Grand HSC EXAMINATION SET - A PHYSICS Test HINT & …2. a) Formation of Stationary waves on strings...

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HSC EXAMINATION SET - A 2016 PHYSICS

HINT & SOLUTION

Grand Test

SOLUTIONS

Q.1.

1. B. 2.B. 3.A. 4.A. 5.D. 6.B. 7.B.

Q.2.

1. a) The substances that does not transmit any incident heat radiations are called athermanous substances

b) The substances through which heat radiations can pass are called diathermanoussubstances

2. Given: 54 waves in one minute, 𝜆 = 10 𝑚𝑡𝑠

To find v = ?

Formula 𝑣 = 𝑛𝜆

Solutionn

𝑛 =54 𝑤𝑎𝑣𝑒𝑠1 𝑚𝑖𝑛𝑢𝑡𝑒

= 5460 𝑠𝑒𝑐−1

We have 𝑣 = 𝑛𝜆 = 5460

× 10 = 9 𝑚/𝑠

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3. a) The maximum distance between two molecules upto which intermolecular forces are effective is called range of molecular attraction

b) The layer of surface liquid whose thickness is about to range of molecularattraction is called surface film.

4. Given :

∆𝜃 = 40− 20 = 200𝐶

𝐴 = 1𝑚𝑚2 = 10−6 𝑚2

𝛼 = 1.15 × 10−5 0 𝐶

𝑌𝑠𝑡𝑒𝑒𝑙 = 2 × 1011𝑁/𝑚2

Formula :

𝐹 = 𝑌 𝐴 𝛼 ∆𝜃 = 2 × 1011 × 10−6 × 1.15 × 10−5 × 20 = 44𝑁

5. a) An ideal simple pendulum consists of a point mass suspended by a weightless inextensible string from a firm or rigid support.

b) An practical simple pendulum consists of a small heavy spherical metallic bobsuspended by a light inextensible string from a firm support.

6. Given 𝐾0 = 10 𝑐𝑚, 𝑟 = 6 𝑐𝑚,𝑇𝑜 𝑓𝑖𝑛𝑑 𝐾𝐶

Formula :

𝐼0 = 𝐼𝐶 + 𝑀𝑟2

𝐾02 = 𝐾𝐶2 + 𝑟2

𝐾𝐶2 = 𝐾02 − 𝑟2

(10)2 − (6)2 = 64

∴ 𝐾𝐶 = 8 𝑐𝑚𝑠


7. a) The space surrounding the material body in which its attraction (gravitational) can be experienced is called gravitational field.

b) The force experienced by unit mass placed at any point in the gravitational fieldis called intensity of gravitational field.

8. Given 𝑚 = 0.4 𝑘𝑔, 𝑟 = 2 𝑚𝑡𝑠, 𝑛 = 60 𝑟𝑒𝑣/𝑚𝑖𝑛, 1𝑟𝑒𝑣/𝑠𝑒𝑐

To find F = ?

Formula : F= 𝑚𝑟𝜔2

Solution : We have

F= 𝑚𝑟𝜔2 = F = 𝑚𝑟(2𝜋𝑛)2

0.4 × 2 × (2 × 3.142 × 1)2 = 31.53𝑁

Q. 3.

1) Solution :

Let 𝑣1 𝑎𝑛𝑑 𝑣2 be the velocities of particles at distances of 𝑥1 𝑎𝑛𝑑 𝑥2 respectively.

𝑣1 = 𝜔2�𝑎2 − 𝑥1 2 … … … … … … (1)

𝑣2 = 𝜔2�𝑎2 − 𝑥2 2 … … … … … … (2)

Squaring and subtracting (2) from (1), we get

𝑣1 2 − 𝑣2 2 = 𝜔2(𝑎2 − 𝑥1 2) −𝜔2(𝑎2 − 𝑥2 2)

𝑣1 2 − 𝑣2 2 = 𝜔2(𝑎2 − 𝑥1 2 − 𝑎2 + 𝑥2 2)

𝑣1 2 − 𝑣2 2 = 𝜔2( 𝑥2 2 − 𝑥1 2)

𝜔2 = 𝑣1 2 − 𝑣2 2

𝑥2 2 − 𝑥1 2


Taking square root of both sides we get

𝑥2 − 𝑥1 𝜔 = �

𝑣1 2

2− 𝑣2 2

2

Period of S.H.M is

𝑇 =2𝜋𝜔 =

2𝜋

� 𝑣1 2

2− 𝑣2

22

𝑥2 − 𝑥1

𝑇 = 2𝜋 � 𝑥2 2 − 𝑥1 2

𝑣1 2 − 𝑣2 2

2) Solution :

Let 𝑙1 𝑎𝑛𝑑 𝑙2 be the lengths of closed and open organ pipes respectively. The frequency of the first overtone of the closed pipe.

