Wave Optics
Maharashtra BoardClass12thPhysicsChapter7
SolutionPart1
Solution : Part1

Question 1.
Choose the correct option.
i) Which of the following phenomenon proves that light is a transverse wave?
(A) reflection
(B) interference
(C) diffraction
(D) polarization
(D) polarization
ii) Which property of light does not change when it travels from one medium to another?
(A) velocity
(B) wavelength
(C) amplitude
(D) frequency
(D) frequency
iii) When unpolarized light is passed through a polarizer, its intensity
(A) increases
(B) decreases
(C) remains unchanged
(D) depends on the orientation of the polarizer
(B) decreases
iv) In Young’s double slit experiment, the two coherent sources have different intensities. If the ratio of maximum intensity to the minimum intensity in the interference pattern produced is 25:1. What was the ratio of intensities of the two sources?
(A) 5:1 (B) 25:1 (C) 3:2 (D) 9:4
(D) 9:4
v) In Young’s double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will satisfy.
(A) The interference pattern will remain unchanged
(B) The fringe width will decrease
(C) The fringe width will increase
(D) The fringes will shift.
(A) The interference pattern will remain unchanged
Question 2.
Answer in brief.
i) What are primary and secondary sources of light?
The Sun, moon, stars, light bulb, etc. are the several sources of light around us, These can be classified into primary and secondary sources of light.
Primaly sources of light :The sources that emit light on their own are called primary sources. This emission of light may be due to
 the high temperature of the source, e.g., the Sun, the stars, objects heated to high temperature, a ﬂame, etc.
 the effect of current being passed through the source, e.g., tubelight, TV, etc.
 chemical or nuclear reactions taking place in the source, e.g., ﬁrecrackers, nuclear energy generators, etc.
Secondary sources of light : Some sources are not self luminous, i.e., they do not emit light on their own, but reflect or scatter the light incident on them. Such sources of light are called secondary sources, e.g. the moon, the planets, objects such as humans, animals, plants, etc. These objects are visible due to reﬂected light.
Many of the sources that we see around are secondary sources and most of them are extended sources.
ii) What is a wavefront? How is it related to rays of light? What is the shape of the wavefront at a point far away from the source of light?
Wavefront or wave surface : The locus of all points where waves starting simultaneously from a source reach at the same instant of time and hence the particles at the points oscillate with the same phase, is called a wavefront or wave surface.
Wavefront related to rays of light :
Consider a point source of light O in a homogeneous isotropic medium in which the speed of light is v. The source emits light in all directions. In time t, the disturbance (light energy) from the source, covers a distance vt in all directions, i.e, it reaches out to all points which are at a distance vt from the point source. The locus of these points which are in the same phase is the surface of a sphere with the centre O and radius vt. It is a spherical wavefront.
In a given medium, a set of straight lines can be drawn which are perpendicular to the wavefront.
According to Huygens, these straight lines are the rays of light. Thus, rays are always normal to the wavefront. In the case of a spherical wavefront, the rays are radial.
Shape of the wavefront at a point far away from the source of light :
If a wavefront has travelled a large distance away from the source, a small portion of this wavefront appears to be plane. This part is a plane wavefront.
iii) Why are multiple colours observed over a thin film of oil floating on water? Explain with the help of a diagram.
Interference due to thin films:
The brilliant colours of thin oil films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film.
The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in the figure.
 The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC.
 At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence θ_{1} , i.e., the angle of incidence at the top surface, which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it.
 The two rays AE and CF interfere, producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.
iv) In Young's double slit experiment what will we observe on the screen when white light is incident on the slits but one slit is covered with a red filter and the other with a violet filter? Give reasons for your answer.
 In Young's doubleslit experiment, when white light is incident on the slits and one of the slit is covered with a red ﬁlter, the light passing through this slit will emerge as the light having red colour.
 The other slit which is covered with a violet ﬁlter, will give light having violet colour as emergent light.
 The interference fringes will involve mixing of red and violet colours.
 At some points, fringes will be red if constructive interference occurs for red colour and destructive interference occurs for violet colour.
 At some points, fringes will be violet if constructive interference occurs for violet colour and destructive interference occurs for red colour.
 The central fringe will be bright with mixing of red and violet colours.
v) Explain, what is optical path length ?. How is it different from actual path length?
