Ch-10 Wave Optics - For HW

8/17/2019 Ch-10 Wave Optics - For HW

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PU12 Physics Question Bank – Wave Optics

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Note: Question which are highlighted with yellow colour are important

Introduction, Huygens’ Principle , Reflection and refraction of plane waves using Huygens’

Principle.

ONE MARKS QUESTIONS 1.

Name the scientist who first proposed the wave theory of light.

Sol: Christian Huygens.

2.

What is the reason to believe that light is a wave motion?

Sol: Light undergoes interference, diffraction and polarisation which can be explained on the

basis of wave nature of light.

3. Define the term wavefront.

Sol: A wave front is the locus of all points (particles) in a medium which are vibrating in same

phase. It is a surface of constant phase.4. How is the direction of a ray related to a wave front?

Sol: The direction of a ray is normal to the corresponding wavefront.

5.

What is the phase difference between two points on a wavefront?

Sol: Zero (There is no phase difference)

6.

What is the geometrical shape of the wavefront of light diverging from a point source?

Sol: Spherical

7. What type of wave front is received if light is coming from a very distant point source?

Sol: Plane wave front8. What type of wavefront is obtained by a light source in the form of a narrow slit?

Sol: Cylindrical wavefront

9. What type of wavefront is realised when aplane wavefront (plane wave) passes through a prism?

Sol: Spherical wavefront.

10.

What is the shape of the wavefront when a plane wave gets reflected from a convex mirror?

Sol: Spherical

11.

What is the shape of the wavefront when a plane wave gets reflected from a concave mirror?

Sol: Spherical12.

Which type of source of light gives rise to cylindrical wavefront?

Sol: A linear source of light gives rise to cylindrical wavefront.

13. What are secondarywavelets?

Sol: These are wavelets emanating from the secondary sources of disturbance on the wavefront.

14. Who predicted the existence of electromagnetic waves?

Sol: James clerk Maxwell.


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TWO MARKS QUESTIONS

1.

State Huygens principle.

Sol: According to Huygens principle each point on a wavefront is a source of secondarydisturbance and the secondary wavelets emanating from these points spread out in all directions

with the speed of light. The forward envelope of the secondary wavelets at a later time gives the

new wavefront (new position at that time).

2. A light wave travels from air to glass. How will the following be affected (i) energy of the wave?

(ii) frequency of the wave?

Sol: (i) Energy of the wave decrease as part of the light wave is reflected (back into air)

(ii) Frequency of the wave remains the same.

3.

When a monochromatic light incident on a surface separating two media both the reflected and

reflected and refracted light will have the same frequency as the incident light frequency. Explain

why.

Sol: Reflection and refraction arise through interaction of incident light with the atoms of the

medium. These can be treated as oscillators which take up the frequency of the light incident and

undergo forced oscillations. The frequency of the light emitted by a charged oscillater equals its

frequency of the reflected and refracted light equals that of the incident light.

4. When monochromatic light travels from one medium to another its wavelength changes but

frequency remains the same. Explain.

Sol: Frequency is the characteristic of the source while wavelength is characteristic of the mediu.

When a monochromatic light travels from one medium to another, its speed changes and

correspondingly its wavelength monochromatic light travels from one medium to another, its

speed changes and correspondingly its wavelengthC

V

also changes but the frequency V

remains the same.

5.

Who proposed electromagnetic wave theory. Who first detected electromagnetic waves?

Sol: James clerk Maxwell proposed electromagnetic wave theory. Heinrich Hertz producted anddetected electromagnetic waves (radio waves) for the first time.


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THREE MARKS QUESTIONS

1. What is a wavefront? Draw the geometrical shape of the wavefronts when (i) light diverges from

a point source and (ii) light emerges out of a convex lens when a point source is placed at its

focus.Sol: A wavefront is the locus of all points (particles) which are in phase at any instant. It is a

surface of constant phase.

2.

State Hyygen’s principle and using the same explain the geometrical construction of spherical

wavefront.

Sol: Huygen’s principle: Each point on a wavefront is a source of secondary disturbance. The

secondary wavelets starting from these points spread out in all directions with the speed of the

wave (light) the forward envelope of these secondary wavelets gives the new wavefront at a later

time.

Consider a spherical wave front with O as centre moving towards right. Let

PQ be a portion of spherical wavefront at time 0.t Each point on PQ is a

source of secondary disturbance which travels with the speed v of the waves

in the medium. To find the shape of the new wavefront at .t T The new

wavefront RS also is spherical with point O as centre.

