDI I

dCV

dtdV

Cdt

Therefore,

3

6

10500 V/s

2 10DIdV

dt C



Therefore, applying a varying potential difference of 500 V/s would produce a displacement current of desired value.

SECTION C

11. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on aface of the prism. By rotating the prism, the minimum angle of deviation is measured to be 400. What is the refractive index of the prism? If the prism is placed in water (µ = 1.33), predict thenew angle of minimum deviation of the parallel beam. The refracting angle of prism is 600.

Sol.

Here,

0

0

δ 40

60m

A

0 0

0

0

0

sin δ 2As μ

sin 2

sin 60 40 2 μ

sin 60 2

sin 50

sin 300.7660

0.5000

mA

A

When prism is immersed in water, we have to take refractive index of glass w.r.t. water;

gg

μμ

μ

1.5321.33

1.149

aw

aw

If 'δm is angle of minimum deviation in water, then

'sin δ 2μ

sin 2mw

g

A

A



'

0

sin δ 2 μ sin 2

1.149 sin 301.149

20.5745

wm gA A

' 1

0 '

' 0 '

' 0 '

0 ' 0

0 '

sin δ 2 sin 0.5745

35 4

δ 70 8

δ 70 8

70 8 60

10 8

m

m

m

A

A

A

The new angle of minimum deviation of the parallel beam is 1008’.

12. A cell of emf 'E' and internal resistance 'r' is connected across a variable resistor 'R'. Plot agraph showing variation of terminal voltage 'V' of the cell versus the current 'I'. Using the plot, show how the emf of the cell and its internal resistance can be determined.

Sol.

Terminal Voltage V of the cell is, V E Ir

where E is the EMF of the cell, r is the internal resistance and I the current through the circuit.

So, V Ir E

On comparison with the Equation of a line, y mx c

, , ,y V m r x I c E

On plotting the graph between terminal voltage‘V’ and the current ‘I’

EMF of Cell = V-axis intercept

Internal Resistance = Slope of the Line



13. Explain, with the help of a circuit diagram, the working of a p-n junction diode as a half-waverectifier.

Sol.

The circuit diagram for half wave rectifier is as shown below:

Working: During the positive half cycle of the input a.c. The p-n junction is forward biased i.e theforward current flows from p to n, the diode offers a low resistance path to the current. Thus weget output across-load i.e. a.c input will be obtained as d.c output

During the negative half cycle of the input a.c. The p-n junction is reversed biased i.e the reversecurrent flows from n to p, the diode offers a high resistance path to the current. Thus we get no output across the load.

This principle is explained in the diagram given below.

14. Two particles with charge ratio 1 : 2 and mass ratio 1 : 4 have been accelerated through samepotential. Find the ratio of

(i) Their momenta.

(ii) de Broglie wavelengths. The particles are initially at rest.



Sol.

(i) Let the charges on the two particles be q and 2q and mass be m and 4m

We have

1

2

1 2

acquired charge

and 2

: 1: 2.

ii de Broglie wavelength

KE V

K qV

K q V

K K

hm

1 11

2 2 2

2

/ 2

/ 2

hm K

h m K

h m K

2 2

1 1

4 2

1

2 2 :1.

m Km K

mm

15. A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive index 1.65, (ii) a medium of refractive index 1.33.

(a) Will it behave as a converging or a diverging lens in the two cases?

(b) How will its focal length change in the two media?

Sol.

(a) Let airf be the focal length of lens in air then,

1 2

1 2

1 1 11

1 1 1 2 ...(1)

1

gair

airair g

nf R R

R R ff n



(i) When lens is dipped in a medium A here 1.65An

Focal length be Af when dipped in a medium A then 1 2

1 1 11A

A

nf R R

Using equation (1) we have,

1 1.5 2 1

1.6 5 1 5.5

5.5A air air

A air

f f f

f f

As sign of Af is opposite to that of airf ,

Therefore, the lens will behave as diverging lens.

