HSC-Physics-Paper-2-Target.pdf

Written according to the New Text bo ok (2012-2013) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

Std. XII Sci.

Perfect Physics - II Prof. Mrs. Jyot i D. Deshpande

Prof. Umakant N. Kon dapure

(M.Sc., D.H.E.

(M.Sc., B.Ed., Solapur)

H.O.D., R. Jhunjhunwala College)

Salient Features:       

Exhaustive coverage of syllabus in Question Answer Format. Covers answers to all Textual Questions and numericals. Covers relevant NCERT questions. Simple and Lucid language. Neat, Labelled and authentic diagrams. Solved & Practice numericals. Includes Board Question Paper of February 2013.

Target PUBLICATIONS PVT. LTD. Mumbai, Maharashtra Tel: 022 – 6551 6551 Website : www.targetpublications.org email : [email protected]

Std. XII Sci. Perfect Physics - II

Target Publications Pvt Ltd.

Printed at: Spark Offset Nerul Navi Mumbai Published by

Target PUBLICATIONS PVT. LTD. Sixth Edit ion : March 2013

Price :

180/-

Shiv Mandir Sabhagriha, Mhatre Nagar, Near LIC Colony, Mithagar Road, Mulund (E), Mumbai - 400 081 Off.Tel: 022 – 6551 6551 email: [email protected]

PREFACE In the case of good books, the point is not how many of them you can get through, but rather how many can get through to you.

Physics is the study of matter and energy and the interaction between them. It is an intrinsic science providing the indepth information of light, motion, force, magnetism, mechanism, current etc. It also reveals the magic behind the wonderful existence of natural phenomenon like planets, galaxies and stars. Hitech gadgets, modern machinery, gigantic skyscrapers, speedy trains, superior infrastructure are some of the marvels of physics. It not only transforms the life of one who are involved in its study but also benefit the future generation. In order to study such a vast science and to master it, one needs to understand and grasp each and every concept thoroughly. For this we bring to you “Std XII : PERFECT PHYSICS - II” a complete and thorough book which analyses and extensively boost confidence of the student. Topic wise classified “question and answer” format of this book helps the student to understand each and every concept thoroughly. Significant formulas, summary, laws, definitions and statements are also given in systematic representation. Solved problems are also provided to understand the application of different concepts and formulae. Practice problems and multiple choice question help the students, to test their range of preparation and the amount of knowledge of each topic. And lastly, I would like to thank all those people who have helped me in preparing this exclusive guide for all students. There is always room for improvement and hence we welcome all suggestions and regret any errors that may have occurred in the making of this book. A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants! Yours faithfully Publisher

TARGET Publication s

Paper Pattern •

There will be one single paper of 70 Marks in Physics.

•

Duration of the paper will be 3 hours.

•

Physics paper will consist of two parts viz: Part-I and Part-II.

•

Each part will be of 35 Marks.

•

Same Answer Sheet will be used for both the parts.

•

Each Part will consist of 4 Questions.

•

The paper pattern for Part I and Part II will be as follows: −

−

Question 1:

(7 Marks)

This Question will be based on Multiple Choice Questions. There will be 7 MCQs, each carrying one mark. One Question will be based on calculations. Students will have to attempt all these questions. Question 2:

(12 Marks)

This Question will contain 8 Questions, each carrying 2 marks. Students will have to answer any 6 out of the given 8 Questions. In this question, 4 marks will be based on calculations. Question 3:

(9 Marks)

This Question will contain 4 Questions, each carrying 3 marks. Students will have to answer any 3 out of the given 4 Questions. In this question, 2 marks will be based on calculations. Question 4:

(7 Marks)

This Question will contain 2 Questions, each carrying 7 marks. Students will have to answer any 1 out of the given 2 Questions. In this question, 2/3 marks will be based on calculations.

Distribu tion o f Marks Acco rding to Type of Questions Type of Questions

Marks

Marks with option

Percentage (%)

Objectives

14

14

20

Short Answers

42

56

60

Brief Answers

14

28

20

Total

70

98

100

Sr. No.

Unit

Page No.

Maximum Marks

Marks with option

10

Wave Theory of light

1

03

04

11

Interference and Diffraction

30

04

06

12

Electrostatics

70

03

04

13

Current Electricity

107

03

04

14

Magnetic Effects of Electric Current

139

03

04

15

Magnetism

165

03

04

16

Electromagnetic Induction

181

04

06

17

Electrons and Photons

223

03

04

18

Atoms, Molecules and Nuclei

243

04

06

19

Semiconductors

278

03

04

20

Communication Systems

310

02

03

21

Board paper – 2013

323

-

-

Note: 1. All the Textual questions are represented by * mark. 2. Answers of Intext Questions are represented by # mark.

td. XII c .: Per ect Phys cs - II

TARGET Publications

10

Wave theory of light b.