3.

(a) Use Hyygens geometrical construction to show the behaviour of plane wavefront

(i) passing through a prism. (ii) reflecting at a concave mirror.

(b) Name the theory of light which predicated that light travels faster in denser medium than in

rarer medium.

(b) Newton’s corpuscular theory predicted that the velocity of light in rarer medium is lesser than

in denser medium.


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4.

Sketch the geometrical shape of the wavefront (i) emerging from a linear slit (rectangular slit) (ii)

corresponding to parallel beam of light. (iii) converging beam of light.

Sol:

5.

Derive the law of reflection of light on the basis of Huygens wave theory of light.

Sol: Consider a plane wavefront AB incident on a plane reflecting surface . MN If V be the speed

of light in the medium and t be the time taken by the wavefront to advance from point B to ,C

then . BC vt

By the time the disturbance from B reaches ,C the secondary wavelets from A would have

spread over a hemisphere of radius . AD BC vt

The tangent plane CD drawn from point C over this hemisphere of radius vt will be the new

reflected wavefront.

Let i be the angle of incidence and r be the angle of

reflection.

From similar triangles ADC and CBA

AD BC

Also AB CD

Hence ˆ ˆ BAC DCA

i.e., i r

angle of incidence angle of reflection.

This is the law of reflection.

FIVE MARKS QUESTIONS

1. State Huygens’ principle. Using the geometrical construction of secondary wavelets, explain the

refraction of a plane wavefront incident on a plane surface. Hence verify Snell’s law of

refraction.


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Sol: Huygens’ principle: Each point on a wavefront is a source of secondary disturbance and the

secondary wavelets emanating from these points spread out in all directions with the speed of the

wave. The forward envelope of the secondary wavelets at a later time will give the new

wavefront at that time.

Consider XY as surface of separation between a medium 1 of refractive index 1n and medium 2

of refractive index 2n such that 2 1.n n

Let 1v and 2v be the speeds of light in the media 1 and 2 respectively such that 1 2.v v

Consider XY as surface of separation between a medium 1 and 2 respectively such that 1 2.v v

Consider a plane wavefront AB propagating in the direction PA (or RB ) incident on XY at an

angle of incidence .

i. If t is the timetaken by the wavefront to travel the distance BS then 1 BS v t

During the time the disturbance from B reaches the point ,S the secondary wavelets from A

would have spread over a hemisphere of radius 2 AQ V t in medium 2. The tangent plane SQ

drawn from S will be the new refracted wavefront. Let r be the angle of refraction.

From the triangle ABS

sin BS

i AS

From the triangle AQS sin AQ

r AS

1

2

sin

sin

v t i BS

r AQ v t 1

2

sin

sin

vi

r v

But 1 2 1 22 1

v nn

v n

1 2sin

sin

in

r

This is Snell’s law of refraction 1 2sin sin .n i n r

1

1

C n

V

22

C n

V

:C speed of light in vacuum.


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COHERENT AND INCOHERENT ADDITION OF WAVES

INTERFERENCE OF LIGHT WAVES AND YOUNG’S EXPERIMENT

ONE MARKS QUESTIONS 1.

What are coherent sources?

Sol: Sources are said to be coherent if the waves emanating from them have a constant phase

relation (zero phase difference or constant phase difference) and same frequency.

2.

State the principle of superposition of waves.

Sol: Principle of superposition of waves states that at a particular point in a medium, the resultant

displacement produced by a number of waves is the vector sum of the displacement produced by

each of the waves.

3.

What are incoherent sources?Sol: Two sources which emit waves which do not have a constant phase difference are called

incoherent sources.

4.

State the conditions for two light waves to be coherent.

Sol: The light waves must be of same frequency and they should have zero phase phase

difference or a constant phase difference.

5.

What is interference of light?

Sol: The modification in the distribution of light energy (intensity) due to super position of

coherent light waves travelling in almost the same direction is called interference of light.6. Give a relation between path difference and wavelength for constructive interference between

tow light waves.

Sol: The path difference between the two light waves must be an integral of multiple of

wavelength for constructive interference i.e.., 0,1, 2,3.......n n

7. State the path difference between two waves for destructive interference.

Sol: For destructive interference the path difference between the two waves must be odd multiple

of .

2

( wave length of light) i.e., 2 1

2

n

0,1, 2,3.......n

8.

How does the fringe width in a young’s double slit experiment change when the distance

between the slits and the screen is doubled?