(ii) When lens is dipped in Medium B 1.33Bn

Let Bf be the focal length of lens when dipped in medium B.

Then,

1 2 1 2

1 1 1 1 11 1

1 1.5 2 0.341

1.33 1.33

3.91

gg

B B

B air air

B air

nn

f R R n R R

f f f

f f

As the Sign of Bf is same as that of airf the lens will behave as a converging lens.

Change in Focal length:

(i) Focal length will become negative and its magnitude would increase by half.

(ii) Focal length increases.

16. Write briefly any two factors which demonstrate the need for modulating a signal. Draw asuitable diagram to show amplitude modulation using a sinusoidal signal as the modulating signal?



Sol.

The need for modulation can be summarized as follows:

For transmitting a signal, we need an antenna or an aerial. This antenna should have a sizecomparable to the wavelength of the signal (at least λ/4 in dimension) so that the antenna properlysenses the time variation of the signal. For an electromagnetic wave of frequency 20 kHz, the wavelength λ is 15 km. Obviously, such a long antenna is not possible to construct and operate. Hence direct transmission of such baseband signals is not practical.

A theoretical study of radiation from a linear antenna (length l) shows that the power radiated is proportional to (l/λ) 2. This implies that for the same antenna length, the power radiated increases with decreasing λ, i.e., increasing frequency. Hence, the effective power radiated by a longwavelength baseband signal would be small. For a good transmission, we need high powers and hence this also points out to the need of using high frequency transmission.

Because of these reasons, we use the technology of modulation, for transmitting message signals effectively for long distances.

17. Show diagrammatically the behaviour of magnetic field lines in the presence of (i)paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature?

Sol.

(i) The behaviour of magnetic field lines in the presence of a diamagnetic substance:



(ii) The behaviour of magnetic field lines in the presence of a paramagnetic substance:

The differnce in features is because of the difference in the relative permeabilities. The relative permeability of the diamagnetic substance is less than 1; so, the magnetic lines of force do not prefer passing through the substance. The relative permeability of a paramagnetic substance is greater than 1; so, the magnetic lines of force prefer passing through the substance.

18. Draw a circuit diagram of n-p-n transistor amplifier in CE configuration. Under what condition does the transistor act as an amplifier?

Sol.

The circuit diagram of an NPN transistor amplifier in CE configuration is given below:

The transistor acts as an amplifier when the input circuit (emitter–base) is forward biased with low voltage VBB and the output circuit ( collector–base) is reverse biased with high voltage VCC .



19. Two Polaroids A and B are kept in a crossed position. How should a third Polaroid C be placed

between them so that the intensity of polarised light transmitted by Polaroid B reduces to18th of the

intensity of unpolarised light incident on A?

Sol.

Let the third Polaroid C be kept at with the axis of Polaroid A.

Intensity of polarised light from C,

21 0 0cos , where is intensity of polarised light from AI I I

Angle between axis of B and C o90 θ

Intensity of polarised light from B,

2 o 22 1 1cos 90 θ sin θI I I

Using (i), 2 2 02 0

2cos θsin θ

8I

I I

Note that 2I0 is the intensity of unpolarised light incident on A,

2 2

2

2

o

oo

4cos sin 1

2sin cos 1

or sin 2 1

sin 2 1 sin 90

9045

2

20. A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series witha resistor R = 1 MΩ across 2 V battery. Calculate the magnetic field at a point P, half way between the centreand the periphery of the plates as shown in figure after 10–3 sec. (You may start by assuming that the charge on the capacitor at time t is, q (t) = CV [1 – exp. (–t/τ)], where the time constant τ is equal to CR).



Sol.

The time constant of the C-R circuit is given by: 9 6 310 10 10 sCR

20 0

Charge on capacitor at any time during charging is given by: 1

10

The electric field inbetween the plates at time is,π

t

t t

t q CV e

dq CVCV e e

dtq q

t EA r

Consider a circular loop of radius 1 2mr , parallel to the plates passing through P. The

magnetic field at all points on this loop is of the same value and acting tangentially to the loop.