10.0 Introduction Q.1. State the postulates corpuscular theory.

of

Newton’s

Ans: Postulates of Newton’s corpuscular theory:

i.

ii. iii. iv. v.

Every source of light emits large number of tiny particles known as ‘corpuscles’ in a medium surrounding the source. These corpuscles are perfectly elastic, rigid and weightless. The corpuscles travel in a straight line with very high speeds, which are different in different media. One gets a sensation of light when the corpuscles fall on the retina. Different colours of light is due to different sizes of corpuscles.

Q.2. State the drawbacks of Newton’s corpuscular theory. Ans: Drawbacks of Newton’s corpuscular theory:

i. ii. iii.

iv.

It could not explain partial reflection and refraction at the surface of a transparent medium. It was unable to explain phenomenon such as interference, diffraction, polarisation etc. This theory predicted that speed of light in a denser medium is more than the speed of light in a rarer medium which was experimentally proved wrong by Focault. Hence the Newton’s corpuscular theory was rejected. When particles are emitted from the source of light, the mass of the source of light must decrease but several experiments showed that there is no change in the mass of the source of light.

Q.3. Explain i. Maxwell’s electromagnetic theory ii. Ans : i.

Planck’squantum theory of light. Maxwell’s electromagnetic theory of light:

a.

Maxwell postulated the existence of electromagnetic waves.

Wave Theory of Lig ht

ii.

Light is an electromagnetic wave which requires no material medium for its propagation. So light can travel through a medium where there is no atmosphere i.e. in vacuum.

Planck’s quantum theory:

a.

b.

According to Planck’s quantum theory, light is propagated in the form of packets of light energy called quanta. Each quantum of light (photon) has energy E = h ν where, h = Planck’s constant = 6.63 × 10−34 Js ν = frequency of light

10.1 Wave theory of light *Q.4. Give a brief account of Huygen’s wave [Oct 01, 04] theory of light. Ans: Huygen’s wave theory of light:

In 1678, Dutch physicist Christian Huygen proposed a theory to explain the wave nature of light. This theory is called Huygen’s wave theory of light. Main postulates of Huygen’s wave theory: i. Light energy from a source is propagated in the form of waves: The

particles of the medium vibrate about their mean position in the form of simple harmonic motion. Thus the particles transfer energy from one particle to its neighbouring particle and reach the observer. ii.

In homogeneous isotropic medium the velocity of wave remains constant:

Speed of the wave is not affected because density and temperature of isotropic medium is same. iii.

Different colours of light waves are due to different wavelengths of light waves:

Each wave has its own wavelength. As the wavelength changes, its colour and frequency also changes. This is indicated by change in the colour. 1


iv.

The material medium is necessary for the propagation of wave: Periodic

disturbance is created in the medium at one place which is propagated from that place to another place. The medium only carries disturbance and handover it to the next particle. Hence it is assumed that luminiferous ether (hypothetical medium) is present everywhere and even in vacuum which possess the property of elasticity and inertia. Q.5. State the merits of Huygen’s wave theory of light. Ans: Merits of Huygen’s wave theory of light:

i.

ii. iii. iv.

It gives satisfactory explanation for laws of reflection, refraction and double refraction of light assuming transverse nature of the light waves. It also explains the theory of interference and diffraction. It explains the phenomenon of polarisation of light. It experimentally proved that velocity of light in rarer medium is greater than that in a denser medium.

Q.6. State demerits of Huygen’s wave theory of [Oct 01] light. Ans: Demerits of Huygen’s wave theory of light:

i. ii. iii.

iv.

This theory could not explain rectilinear propagation of light. It could not explain Compton effect, photoelectric effect, Raman effect etc. It could not explain properly the propagation of light through vacuum. This is because ether has high elastic constant and zero density which gives contradictory results. According to Huygen’s wave theory, luminiferous ether medium exists everywhere in the universe even in vacuum which are treated as material medium for propagation of light wave. However Michelson’s and Morley’s theory disapproved the existence of ether medium.

TARGET Publications

2.

Electromagnetic nature of light was experimentally proved by Maxwell in 1873.

3.

Light wave is assumed to be transverse. Its speed in a hypothetical medium is given by v =

According to wave theory of light, a source of light sends out disturbance in all directions. When these waves carrying energy reach the eye, they excite the optic nerves and the sensation of vision is produced. 2

ρ

where E and ρ are elasticity and

density of the medium. 4.

Huygen’s theory was not accepted immediately due to following reasons: i. If light were waves, they should bend around the sharp corners in the same manner as the sound waves. ii. If light were waves, they could not travel through vacuum. This difficulty was overcome by assuming the existence of a hypothetical medium (ether) which was assumed to fill the whole space.

10.2 Wavefront and wave normal Q.7. Explain the concept of wavefront. Ans: Concept of wavefront:

i.

ii.

iii.