Sol: The fringe width gets doubled D

9. What is the effect on the interference fringes in young’s double slit experiment if the

monochromatic source is replaced by another of shorter wavelength?

Sol: The fringe width decreases


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10.

In a young’s double slit experiment the separation between the two slits decreased. What will

happen to fringe width?

Sol: The fringe width increases.1

d

11.

In a young’s double slit experiment, the interference pattern is not detected when the two

coherent sources are for apart, why?

Sol: The fringe width becomes too small to be detected.1

d

12.

How will the intensity of maximum and minima change in a young’s double slit experiment if

one of the slits is covered by a transparent paper which transmits only half of the light intensity?

Sol: The intensity of maxima decreases and the intensity of minima increases.

13.

What is the ratio of the fringe width of bright and dark fringes in the interference pattern inyoung’s double slit experiment?

Sol: 1: 1

14. Does interference phenomenon violate the law of conservation of energy?

Sol: No There is only redistribution of energy.

15. Give on example of interference of light observed in daily life.

Sol: Colours produced on a soap bubble is an example of interference of light at thin films.

16.

What is the principle of interference in young’s double slit experiment.

Sol: The principle is division of wavefront.17.

State the conditions for constructive interference in terms of phase difference between two light

waves.

Sol: The phase difference between the light waves must be an even multiple of . i.e. 2n

0,1, 2,......n

TWO MARKS QUESTIONS

1. Two independent sources of light cannot be considered as coherent sources. Explain why?

Sol: Each source of light contains a very large number of atoms and light is emitted by theseatoms independently and randomly. There will be no constant phase relation between the

emitted waves. The phase difference changes rapidly in unit time. Thus independent sources

cannot be coherent.

2. What happens to the fringe width and the angular separation of the fringes in Young’s double slit

experiment when the screen is moved away from the plane of the slits?

Sol: The fringe width is given by D

d

As D increases' ' also increases –fringe width increases.


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The angular separation of the fringes is given byd

and hence the angular separation remains

same.

3.

In a young’s double slit experiment, the source slit is moved closer to the double slit plane. Whatis the effect on interference fringes? Explain.

Sol: Suppose s be the size of the source and S be the separation between source and the double

slit plane. The condition for interference fringes to be seen is s

S d

: wavelength of light

:d double slit separation

When source slit is moved closer to double slit plane, S decreases. The interference patterns

produced by different parts of the source, overlap the pattern gets less sharp and may even

disappear.

4. What will be the effect on the interference fringes in young’s double slit experiment when the

monochromatic source is replaced by a source of white light?

Sol: When white light is used instead of a monochromatic light the interference due to different

component colours of white light overlap. At the centre of the screen the path difference is zero

for all the wavelength. Hence the central fringe is white. On either side of the central fringe few

coloured fringes will be seen. The fringe closest on either side of the central while fringe is blue

(or violet) as the wavelength of blue (or violet ) as the wavelength of blue is smaller than that ofred while the farthest fringe is red.

5. What are coherent and incoherent sources of light?

Sol: Two sources of light which continuously emit light waves of same frequency with zero or

constant phase difference between them are called coherent sources.

Two sources of light which do not emit light waves with a constant phase difference are called

incoherent sources.

6.

Show that the law of conservation of energy is obeyed during interference of light.

Sol: Let 1a and 2a be the amplitudes of the light waves from two coherent sources. In theinterference pattern the intensities of maxima and minima points will be respectively

2

max 1 2 I a a and

2

min 1 2 I a a

The average value will be

2 2

1 2 1 2max min

2 2av

a a a a I I I

2 21 2av I a a

If there is no interference between the light waves from the two sources then the intensity at

every point would be the same and equal to I


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2 21 2 1 2 I I I a a

This is the same asav

I in the interference pattern.

Hence the law of conservation of energy is obeyed.

THREE MARKS QUESTIONS

1.

What is a sustained interference pattern? State the necessary conditions to obtain sustained

interference pattern.

Sol: The interference pattern in which the positions of maxima and minima of intensity (i.e.,

bright and dark fringes) do not change with time is called sustained interference pattern.

1.

The two sources of light must be coherent – emitting light of same phase or constant phase

difference.

2. The sources must be monochromatic – having same wavelength and frequency.

3. The amplitude of interfering waves must be same for better contrast.

4.

The sources must be narrow and close to each other.

5.

The interfering waves must travel nearly along the same direction.

6.

The interfering waves should be in the same state of (or plane of ) polarisation.

2.

