Bihar Board - Class 12 Physics - Chapter 10: Wave Optics Long Answer Question

Long Question Answer

1. What is Wavefront ? Explain types of Wavefront .

Ans: - Wavefront :-  A wavefront is a continuous locus of all those points in a medium that oscillate in the same phase. The physical view of a wavefront is the ripples on a water surface. When a stone is dropped in a water pond, the disturbance travels radially outward and what you see are circular wave-fronts traveling outward.

Types of Wavefront :-

Depending on the mode of propagation of light or the source, the wave fronts can be converging, diverging, or plane. For a point source, spherical diverging wavefronts are formed. For an extended line source, cylindrical diverging wavefronts are formed. For a source at infinity, a plane wavefront exists.

2.  Explain Huygens’ Principle.

Ans: - Huygens’ Principle:-  According to this principle,

Each point of a wavefront is a source of secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave.

The envelope of these wavelets gives the shape of the new wavefront.

If we draw a common tangent to all these spheres, then we obtain an envelope which is again a sphere centered at the point source.

3. Find the difference between Diffraction and Interference.

Ans : - Difference between Diffraction and Interference:-

4. Explain the reflection of a Plane Wave.

Ans : - Reflection of a Plane Wave:-

Let us consider a plane wave AB incident at an angle 'i' on a reflecting surface MN.

Time taken by the wave to advance to point C from point B will be t.

Hence BC = vt

Let EC represent a tangent drawn from C to wavefront from E to the spherical wavefront.

AE = vt

Consider, ΔAEC and ΔABC

AC = AC (Common side)

∠AEC = ∠ABC (Each 90°)

AE = BC (Each vt)

Hence, ΔAEC ≅ ΔABC

∠i = ∠r

which proves the law of reflection.

5. Explain the refraction of a Plane Wave.

Refraction of a Plane Wave:-

Let v₁ and v₂ represent the speed of light in medium-1 and medium-2 respectively. Consider a plane wavefront AB propagating in the direction AA', incident on the medium boundary at point A at an angle of incidence 'i'. Let t be the time taken to travel from B to C.

BC = v₁t

From point A, draw a sphere of radius v₂t, let CE represent the forward tangent plane. It is a refracted wavefront at t.

AE = v₂t

From ΔABC, 

                                    sini = BCAC = v₁tAC                   .....(i)

From ΔAEC, 

                                    sinr = AEAC = v₂tAC              .....(ii)

Dividing (i) by (ii), we have

                                       sinisinr = v₁tv₂t

                                       sinisinr = v₁v₂

which proves the law of refraction.

6. What are Coherent and Incoherent Sources of Light? Explain Interference of Light Wave and conditions of interference.

Ans (i). Coherent sources

Two sources of light that continuously emit light waves of the same frequency (or wavelength) with a zero or constant phase difference between them are called coherent sources. 

Ex- LASER.

(ii). Incoherent sources

Two sources of light that do not emit light waves with a constant phase difference are called incoherent sources. 

Ex- Two different light sources produce incoherent waves.

Interference of Light Wave:-

Interference is the phenomenon in which two waves superpose to form the resultant wave of the lower, higher, or same amplitude. When the crest of one wave falls on the crest of another wave such that the amplitude is maximum then interference is called constructive interference. When the crest of one wave falls on the trough of another wave such that the amplitude is minimum then interference is called destructive interference.

Conditions for interference:-

Two sources of light must be coherent.

The frequencies (or wavelength) of the two waves should be equal.

The light must be monochromatic.

The amplitudes of the interfering waves must be equal or nearly equal.

The two sources must be narrow.

7. What is Polarization? Explain Polarization by reflection.

Polarization:  If the vibrations of a wave are present in just one direction in a plane perpendicular to the direction of propagation, the wave is said to be polarized or plane polarized. The phenomenon of restricting the oscillations of a wave to just one direction in the transverse plane is called the polarization of waves.