According to Huygen’s theory, light travels in the form of waves which are emitted from the source. Consider a point source of light S situated in air or vacuum. Light waves spread out in all possible directions from the source of light with same speed c. After time t seconds, each light wave covers a distance equal to ct. E

A

D S C iv.

Note:

1.

E

v.

ct

B

Draw a spherical surface by considering radius ct and S as its centre. This surface cuts waves of light at different points A, B, C, D, E etc. All the points on this surface are in the same phase. It is equiphase surface. Such a surface is called spherical wave surface. Wave Theory of Light


TARGET Publications

*Q.8. Define i. Wavefront ii. Wave normal iii. Wave surface Ans: i.

ii.

ii.

A wavefront originating from a point source of light at infinite distance is called plane wavefront.

Example: The light from the Sun reaches the surface of the Earth in the form of plane wavefront.

Wavefront: A locus of all the points of the medium to which waves reach simultaneously so that all the points are in the same phase is called wavefront. Wave normal: A perpendicular drawn to the surface of a wavefront at any point of a wavefront in the direction of propagation of light waves is called a wave normal.

P′

P′′

Ray of light (Wave normal) Source at infinity Plane wavefront

iii.

N1

P

N2 wave normal Q′ Q′′

Example: A tube light emits cylindrical wavefront.

N3

In the figure curve PQ, P ′Q′ and P′′ Q′′ represent wavefront. SN1, SN2 and SN3 represent wave normal. iii.

Ray of light (Wave normal)

Wave surface: The surface of sphere with source as centre and distance travelled by light wave as radius where each wave arrive simultaneously is called wave surface.

Q.9. State different types of wavefronts with examples. Ans: Depending upon the source of light,

Cylindrical wavefront

Q.10. State the main characteristics of wavefront. Ans: Characteristics of wavefront:

i. ii.

wavefronts are classified into three types. i.

Cylindrical wavefront: A wavefront originating from a linear source (slit) of light at a finite distance is called cylindrical wavefront.

wavefront

S Q

Plane wavefront:

Spherical wavefront: A wavefront originating from a point source of light at finite distance is called spherical wavefront.

Example: Candle spherical wavefront P S

P1

flame

iii.

produces

P2

Q Q1 Q2 R R 1 R 2

iv.

Wavefronts travel with the speed of light in all directions in an isotropic medium. The phase difference between any two points in the same phase on the two consecutive wavefront is 2 π. So, if the phase at one crest is 2π then phase at next consecutive crest = 4 π and so on. It always travels in the forward direction. During the propagation of spherical wavefront from a source, wave become weaker. It is so because same energy is distributed over circumference of larger circles of increasing radii. In an isotropic medium, it travels with different velocities in different directions due to different densities of the medium.

Spherical wavefront


3

td. XII c .: Per ect Phys cs - II Q.11. State the main characteristics of wave normal. Ans: Characteristics of wave normal:

i. ii. iii.

iv.

It gives the direction of propagation of wave. It is perpendicular to wavefront. In a homogeneous isotropic medium wave normal is same as direction of ray of light. It is drawn from the point of generation of wavefront.

TARGET Publications

10.4 Construction of plane and spherical wavefront Q.14.What is the shape of the wavefront in each of the following cases? (NCERT) i.

Light diverging from a point source.

ii.

Light emerging out of a convex lens when a point source is placed at its focus.

Ans: i.

ii.

Spherical wavefront. Plane wavefront.

*Q.15.Explain the Huygen’s construction of plane [Oct 99] wavefront. OR

10.3 Huygen’s principle *Q.12.State Huygen’s principle. [Oct 99, 04, 08] Ans: It is the geometrical construction to determine

new position of a wavefront at later instant from its position at any instant.

Using Huygen’s principle explain the propagation of a plane wavefront. [Feb 06] Ans: Huygen’s construction of plane wavefront:

i.

Statement:

i.

Every point on the primary wavefront acts as a secondary source of light and sends out secondary waves(wavelets) in all possible directions.

ii.

The new secondary wavelets are more effective in the forward direction only (i.e. direction of propagation of wave front).

iii.

i.

Primary source of light

It is a real source of light. ii. It sends out primary waves in all possible directions. iii. Primary wave is effective at every point on its surface. iv.

4

iii.

The resultant wavefront at any position is given by the tangent to all the secondary wavelets at that instant.

Q.13. Distinguish between primary source of light and secondary source of light. Ans: No.

ii.

iv.

A plane wavefront is formed when point of observation is very far away from the primary source. Let PQR represents a plane wavefront at any instant. According to Huygen’s principle, all the points on this wavefront will act as secondary source of light sending out secondary wavelets in the forward direction. Draw hemispheres with P, Q, R…. as centres and ‘ct’ as radius. The surface of tangency of all such hemispheres is P1Q1R 1…. at instant ‘t’. It is a new wavefront at time ‘t’. The plane wavefront is propagated as a plane wave in homogeneous isotropic medium. They are parallel to each other.