State two conditions for sustained interference of light. Draw the variation of intensity with

position in the intensity pattern of young’s double slit experiment.

Sol:

The intensity is maximum at the centre ' 'O then at points where x is odd multiple of2

D

d

the

intensity becomes zero indicating dark fringe and at points where x is an integral multiple of

, D

d

the intensity becomes maximum indicating bright fringes.

3. What is the effect on the interference pattern in a Young’s double slit experiment when

(i)

A light of smaller frequency is used?

(ii)

The separation between the double slits is increased?

(iii)

The entire apparatus is immersed in water

Sol: (i) The fringe width is given by D

d

as frequency becomes lesser, wavelength

increases hence the fringe width increases.


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(ii)1

,d

As increases decreases

(iv)

In water the wavelength decreases.

Since ,

the fringe width ' ' also decreases i.e., the fringes become narrover.

FIVE MARKS QUESTIONS

1. Give the theory of interference of light and obtain the condition for constructive and destructive

interference.

Sol: Let1

sin y a t and 2 sin y b t

Where1

Y and2

Y are the instantaneous values of the electric field vectors of light at a point in the

region of superposition at any instant of time .t a and b are the amplitudes of the two waves

while is the angular frequency and is the phase difference between the two waves at the

point considered.

From the principle of superposition, the resultant electric field vector (displacement) at the point

is

1 2 y y y

sin sin sin sin cos cos sin y a t b t a t b t b t

cos sin sin cos y a b t b t

Let cos cosa b R

sin sinb R

Then cos sin sin cos cotY r t R

sin y R t

The resultant wave is also a harmonic wave of amplitude R and leads the first wave by angle .

Also 22 2 2 2 2 2

cos sin cos sin R R a b b

2 2 22 cos R a ab b

2 22 cos R a b ab

For constructive interference:

The amplitude R and hence intensity at a point of superposition will be be maximum when

cos 1

0,2 ,4 ,......2n

Phase difference 2n n

0,1, 2,........n n

Corresponding path difference: 2 12 2

n


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2 12

n

0,1, 2......n

Hence for destructive interference the phase difference between the interfering waves must be

odd multiple of

while path difference must be an integral multiple of wavelength

of lightwave.

For destructive interference:

The resultant amplitude R and intensity at the point will be minimum when

cos 1 or ,3 ,5 ,......... 2 1n

2 1 0,1, 2, 3,.........n n

Corresponding path difference: 2 12 2

n

2 1 0,1, 2,......2

n n

Hence for destructive interference the phase difference between the interfering waves must be

odd multiple of while path difference must be odd multiple of .2

Alternatively

Consider two coherent sources1S and

2.S Let p be a point equidistant from these two sources

such that1 2

.S P S P The waves from1S and

2S arrive at point P in phase. The displacement

produced by source 1S at P is 1 cosY a t

:a amplitude of the wave.

The displacement by2

S at . P

2 cosY a t

The resultant displacement by2

S at , P

1 2 2 cos y y y a t

Intensity is proportional to the square of the amplitude. Hence resultant intensity is

0

4 I I

Where0

I is the intensity due to each of the source.

2

0 I a

This is called constructive interference.

Consider a point Q for which

2 1 2S Q S Q

i.e., the path difference between the two wave from1S and

2S reaching Q is 2 and

corresponding phase difference is 4 . Then

1 cos y a t


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2 cos 4 cos y a t a t

Thus these two displacement are in phase and intensity will be again0

4 I giving rise to

constructive interference. In general, for constructive interference the path difference between the

waves from the coherent sources must be an integral multiple of wavelength .

i.e.1 2

S p S p n

0,1, 2,......n

Suppose the path difference between the waves from1S and

2S at some other point R be 2.5

i.e.,2 1

2.5S R S R

2

2.5

5

The net displacement cos cos 0 y a t a t

This results in zero intensity and corresponds to destructive interference.

In general for destructive interference the path difference between the waves from source must

be an odd multiple of2

1 2 2 1 0,1, 2...........2

S P S P n n

In general if1

cos y a wt due to1S

2 cos y a t due to 2S at point , A then

The resultant displacement at A is

1 2 cos cos y y y a wt wt

2 cos cos

2 2

wt wt wt wt y a

2 cos cos

2 2 y a wt

cos2

y A wt

where amplitude of the resultant displacement is 2 cos2

A a

The intensity at the point is

2 2 2

04 cos 4 cos

2 2 I a I

for constructive interference leading to maximum intensity.