𝑛1 = 3 �𝑣

4 𝑙1 �

The frequency of the third overtone of the open pipe is.

𝑛2 = 4 �𝑣

4 𝑙2 �

For resonance with a given fork

𝑛1 = 𝑛2

∴ 3 �𝑣

4 𝑙1 �= 4 �

𝑣4 𝑙2 �

∴ � 𝑙1𝑙2 � = �

38 �


3) Maxwell Distribution :

For a given mass of gas, the velocities of all molecules are not the same, even when bulk parameters like pressure, volume and temperature are fixed. Collisions change the directions and speed of molecules, but in a state of equilibrium the distribution of speed is constant.

1) Maxwell on purely statistical considerations shows that the distribution ofmolecular velocities in a gas takes place according to a definite law.

2) This law is known as Maxwell’s law of distribution of molecular velocities.3) The figure shows the curve representing the distribution of velocity in a gas.4) Conclusions from the graph.

a) At C = 0, N (C) = 0 i.e. the molecules of zero velocity are not present inthe gas.

b) The velocity corresponding to the peak of the curve is known as mostprobable velocity (𝐶𝑃)

c) The curve is not symmetrical i.e. asymmetrical.d) The number of molecules having very high and very low velocities is

very low in the gas.e) The shaded region of the curve represents the number of molecules of the

gas between the region C and (𝐶 + ∆𝐶 )

4) Moment of Inertia of a ring about its diameter


For uniform ring, the moment of inertia of the ring about any diameter is the same. We want to find the moment of inertia of a ring about a diameter as shown in the figure. If 𝐼𝑍 is the moment of inertia of a ring about Z axis passing through centre of mass of the ring and perpendicular to its plane then 𝐼𝑍 = 𝐼𝐶. Let 𝐼𝑋 𝑎𝑛𝑑 𝐼𝑦 be the moment of inertia of the ring about the diameter along x and y axes respectively.

Using perpendicular axes theorem

𝐼𝑍 = 𝐼𝑋 + 𝐼𝑦 … … … … … … … (1)

But 𝐼𝑍 = 𝐼𝐶 . And for a ring 𝐼𝑋 = 𝐼𝑦

Equation (1) becomes

𝐼𝐶 = 2𝐼𝑋 = 2𝐼𝑦

𝐿𝑒𝑡 𝐼𝑋 = 𝐼𝑦 = 𝐼𝐷 𝑤ℎ𝑒𝑟𝑒 𝐼𝐷

Is the moment of inertia of a ring about its diameter

∴ 𝐼𝐶 = 2𝐼𝐷

But 𝐼𝐶 = 𝑀𝑅2

∴ 𝐼𝐷 =12𝑀𝑅

2


Hence the moment of inertia of a thin uniform ring about its diameter is given by

𝐼𝐷 =12𝑀𝑅

2

Q. 4

1. a) Conical Pendulum

Conical Pendulum is a simple pendulum which is given such a motion that a bob

describes a horizontal circle and the string describes a cone.

S ------ rigid support

T -------tension in the string

l---------length of the string

h--------axial height of the cone


v--------velocity

r---------radius of horizontal circle

𝜃--------semi vertical angle of cone

mg------weight of the sphere (bob)

Consider a bob of mass m revolving along a horizontal circle of radius r with

velocity v. Let 𝜃 be semi vertical angle of cone. The forces acting on bob at

position A are,

(1) Weight mg acting vertically downward.

(2) Tension T in upwards along the string. Tension (T) in the string can be

resolved into,

a) 𝑇 𝑐𝑜𝑠𝜃 vertically upward.

b) 𝑇 𝑠𝑖𝑛 𝜃 horizontal and directed towards the center of circle.