Secondary source of light

It is a fictitious source of light. It sends out secondary waves only in the forward direction. Secondary wave is effective only at the point where it touches the envelope. Primary source is Secondary source situated in air. is situated on a wavefront.

P

ct

P1

N1

Q1

N2

R 1

N3

ct Q ct R

PQR: Plane wavefront at any instant, P1Q1R1 : Plane wavefront after time ‘t’, PP1N1, QQ1N2, RR1N3 : wave normals at PQR

Wave Theory of Light


TARGET Publications

v.

vi.

PP1 N1, QQ1 N2, RR 1 N3 are the wave normals at P, Q, R. These wave normals show the direction of propagation of plane wavefront. The new wavefront P1Q1R 1 is parallel to primary wavefront PQR.

*Q.16. Explain the Huygen’s construction of spherical wavefront. Ans: Huygen’s construction of spherical wavefront:

i. ii.

Spherical wavefront is formed when source of light is at a finite distance from point of observation. Let S be the point source of light in air. PQR represents spherical wavefront at any instant. The wavefront PQR acts as a primary wave which is propagated in air. N1 ct

P1

vii.

These wave normals show the direction of propagation of spherical wavefront. viii. The new wavefront P1Q1R 1 is parallel to PQR at every instant. Note:

The intensity of secondary waves varies from maximum in forward direction to zero in backward direction. This indicates that secondary waves are effective only in forward direction. 10.5 Reflection at a plane surface *Q.17 With the help of a neat diagram, explain the reflection of light from a plane reflecting surface on the basis of wave theory of light. OR On the basis of wave theory of light explain the laws of reflection. [Oct 96] Ans: Reflection of plane wavefront from plane reflecting surface:

According to laws of reflection: i. The incident rays, reflected rays and normal to the reflecting surface at the point of incidence, all lie in the same plane. ii. The incident rays and the reflected rays lie on opposite sides of normal. iii. The angle of incidence is equal to angle of reflection. i.e. ∠ i = ∠ r.

P S

Q R

ct Q1

N2

ct R 1 N3

PQR : Primary wavefront, P1Q1R1 : Secondary wavefront after time t, SPN1, SQN2, SRN3 : Wave normals at P, Q, R

Explanation: B

T

M

iii.

iv.

v.

vi.

According to Huygen’s principle, all the points on PQR will act as secondary source of light and send secondary wavelets with same velocity ‘c’ in air. To find out new wavefront at later instant ‘t’, draw hemispheres with P, Q, R…. as centres and ‘ct’ as radius in the forward direction. The surface tangential of all such hemispheres is an envelope at that instant ‘t’. Such a surface is passing through the points P1, Q1, R 1…. on the hemispheres and touching all the hemispheres. This surface is the new wavefront at that instant ‘t’. SPN1, SQN2, SRN3 are the wave normals at P, Q, R.


N

A

Q

R

P

i r

X

i r

A1

Y

B1

Reflection of light XY : Plane reflecting surface AB : Plane wavefront RB1 : Reflecting wavefront A1M, B1N : Normal to the plane AA1M =

BB1N =

i = Angle of incidence

TA1M =

QB1N =

r = Angle of reflection 5


i. ii.

iii.

iv.

v.

vi.

Let a plane wavefront AB is advancing obliquely towards plane reflecting surface XY. AA1 and BB1 are incident rays. When ‘A’ reaches XY at A1, then ray at ‘B’ reaches point ‘P’ and it has to cover distance PB1 to reach the reflecting surface XY. Let ‘t’ be the time required to cover distance PB1. During this time interval secondary wavelets are emitted from A 1 and will spread over a hemisphere of radius A1R. Distance covered by secondary waves to reach from A1 to R in time t is same as the distance covered by primary waves to reach from P to B1. Thus A1R = PB1 = ct. All other rays between AA1 and BB1 will reach XY after A1 and before B1. Hence they also emit secondary wavelets of decreasing radii. The surface touching all such hemispheres is RB1 which is reflected wavefront bounded by reflected rays A1R and B1Q.

Note:

1.

Frequency, wavelength and speed of light does not change after reflection. If reflection takes place from a denser medium then phase changes by π radian.

2.

Q.18. Draw neat labelled ray diagram of reflection of light from a plane reflecting surface using plane wavefront. [Mar 96, Oct 99, 04] Ans: Refer Q.17 (diagram) 10.6 Refraction at plane surface *Q.19.Explain refraction of light on the basis of wave theory. Hence prove laws of refraction. [Mar 96, Oct 08] OR Prove the laws of refraction on the basis of wave theory light. [Feb 02, 03, 05, Oct 03, 05, 06] Ans: Laws of refraction:

i.

ii.

Draw A1M ⊥ XY and B1 N ⊥ XY. Thus, angle of incidence is ∠AA1M =∠BB1N = i and Angle of reflection is ∠MA1R =∠NB1Q = r .