Optical path length : Consider, a light wave of angular frequency ω and wave vector k travelling through vacuum along the xdirection. The phase of this wave is (kx—ωt) The speed of light in vacuum is c and that in medium is v
k = \(\frac{2π}{λ}=\frac{2πν}{νλ}=\frac{ω}{v} \) as ω = 2πν and v = νλ, where ν is the frequency of light.
If the wave travels a distance Δx, its phase, changes by ΔΦ = kΔx = ωΔx/v.
Similarly, if the wave is travelling in vacuum,
K = ω/c and ΔΦ = ωΔx/c
Now, consider a wave travelling a distance Δx in the medium, the phase difference generated is,
ΔΦ’=k'Δx = ωnΔx/c = ωΔx’/c …...(1)
where Δx’ = nΔx ... (2)
The distance nΔx is called the optical path length of the light in the medium; it is the distance the light would have travelled in the same time t in vacuum (with the speed c).
The optical path length in a medium is the corresponding path in vacuum that the light travels in the same time as it takes in the given medium.
Now, speed = distance/time
∴ time = distance/speed
∴ t = d_{medium}/v_{medium }= d_{vaccum}/v_{vaccum}
Hence the optical path = d_{vaccum} = \(\frac{v_{vaccum}}{v_{mediuam}}×d_{medium}\) = n x d_{medium}
Thus, a distance d travelled in a medium of refractive index n introduces a path difference = nd — d = d(n — 1) over a ray travelling equal distance through vacuum.
Two waves interfere constructively when their optical path lengths are equal or differ by integral multiples of the wavelength.
Question 3.
Derive the laws of reflection of light using Huygens’ principle.
Consider a plane wavefront AB of monochromatic light incident obliquely at an angle i on a plane mirror MN. The wavefront AB touches the reﬂecting surface MN at A at time t = 0. Let v be the speed of light in the medium.
When wavefront AB is incident on the mirror, at ﬁrst, point A becomes a secondary source and emits secondary waves in the same medium. If T is the time taken by the incident wavefront to travel from B to C, then BC = vT. During this time, the secondary wave originating at A covers the same distance, so that the secondary spherical wavelet has a radius vT at time T.
To construct the reﬂected wavefront, a hemisphere of radius vT is drawn from point Draw a tangent EC to the secondary wavelet.
As all points on EC are in the same phase of wave motion, EC represents the reﬂected wavefront.
The arrow AE shows the direction of propagation of the reﬂected wave.
AP is the normal to MN at A,
∠ RAP = i = angle of incidence and
∠ PAE = r = angle of reﬂection
In Δ ABC and Δ AEC,
AC is common,
AE =BC and ∠ ABC = ∠ AEC = 90°
Δ ABC and Δ AEC are congruent.
∠ ACE = ∠ BAC = i …….(1)
Also, as AE is perpendicular to CE and AP is perpendicular to AC,
∠ ACE = ∠ PAE =r ...(2)
From Eqs (1) and (2),
i = r
Thus, the angle of incidence is equal to the angle of reﬂection. This is the ﬁrst law of reﬂection. Also, it can be seen from the ﬁgure that the incident ray and reﬂected ray lie on the opposite sides of the normal to the reﬂecting surface at the point of incidence and all of them lie in the same plane. This is the second law of reﬂection. Thus, the laws of reﬂection of light can be deduced by Huygens’ construction of a plane wavefront.
Question 4.
Derive the laws of refraction of light using Huygens’ principle.
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle i on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let v_{1} and v_{2} be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser medium) respectively.
When the wavefront reaches MN at point A at t = 0, A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time t = T. Thus, BC = v_{1}T. During the time T, the secondary wavelet originating at A covers a distance AE in the denser medium with radius v_{2}T.
As all the points on CE are in the same phase of wave motion, CE represents the refracted wavefront in the denser medium. CE is the tangent to the secondary wavelet starting from A. It is also a common tangent to all the secondary wavelets emitted by points between A and C. PP’ is the normal to the boundary at A.
∠ A’AP = ∠ BAC = the angle of incidence (i) and
∠ P’AE = ∠ ACE = the angle of refraction (r).
