Unit 6ab tủ tài liệu training

In Young’s double slit experiment, one of the two slits is covered with a cellophane sheet whichabsorbs half the light intensity.. How will the width of the central maximum change if a s

VERY SHORT AND SHORT-ANSWERS QUESTIONS

51 What are red and blue shifts ?

52 What is Fresnel distance ? Write an expression for it.

53 Write an expression for the resolving power of a telescope.

54 What is the limit of resolution of a microscope ? Write an expression for it.

55 What is optical activity ?

56 Give an expreiment to show that light waves are transverse in nature (AISSCE 1994)

57 Write some uses of polaroids.

58 Write the conditions for maxima and minima of diffraction pattern.

59 The refractive index of glass w.r.t air is 5/3, and that of water is 4/3 Find the refractive

index of glass w.r.t water

60 The refractive indices of two media X and Y are 5/3 and 5/4 respectively Which medium is

denser ?

61 What happens to the path when a ray passes from a rarer to a denser medium ?

62 The wavelength of light is 0.00006 cm Express this wavelength in micron and in angstrom.

63 In Young’s double slit experiment, violet, yellow and red lights are successively used For

which colour will the fringe width be maximum ?

64 What is the phase difference corresponding to a path difference of 2 λ ?

65 What is the shape of the wave front for a point source ?

66 Two slits are separated by a distance less than the wavelength of light used Can you observe

interference pattern on the screen ?

67 A Young’s double slit arrangement is shifted from air to inside water What happens to the

fringe width ?

68 In Young’s double slit experiment, one of the two slits is covered with a cellophane sheet which

absorbs half the light intensity What will happen to the fringes ?

69 What is the ratio of the fringe widths for bright and dark fringes in Young’s double slit experiment.

70 Can we obtain a interference pattern on a screen by using two identical sodium lamps.

71 Do we get any information about the longitudinal and transverse nature of light from the

phenomenon of diffraction

72 How will the width of the central maximum change if a single slit diffraction set up is

im-mersed completely in water without changing any other factor ?

73 When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot

is seen at the centre of the shadow of the obstacle Explain Why ?

74 Name one phenomenon which is exhibited by light waves but not by sound waves.

75 Can electromagnetic waves be polarised ?

76 Does the value of Brewster’s angle depend on the colour of light ?

77 What is Doppler shift.

78 A star is accelerating towards the earth It is emitting orange colour Whether its colour, as

seen from the earth, will turn gradually yellow or red ?

79 State Malus law.

80 Light of wavelength 5000 Å in air enters a medium of refractive index 1.5 What will be its

frequency in the medium ?

81 A and B are two points on a water surface where waves are generated What is the phase

difference if (a) A and B are on the same wavefront but separated by one wavelength, (b) A

and B are on successive crests

82 In Young’s double slit experiment, if the distance between the two slits is halved and the

distance between the slits and the screen is made four times, then by what factor will the fringe

83 Widths of the two slits in Young’s experiment are in the ratio 9 : 1 What is the ratio of the

amplitudes of light waves coming through them

84 The phase difference between two waves reaching a point is π/2 What is the resultant ampli-tude if the ampliampli-tudes of the two waves are 3 mm and 4 mm

85 A ray of light strikes a transparent medium of refractive index 3 If the reflected and re-fracted rays are multually perpendicular, what is the angle of incidence ?

86 Two coherent sources have intensity ratio 64 :1 Calculate their amplitude ratio.

87 Refractive index of glass for lights of yellow, green and red colours are µy, µg, and µr respec-tively Rearrange these symbols in an increasing order of values (AISSCE 1997)

88 Deduce the laws of reflection on the basis of Huygens’ principle (AISSCE 1992)

89 Using Huygens’ principle, derive Snell’s law (AISSCE 1993)

90 Derive an expression for the angular width of the central maximum of the diffraction pattern produced by a single slit, illuminated with monochromatic light (AISSCE Delhi (comp) 1991)

91 In Young’s double slit experiment derive an expression for fringe width.

92 Explain polarisation by reflection and obtain Brewster’s law.

93 State and explain Huygens’ principle.

94 Explain the terms Fresnel distance and Fresnel zone.

95 The polarizing angle of a medium is 60º What is the refractive index of the medium ?

(AISSCE Delhi 1999)

96 What is the polarizing angle of a medium of refractive index 3? (AISSCE 1999)

97 What changes in the interference pattern in Young’s double slit experiment will be observed

when

(i) the distance between the slits is reduced, and

(ii) the apparatus is immersed in water ?

98 State Huygens’ postulates of wave theory Sketch the wavefront emerging from (i) a point

source of light and (ii) a linear source of light like slit. (AISSCE Delhi 2000)

99 In a single slit diffraction experiment, if the width of the slit is doubled, how does the (i)

intensity of light (ii) width of the central maximum change? Give reasons for your answer.