∠ RA1B1 = 90 − r PB1A1 = 90 − i vii. In ∆A1RB1 and ∆A1PB1 ∠A1RB1 ≅ ∠A1PB1 A1R = PB1 (Reflected waves travel equal distance in same medium in equal time). A1B1 = A1B1 (common side) ∴ ∆A1RB1 ≅ ∆A1PB1 ∴ ∠ R 1AB1 = ∠PB1A1 ∴ 90 − r = 90 − i ∴ i=r viii. Also from the figure it is clear that incident ray, reflected ray and normal lie in the same plane. ix. This explains laws of reflection of light from plane reflecting surface on the basis of Huygen’s wave theory. 6

TARGET Publications

iii.

Ratio of velocity of light in rarer medium to velocity of light in denser medium is a constant called refractive index of denser medium w.r.t. rarer medium. The incident rays, refracted rays and normal lie in the same plane. Incident ray and refracted ray lie on opposite sides of normal.

Explanation:

Phenomenon of refraction can be explained on the basis of wave theory of light. B

M M1

P A X

i

A1 r N

i

air (µ1)

r r

R

N1

B1 glass Y (µ2) R 1

Refraction of light XY : plane refracting surface AB : incident plane wavefront B1R : refracted wavefront AA1, BB1 : incident rays A1 R, B1R1 : refracted rays AA1M = BB1M1 = i : angle of incidence RA1N = R1B1N1 = r : angle of refraction



TARGET Publications

i.

Let XY be the plane refracting surface separating two media air and glass of refractive indices µ1 and µ2. ii. A plane wavefront AB is advancing obliquely towards XY from air. It is bounded by rays AA 1 and BB1 which are incident rays. iii. When ‘A’ reaches at ‘A1’ then ‘B’ will be at ‘P’. It still has to cover distance PB1 to reach XY. iv. According to Huygen’s principle, secondary wavelets will originate from A1 and it will spread over a hemisphere in glass. v. All the rays between AA1 and BB1 will reach XY and spread over the hemispheres of increasing radii in glass. The surface of tangency of all such hemispheres is RB1. This gives rise to refracted wavefront B1R in glass. vi. A1R and B1R 1 are refracted rays. vii. Let c1 and c2 be the velocities of light in air and glass respectively. viii. At any instant of time t distance covered by incident wavefront from P to B 1 = PB1 = c1t Distance covered by secondary wave from A1 to R = A1R = c2t.

iv.

sin r =

v.

ii.

iii.

From fig. ∠AA1M + ∠MA1P = 90° and ∠MA1P + ∠PA1B1 = 90° From equations (i) and (ii) ∠AA1M = ∠PA1B1 = i Similarly, ∠ NA1R = ∠ N1B1R 1 = r We have ∠ N1B1R 1 + ∠ N1B1R = 90° and ∠ N1B1R + ∠A1B1R = 90° From equations (iii) and (iv) ∠ N1B1R 1 = ∠A1B1R = r In ∆ A1PB1 sin i = PB1 = c1t A1B1 A1B1


…(i) …(ii)

∴

….(v)

….(vi)

Dividing equation (v) by (vi), we have

sin i c1 = sin r c2 Also

vi.

….(vii)

c1 µ 2 1 = = µ2 c 2 µ1

….(viii)

Where 1µ2 = R.I. of glass w.r.t air. From the explanation, it is clear that incident rays AA1, BB1, refracted rays A1R, B1R 1 and normal MN and M1 N1 lie on the same plane XY. Also incident ray AA1 and refracted ray A 1R lie opposite side of normal MN. Hence laws of refraction can be explained.

Q.20. Show that velocity of light in rarer medium is greater than velocity in denser medium. [Oct 08] Ans : i.

∴ ∴

To show velocity of light in rarer medium is greater than velocity in denser medium, we have to prove c1 > c2. From figure ∠i > ∠r sin i > sin r sini > 1 sinr

∴

µ2 >1 µ1

ii.

Since ,

…(iii) …(iv)

A1R c t = 2 A1B1 A1B1

sin i c1t / A1B1 c1 t = = sin r c2 t / A1B1 c2 t

Proof of laws of refraction:

i.

In ∆ A1RB1

But

c1 µ 2 = c 2 µ1

µ2 >1 µ1

∴

c1 > 1 c2

∴

c1 > c2 Hence velocity of light in rarer medium is greater than velocity in denser medium. 7


TARGET Publications

It is given by 1 ν =

λ

ii. iii.

−1

Fig. (a) S1

S2 B

A

S1

−1

Unit: m in SI system and cm in CGS system. Dimension: [M0L−1T0]

B

A

OR Reciprocal of wavelength of the light is called wave number.

S2

S1

Q.21. Define wave number. Write down its unit and dimension. Ans: i. Definition: Wave number is defined as number of waves per unit distance.

Fig. (b) S2 B

A

Note:

1.