Polarization by reflection

When a light wave is incident on a boundary of a medium, a part of a light wave is reflected back into the medium from which it is incident and a part of the wave is refracted into the other medium.

When unpolarized light is incident on the boundary between two transparent mediums, for an angle of incidence in which the reflected wave travels at a right angle to the refracted wave, the reflected light is polarized while the refracted light is partially polarized.

Brewster’s Law :-

According to Brewster’s law, When unpolarized light is incident on a transparent substance surface, it experiences maximum plan polarization at the angle of incidence whose tangent is the refractive index of the substance for the wavelength.

                                 n = tani        (where i = incident angle)

8. Answer the following questions :
(i) In what way is diffraction from each slit related to the interference pattern in a double slit experiment?
(ii) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle. Explain why.

Answer:
(i) Diffraction from each slit is related to interference pattern in a double slit experiment in the following ways :

The intensity of minima for diffraction is never zero, while for interference it is generally zero.

All bright fringes for diffraction are not of uniform intensity, while for interference, these are of uniform intensity

(ii) Waves from the distant source are diffracted by the edge of the circular obstacle and these diffracted waves interfere constructively at the center of the obstacle’s shadow producing a bright spot.

9. Explain, polarization affords convincing evidence of the transverse nature of light.

Answer: The phenomenon of interference and diffraction establishes the wave nature of light. Both these phenomena are exhibited by both the types of wave motion i.e., longitudinal and transverse.

In transverse wave motion the vibrations of particles take place in a direction perpendicular to the direction of propagation of the wave. Hence the vibrations in ordinary light are distributed symmetrically in all directions in a plane perpendicular to the direction of propagation. Such light is called unpolarised light.

Light waves which exhibit different properties in different directions are called polarized light waves. This can be demonstrated by using a pair of tourmaline crystals in the path of light waves. 

10. What is Polarization? Explain Polarization by Scattering.

Polarization:-  If the vibrations of a wave are present in just one direction in a plane perpendicular to the direction of propagation, the wave is said to be polarized or plane polarized. The phenomenon of restricting the oscillations of a wave to just one direction in the transverse plane is called the polarization of waves.

Polarization by Scattering :-

When light is incident on the small particles of the atmosphere such as dust, and air molecules it is absorbed by the electrons in the molecules, hence electrons start vibrating. These vibrating electrons emit radiation in all directions except in their own line of vibration. The emitted radiations (light) scattered in a direction perpendicular to the direction of incident light are plane polarized. The light in all other directions is partially polarized.

11. (a) Write the conditions under which light sources can be said to be coherent.
(b) Why is it necessary to have coherent sources in order to produce an interference pattern? 

Answer:
(a) Coherent sources of light: The sources of light, which emit continuously light waves of the same wavelength, same frequency and in the same phase are called Coherent sources of light.
Interference pattern is not obtained. This is because the phase difference between the light waves emitted from two different sodium lamps will change continuously.

(b) Conditions for interference: The important conditions for obtaining interference of light are :

1. The two sources of light must be coherent. i.e. they should exist continuous waves of the same wavelength or frequency.
2. The two sources should be monochromatic.
3. The phase difference of waves from two sources should be constant.
4. The amplitude of waves from two sources should be equal.
5. The coherent sources must be very close to each other.

12. Draw the intensity pattern for single slit diffraction and double slit interference. Hence, state two differences between interference and diffraction patterns. 

Answer:
(i) Intensity distribution in the diffraction due to single slit
                               

(ii) Intensity pattern for double slit interference.
                         

(iii) Difference between Interference and Difference Patterns:
The diagram, given here, shows several fringes, due to double slit interference, ‘contained’ in a broad diffraction peak. When the separation between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.
                                         
Basic features of distinction between interference and diffraction patterns :
(i) The interference pattern has a number of equally spaced bright and dark bands while the diffraction pattern has a central bright maximum which is twice as wide as the other maxima.
(ii) Interference pattern is the superimposition of two wave slits originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
(iii) For a single slit of width ‘a’ the first null of diffraction pattern occurs at an angle of a. At the same angle of a , we get a maxima for two narrow slits separated by a distance ‘a’.