The weight mg is balanced by vertical component 𝑇 𝑐𝑜𝑠𝜃

∴ 𝑇 𝑐𝑜𝑠𝜃 = 𝑚𝑔… … … . . (1)

The horizontal component 𝑇 𝑠𝑖𝑛 𝜃 provides necessary centripetal force directed

towards O for uniform circular motion of bob.

𝑇 𝑠𝑖𝑛 𝜃 = 𝑚𝑣2

𝑟...........(2)

Dividing (2) by (1) we get

𝑡𝑎𝑛 𝜃 =𝑣2

𝑟𝑔

∴ 𝑣 = �𝑟𝑔𝑡𝑎𝑛 𝜃

This relation gives the speed of the bob of conical pendulum.

But 𝑣 = 𝑟𝜔


∴ 𝑟𝜔 = �𝑟𝑔𝑡𝑎𝑛 𝜃

𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑡𝑎𝑛 𝜃 =𝑟ℎ

𝑟 2𝜔2 = 𝑟𝑔 �𝑟ℎ�

𝜔2 = �𝑔ℎ�

𝑔∴ 𝜔 = � … … … … (3) ℎ

Equation (3) gives the angular speed of bob of a conical pendulum.

Periodic time of conical pendulum

𝑇 =2𝜋𝑟𝑣

∴ 𝑇 =2𝜋𝑟

�𝑟𝑔𝑡𝑎𝑛 𝜃

∴ 𝑇 = 2𝜋�𝑟

𝑔𝑡𝑎𝑛 𝜃… … … … … … … … … (4)

𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑠𝑖𝑛 𝜃 =𝑟𝑙

∴ 𝑟 = 𝑙 𝑠𝑖𝑛 𝜃

Substituting this value in (4) we get

∴ 𝑇 = 2𝜋�𝑙 𝑠𝑖𝑛 𝜃𝑔𝑡𝑎𝑛 𝜃


∴ 𝑇 = 2𝜋�𝑙 𝑐𝑜𝑠 𝜃𝑔

1. b) Data : 𝑔 = 9.8 𝑚/𝑠2, R = 6400 km, d = 200km = 2 × 106 𝑚

Solution :

The acceleration due to gravity at a point at a depth d from the earth’s surface is given by

𝑔𝑑 = 𝑔 �1 −𝑑𝑅�

∴ 𝑔𝑑 = 9.8 �1 −2 × 106

6.4 × 106�

∴ 𝑔𝑑 = �9.8 × 4.4

6.4 �

∴ 𝑔𝑑 = 6.737𝑚/𝑠2

2. a) Formation of Stationary waves on strings When two identical progressive waves both (transverse or longitudinal)

travelling along the same path in opposite directions, interfere with each other, by superposition of waves resultant wave obtained in the form of loops, is

called a stationary wave. Consider two simple harmonic progressive waves of equal amplitude and frequency, propagating on a long uniform string in opposite directions

If wave (frequency ‘n’ and wavelength 𝜆) is travelling along positive direction of X-axis then

𝑌1 = 𝑎 sin 2𝜋 �𝑛𝑡 −𝑥𝜆�… … … (1)


If wave (frequency ‘n’ and wavelength 𝜆) is travelling along negative direction of X-axis then

𝑌2 = 𝑎 sin 2𝜋 �𝑛𝑡 +𝑥𝜆�… … … (2)

These waves interfere to produce stationary wave. The resultant displacement of stationary wave is given by the principle of superposition of waves

𝑌 = 𝑌2 + 𝑌1

𝑌 = 𝑎 sin 2𝜋 �𝑛𝑡 +𝑥𝜆� + 𝑎 sin 2𝜋 �𝑛𝑡 −

𝑥𝜆�

𝑈𝑠𝑖𝑛𝑔 𝑠𝑖𝑛𝐶 + 𝑠𝑖𝑛𝐷

= 2𝑠𝑖𝑛 �𝐶 + 𝐷

2 � 𝑐𝑜𝑠 �𝐶 − 𝐷

2 � 𝑎𝑛𝑑 cos(−𝜃) = 𝑐𝑜𝑠𝜃 𝑤𝑒 𝑔𝑒𝑡

𝑌 = 2𝑎 �sin(2𝜋𝑛𝑡)𝑐𝑜𝑠 �2𝜋𝑥𝜆 ��

𝑌 = �2𝑎 𝑐𝑜𝑠2𝜋𝑥𝜆 � sin(2𝜋𝑛𝑡)