From Δ ABC and Δ AEC,
sin i = \(\frac{BC}{AC}\) and sin r = \(\frac{AE}{AC}\)
∴ \(\frac{sin\,i}{sin\,r}=\frac{BC/AC}{AE/AC}=\frac{BC}{AE}=\frac{v_1T}{v_2T}=\frac{v_1}{v_2}\)
By deﬁnition, the refractive index of medium 2 with respect to medium 1,
_{1}n_{2} = \(\frac{n_2}{n_1}=\frac{v_1}{v_2}\)
∴ \(\frac{n_2}{n_1}=\frac{sin\,i}{sin\,r}\)
n_{1} sin i = n_{2} sin r ….. (1)
Here, n_{1} and n_{2} are the absolute refractive indices of medium 1 and medium 2 respectively. Eq. (1) is Snell's law of refraction. Also, it can be seen from the ﬁgure, that the incident ray and the refracted ray lie on the opposite sides of the normal and all three of them lie in the same plane.
Thus, the laws of refraction of light can be deduced by Huygens’ construction of a plane wavefront.
If v_{1} > v_{2}, i.e. n_{1} < n_{2}, then r < i (bending of the refracted ray towards the normal).
Question 5.
Explain what is meant by polarization and derive Malus’ law.
According to the electromagnetic theory of light, a light wave consists of electric and magnetic ﬁelds vibrating at right angles to each other and to the direction of propagation of the wave. If the vibrations of in a light wave are in all directions perpendicular to the direction of propagation of light, the wave is said to be unpolarized.
If the vibrations of the electric ﬁeld in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric ﬁeld is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be planepolarized or linearly polarized.
This phenomenon of restricting the vibrations of light, i.e., of the electric field vector in a particular direction, which is perpendicular to the direction of the propagation of the wave is called polarization of light.
Consider an unpolarized light wave travelling along the xdirection. Let c, n and λ be the speed, frequency and wavelength, respectively, of the wave. The magnitude of its electric ﬁeld (\(\vec{E}\) ) is,
E = E_{0} sin (kx − ωt), where E_{0} = E_{max} = amplitude of the wave, ω = 2pn = angular frequency of the wave and k = 2p/λ = magnitude of the wave vector or propagation vector.
The intensity of the wave is proportional to E_{0}^{2}.
The direction of the electric ﬁeld can be anywhere in the yz plane. This wave is passed through two identical polarizers as shown in below Fig.
When a wave with its electric ﬁeld inclined at an angle Φ to the axis of the ﬁrst polarizer is passed through the polarizer, the component E_{0} cos Φ will pass through it. The other component E_{0} sin Φ which is perpendicular to it will be blocked.
Now, after passing through this polarizer, the intensity of this wave will be proportional to the square of its amplitude, i.e., proportional to E_{0} cos Φ^{2}.
The intensity of the planepolarized wave emerging from the ﬁrst polarizer can be obtained by averaging E_{0} cos Φ^{2 }over all values of Φ between 0 and 180°. The intensity of the wave will be proportional to ½E_{0}^{2 } as the average value of cos^{2}Φ over this range is Thus the intensity of an unpolarized wave reduces by half after passing through a polarizer.
When the planepolarized wave emerges from the ﬁrst polarizer, let us assume that its electric ﬁeld ( ) is along the ydirection. See Fig . Thus, this electric ﬁeld is,
\(\vec{E}\) =\(\hat{j}\) E_{10} sin (kx—ωt) …..(1)
where, E_{10} is the amplitude of this polarized wave.
The intensity of the polarized wave,
I_{1} ∝ lE_{10}l^{2} …….(2)
Now this wave passes through the second polarizer whose polarization axis (transmission axis) makes an angle θ with the ydirection. This allows only the component E_{10 }cos θ to pass through it.
Thus, the amplitude of the wave which passes through the second polarizer is
E_{20} = E_{10} cos θ and its intensity,
I_{2} ∝ E_{20}^{2}
∴ I_{2} ∝ E_{20}^{2} cos^{2}θ
∴ I_{2} = I_{1 }cos^{2}θ ….(3)
Thus, when planepolarized light of intensity I_{1} is incident on the second identical polarizer, the intensity of light transmitted by the second polarizer varies as cos^{2}θ, i.e., I_{2} = I_{1 }cos^{2}θ, where θ is the angle between the transmission axes of the two polarizers. This is known as Malus’ law.