The phase difference must be

0, 2 , 4 , ..... 2n

0,1, 2,........n


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Corresponding path difference 2 12

n

0,1, 2,..........n

2.

Define fringe Obtain an expression for the frings width in the interference pattern in Young’sdoubts slit experiment.

Sol: The distance between two consecutive bright fringes or two consecutive dark fringes is called

fringe width. Consider two slits A and B with a separation .d These two act as coherent

sources of light.

Let D be the distance of the screen from the plane of the slits. Let be the wavelength of the

light passing through the slits. At O the centre of the fringe pattern a bright fringe is formed.

Let P be a point on the screen at a small distance x from ,O path difference between the waves

from A and B arising at P is BP AP

From2

2 2 2 2:

2

le d BNP BP BN PN D x

le 2

2 2 2 2:

2

d AMP AP AM PM D x

2 2

2 2 2 22

4 4

d d BP AP x xd x xd xd

2 BP AP BP AP xd

BP AP also BP AP D .

As d is very small compared to D as P is very close to .O

2 D D xd

2 2 D xd

Path difference: xd

D

For a bright fringe to be formed at , P n

i.e., xd n D

n D x

d

Similarly the distance of the 1 th

n bright fringe from O is

1

1n

n D x

d

By definition, fringe width is1n n x x

1 1n D n D D n nd d d


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D

d

Similarly for a dark fringe to be formed at P1

2 12

n and 2 12

n n D xd

It can be shown that in case of dark fringes also D

d

The fringe width in interference pattern is D

d


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Diffraction

One mark Questions

1.

What is diffraction of light?

Sol: The phenomenon of bending of light around the corners of (edges of) small obstacles or

apertures and its spreading into geometrical shadow region is called diffraction of light.

2. State the essential condition for diffraction of light to occur.

Sol: Diffraction of light occurs when the size of the obstacle or aperture causing diffraction is

comparable to the wavelength of light.

3. What should be the order of the size of the obstacle or aperture for diffraction of light?

Sol: The size of the obstacle or aperture should be of the order of the wavelength of light used.

4. Give one example of the diffraction effect of light.

Sol: The colour pattern we see when viewing a CD (Compact disc) is an example of diffraction

effect of light.

5.

Who discovered diffraction of light?

Sol: Grimaldi discovered diffraction of light.

6.

Which nature of light explains diffraction phenomenon.

Sol: Wave nature of light explains the phenomenon of diffraction.

7. What is the condition for first minimum in case of diffraction at a single slit?

Sol: asin

Where a: width of the slit. wavelength of light

- angular position of the minimum.

8. Can longitudinal waves undergo diffraction?

Sol: Yes. All waves exhibit diffraction.

9. Why the centre of the diffraction pattern due to a single slit is of maximum intensity?

Sol: The wavelets from all the points of the diffracted wavefront interfere constructively as they

arrive in phase to give rise to maximum intensity.

10.

State the condition for secondary maxima for diffraction of light at a single slit.

Sol: sin 2 1 1, 2, 3...2

a n n

a = width of the slit. wavelength of light.

11.

How does the width of cental maximum vary if the wavelength of light is reduced?

Sol: The width of the central maximum decreases.

12. Is the width of all secondary maxima same?

Sol: No. the width of secondary maxima decreases with increase in the order.

13. What is resolving power of an optical instrument?


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Sol: Resolving power of an optical instrument is its ability to resolve two closely lying point

objects.

14. Define limit of resolution of a microscope.

Sol: Limit of resolution of a microscope is the minimum distance of separation between two point

objects at which they can just be seen as separate points.

15. Define resolving power of a microscope.

Sol: It is the reciprocal of the minimum distance of separation between two point objects at which

they can just be resolved.

16.

Define resolving power of a telescope.

Sol: The resolving power of a telescope is defined as the reciprocal of the smallest angular

separation between two point objects at which they can seen be as distinct.

17.

How can the resolving power of a telescope be increased?

Sol: Resolving power of a telescope can be increased by increasing the diameter of the objective of

the telescope.

18. What is Fresnel distance?

Sol: It is the distance from an obstacle or aperature causing diffraction at which the diffraction

spread of the beam becomes comparable with the size of the aperature.

19.

Give the expression for Fresnel distance.

Sol:2

f a

Z

a: size of the aperature : wavelength of light.

20. In a young’s double slit experiment light of wavelength comparable to the size of the slits is used

what happens if one of the slits is closed?

Sol: A single slit diffraction pattern may result instead of interference pattern.

Two marks questions

1.