The electric flux through this loop is, 2

2 22 2

0 0

π ππE

q q rE r r

r r

The displacement current, 2 2 2

0 0 2 2 20

tED

d d q r r dq r CVi e

dt dt r r dt r

When, t = 10–3 s, 3 3

2 9 610 10

2 3

1 2 10 2 0.5 10A

1 10Di ee

Applying Ampere Maxwell’s law to the loop, we have

0 0 0

60

6 7 6 13130

. 0

0.5 10or 2π

0.5 10 2 10 0.5 10 2 10or 0.74 10 T

2π 1 2 1 2 2.718 2.718

D D DB dl i i i i

B re

Be

21. Draw a labeled diagram of a full wave rectifier circuit. State its working principle. Show theinput-output waveforms?

Sol.

To get an output voltage for both half cycles of the input signal, we use full wave rectifiers. Thecommonly used full wave rectifier circuits are center-tap rectifier and bridge rectifier. The figurebelow shows the center-tap rectifier circuit.



Now consider the circuit. The P-side of the diodes D1 and D2 are connected to the secondaryterminals of the transformer. The N-sides of the diodes are connected together. The load is connected between this point and the midpoint of the transformer. When the input signal to diode D1 is positive, it conducts and load current flows. During this time, the input to diode D2 is negative with respect to the midpoint. During the negative half cycle of the input signal, thevoltage at D1 is negative and that at D2 is positive. So D2 conducts during this time period. Thus, we get output voltage during both the half cycles. As the full wave rectifier rectifies both the half cycles, it is more efficient than the half wave rectifier. The waveforms are given below:

OR

Write any two distinguishing features between conductors, semiconductors and insulators on thebasis of energy band diagrams.

Sol.

Conductors:



(i) For conductors, the valence band is completely filled and the conduction band can have twopossibilities—either it is partially filled with an extremely small energy gap between the valenceand conduction bands or it is empty, with the two bands overlapping each other, as shown.

(ii) On applying an even small electric field, conductors can conduct electricity.

Insulators:

(i) For insulators, the energy gap between the conduction and valence bands is very large. Also, theconduction band is practically empty, as shown.

(ii) When an electric field is applied across such a solid, the electrons find it difficult to acquiresuch a large amount of energy to reach the conduction band. Thus, the conduction band continuesto be empty. That is why no current flows through insulators.

Semiconductors:

(i) The energy band structure of semiconductors is similar to that of insulators, but in their case, the size of forbidden energy gap is much smaller than that of the insulators, as shown.

(ii) When an electric field is applied to a semiconductor, the electrons in the valence band find itcomparatively easier to shift to the conduction band. So, the conductivity of semiconductors liesbetween the conductivity of conductors and insulators.



22. A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm apart. A point object lies 60 cm in front of the convexlens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.

Sol.

For convex lens: 20 cm; 60 cmf u OC

As1 1 1v u f

Therefore,1 1 1v f u

1 120 603 1 260 60

6030 cm

2v

If image by lens alone is formed at I1, then CI1 = 30 cm.

For the convex mirror, I1 acts as a virtual object.

Therefore, 1 1 30 15 15 cm; 20 cm, ?u PI CI CP R v

From,1 1 2v u R

1 2 1

2 1 120 15 3030 cm

v R u

v



Therefore, the final image by the combination is formed at I, where PI = 30 cm.

SECTION D

23. For the past some time, Aarti had been observing some erratic body movement, unsteadiness and lack of coordination in the activities of her sister Radha, who also used to complain of severeheadache occasionally. Aarti suggested to her parents to get a medical check-up of Radha. Thedoctor thoroughly examined Radha and diagnosed that she has a brain tumour.

(a) What, according to you, are the values displayed by Aarti?