Question 6.
What is Brewster’s law? Derive the formula for Brewster angle.
Brewster's law : The tangent of the polarizing angle is equal to the refractive index of the reflecting medium with respect to the surrounding (_{1}n_{2}).
If θ_{B} is the polarizing angle,
tan θ_{B} = _{1}n_{2} = \(\frac{n_2}{n_1}\)
Here n_{1} is the absolute refractive index of the surrounding and n2 is that of the reﬂecting medium.
The angle θ_{B} is called the Brewster angle.
Derivation of formula for Brewster angle
Consider a ray of unpolarized monochromatic light incident at an angle θ_{B} on a boundary between two transparent media as shown in Fig. Medium 1 is a rarer medium with refractive index n_{1} and medium 2 is a denser medium with refractive index n_{2}. Part of incident light gets refracted and the rest gets reﬂected. The degree of polarization of the reﬂected ray varies with the angle of incidence.
The electric ﬁeld of the incident wave is in the plane perpendicular to the direction of propagation of incident light. This electric ﬁeld can be resolved into a component parallel to the plane of the paper, shown by double arrows, and a component perpendicular to the plane of the paper shown by dots, both having equal magnitude. Generally, the reﬂected and refracted rays are partially polarized, i.e., the two components do not have equal magnitude.
In 1812, Sir David Brewster discovered that for a particular angle of incidence. θ_{B}, the reﬂected wave is completely planepolarized with its electric ﬁeld perpendicular to the plane of the paper while the refracted wave is partially polarized. This particular angle of incidence (θ_{B}) is called the Brewster angle.
For this angle of incidence, the refracted and reﬂected rays are perpendicular to each other.
For angle of refraction θ_{r},
θ_{B} + θ_{r }= 90° ...(1)
From Snell's law of refraction,
n_{1} sin θ_{B} = n_{2} sin θ_{r} ….(2)
From Eqs. (1) and (2), we have,
n_{1} sin θ_{B} = n_{2} sin(90°— θ_{B}) = n_{2} cos θ_{B}
∴ θ_{B} = tan^{1}\(\frac{n_2}{n_1}\)
This is called Brewster's law.
Question 7.
Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.
Young’s Double Slit Experiment:
In this experiment, a plane wavefront is made to fall on an opaque screen AB having two similar narrow slits S_{1} and S_{2}.
 A plane wavefront is obtained by placing a linear source S of monochromatic light at the focus of a convex lens.
 It is then made to pass through an opaque screen AB having two narrow and similar slits S_{1} and S_{2}.
 S_{1} and S_{2 }are equidistant from S so that the wavefronts starting simultaneously from S and reaching S_{1} and S_{2 }at the same time are in phase.
 A screen PQ is placed at some distance from screen AB as shown in Fig.
 S_{1} and S_{2 }act as secondary sources. The crests/troughs of the secondary wavelets superpose and interfere constructively along straight lines joining the black dots shown in Fig. The point where these lines meet the screen have high intensity and are bright.
 Similarly, there are points shown with red dots where the crest of one wave coincides with the trough of the other. The corresponding points on the screen are dark due to destructive interference.
 These dark and bright regions are called fringes or bands and the whole pattern is called interference pattern.
Conditions for occurrence of dark and bright fringes on the screen :
(Consider Young's double—slit experimental set up.)
Two narrow coherent light sources are obtained by wavefront splitting as monochromatic light of wavelength A emerges out of two narrow and closely spaced, parallel slits S_{1} and S_{2 }of equal widths. The separation S_{1}S_{2} = d is very small. The interference pattern is observed on a screen placed parallel to the plane of S_{1}S_{2} and at considerable distance D (D >> d) from the slits. OO’ is the perpendicular bisector of segment S_{1}S_{2}, Fig.
Consider, a point P on the screen at a distance y from O’ (y << D). The two light waves from S_{1} and S_{2 }reach P along paths S_{1}P and S_{2}P, respectively.
If the path difference (Δl) between S_{1}P and SZP is an integral multiple of λ, the two waves arriving there will interfere constructively producing a bright fringe at P. On the contrary, if the path difference between S_{1}P and S_{2}P is half integral multiple of λ, there will be destructive interference and a dark fringe will be produced at P.