Do the phenomina of interference and diffraction obey the principle of conservation of energy.

Explain.

Sol: Yes. The phenomena of interference and diffraction obey the principle of conservation of

energy. In interference and diffraction the light energy is redistributed. If the energy reduces in

one region producing a bright fringe. There is no loss or gain of energy.

2. Diffraction is commonly observed in case of sound waves while it is not easily observed for light

waves. Why? Explain.

Sol: Diffraction effect is observable only when the size of the obstracle or aperature is comparable

to the wavelength of the waves.

In case of sound waves, their wavelength is of the order of size of the objects around us. Hence

they get easily diffracted. But the wavelength of light is much smaller then the size of the objects

around us and thus diffraction of light is not easily observed.


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3.

State with reason, how would the linear width of the Central maximum change if (i)

monochromatic yellow light is replaced with red light and (ii) the distance between the slit and

the screen is increased.

Sol: The linear width of the central maximum is given by2 D

W a

wavelength of light.

D = distance between the screen and the slits.

a: width of the slit.

(i) since red yellow and W . The linear width of the central maximum increases.

(ii) W D

As D increases, the linear width increases.

4. Write the expression for the resolving power of a microscope and explain the symbols.

Sol:2 sin

.1.22

micn

R P

wavelength of light

semivertical angle.

n refractive index of the medium between the object and the objective.

5. Write the expression for the resolving power of an astronomical telescope and explain the

symbols.

Sol: .1.22

tel d

R P

Where d: diameter of the objective lens.

: wavelength of light

6. How does the resolving power of a microscope change upon (i) decreasing the wavelength of

light ? (ii) decreasing the diameter of the objective lens?

Sol:2 sin

.

1.22

n R P

(i) Resolving power1

. R P

. Resolving power increases as the wavelength of light is decreased.

(ii) . sin R P

On decreasing the diameter of the objective lens the semivertical angle decreases and hence the

resolving power also decreases.

7. Astronomers presser to use telescopes with large diameter objectives to observe astronomical

objects. Explain Why.


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Sol: Objective with large diameter has a greater capacity to gather light and to form bright image

of distant, faint stars. Also larger diameter implies greater resolving power since resolving power

R.P. is directly proportional to diameter of the objective

8.

Give the basic difference between interference and diffraction.

Sol: Interference is the result of superposition of secondary waves starting from two wavefronts

originating from two coherent sources.

Three marks question.

1.

Draw a graph showing the variation of intensity with diffraction angle in a single slight

diffraction experiment. Explain the distribution.

Sol: This graph represents the intensity distribution for

maxima and minima in the diffraction pattern with

diffraction angle

It has a broad central maximum at 0 of the incident

light. On either side, it has secondary maxima of

decreasing intensity at positions.

2 12

na

a: width of the slit.

and minima at positions

a

2.

Mention any three differences between interference fringe pattern and diffraction pattern.

Sol: Interference pattern

It has a number of equally spaced a alternate bright and dark bands (fringes)

All bright fringes are of same intensity.

Regions of dark fringes are perfectly dark providing a good contrast. Diffraction patter

It has a central bright maximum (which is twice as large as other maxima) followed on either

side by alternate dark and lesser bright bands.

Intensity of bright fringes decreases as we go around from central maximum.

Dark frings are not perfectly dark providing a poor contrast.

Five marks questions

1. Explain the intensity distribution in case of diffraction at a single slit and draw the intensity

distribution curve.

Sol: Consider a rectangular slit AB of width a illuminated by light of wavelength

Let a plane wavefront be incident on AB as shown. According to

Huygen’s principle all the points on the wavefront between A

and B acts as sources of secondary wavelets which move in the


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direction shown. The intensity at a point on the screen is the resultant effect produced by all these

secondary wavelets. The path difference between waves from A and B.

Reaching P is sin BN BN AP g If is small sin then BN a

Intensity of the central maximum:

The secondary wavelets from all the points between AB reach central point O in phase resulting

in maximum intensity. There will be reinforcement of infinite number of secondary wavelets. The

intensity has a central maximum at 0

Intensity of minima:

If the point P is such that the path difference between wavelets from A and B is . i.e.

asin BN , then the wavelets from A nad C and those from C and B suffer a path difference

of ,2

C

being the midpoint of AB. These pairs of wavelets and all other pairs of wavelets from

the corresponding points in the upper and lower pair of parts will suffer a path difference2

and

hence they interfere destructively resulting in zero insensity. Thus the second minimum is formed

at P. In general, the condition for minima (dark bands) is sina n or a n where

1, 2, 3...n

Intensity of secondary maxima

If the path difference3

sin ,2

a q

then AB can be considered to be made up of three equal parts.