(b) How can radioisotopes help a doctor to diagnose brain tumour?

Sol.

(a) Aarti has displayed awareness and care towards the health of her sister.

(b) During the intake of different elements and compounds, the biological organisms absorb them differently. Also, the exact distribution of the elements and their function in the various parts oforganisms cannot be known clearly. For this, a radioisotope is made to enter the organism alongwith the elements and compounds, whose absorption, functioning and distribution to the brain has to be studied. The radioisotope acts as a tag of label for the element or compound under study. Bydetecting the radiation emitted by the isotope from the brain, the details regarding the absorptionand function of the compounds by the organisms are found out. In this way, radioisotopes help adoctor to diagnose brain tumour.

SECTION E

24. a) With the help of the neat and labelled diagram, discuss the working of common emitter n-p-n transistor amplifier.

b) Explain the phase relationship between input and output signal voltage and its voltage gain and current gain.

Sol.

Here emitter is common to both the input and the output circuits, The input (emitter base) circuit is forward biased with battery VBB of voltage VEB, and the output (collector-emitter) circuit is reversed biased with battery VCC of voltage VCEo Due to this, the resistance of input circuit is low



and that of output circuit is high. Re is a load resistance connected in the collector circuit. The low input a.c. voltage signal is applied across the base-emitter circuit and the amplified a.c. voltagesignal (i.e., output) is obtained as the change in collector voltage. In circuit diagram arrows represent the direction of conventional current or hole current, which is opposite to the direction ofelectronic current.

When no a.c. signal voltage is applied to the input circuit, but the emitter base circuit is closed, let us consider, that Ie ,Ib and Ic be the emitter current, base current and collector current respectively. Then, according to Kirchhoff's first law,

Ie = Ib + Ic

In this case, the output signal voltage obtained across collector is 1800 out of phase with the inputvoltage signal.

Phase relationship between input and output voltages:

a) When the positive half cycle of input a.c. signal voltage comes, it opposes the forward biasingof the emitter base circuit. Due to it, the emitter current decreases and hence collector current decreases; consequently the collector voltage Vc increases. Since the collector is connected to the negative terminal of a VCC battery of voltage VCE, therefore, the increase in collector voltagemeans, the collector will become more negative. This indicates that during the positive half cycleof input a.c. signal voltage, the output signal voltage at the collector varies through the negativehalf cycle, i.e., 1800 out of phase.

b) During the negative half cycle of input a.c. signal voltage, it supports the forward biasing of theemitter-base circuit, due to it; the emitter current increases and hence collector current increases; consequently the collector voltage Vc decreases i.e., the collector becomes less negative. Thus, the output signal voltage at the collector varies through the positive half cycle, i.e., 1800 out of phase.

Hence, in common emitter transistor amplifier circuit, the input signal voltage and the output collector voltage are 1800 out of phase.



Current Gain: It is defined as ratio of change in collector current (ΔIC) to the change in basecurrent (ΔIb) at constant collector voltage. It is denoted by βa.c.

a.c.Therefore, β c

b

II

Voltage Gain: It is defined as the ratio of the change in output voltage (ΔVC) to the change in input voltage (ΔVi).

V

. .

. . Resistance Gain

C

i

C o

b i

oa c

i

a c

VA

V

I RI R

RR

Where Ro and Ri are the output and input resistances of the circuit.

Here βa.c. > > αa.c., but resistance gain (Ro / Ri) is less than that in case of a common base transistor amplifier, hence the a.c. voltage gain in a common emitter amplifier is greater as compared to that of the common base transistor amplifier.

From the above equation:

V . .o

a ci

m o

RA

R

g R

Actually, V m oA g R ,here the negative sign indicates phase reversal of output.

OR

a) Draw the circuit arrangement for studying the input and the output characteristic of an n-p-ntransistor in CE configuration. With the help of these characteristics define:

i) Input resistance

ii) Current Amplification factor

b) Describe briefly with the help of a circuitdiagram how an n-p-n transistor is used to produce self sustained oscillations?