During refraction, speed and wavelength of light changes but frequency remains same. Change in wavelength is due to change in speed of light as it travels from one medium to another. More the denser medium less is the wavelength. Phase of light does not change during refraction.

2. 3. 4.

Fig. (c)

iv.

v.

10.7 Polarisation *Q.22.What do you mean by polarisation? Explain the concept of polarisation by considering transverse wave on a rope. [Oct 09] Ans: Polarisation: The phenomenon of restriction of the vibration of light waves in a particular plane perpendicular to direction of wave motion is called polarisation of light. Concept of Polarisation:

i.

ii.

iii.

8

Consider two slits S1 and S2 which are kept parallel to each other. A string AB is passed through both the slits. One end of the string A is in our hand and the other end B is fixed to a rigid support as shown in fig. (a). Now, A is given a jerk up and down so that transverse wave is formed in the string. It is observed that, transverse wave pass through both the parallel slits without loss in amplitude of vibrations as shown in fig. (b). Now the slit S2 is kept perpendicular to slit S1. In this case transverse wave travels up to slit S2 but there are no vibrations in the string through S2 as shown in fig. (c). This means slit S2 does not allow the transverse wave to pass through it. In this case amplitude of vibrations reduces to zero.

Instead of transverse vibration, if we produce longitudinal vibration then it will pass through slit without change in amplitude of vibrations even the slits may parallel or at right angles to each other. From the above experiment it is concluded that transverse vibrations can pass through the slits only in certain conditions. i.e. vibrations are restricted in certain plane. This phenomenon is called polarisation.

Note:

1. 2.

There is no effect of position of slit on the propagation of longitudinal waves. This means longitudinal wave cannot be polarised. There is effect of position of slit on the propagation of transverse waves. This means transverse waves can be polarised. So, polarisation is the property of transverse waves only.

Q.23. Explain in brief the transverse nature of light. Ans: Explanation of transverse nature of light:

i. ii. iii.

Consider a tourmaline crystal C1 with its crystallographic axis perpendicular to the direction of propagation of light. Ordinary light (unpolarised light) is made to incident on crystal C1 as shown in fig. (a). The components of electric field vector which are in the plane of paper pass through the crystal and the components of electric field vector which are perpendicular to plane of paper are blocked. Wave Theory of Light


TARGET Publications

23.

The angle between the original direction of incident ray and reflected ray is (A) angle of deviation due to reflection (B) angle of emergence (C) angle of reflection (D) angle of refraction

24.

When wavefront strikes a reflecting surface, (A) it comes to rest (B) it penetrate the reflecting surface (C) the surface bends (D) the points on the surface becomes source of secondary wavelets

25.

A rays of light of frequency 4 × 1014 Hz is refracted through glass of R.I 1.5. If c = 3 × 108 m/sec in air, the percentage change in the wavelength from air to glass is (A) 50% (B) 25% (C) 20% (D) 33%

26.

The frequency of a beam of light in air is 8 × 10 14 Hz. The wave number of the beam of light in air is (A) 1.67 × 106 m−1 (B) 2.67 × 106 m−1 (C) 3.67 × 106 m−1 (D) 4.67 × 106 m−1

27.

In case of refraction of light for normal incidence, there is no deviation because (A) i = 90° then r = 0 ° (B) i = 0° then r = 0° (C) i = 0° then r = 90° (D) i = 90° then r = 90 °

32.

Which of the following is correct? c sini (A) 1µ2 = (B) 1µ2 = 2 sinr c1 (C)

⎛µ⎞

2 cos−1 ⎜ ⎟ ⎝2⎠

(D)

⎛µ⎞

2 sin−1 ⎜ ⎟ ⎝ 2⎠

28.

The velocity of light in air is c. Its velocity in a medium of refractive index 1.4 will be c (A) c (B) 1.4 (C) c × 1.4 (D) c + 1.4

29.

The refractive index of glass is 1.68 and that of an oil is 1.2. When a light ray passes from oil to glass, its velocity will change by a factor (A) 1/1.2 (B) 1.68 × 1.2 1 1 (C) (D) 1.4 1.68 × 1.2

30.

The laws of refraction of light are valid for (A) plane mirror (B) concave mirror (C) convex mirror (D) glass lens

1

µ2 =

µ1 µ2

(D)

1

µ2 =

sinr sini

33.

A monochromatic beam of light is refracted into water and then into glass. If λa, λw and λg are its wavelengths in air, water and glass respectively, then (A) λa = λw = λg (B) λa > λw > λg (C) λa < λw < λg (D) λa >λw or λg & λw = λg

34.

The velocity of light in vacuum is 3 × 108 m/s. Determine the velocity, wavelength and frequency of, green light of wavelength 5270 A.U. in glass. Refractive index of glass is 1.5. (A) 2 × 108 m/s, 3513 AU, 5.7 × 1014 Hz (B) 3 × 108 m/s, 4513 AU, 6.7 × 1014 Hz (C) 4 × 108 m/s, 8900 AU, 7.7 × 1014 Hz (D) 5 × 108 m/s, 7000 AU, 8.7 × 1014 Hz

35.