The waves from the corresponding points on the first two parts will have a path difference of .2

Hence they interfere destructively. Thus the intensity at the point considered due to first two

parts of the slit is zero. However the wavelets from the third part will produce some intensity at

the point giving rise to the first secondary maximum. In general sin 2 12

a n

0.1.2..n is the condition for secondary maxima.

The diffraction pattern consists of central maximum followed by alternte minima and secondary

maxima on either side.


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Polarisation

One mark Question

1. What is polarisation of light?

Sol: The phenomenon of restricting the vibrations of the electric field vector of a light wave to just

one direction perpendicular to the direction of wave propogation is called polarisation of light.

2. Which type of waves show the property of polarisation?

Sol: Transverse waves show the property of polarisation.

3.

Can longitudianal waves be polarised?

Sol: No. longitudinal waves cannot be polarised.

4.

Do ultrasonic waves undergo polarisation?

Sol: Ultrasonic waves do not undergo polarisation as they are longitudinal.

5.

What is plane polarised light?

Sol: The light wave in which the electric field vectors are restricted to a single direction (plane)

perpendicular to the direction of wave propagation is called plane polarised or linearly polarised

light.

6. Which special characteristic of light is demonstrated only by the phenomenon of polarisation?

Sol: Transverse wave nature of light is demonstrated by the phenomenon of polarisation.

7.

Which among the infrared waves, sound waves and radiowaves can be polarised?

Sol: Only infrared waves and radio waves can be polarised.

8.

Is the sun light scattered by the molecules of the atmosphere polarised?

Sol: Yes, the scattered sun light is partially polarised.

9.

Mention a method to produce plane polarised light.

Sol: Polarisation by reflection.

10. Name three properties which are mutually perpendicular in a plane polarised light.

Sol: Electric field vector, magnetic field vector and the direction of propogation of light wave.

Or

Plane of vibration, plane of polarisation and direction of propagation of the light wave.

11.

What is unpolarised light?

Sol: The light wave in which the electric field vectors (vibrations) are present in all possible

directions, in a plane perpendicular to the direction of propagation is called unpolarised light.

12.

Give the representation of unpolarised light.

Sol:


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13.

Give the representation of plane polarised light

Sol:

Or

14. State Brewster’s law

Sol: Brewster’s law stares that, for transparent medium the refractive index of the medium is

equal to the tangent of the polarising angle of incidence.

15.

What is meant by polarising angle of Brewster’s angle?

Sol: It is that angle of incidence on a medium for which the reflected light is completely plane

polarise.

16.

Unpolarised light is incident on a plane surface of glass of refractive index n at an angle i. If the

reflected light gets totally polarised, write the relation between the angle i and refractive index n

Sol: tann i

17.

A ray of light is incident on a medium at ;polarising angle. What is the angle between the

reflected and refracted rays?

Sol: The angle between reflected and refracted rays is 90 .

18.

Does the value of the polarising angle of incidence depend on the colour of light?

Sol: Yes

19. What are polaroids?

Sol: These are thin sheets of plastic containing long chain molecules which are used to get plane

polarised light by selective absorption.

20.

Unpolarised light of intensity 0 I is passed through a polaroid. What is the intensity of the light

transmitted by the polaroid?

Sol: 02

I

Two marks questions

1. Explain why longitudinal waves cannot be polarised.

Sol: In polarisation, vibrations (eg.: electric field vectors) perpendicular to the direction of

propagation are restricted to only one direction. But in longitudinal waves vibrations occur only

along the direction of propagation but not perpendicular to it. Hence their polarisation is not

possible.

2.

Light waves can be polarised while sound waves cannot be polarised. Explain why.

Sol: Light waves are transverse waves with vibrations perpendicular to the propagation direction

and hence they can be polarised.

Sound waves are longitudinal waves with vibrations only along the direction of propagation and

hence they cannot be polarised.


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3.

Which of the following can be polarised (i) microwaves (ii) ultrasound waves? Give reasons.

Sol: Microwaves can be polarised as they are transverse electromagnetic waves.

Ultrasound waves cannot be polarised as they are longitudinal waves.

4.

Distinguish between unpolarised light and plane polarised light.

Sol: Unpolarised light:

The light in which the electric field vibrations are present in all

directions in a plane.