𝑌 = 𝐴sin(2𝜋𝑛𝑡)

𝐵𝑢𝑡 𝜔 = (2𝜋𝑛)

𝑌 = 𝐴sin𝜔𝑡 𝑤ℎ𝑒𝑟𝑒 𝐴 = �2𝑎 𝑐𝑜𝑠2𝜋𝑥𝜆 �

A is amplitude of resultant stationary wave, i.e. amplitude is periodic in space, hence we can see loops in case of transverse waves forming stationary wave on string.

Thus in stationary wave, frequency is same as that of progressive waves but amplitude varies with position of particle.


The above equation represents the resultant displacement of two simple harmonic progressive waves which is not moving because of absence of term x in the equation, therefore it is called stationary wave.

b) Data :

𝐸 = 5.67 × 104𝑤𝑎𝑡𝑡/𝑚2,𝜎 = 5.67 × 10−8𝑤𝑎𝑡𝑡/𝑚2𝐾4

T = ?

Solution

𝐸 = 𝜎𝑇4

∴ 𝑇4 =𝐸𝜎

∴ 𝑇 = �𝐸�1�4

∴ 𝑇 = �

𝜎

5.67 × 104

5.67 × 10−8�1�4

∴ 𝑇 = [1012]1�4


∴ 𝑇 = [103]

∴ 𝑇 = 1000 𝐾

Q. 5.

1. . 2. . 3. . 4. . 5.C. 6.D. 7.D.

Q. 6.

1) Given 𝑖𝑃 = 56040 ′ Find the value of a𝜇𝑔

Formula 𝑡𝑎𝑛 𝑖𝑃 =a𝜇𝑔

Solution :

According to Brewster’s law, if 𝑖𝑃 the polarising angle then 𝑡𝑎𝑛 𝑖𝑃 =a𝜇𝑔

𝑡𝑎𝑛 ( 56040 ′ ) =a𝜇𝑔

a𝜇𝑔 = 1.5204

2) The various energy losses in a transformer are

• Flux losses• Copper losses• Iron losses• Hysteresis losses• Magnetostriction

3) Thomson’s Atomic Model (1904)

1. The first atom model was suggested by Sir J.J. Thomson in 1903.


2. According to this model, an atom is a uniform sphere of positively chargedmaterial having a radius of the order of 10−10 m in which electrons areembedded in number such as to balance the positive charge of the atom.

3. Thomson’s atom model is also called as plum – pudding model.4. The number of positive charges is equal to negative charges, so atom is

electrically neutral.5. Thomson’s Atomic Model was able to explain.a) The Thermionic emissionb) Photo – emissionc) Emission and absorption of spectrum of single wave length

Drawbacks

1. This model could not explain scattering 𝛼 − 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 by a thin foil of metal.2. Thomson’s could not explain the stable structure of atom and atomic spectra.

4) Solar cells

d) It is semi conductor device that works on the phenomenon of photo electriceffect.

e) It coverts light energy from the sun (i.e. solar energy) into electric energy orelectric current.

f) Its working is based on photo voltaic effect (i.e when light falls on the opencircuit of a p-n junction diode silicon or germanium, then e.m.f. is generatedacross its terminals).

Advantages of solar cells

1. They are portable2. They require little maintenance3. No fuel or chemical is required for its operation.

Materials used in solar cells

1. For solar cells Si and Ge are most widely used as semi conducting materials2. Gallium Arsenide (GaAs), Indium Arsenide (InAs) and Cadmium Arsenide

(CdAs) are also used to prepare the solar cells.


5) Since no current flows through the 5Ω the circuit represents a balancedWheatstone’s bridge.