Sol.

a) The study of the characteristics of a transistor when grounded emitter is kept as a common terminal is as shown in the figure. The base is the input terminal and collector is the outputterminal as shown. The various currents are marked keeping in view the condition

e b cI I I

Input Characteristic:

The input characteristics of the transistor represent the variation of the base current Ib

with base emitter voltage VBE, keeping VCE

fixed. Their shape is shown in figure. Thecurrent is small as long as VBE is less than the barrier voltage. When VBE is greater than the barrier voltage, the curves look similar to that of a forward biased diode. More than 95% of emitter electrons (in npn transistor) and emitter holes (in pnp transistor) goes to the collector to form the collector current. That is why Ib is much smaller (in micro- ampere).

As long as the collector- emitter junction is reverse biased, the input characteristics arenot affected much by small changes in VCE.

Output characteristic:

The output characteristic is obtained byobserving the variation of Ic as VCE is varied keeping Ib constant. It is obvious that if VBE

is increased by a small amount, both holecurrent from the emitter region and the electron current from the base region will increase. As a consequence both Ib and Ic

will increase proportionately. This shows that when Ib increases Ic also increases. The plot of Ic

versus VCE for different fixed values of Ib gives one output characteristic. So there will be different output characteristics corresponding to different values of Ib as shown in Figure.

i) Input resistance (ri): This is defined as the ratio of change in base-emitter voltage (ΔVBE) to theresulting change in base current (ΔIb) at constant collector-emitter voltage (VCE). This is dynamic(ac resistance) and as can be seen from the input characteristic, its value varies with the operatingcurrent in the transistor:



CE

BEi

B V

Vr

I

The value of ri can be anything from a few hundreds to a few thousand ohms.

ii) Current amplification factor (β): This is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage (VCE) when the transistor is in active state.

CE

cac

b V

II

This is also known as small signal current gain and its value is very large.

Also the ratio of Ic and Ib gives dc β of the transistor. Hence,

cdc

b

II

Since Ic increases with Ib almost linearly and Ic = 0 when Ib = 0, the values of both βdc and βac arenearly equal. So, for most calculations βdc can be used. Both βac and βdc vary with VCE and Ib (or Ic) slightly.

b) For transistor as an oscillator, the L-C circuit is inserted in the emitter base circuitof transistor which is forward biased with battery VBB. The collector emitter circuit is reverse biased with battery Vcc. A coil L1 is inserted in collector emitter circuit. It is coupled with L in such a way that if increasing magnetic flux is linked with L, it will support the forward bias of emitter basecircuit and if decreasing magnetic flux is linked with L, it will oppose the forward bias of the emitter base circuit. Hence, the oscillator willproduce self sustained oscillations.

25. (a) In Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.

(b) The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9 :25. Find the ratio of the widths of the two slits.

Sol.



(a) In Young's double-slit experiment, the wavefronts from the two illuminated slits superpose on the screen. This leads to formation of alternate dark and bright fringes due to constructive and destructive interference, respectively. At the centre C of the screen, the intensity of light is maximum and it is called central maxima.

Let S1 and S2 be two slits separated by a distance d. GG′is the screen at a distance D from the slits S1 and S2. Point C is equidistant from both the slits. The intensity of light will be maximum at this point because the path difference of the waves reaching this point will be zero.

At point P, the path difference between the rays coming from the slits S1 and S2 is S2P – S1P.

Now, S1S2 = d, EF = d, and S2F= D

∴In Δ S2PF,

1 22 22 2

1 22

1 222

2 2

212

S P S F PF

dx

dS P D x D

D

Similarly, in ΔS1PE,

1 22

1 2

2 2

2 1 2 2

21

1 12 21 12 2

dx

S P DD

d dx x

S P S P D DD D

On expanding it binomially,

2 1

14

2 2d xd

S P S P xD D

For bright fringes (constructive interference), the path difference is an integral multiple of wavelengths, i.e. path difference is nλ.