(AISSCE Delhi 2000)

100 What is the effect on the interference pattern observed in a Young’s double slit experiment in

the following cases:

(i) Screen is moved away from the plane of the slits

(ii) Seperation between the slits is reduced

(iii) Widths of the slits are doubled.

ANSWERS

50 The phenomenon of apparent change in the frequency of light due to relative motion of source

and observer is called Doppler effect.

51 Due to Doppler effect, a wavelength in the middle of the visible spectrum will be shifted

towards red if the source and observer move away from each other, and towards blue if they

approach each other The terms red shift and blue shift are used for wavelength increase and

wavelength decrease, respectively, even when the wavelength is not in the visible region

52 Fresnel distance (D F) is the distance of the screen from the slit at which the spreading of light due to diffraction becomes equal to the size of the slit It is given by F a2

D = λ

53 Resolving power of a telescope = 1.22aλ,

where a is the diameter of the aperture of the objective.

54 The limit of resolution of a microscope is defined as the smallest distance between two objects

so that they can be seen distincity It is given by

2µ sin

θ

where µ is the refractive index of the medium between the object and the objective of the microscope and θ is the semi vertical angle

55 Certain crystals or compounds in solution have the ability to rotate the plane of polarization of plane polarized light This property is called optical activity.

56 A tourmaline crystal cut parallal to its axis behaves like a very fine slit When unpolarised

light is incident normally on a pair of such tourmaline crystal, placed with their axes parallal, the intensity of the emergent light is maximum Now we place the two crystals with their axes perpendicular to each other The intensity of outcoming light in this case is minimum This experiment shows that light waves are transverse in nature

57 Polaroids are used in

(a) sun glasses, (b) window panes, (c) photography as filters.

58. For maxima sin = (2 1) 1, 2,3,

2

For minima asin =θ nλ, n=1, 2,3,

Symbols have their usual meanings

59. µg 5, µ 4

µ

a g

w g

a w

60 X is denser than Y.

61 When a ray of light passes from a rarer to a denser medium it bends towards the normal.

62 0.00006 cm = 0.6 micron = 6000 Å.

63 Fringe width β ∝ λ

The wavelength is maximum for red, so the fringe with will be maximum for red

64. Phase difference = 2π (path difference)

λ

2π 2

λ

= 4π radians

65 Spherical.

66 No In this case the fringe width will become very large.

67 In water, wavelength λ will decrease and so the fringe width will also decrease β = D

d

λ

68 Bright fringes will be less bright and dark fringes will be less dark Thus the contrast between

the fringes will decrease

69 1 : 1

70 No

71 Diffraction does not give any information about the transverse or longitudinal nature of light.

72 The width of the central maximum in the single - slit diffraction pattern is 2λD/a Since the

wavelength λ decreases in water, the width of the central maximum will decrease on

immers-ing the set-up in water

73 At the centre, the light waves diffracted from the edge of the obstacle interfere constructively.

So a bright spot is seen at the centre of the shadow

74 Polarisation is exhibited by light waves but not by sound waves.

75 Electromagnetic waves are transverse in nature So they can be polarised.

76 Yes Brewster’s angle is tan–1(µ), and µ is different for different colours

77 The apparent change in the frequency of waves due to relative motion of observer and source

is called Doppler shift.

78 Since the star is accelerating towards the earth, the wavelength of the light emitted by it will go

on decreasing Therefore its colour will gradually turn yellow.

79 When plane polarised light is passed through an analyser, the intensity of the outcoming light

changes as the analyser is rotated in its own plane It is given by 2

0

= cos θ,

where I0 is the maximum intensity of the transmitted light and θ is the angle between the axes

of the polariser and the analyser This equation is called Malus law.

–10

3 10

5000 10

Frequency does not change in going from one medium to another

81 (a) If points A and B are on the same wavefront, the phase difference

bet-ween them is zero.

(b) If A and B are on successive crest, then the path difference between

them is λ and so the phase difference is 2π

82. β = D

d

λ

β = / 2

8

λ

′

83 Let d1and d2be the width of two slits, then

9 3 1

84 We have A = 2 2

1 2 2 1 2cos θ

Here A1= 3 mm, A2= 4mm, φ = π/2

A

=5mm

85 When the refracted and reflected rays are mutually perpendicular, the angle of incidence is

given by i = tan–1 (µ) = tan–1 ( 3)= 60º

86 Amplitude ratio 1 1

2

1

2

64 Here

1 64 1

I

I

a

a

=

1

87. µr< µy< µg

88 Let XY be a reflecting surface and AB be a plane wavefront just incident on the surface at A.

The normals LA and MB to the wavefront represent incident rays If AN is the normal to the surface at A, then ∠ LAN = i,

C Q

A

L

i

N P

r

R N'

r

B M

Fig 6 A.2

the angle of incidence CB′ is the reflected wavefront and r is the angle of reflection.