𝑋 18 =

26

∴ 𝑋 =2 × 18

6 = 6Ω

6) 𝐺𝑖𝑣𝑒𝑛 𝑛1 = 10,𝑇𝑜 𝑓𝑖𝑛𝑑 𝑛2

Formula 𝑀 = 𝑚 × 2𝑙, 𝑛 = 12𝜋 𝐼�𝑀𝐵

Solution

𝑀1 = 𝑚 × 2𝑙, 𝑀2 = 3𝑚 × 2𝑙 = 3𝑀

We have 𝑛 = 12𝜋 𝐼�𝑀𝐵

𝑛2𝑛1

= �𝑀2

𝑀1= √3

∴ 𝑛2 = 𝑛1√3 = 10√3 = 17.32

∴ 𝑛2 = 17.32 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛𝑠/𝑚𝑖𝑛

7) If q is the net charge inside the closed surface, the net electric flux leavingthe closed surface = 𝜙2 − 𝜙1

If q is the net charge inside the closed surface, then according to Gauss’s theorem

𝜙2 − 𝜙1 =𝑞 𝜖0


𝑞 = 𝜖0(𝜙2 − 𝜙1)

8) Given 𝜆1 = 25000𝐴, 𝜆2 = 35000𝐴

To find 𝑛1 =? 𝑛2

Formula 𝑛𝜆 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑛1𝜆1 = 𝑛2𝜆2

𝑛1𝑛2

=𝜆2𝜆1

=35002500 =

75

Thus the seventh order fringe of wave length 25000𝐴 and fifth order fringe of wave length 35000𝐴 will co-incide

Q. 7.

1) Resolving power of a microscope

1. It is defined as the reciprocal of the least separation between two closeobjects so that they appear just separated, when seen through the microscope

2. Limit of resolution of the microscope = 𝑑 = 𝜆2𝜇 𝑠𝑖𝑛𝜃

3. The resolving power of a microscope increases with the increase in the valueof the medium between its objective and the object. For the reason, oilimmersion objective microscopes are used to achieve high resolving power.

4. The resolving power of a microscope increases with the decrease in thevalue of wavelength of the light used to illuminate the object.

5. Since the wave length of ultra violet light is less than that of the visiblelight, the microscopes employing ultra violet light for illuminating theobjects are used to achieve high resolving power. Such microscopes arecalled ultra microscopes.

6. Still higher resolving power is obtained in electron microscope.


7. The resolving power of an electron microscope is 1000 times greater thanthat of a microscope using visible light.

2) Data k = 5

E = 500 V /m

l = 1 cm = 10−2 𝑚𝑡𝑠

𝑣 = (1 𝑐𝑚)3 = 10−6 𝑚3

Energy stored per unit volume

𝑈 =12𝑘 𝜖0𝐸2

𝑈 =12 × 5 × 8.85 × 10−12 × (500)2

𝑈 = 553.1 × 10−8 𝐽 /𝑚3

Energy contained in the cube

= 𝑈 × 𝑉

= 553.1 × 10−8 × 10−6

= 553.1 × 1014 𝐽 /𝑚3

3) Resonant Frequency

A given A.C. circuit with constant value of inductance (L) and capacitance (C) can be in resonance only for a particular frequency.

The frequency of A.C. for which resonance takes place and maximum current (r.m.s.) flows through the circuit, is called resonant frequency (𝑓𝑟).


The condition for resonance is (𝑍 = 𝑍𝑀𝐼𝑁)

𝑋𝐿 = 𝑋𝐶

𝜔𝐿 = 1𝜔𝐶

∴ 𝜔2 = 𝐿𝐶 ∴ 𝜔 = 1

√𝐿𝐶

𝐵𝑢𝑡 𝜔 = 2 𝜋𝑓𝑟

∴ 2 𝜋𝑓𝑟 = 1√𝐿𝐶

∴ 𝑓𝑟 = 12 𝜋√𝐿𝐶

………………………(1)

The variation of r.m.s. current with frequency of A.C. is as shown in the figure. The curve is called the series resonance curve. At resonance the r.m.s. current becomes maximum.

This circuit at resonant condition is very useful in radio receivers T.V. receivesr for tuning the signal from a desired transmitting channel or station.

We, know that when alternating current of different frequencies are sent through series resonant circuit it offers minimum impedance to the current of resonant frequency and high impedance to the other frequencies current. In other words the circuit only accepts only current of the resonant frequency and rejects current of other frequencies. For this reason it is called an acceptor circuit.