Therefore,xd

nD

n Dx

d

, where n = 0, 1, 2, 3, 4, …

For n = 0, xo = 0

1

2

3

1,

22,

33,

Dn x

dD

n xdD

n xd

.

.

, n

n Dn n x

d

Fringe width (β):

The separation between the centres of two consecutive bright fringes is called the width of a dark fringe.

1 1n n

Dx x

d

Similarly, for dark fringes,

2 12n

Dx n

d

For 11,2D

n xd

For 2

32,

2D

n xd

The separation between the centres of two consecutive dark interference fringes is the width of abright fringe.

Therefore, 2 1n n

Dx x

d



Therefore, β1 = β2

All the bright and dark fringes are of equal width as β1 = β2.

(b) Let W, A, I represent the slit width, amplitude and intensity.

21 2min

2max 1 2

1 2

1 2

925

35

A AII A A

A A

A A

or 1

2

41

AA

211

22 2

161

AWW A

OR

(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slitilluminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.

(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 × 10−6 m. The distance between the slit and thescreen is 1·5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.

Sol.

(a) The phenomenon of bending of light round the sharp corners of an obstacle and spreading into the regions of the geometrical shadow is called diffraction.

\

Expression for Fringe Width:



Consider a parallel beam of light from a lens falling on a slit AB. As diffraction occurs, the pattern is focused on the screen XY with the help of lens L2. We will obtain a diffraction pattern that is a central maximum at the centre O flanked by a number of dark and bright fringes called secondarymaxima and minima.

Central Maximum: Each point on the plane wave front AB sends out secondary wavelets in alldirections. The waves from points equidistant from the centre C lying on the upper and lower half reach point O with zero path difference and hence, reinforce each other, producing maximum intensity at point O.

Positions and Widths of Secondary Maxima and Minima

Consider a point P on the screen at which wavelets travelling in a direction making angle θ with CO are brought to focus by the lens. The wavelets from points A and B will have a path differenceequal to BN.

From the right-angled Δ ANB, we have:

BN = AB sin θ

BN = a sin θ … (i)

Suppose, BN = λ and θ = θ1

Then, the above equation gives

λ = a sin θ1

1sina ... (ii)

Such a point on the screen will be the position of first secondary minimum.

If BN = 2λ and θ = θ2, then



2λ = a sin θ2

2

2sin

a ... (iii)

Such a point on the screen will be the position of second secondary minimum.

In general, for nth minimum at point P,

sin n

na ... (iv)

If yn is the distance of the nth minimum from the centre of the screen, from right-angled Δ COP, wehave:

tan n

OPCO

tan nn

yD

... (v)

In case θn is small, sin θn ≈ tan θn

Therefore, Equations (iv) and (v) give

n

n

y nD a

n Dy

a

Width of the secondary maximum,

1

1n n

n DnDy y

a a

Da ... (vi)

Since β is independent of n, all the secondary maxima are of the same width β.

If32

BN

and 1θ θ , from equation (i), we have:

1

3sinθ

2a



Such a point on the screen will be the position of the first secondary maximum.

Corresponding to path difference,

If52

BN

and 2θ θ , the second secondary maximum is produced

In general, for the nth maximum at point P,

2 1sinθ

2n

n

a

... (vii)

If ny is the distance of nth maximum from the centre of the screen, then the angular position of the nth maximum is given by

tanθ nn

yD

... (viii)

In case θn is small,

sinθ tanθ

2 12

n n

n

n Dy

a

Width of the secondary minimum,

1

1n n

n DnDy y

a a

Da ... (ix)

Since is independent of n, all the secondary minima are of the same width .

(b) For first maxima of the diffraction pattern we know3

sinθ2a

where a is aperture of slit.