From above Fig.
(S_{2}P)^{2} = (S_{2}S_{2}’)^{2} + (PS_{2}’)^{2}
= (S_{2}S_{2}’)^{2} + (PO’ + O’ S_{2}’)^{2}
= D^{2} + \((y+\frac{d}{2})^2\) ……. (1)
and
(S_{1}P)^{2} =(S_{1}S_{1}’)^{2} + (PS_{1}’)^{2}
=(S_{1}S_{1}’)^{2} + (PO’—O’ S_{1}’)^{2}
= D^{2 }+ \((y\frac{d}{2})^2\) …… (2)
(S_{2}P)^{2} − (S_{1}P)^{2} = \([D^2 + (y\frac{d}{2})^2]\) – \([D^2 +(y\frac{d}{2})^2]\)
∴ (S_{2}P + S_{1}P)(S_{2}P − S_{1}P) = [D^{2} + y^{2} + \(\frac{d^2}{2}\) + yd] – [D^{2} + y^{2} + \(\frac{d^2}{2}\) – yd] = 2yd
∴ S_{2}P − S_{1}P = Δl = 2yd/(S_{2}P + S_{1}P)
In practice, D >> y and D >> d
S_{2}P + S_{1}P ≈ 2D
Path difference,
Δl = (S_{2}P − S_{1}P) ≈ \(2\frac{yd}{2D} = y\frac{d}{D}\) ….. (3)
Expression for the fringe width (or band width) ;
The distance between consecutive bright (or dark) fringes is called the fringe width (or band width) W, Point P will be bright (maximum intensity), if the path difference, Δl = y_{n} \(\frac{d}{D}\) = n λ where n = 0, 1, 2, 3...,
Point P will be dark (minimum intensity equal to zero), if y_{m} \(\frac{d}{D}\) = (2m — 1)\(\frac{λ}{2}\) , where, m = 1, 2, 3...,
Thus, for bright fringes (or bands),
y_{n} = 0, \(\frac{λD}{d}\) , \(\frac{2λD}{d}\) ….
and for dark fringes (or bands),
y_{m} = \(\frac{λD}{2d}\) , \(\frac{3λD}{2d}\), \(\frac{5λD}{2d}\) ….
These conditions show that the bright and dark fringes (or bands) occur alternately and are equally spaced. For Point O’, the path difference (S_{2}O’ — S_{1}O’) = 0. Hence, point O’ will be bright.
It corresponds to the centre of the central bright fringe (or band). On both sides of O’, the interference pattern consists of alternate dark and bright fringes (or band) parallel to the slit.
Let y_{n} and y_{n +1}, be the distances of the nth and (n + 1)th bright fringes from the central bright fringe. .
∴ y_{n} \(\frac{d}{D}\) = n λ , ∴ y_{n} = \(\frac{nλD}{d}\) ….. (4)
And \(\frac{y_{n+1}d}{D}\) = (n+1)λ, ∴ y_{n+1} = \(\frac{(n+1)λD}{d}\) ….. (5)
The distance between consecutive bright fringes
= y_{n+1} − y_{n} = \(\frac{λD}{d}[(n+1)n]\) = \(\frac{λD}{d}\) …….(6)
Hence, the fringe width,
∴ W = Δy = y_{n+1} − y_{n} = \(\frac{λD}{d}\) (fo rbright fringes) ...(7)
Alternately, let y_{m} and y_{m +1 }be the distances of the m th and (m + 1)th dark fringes respectively from the central bright fringe.
∴ \(\frac{y_md}{D}\) = (2m—1)\(\frac{λ}{2}\) and
\(\frac{y_{m+1}d}{D}\)= [2(m+1)—1]\(\frac{λ}{2}\) = (2m+1)\(\frac{λ}{2}\) …. (8)
∴ y_{m} = (2m—1) \(\frac{λD}{2d}\) and y_{m +1} = (2m+1)\(\frac{λD}{2d}\)
The distance between consecutive dark fringes,
y_{m+1}—y_{m} = \(\frac{λD}{2d}\)[(2m+1)(2m1)] = \(\frac{λD}{d}\)...(10)
∴ W = y_{m+1} − y_{m} = \(\frac{λD}{d}\)(for dark fringes) …..(11)
Eqs. (7) and (11) show that the fringe width is the same for bright and dark fringes.