4) Data

𝑚 = 9 × 10−31 𝑘𝑔

𝑒 = 1.6 × 10−19𝐶

ℎ = 6. 63 × 10−34𝐽𝑠, 𝜖0 = 8.85 × 10−12𝐶2/𝑁𝑚2

Solution

For the first orbit


n = 1

ℎ2𝜖0

𝑟1 =

𝑟1 = �𝜋𝑚𝑒2 �

(6. 63 × 10−34)2 × 8.85 × 10−12 3.142 × 9 × 10−31 × (1. 6 × 10−19)2

𝑟1 = 5.34 × 10−11 𝑚𝑡𝑠

𝑟1 = 0.53740𝐴

In general 𝑟𝑛 = 𝑟1𝑛2

𝑟3 = 𝑟1 × (3) 2

𝑟3 = 0.537 × 9

𝑟3 = 4.8370𝐴

Q. 8.

1) Space Communication

a) The process of transfer of information (speech, pictures or both) from sourceto large distances without wires (i.e. through space as communicationchannel) is called space communication.

Examples : Radio communication, Television communication and Satellite communication

b) Messages are in the form of sound wavesc) Waves can be send from one place to another by superimposing it on

undamped electromagnetic waves.d) The radio waves can propagate from transmitting antenna to the receiving

antenna by three types of propagation, which are as follows.1) Ground wave propagation2) Sky wave propagation3) Space wave propagation.


2) a) Inductive reactance (𝑋𝐿)

1) The non ohmic or non resistive opposition offered by an inductor coil to theflow of varying current or A.C. current through it is called as Inductivereactance (𝑋𝐿)

2) Inductive reactance can also be defined as the ratio of effective (r.m.s.)voltage across the inductance to the effective (r.m.s.) current through it.

3) Note that the inductor coil produces leading of applied A.C. voltage withthe resulting A.C. current. So it is always taken as + 𝑋𝐿

4) S.I. unit of (𝑋𝐿) is ohms5) It is given by (𝑋𝐿) = 𝜔𝐿 = 2𝜋𝑓𝐿

b) Capacitive reactance (𝑋𝐶)

1) The non ohmic or non resistive opposition offered by a capacitor to the flowof varying current or A.C. current through it is called as Capacitivereactance (𝑋𝐶)

2) Capacitive reactance can also be defined as the ratio of effective (r.m.s.)voltage across the capacitance to the effective (r.m.s.) current through it.

3) Note that the capacitor produces leading of the resulting A.C. current withthe applied A.C. voltage. So it is always taken as + 𝑋𝐶

4) S.I. unit of (𝑋𝐶) is ohms5) It is given by (𝑋𝐶) = 1

𝜔𝐶= 1

2𝜋𝑓𝐶

3. a)

Energy of Bohr’s orbit

1) Energy of electrons increases as the number of orbit increases. Thus, energyis minimum in the innermost orbit and maximum in the outermost orbits. i.eelectrons in the innermost orbit are tightly bound to the nucleus.

2) The energy of an electron in a Bohr orbit is inversely proportional to thesquare of the principal quantum number of that orbit.

𝐸 = 1𝑛2 𝑎𝑛𝑑 𝐸𝑛 =

𝐸1𝑛2 =

−13.6𝑛2 𝑒𝑉


3) Negative sign in the equation of total energy indicates that the electron isbound to the nucleus. Thus negative sign is a measure of stability of electronin an orbit.

4) The total energy of electron in terms of Rydberg’s constant is given by

𝐸𝑛 = − Rch𝑛2

b) Diffraction of light

1) The phenomenon of bending of light waves around corners and theirspreading into the regions of the geometrical shadow of an object is calleddiffraction

2) There are two kinds of diffraction. They are (a) Frensel’s diffractions and(b)Fraunhofer diffraction

3) Frensel’s diffractions : In this case either the source or the screen or both areat finite distances from the diffracting aperture or the obstacle. No lenses areused to make the rays parallel or convergent. The wave fronts are eitherspherical or cylindrical.

4) Fraunhofer diffraction : In this case either the source and the screen are atinfinite distances from the diffracting aperture. This is achieved by placingthe source and the screen in the focal planes of the two lenses.

a) The diffraction is readily observed in case of radio waves and sound waves.b) Diffraction effect is observed in principle of all types of wave motion of

which is due to interference of secondary wavelets.5) Wavefront obtained due to this may be either spherical or cylindrical.

Grand HSC EXAMINATION SET - A PHYSICS Test HINT & …2. a) Formation of Stationary waves on strings...

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