To prove the laws of reflection, consider any point P on the incident wave front Starting from

P we consider a ray PQR such that PQ is perpendicular to AB and QR is perpendicular to B′C

Let c be the velocity of light The distance travelled by the ray from P to R = PQ + QR.

Total time taken

=

t

c

c

+

= +

=

+

=

This time is to be the same for all wavelets on the incident wavefront., So it must be independent

of AQ.

i.e., angle of incidence = angle of reflection This proves the first law of reflection Also, the

incident wavefront, the reflected wavefront and the normal to the surface, all lie in the same plane i.e., the plane of the paper This is the second law

89 Let XY be a plane surface separating two media and let PA be a plane wavefront just incident

on it Normals LA and MP to the incident wavefront represent incident rays AN is normal

to the surface at the point A ∠ NAL = i, the angle of incidence.

The wavefront first strikes at A, so secondary wavelets will start first from A, which travel

with velocity c2 in the second medium During time t the disturbance reaches from P to P′ In

this time the secondary wavelet from A will travel a distance c2 t in the second medium With

A as centre and c2t as radius, we draw an arc and from P′ we draw a tangent P′A ′ P′A′is the refracted wavefront

A X

L

N

M

P

P'

Y

r r

Q'

Q

K

i

i

A'

Fig 6 A.3

Consider a ray QKQ′such that QK is perpendicular to AP and KQ′is perpendicular to A P ′ ′ The time taken by the ray to travel a distance QK + KQ′is

1 2

sin sin – sin

t

t

AK

′

′ +

′

′

The rays from different points on the incident wavefront will take the same time to reach the

corresponding points on the refracted wavefront It is possible only if t is independent of AK.

1

1 2 2

sin

sin

c i

= =

1µ2 is called the refractive index of the second medium with respect to the first medium and the above equation is called Snell’s law

90 Fig 6 A.4 shows the diffraction of a monochromatic wavefront from a single slit (AB) The

secondary wavelets that go straight across the slit arrive at the lens in same phase and are brought to focus on the screen at O Thus the intensity at O is maximum

P

O

P' B

N C

Fig 6 A.4(a)

a

a sin θ θ

θ

O

Central Maximum

–

——

a

–—

a

↑

I

Fig 6 A.4(b)

Consider now the rays making an angle θ with the direction of normal to the slit, such that a sin

θ = λ

In this case, if we imagine the slit to be divided into two equal parts AC and CB, then for every point in AC, there is a point in CB such that the path difference between the rays from these points is λ/2 Thus the points P and P′at which these rays converge will be minima Since θ is very small, this gives θ = λ/a.

Angular width of central maximum 2θ = 2

a

λ

=

91 S is a source of monochromatic light illuminating two slits S1and S2 S1and S2are very close

to each other and equidistant from source S S1and S2 will act as coherent sources of light and wavefronts from them will spread in all directions Due to their superposition, alternate bright and dark bands are obtained on the screen O is equidistant from S1and S2, so maximum is obtained at O Let y be the distance of a point P on the screen from the centre O

A P

O

B

Fig 6 A.5

y

D

S 2 C

S 1

d

The path difference between the rays reaching point P from S1and S2 is given by

2

2

2 and

2

d

d

= + = +  

= + + 

= +   

We can take S2 P + S1P = 2D

Path difference –

2 For maxima , 0,1, 2,

yd

D

n D d

λ

This is the position of the nth maximum on the screen

Fringe width β = Distance between two successive maxima = yn+1 – yn

–

or

+ λ λ

=

β =λD

d

92 Unpolarized light, on reflection from a

transpar-ent surface, gets polarised The degree of

polar-ization depends on the angle of incidence At a

particular angle of incidence the reflected beam

becomes completely polarised This angle is

called polarising angle or Brewster’s angle (θp)

The plane polarised light has vibrations

perpendicu-lar to the plane of the paper The refracted light is

partially polarised It is found that when light is

in-cident at polarising angle, the reflected and the

re-fracted rays are perpendicular to each other

Brewster’s law

In the figure

θp+ r = 90º

or r = 90 – θp

According to Snell’s law

sin µ

sin

i

p p p p p

sin

sin (90 – )

sin

cos

θ

= θ θ

= θ

θ =

This is Brewster’s law

B

r

90° Medium

Air

θp

θp

N N

Fig 6 A.6

93 Huygens’ principle is a geometrical construction which enables us to find the new position of

the wavefront at a later time from its given position at any instant It is based on the following assumptions:

(a) Each point on a given wavefront acts as a source of disturbance called secondary wavelets,

which travel in all the directions with the speed of light in that medium

A

B

A ′

B ′

A"

B"

Fig 6 A.7(a)

A

B

A ′

B ′

A"

B"

a b c d e f

Fig 6 A.7(b)

a b

c

d

e f g