For small values of θ, sinθ tanθyD

Where y is the distance of first minima from central line and D is the distance between the slit and the screen.



So,32

y Da

For 590 nm,9

1 6

1

3 590 101.5

2 2 100.66375 m

y

y

For 596 nm,

9

2 6

2

3 596 101.5

2 2 100.6075 m

y

y

Hence, Separation between the positions of first maxima 2 1 0.00675 my y

26. (a) Deduce an expression for the frequency of revolution of a charged particle in a magneticfield and show that it is independent of velocity or energy of the particle.

(b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles.

Sol.

(a) When a charged particle with charge q moves inside a magnetic field B

with velocity v, itexperiences a force, which is given by:

F q v B

Here, v

is perpendicular to B

, F

is the force on the charged particle which acts as the centripetal force and makes it move along a circular path.



Let m be the mass of the charged particle and r be the radius of the circular path.

2mv

q v Br

v and B are at right angles:

2mvqvB

rmv

rBq

Time period of circular motion of the charged particle can be calculated as shown below:

2

2

2

rT

vmv

v Bqm

TBq

Therefore, Angular frequency is

2TBqm

Therefore, the frequency of the revolution of the charged particle is independent of the velocity or the energy of the particle.



(b) It consists of two D-shaped hollow semicircular metal chambers D1 and D2, which are called dees. The two dees are placed horizontally with a small gap separating them. The dees areconnected to the source of high frequency electric field. The dees are enclosed in a metal boxcontaining a gas at a low pressure of the order of 10–3 mm mercury. The whole apparatus is placed between two poles of an electromagnet. The magnetic field acts perpendicular to the dees. Thepositive ions are produced in the gap between the two dees by ionisation of the gas.

A cyclotron involves the use of an electric field to accelerate charged particles across the gap between the two D-shaped magnetic field regions. The magnetic field is perpendicular to the paths of the charged particles that make them follow in circular paths within the two dees. An alternatingvoltage accelerates the charged particles each time they cross the dees. The radius of each particle’s path increases with its speed. So, the accelerated particles spiral toward the outer wall of the cyclotron.

Square wave electric fields are used to accelerate the charged particles in a cyclotron.

The accelerating electric field reverses just at the time the charged particle finishes its half circle, so that it gets accelerated across the gap between the dees.

The particle gets accelerated again and again, and its velocity increases. Therefore, it attains high kinetic energy.

OR

(a) Using Ampere's circuital law, obtain the expression for the magnetic field due to a longsolenoid at a point inside the solenoid on its axis.

(b) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.

(c) How is the magnetic field inside a given solenoid made strong?

Sol.

a) Solenoid:

It consists of an insulating long wire closely wound in the form of helix.



Its length is large as compared to its diameter.

Magnetic field due to RQ and SP path is zero because they are perpendicular to the axis of solenoid. Since SR is outside the solenoid, the magnetic field is zero.

The line integral of magnetic field induction B

over the closed path PQRS is

PQRS PQ. .B dl B dl BL

0PQRS

0

0

0

From Ampere?s circuital law,

. Total current through rectangle PQRS

BL Number of turns in rectangle Current

BL

B dl

nLI

B nI

b) Toroid:

It is a hollow circular ring on which a large number of turns of a wire are closely wound. Magnetic field along loop 1 is zero because the loop encloses no current.

Three Amperian loops (1, 2, and 3) are shown by dotted lines. Magnetic field along loop 3 is zero because the current coming out of the paper is cancelled

exactly by the current going out of it. Magnetic field at S (along loop 2):



0

0

From Ampere's Law,(2 )

Where,Magnetic FieldRadiusCurrentNumber of turns of toroidal coil

2

B r NI

B

r

I

N

NIB

r

These are the Magnetic field diagrams.

c) The magnetic field lines inside a solenoid can be made strong by:

Inserting a ferromagnetic core. Increasing the number of turns of the solenoid. Increasing the current passing through the solenoid.

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