Perpendicular to the direction of propagation of the wave, is called

unpolarised light. Plane polarised light.:

The light in which the electric field vibrations are restricted to a

single direction perpendicular to the propagation direction is

called plane polarised light.

5.

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

Sol: Brewster’s angle depends on the colour of light.

We have tan pn i

pi depends on the refractive index of the medium (n) and n depends on wavelength of light or

colour. Hence pi depends on colour of light.

6. Indicate n a diagram the plane of vibration and plane of polarisation.

Sol:

Three Marks questions

1.

What is meant by plane polarised light? What type of waves show the property of polarisation?Describe a method for producing a beam of plane polarised light.

Sol: The light in which the electric field vector of light waves vibrate in only one direction (plane)

perpendicular to the direction of propagation is called plane polarised light. Only transverse

waves exhibit the property of polarisation. A beam of plane polarised light can be produced by

using a polaroid.

When unpolarised light falls on a polaroid, only the vibrations parallel to the transmission plane

get transmitted and perpendicular vibrations are selectively absorbed. Hence the emergent beam

form the polaroid is plane polarised.


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2.

What is a polaroid? How is plane polarised light obtained from it? How is it used to distinguish

between unpolarised light and plane polarised light?

Sol: A polaroid is a tin plastic like sheet containing long chain molecules which produces plane

polarised light by selective absorption.

When unpolarised light is incident on a polaroid the electric vectors (vibrations) along the

direction of the aligned molecules get selectively absorbed. The direction perpendicular to the

transmission plane get transmitted. Hence the emergent beam is plane polarised (linearly

polarised)

When unpolarised light is seen through a polaroid the intensity of light reduces by half due to

polarisation. But when the polaroid is rotated there will be no change in intensity of emergent

light.

When linearly polarised light is seen through a polaroid and the polaroid is rotated then the

intensity becomes maximum twice and zero twice in each rotation. This change in intensity helps

in distinguishing between unpolarised and plane polarised light.

3. Explain with the help of a ray diagram how an unpolarised light can be polarised by reflection

from a transparent medium? Write the expression for Brewster’s angle in terms of the refractive

index of denser medium.

Sol: An unpolarised light has two vibrations – one parallel to the plane of incidence, (lines) and

the other component perpendicular to the plane of incidence

(dots).

PQ: incident, unpolarised light. OR: reflected, plane polarised

light.

QS : refracted, partially plane polarised light.

pi : polarising angle of incidence (Brewster’s angle)

When unpolarised light incidents (PQ) on the transparent medium of refractive index n at

polarising angle of incidence (Brewster’s angle) pi , the reflected ray (QR) is found to be

completely plane polarised and the reflected ray is perpendicular to the refracted ray (QS). The

electrons oscillating in the medium produce the reflected wave. The vibrations move in two

directions transverse to the refracted wave. As the arrows are parallel to the direction of the

reflected wave they cannot send energy along the reflected light direction. Hence the reflected

light consists of vibration perpendicular to the plane of incidence (dots) only and is completely

plane polarised.

Brewster’s law tan pn i .

N : refractive index of the medium

: pi Brewster’s angle.


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4. State and explain Malus’ Law.

Sol: Malus’ law states that when a beam of completely plane polarised light is passed through an

analyser (polaroid), the intensity of the transmitted light varies directly as the cosine of the angle

between the transmission directions (plane or axis) of polariser and analyser.

i.e. 2cos I

20 cos I I

Where 0 I is the intensity of the plane polarised light

from the polariser and Ao is the corresponding

amplitude. I is the intensity of transmitted light

emerging from the analyser.

Suppose their planes make an angle as shown, then

only the component 0 cos A is transmitted by the

analyser as it is parallel to its own plane of transmission.

Hence 2 2 2

0 0cos cos I K A KA

2 20 0 0cos . I I I KA

Five marks questions:

1. State Brewster’s law. Show that when a ray of light is incident at polarising angle on the surface

of a transparent medium the reflected and refracted rays are perpendicular to each other.

Sol: Brewster’s law states that the refractive index of the medium is equal to the tangent of the

polarising angle of incidence.

From Brewster’s law:

tan pn i

sin

cos

p

p

in

i ….(1)

Applying Snell’s law for refraction at Q,

sin

sin

pin

r …(2)

From equations (1) and (2)

cos sin pi r sin 90 sin pi r

90 p

i r

90 pr i


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From the diagram p I r

180 ip r

180 90

90

Hence reflected and refracted rays are mutually perpendicular

Documents

Ch-10 Wave Optics - For HW