Monochromatic light of wavelength 6870 A.U is refracted through water surface. Determine its wavelength and frequency in water, if its frequency in air is 4.4 × 1014 Hz. Refractive index of water is 1.33. (A) 7100 A.U, 3 × 1014 Hz (B) 4933 A.U, 4 × 1014 Hz (C) 6565 A.U, 6 × 1014 Hz (D) 5165 A.U, 4.4 × 1014 Hz

36.

A monochromatic light of wavelength 4310 A.U is incident on the surface of a glass slab of R.I 1.6. Determine the wavelength and frequency of light in glass, if its frequency in air is 5.8 × 1014 Hz. (A) 5993 A.U, 2 × 1014 Hz (B) 4693 A.U, 4 × 1014 Hz (C) 3000 A.U, 7 × 1014 Hz (D) 2694 A.U, 5.8 × 1014 Hz

A ray of light is made to pass from vacuum into a medium of refractive index µ. Angle of incidence is twice the angle of refraction. Angle of incidence is given by ⎛2⎞ ⎛µ⎞ (A) sin−1 ⎜ ⎟ (B) 2 cos−1 ⎜ ⎟ ⎝ 2⎠ ⎝µ⎠ (C)

26

31.



TARGET Publications

37.

A light wave has a frequency of 4 × 1014 Hz and a wavelength of 5 × 10−7m in a medium. The refractive index of the medium is (A) 1.5 (B) 1.33 (C) 1.0 (D) 0.66

38.

Time taken by the sun light to pass through window of thickness 4 mm, whose refractive index is 1.5, is (A) 2 × 10−1 sec (B) 2 × 108 sec (C) 2 × 10−11 sec (D) 2 × 1011 sec

39.

40.

41.

42.

43.

44.

A rocket is going away from the earth at a speed 0.2c, where c = speed of light. It emits a signal of frequency 4 × 107. What will be the frequency observed by an observer on the earth? (A) 4 × 106 Hz (B) 3.2 × 107 Hz (C) 3 × 106 Hz (D) 5 × 107 Hz The R.I of glass is 3/2 and refractive index of water is 4/3. If the speed of light in water is 2.25 × 108 m/s what will be speed of light in glass? (A) 2.25 × 108 m/s (B) 2 × 108 m/s (C) 3 × 108 m/s (D) 2.98 × 108 m/s When light travels from air to water its speed is retarded by [µW = 4/3] 3 4 (A) × 107 m/s (B) × 107 m/s 4 3 8 (C) 2.25 × 10 m/s (D) 7.5 × 107 m/s The ratio of velocity of light in glass to water if R.I of glass and water with respect to air is 3/2 and 4/3 respectively will be (A) 9 /8 (B) 8/9 (C) 3/4 (D) 2/3 The number of waves of electromagnetic radiation of wavelengths 5000 Å in a path of 4 cm in vacuum is (A) 7.5 × 104 (B) 8 × 104 (C) 9 × 106 (D) 10 × 106 One cannot see through fog because (A) fog absorbs light (B) light is scattered by the droplets in fog (C) light suffers total reflection at the droplets in fog (D) the refractive index of fog is infinity


45.

An unpolarised beam of transverse waves is one whose vibrations (A) are confined to a single plane (B) occur in all directions (C) have not passed through a polarised disc (D) occur in all directions perpendicular to their direction of motion

46.

The transverse nature of light is shown by (A) interference of light (B) refraction of light (C) polarisation of light (D) dispersion of light

47.

The polarisation of an electromagnetic wave is determined by (A) the electric field only (B) the magnetic field only (C) both the electric and magnetic fields (D) the direction of propagation of electromagnetic waves

48.

The plane of vibration and the plane of polarisation of a beam of light (A) are identical to each other (B) are orthogonal to each light (C) make an angle, which depends on the colour of the light (D) rotate with respect of each other along the path of the beam

49.

Polaroids used to control the intensity of light coming through windows of (A) trains and aeroplanes (B) nicol prism (C) biprism (D) ammeter

50.

Which of the following phenomenon is used to test and measure the optical activity of crystal like quartz? (A) Interference (B) Polarisation (C) Diffraction (D) Refraction

51.

Unpolarised light consists of electric field vectors in (A) any one plane (B) plane of paper (C) perpendicular to plane of paper (D) all possible planes

52.

Waves that cannot be polarised are (A) radio waves (B) X-rays (C) visible light (D) sound waves 27


53.

A ray of light strikes a glass plate at an angle of 60°. If reflected and refracted rays are perpendicular to each other, the R.I. of glass is 1 3 (A) (B) 2 2 (C) 2/3 (D) 1.732

54.