13. In a single slit diffraction experiment, when a tiny circular obstacle is placed in the path of light from a distance source, a bright spot is seen at the center of the shadow of the obstacle. Explain why?
                                                                                           or
State two points of difference between the interference pattern obtained in Young’s double slit experiment and the diffraction pattern due to a single slit.
Answer: Wave diffracted from the edge of any circular obstacle undergoes constructive interference to form a bright spot at the center of shadow.

14. (a) Describe briefly, with the help of a suitable diagram, how the transverse nature of light can be demonstrated by the phenomenon of polarization.
(b) When unpolarized light passes from air to a transparent medium, under what condition does the reflected light get polarized?

Answer:
(a) Light from a source S is allowed to fall normally on the flat surface of a thin plate of a tourmaline crystal, cut parallel to its axis. Only a part of this light is transmitted through A. If now the plate A is rotated, the character of transmitted light remains unchanged. Now another similar plate B is placed at some distance from A such that the axis of B is parallel to that of A. If the light transmitted through A is passed through B, the light is almost completely transmitted through B and no change is observed in the light coming out of B.

             
           
If now the crystal A is kept fixed and B is gradually rotated in its own plane, the intensity of light emerging out of B decreases and becomes zero when the axis of B is perpendicular to that of A. If B is further rotated, the intensity begins to increase and becomes maximum when the axes of A and B are again parallel.

Thus, we see that the intensity of light transmitted through B is maximum when axes of A and B are parallel and minimum when they are at right angles.

From this experiment, it is obvious that light waves are transverse and not longitudinal; because, if they were longitudinal, the rotation of crystal B would not produce any change in the intensity of light.

(b) After falling on a transparent medium, unpolarised light will get polarized after reflection only if refracted and reflected rays make a right angle to each other.

15. A parallel beam of monochromatic light falls normally on a narrow slit of width ‘a’ to produce a diffraction pattern on the screen placed parallel to the plane of the slit. Use Huygens’ principle to explain that
(i) the central bright maxima is twice as wide as the other maxima.
(ii) the intensity falls as we move to successive maxima away from the center on either side.

Answer:
(i) In diffraction pattern, intensity will be minimum at an angle θ = nλ/a
∴ There will be a first minimum at an angle
θ = λ/a, on either side of central maximum
∴ Width of central maxima = 2λ/a
∴ The central bright maxima is twice as wide as the other maxima.

(ii) The intensity of maxima decreases as the order (n) or diffraction maxima increases. This is because, on dividing the slit into an odd number of parts, the contributions of the corresponding (outermost) pairs cancel each other, leaving behind the contribution of only the innermost segment.

For example, for first maximum, dividing split into three parts out of these three parts of the slit, the contributions from first two parts cancel each other; only 1/3rd portion of the slit contributes to the maxima of intensity. Similarly for the second maxima, dividing split into five parts, contribution of the first four parts will be zero (as they cancel each other). The remaining 1/5th portion only will contribute for maxima and so on.

16. State Huygens’s principle. Show, with the help of a suitable diagram, how this principle is used to obtain the diffraction pattern by a single slit. Draw a plot of intensity distribution and explain clearly why the secondary maxima becomes weaker with increasing order (n) of the secondary maxima. 

Answer: Huygen’s principle :

“(i) Each point on a wavefront acts as a fresh source of new disturbance called secondary waves or wavelets;

(ii) The secondary wavelets spread out in all directions with the speed of wave in the given medium;

(iii) The new wavefront at any later time is given by the forward envelope (tangential surface in the forward direction) of the secondary wavelets at that time.

The reason is that the intensity of the central maxima is due to the constructive interference of wavelets from all parts of the slit, the first secondary maxima is due to the contribution of wavelets from one third part of the slit (wavelets from remaining two parts interfere destructively), the second secondary maxima is due to the contribution of wavelets from the one fifth part only (the remaining four parts interfere destructively) and so on. Hence, the intensity’ of secondary maxima decreases with the increase in the order n of the maxima.