Question 8.
What are the conditions for obtaining good interference pattern? Give reasons.
The following conditions have to be satisfied for the interference pattern to be steady and clearly visible.
(1) The two sources of light should be coherent.
 The waves emitted by two coherent sources should be always in phase or have a constant phase difference between them at all times.
 If the phases and phase difference vary with time, the positions of maxima and minima will also change with time and the interference pattern will not be steady.
 For this reason, it is preferred that the two secondary sources used in the interference experiment must be derived from a single original source.
(2) The two sources of light must be monochromatic :
 The two sources of light must be monochromatic, otherwise, interference will result in complex coloured bands (fringes) because the separation of successive bright bands (fringes) is different for different colours. It also may produce overlapping bands.
(3) The two interfering waves must have the same amplitude :
 The two light sources should be of equal brightness, i.e., the waves must have the same amplitude.
 The interfering light waves should have the same amplitude. Then, the points where the waves meet in opposite phase will be completely dark (zero intensity). This will increase the contrast of the interference pattern and make it more distinct.
(4) The separation between the two light sources must be small in comparison to the distance between the plane containing the slits and the observing screen:
 This is necessary as only in this case, the width of the fringes will be sufficiently large to be measurable and the fringes are well separated and can be clearly seen.
(5) The two light sources should be narrow :
 If the source apertures are wide in comparison with the light wavelength, each source will be equivalent to multiple narrow sources and the superimposed pattern will consist of bright and less bright fringes. That is, the interference pattern will not be well deﬁned.
(6) The interfering light waves should be in the Same state of polarization :
 Otherwise, the points where the waves meet in opposite phase will not be completely dark and the interference pattern will not be distinct.
Question 9.
What is meant by coherent sources? What are the two methods for obtaining coherent sources in the laboratory?
Coherent sources : Two sources which emit waves of the same frequency having a constant phase difference, independent of time, are called coherent sources.
In the laboratory, coherent sources can be obtained by 'using (1) Lloyd's mirror and (2) Fresnel's biprism.
Lloyd's mirror : A plane polished mirror is kept at some distance from the source of monochromatic light and light is made incident on the mirror at a grazing angle.
 Some light falls directly on the screen as shown by the black lines in Fig. while some light falls on the screen after reﬂection from the mirror as shown by red lines.
 The reﬂected light appears to come from a virtual source and thus two sources can be obtained. These two sources are coherent as they are derived from a single source.
 Superposition of the waves coming from these coherent sources, under appropriate conditions, gives rise to interference pattern consisting of alternate bright and dark bands on the screen as shown in the ﬁgure.
Fresnel’s Biprism : It is a single prism having an obtuse angle of about 178° and the other two angles of about 1° each. The biprism can be considered as made of two thin prisms of very small retracting angle of about 1°.
 The source, in the form of an illuminated narrow slit, is aligned parallel to the refracting edge of the biprism.
 Monochromatic light from the source is made to pass through that narrow slit and fall on the biprism. See Fig.
 Two virtual images S_{1} and S_{2} are formed by the two halves of the biprism. These are coherent sources which are obtained from a single secondary source S.
 The two waves coming from S_{1} and S_{2 }interfere under appropriate conditions and form interference fringes, like those obtained in Young's double—slit experiment, as shown in the ﬁgure in the shaded region. The formula for y is the same as in Young's experiment.
Question 10.
What is diffraction of light? How does it differ from interference? What are Fraunhoffer and Fresnel diffractions?
Phenomenon of diffraction of light : When light passes by the edge of an obstacle or through a small opening or a narrow slit and falls on a screen, the principle of rectilinear propagation of light from geometrical optics predicts a sharp shadow. However, it is found that some of the light deviates from its rectilinear path and penetrates into the region of the geometrical shadow. This is a general characteristic of wave phenomena, which occurs whenever a portion of the wavefront is obstructed in some way. This bending of light waves at an edge into the region of geometrical shadow is called diffraction of light.
Differences between interference and diffraction :
 The term interference is used to characterise the superposition of a few coherent waves (say two) But when the superposition at a Point involves a large number of waves coming from different parts of the same wavefront, the effect is referred to as diffraction.