When unpolarised light is passed through crossed polaroids then light passes through first polaroid is (A) also passes through second polaroid (B) blocked by second polaroid (C) partially passes through second polaroid (D) passes with greater intensity

55.

The critical angle does not depend upon (A) wavelength (B) refractive index (C) temperature (D) frequency

56.

When unpolarised light is incident on a plane glass at Brewster’s angle, then which of the following statements is correct? (A) Reflected and refracted rays are completely polarised with their planes of polarisation parallel to each other (B) Reflected and refracted rays are completely polarised with their planes of polarisation perpendicular to each other (C) Reflected light is plane polarised but transmitted light is partially polarised. (D) Reflected light is partially polarised but refracted light is plane polarised

57.

58.

59.

28

Refractive index of material is equal to tangent of polarising angle. It is called (A) Lambert’s law (B) Bragg’s law (C) Brewster’s law (D) Malus law When a light wave suffers reflection at the interface from air to glass, the change in phase of reflected wave is equal to (A) 0 (B) π (C) π/2 (D) 2π According to Brewster’s law, at polarising angle the reflected and refracted rays are (A) parallel to each other (B) antiparallel to each other (C) perpendicular to each other (D) at 40° to each other

TARGET Publications

60.

It is believed that the universe is expanding and hence the distant stars are receding from us. Light from such a star will show (A) shift in frequency towards longer wavelengths (B) shift in frequency towards shorter wavelength (C) no shift in frequency but a decrease in intensity (D) a shift in frequency sometimes towards longer and sometimes towards shorter wavelengths

61.

Angle of polarisation for a transparent medium (A) does not depend of wave length of light (B) increases as wavelength increases (C) decreases as wavelength increases (D) changes irregularly with increase in wavelength

62.

The refractive index of certain glass is 1.5 for yellow light of wavelength 591 nm in air. The wavelength of the light in the glass will be (A) 591 am (B) 394 nm (C) 886.5 nm (D) 295.5 nm

63.

A ray of light strikes a glass plate at an angle of 60°. If the reflected and refracted rays are perpendicular to each other, the index of refraction of glass is (A) (C)

1 2 3/2

(B) (D)

3 2 1.732

64.

The angle between polariser and analyser is 30°. The ratio of intensity of incident light and transmitted by the analyser is (A) 3 : 4 (B) 4 : 3 (C) (D) 2 : 3 3 :2

65.

For a given medium, the polarising angle is 60°. The critical angle for this medium (A) 47°23′ (B) 60°10′ (C) 23°30′ (D) 35°16′

66.

The angle of incidence is 60° and the angle of refraction is 30°. The polarising angle for the same medium is (A) 55° (B) 45° (C) 30° (D) 60° Wave Theory of Light


TARGET Publications

67.

68.

69.

In a doubly refracting crystal, optic axis is a direction along which (A) plane polarised light does not suffer deviation (B) any beam of light does not suffer deviation (C) double refraction does not take place (D) O-ray and E-ray undergo maximum deviation

Section C

Dichroism is the property where (A) unequal absorption of O-ray and E-ray takes place (B) equal absorption of O-ray and E-ray takes place (C) plane of polarisation rotates (D) unequal reflection of O-ray and E-ray takes place Which of the following polaroid is formed by stretching polyvinyl alcohol by the stress? (A) P-polaroid (B) H-polaroid (C) K-polaroid (D) N-polaroid ANSWERS

Section A

1. 2. 3. 4.

4.5 × 105 m−1 1.5 1.847 6150 Å

5.

53° 3′

6.

58°57′

7.

30°

8. 9.

67° 33′ 1.54

10. 11.

180 km h−1 i. 1.67 iii. 1.25

12.

6.66 × 1014 Hz, 2903 Å

13. 14.

2 × 108 m/s, 2.25 × 108 m/s, 1.125 1.245

15. 16.

37°, 1.327 0.41


ii.

1.

2.143 × 108 m/s, 1.25 × 108 m/s, 1.714

2.

35° 16′, 2 × 108 m/s

3.

16° 22′, 2 × 108 m/s

4.

2500 Å

5.

2667 A.U

6.

2.25 × 108 m/s, 4500 × 10−10 m, 5 × 1014 Hz

7.

24° 44′, 2 × 108 m/s

8.

35° 16′

Section D

1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65. 69.

(B) (A) (A) (B) (B) (C) (D) (C) (B) (A) (D) (D) (A) (D) (C) (C) (D) (B)

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66.

(D) (B) (B) (B) (C) (C) (B) (D) (A) (C) (B) (C) (B) (B) (B) (B) (D)

3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67.

(C) (D) (A) (A) (D) (A) (C) (B) (D) (B) (B) (A) (D) (D) (C) (D) (C)

4. 8. 12. 16. 20 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64. 68.

(C) (B) (D) (B) (B) (D) (B) (A) (D) (B) (B) (B) (D) (C) (A) (B) (A)

1.33

29

HSC-Physics-Paper-2-Target.pdf

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