17. Describe two commonly used devices which use polarized light.

Answer: The two commonly used devices using polarized light are:
1. Liquid crystal display (L.C.D.)
2. Sun glasses (dark glasses).

1. Liquid crystal display:- They are found in many watches and calculators.
Liquid crystals have long molecules whose direction can be controlled by applying electric fields. This is used to modify the light produced by polariser to make its polarization perpendicular to the axis of an analyser. The analyser cuts these lights and produces dark regions. These dark regions are controlled and are used to form numbers and letters.

2. Sun glasses (dark glasses):- These have polaroids. Glare from sunlight reflection from water can be reduced, if the polaroids cut out horizontally polarized light

18. What will be the effect on interference fringes obtained in Young’s double slit experiment if (i) one slit is covered (ii) a source of light of higher wavelength is used. (iii) distance between two slits will be increased to 1 cm and (iv) distance between screen and double slit is increased.

Answer: The fringe width  is given by
                                  =Dd

where λ is the wavelength of light incident, D is the distance between the screen and double slit and d is the distance between the two slits.

On covering one slit, the interference pattern will be replaced by a diffraction pattern due to a single slit. 

On increasing the wavelength of incident light λ, the fringe width will increase.

On increasing d by 1 cm, the fringe will disappear and there will be uniform illumination.

On increasing D, the fringe width will increase.

19. What is the shape of the wavefront in each of the following cases:

(a) Light diverging from a point source.

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

(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Answer:

(a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

20. What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)

Answer: Refractive index of glass,  =1.5 

Brewster angle = θ

Brewster angle is related to refractive index as:

                                    tanθ =

                           =tan^-1(1.5)=56.31▫

Therefore, the Brewster angle for air to glass transition is 56.31°.

21.You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Answer: Let an object at O be placed in front of a plane mirror MO’ at a distance r (as shown in the given figure).

A circle is drawn from the center (O) such that it just touches the plane mirror at point O’. According to Huygens’ Principle, XY is the wavefront of incident light.

If the mirror is absent, then a similar wavefront X’Y’ (as XY) would form behind O’ at distance r (as shown in the given figure).

can be considered as a virtual reflected ray for the plane mirror. Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).

22. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light traveling in a medium?

Answer: No Sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same.

In case of light waves, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.

23.Answer the following questions:

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

(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle. Explain why?

(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.

(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?

Answer:

(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.

(b) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the center of the shadow. This constructive interference produces a bright spot.

(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.

On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.

(e) The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.

24. In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.

Answer: Consider that a single slit of width d is divided into n smaller slits.

∴    Width of each slit,               d'=dn 

Angle of diffraction is given by the relation,

                            =dd'd=d

Now, each of these infinitesimally small slits sends zero intensity in directionθ. Hence, the combination of these slits will give zero intensity.

25. Distinguish between polarized and unpolarised light. Does the intensity of polarized light emitted by a polaroid depend on its orientation? Explain briefly. The vibrations in a beam of polarized light make an angle of 60° with the axis of the polaroid sheet. What percentage of light is transmitted through the sheet?

Answer: A light which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light. The light from the sun, an incandescent bulb or a candle is unpolarised. If the electric field vector of a light wave vibrates just in one direction perpendicular to the direction of wave propagation, then it is said to be polarized or linearly polarized light.

Yes, the intensity of polarized light emitted by a polaroid depends on the orientation of the polaroid. When polarized light is incident on a polaroid, the resultant intensity of transmitted light varies directly as the square of the cosine of the angle between polarization direction of light and the axis of the polaroid.

                                               I ∝ cos2 θ or I = I0 cos2θ
where I0 = maximum intensity of transmitted light;
θ = angle between vibrations in light and the axis of the polaroid sheet.
or                       

                                                  I =I0 cos2 60° = I04

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