 Doubleslit interference fringes are all of equal width. In singleslit diffraction pattern, only the noncentral maxima are of equal width which is half of that of the central maximum.
 In doubleslit interference, the bright and dark fringes are equally spaced. In diffraction, only the noncentral maxima lie approximately halfway between the minima.
 In doubleslit interference, bright fringes are of equal intensity. In diffraction, successive noncentral maxima decrease rapidly in intensity.
Diffraction can be classiﬁed into two types depending on the distances involved in the experimental setup :
 Fraunhofer diffraction : In this class of diffraction, both the source and the screen are at infinite distances from the aperture. This is achieved by placing the source at the focus of a convex lens and the screen at the focal plane of another convex lens.
 Fresnel diffraction: In this case, the distances are much smaller and the incident wavefront is either cylindrical or spherical depending on the source. A lens is not required to observe the diffraction pattern on the screen.
Question 11.
Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.
Fraunhofer diffraction pattern due to a single slit :
When a parallel beam of monochromatic light of wavelength λ illuminates a single slit of ﬁnite width a, we observe on a screen some distance from the slit, a broad pattern of alternate dark and bright fringes. The pattern consists of a central bright fringe, with successive dark and bright fringes of diminishing intensity on both sides. This is called the diffraction pattern of a single slit.
Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance D ( >> a) from the slit and at the focal plane of the convex lens, see Fig.
Location of minima and maxima: We can imagine the single slit as being made up of a large number of Huygens’ sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens’ wavelets
Fig.  Fraunhofer diffraction due to a single slit
Now, imagine the single slit as made up of two adjacent slits, each of width a/2. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase.
They will therefore also in phase at the point P_{0} on the screen, where P_{0} is equidistant from all the Huygens sources. At P_{0}, then, we get the central maximum.
Expressions for the positions of the intensity minima and maxima :
For the ﬁrst minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and O (or O and B) is λ/2, which is the condition for destructive interference.
Suppose, the nodal line OP for the ﬁrst minimum subtends an angle θ at the slit; θ is very small. With P as the centre and PA as radius, strike an arc intersecting PB at C. Since, D >> a, the arc AC can be considered a straight line at right angles to PB.
Then, Δ ABC is a rightangled triangle similar to Δ OP_{0}P.
This means that, ∠ BAC = 0
∴ BC = a sin θ
Difference in path length,
∴ BC = PB − PA = (PB—PO) + (PO—PA) = \(\frac{λ}{2}+\frac{λ}{2}\) = λ
∴ a sin θ = λ
∴ sin θ ≈ θ = λ/a …....(1) ….('.' θ is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the m^{th} minimum (m = ±1, ±2, ±3, ...).
θ_{m} = mλ/a ...(m^{th} minimum) …..(2)
as θ_{m} is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an oddintegral multiple of λ/2
a sin θ_{m} = (2m+1)\(\frac{λ}{2}\) = (m+\(\frac{1}{2}\) ) λ
i.e., at angles given by,
θ_{m} ≈ sin θ_{m} = (m+\(\frac{1}{2}\))\(\frac{λ}{a}\) (m^{th} secondary maximum)
Question 12.
Describe Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.
Rayleigh’s Criterion for Limit of Resolution (or for Resolving Power):
Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the ﬁrst minimum of the other pattern.
The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the ﬁrst minimum. It gives the limit of resolution.
 Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution [above Fig (a)].
 They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution [Fig.(b)].
 They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution [Fig. (c)].
Resolving Power of a Microscope:
 The limit of resolution of a microscope is the least separation between twoclosely spaced points on an object which are just resolved when viewed through the microscope.
 The resolving power of a microscope is deﬁned to be the reciprocal of its limit of resolution.
 In a compound microscope, the objective lens forms a real, magniﬁed image of an object placed just beyond the focal length of the lens. The objective has a short focal length (for greater magniﬁcation) and is held close to the object so that it gathers as much of the light scattered by the object as possible.
Resolving Power of a Telescope:
 Telescopes are normally used to see distant stars. For us, these stars are like luminous point objects, and are far off.
 The resolving power of a telescope is deﬁned as the reciprocal of the angular limit of resolution between two closelyspaced distant objects so that they are just resolved when seen through the telescope.
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