Physics INTERFERENCE OF LIGHT WAVES AND YOUNG’S EXPERIMENT, DIFFRACTION FOR CBSE NCERT -3

### Topic covered

color{blue}{star} INTERFERENCE OF LIGHT WAVES AND YOUNG’S EXPERIMENT
color{blue}{star} DIFFRACTION

### INTERFERENCE OF LIGHT WAVES AND YOUNG’S EXPERIMENT

color{blue} ✍️We will now discuss interference using light waves. If we use two sodium lamps illuminating two pinholes (Fig. 10.11) we will not observe any interference fringes.

color{blue} ✍️This is because of the fact that the light wave emitted from an ordinary source (like a sodium lamp) undergoes abrupt phase changes in times of the order of 10–10 seconds.

color{blue} ✍️Thus the light waves coming out from two independent sources of light will not have any fixed phase relationship and would be incoherent, when this happens, as discussed in the previous section, the intensities on the screen will add up.

color{blue} ✍️The British physicist Thomas Young used an ingenious technique to “lock” the phases of the waves emanating from S_1 and S_2. He made two pinholes S_1 and S_2 (very close to each other) on an opaque screen [Fig. 10.12(a)]. These were illuminated by another pinholes that was in turn, lit by a bright source.

color{blue} ✍️Light waves spread out from S and fall on both S_1 and S_2. S_1 and S_2 then behave like two coherent sources because light waves coming out from S_1 and S_2 are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S_1 and S_2.

color{blue} ✍️Thus, the two sources S_1 and S_2 will be locked in phase; i.e., they will be coherent like the two vibrating needle in our water wave example [Fig. 10.8(a)].

Thus spherical waves emanating from S_1 and S_2 will produce interference fringes on the screen GG′, as shown in Fig. 10.12(b).
The positions of maximum and minimum intensities can be calculated by using the analysis given in Section 10.4 where we had shown that for an arbitrary point P on the line GG′ [Fig. 10.12(b)] to correspond to a maximum, we must have

color {blue}{S_2P – S_1P = nλ; n = 0, 1, 2 }

..............(10.15)

color{purple}{(S_2P)^2 - (S_1P)^2 = [D^2 + (x+d/2)^2 ] - [D^2+ (x - d/2)^2]= 2xd}

where color{purple}{S_1S_2 = d" and "OP = x }. Thus

color {blue}{S_2P – S_1P = (2xd)/(S_2P+S_1P)}

..............(10.16)

color{blue} ✍️If x, d<
color{purple}{S_2P + S_1P = [(100)^2 + (1.05)2]^(½) + [(100)^2 + (0.95)2]^½ ≈ 200.01 cm}

color{blue} ✍️Thus if we replace S_2P + S_1P by 2 D, the error involved is about 0.005%. In this approximation, Eq. (10.16) becomes

color {blue}{S_2P – S_1P ≈(2xd)/200.01}

...........(10.17)

color{blue} ✍️Hence we will have constructive interference resulting in a bright region when

color {blue}{x = x_n = (nλD)/d ; n = 0, ± 1, ± 2, }

............(10.18)

color{blue} ✍️On the other hand, we will have a dark region near

color {blue}{x = x_n = (n+1) (λD)/d ; n= 0, ± 1, ± 2, }

............(10.19)

color{blue} ✍️Thus dark and bright bands appear on the screen, as shown in Fig. 10.13.

color{blue} ✍️Such bands are called fringes. Equations (10.18) and (10.19) show that dark and bright fringes are equally spaced and the distance between two consecutive bright and dark fringes is given by

color {blue}{beta= X_(n+1) -X_n " or "beta= (λD)/d}

.............(10.20)

color{blue} ✍️which is the expression for the fringe width. Obviously, the central point O (in Fig. 10.12) will be bright because S_1O = S_2O and it will correspond to n = 0.

color{blue} ✍️If we consider the line perpendicular to the plane of the paper and passing through O [i.e., along the y-axis] then all points on this line will be equidistant from S_1 and S_2 and we will have a bright central fringe which is a straight line as shown in Fig. 10.13.

color{blue} ✍️ In order to determine the shape of the interference pattern on the screen we note that a particular fringe would correspond to the locus of points with a constant value of S_2P – S_1P. Whenever this constant is an integral multiple of λ, the fringe will be bright and whenever it is an odd integral multiple of λ//2 it will be a dark fringe.

color{blue} ✍️Now, the locus of the point P lying in the x-y plane such that S_2P – S_1P (= Δ) is a constant, is a hyperbola. Thus the fringe pattern will strictly be a hyperbola; however, if the distance D is very large compared to the fringe width, the fringes will be very nearly straight lines as shown in Fig. 10.13.

color{blue} ✍️In the double-slit experiment shown in Fig. 10.12, we have taken the source hole S on the perpendicular bisector of the two slits, which is shown as the line SO. What happens if the source S is slightly away from the perpendicular bisector.

color{blue} ✍️Consider that the source is moved to some new point S′ and suppose that Q is the mid-point of S_1 and S_2. If the angle S′QS is φ, then the central bright fringe occurs at an angle –φ, on the other side.

color{blue} ✍️Thus, if the source S is on the perpendicular bisector, then the central fringe occurs at O, also on the perpendicular bisector. If S is shifted by an angle φ to point S′, then the central fringe appears at a point O′ at an angle –φ, which means that it is shifted by the same angle on the other side of the bisector.

color{blue} ✍️This also means that the source S′, the mid-point Q and the point O′ of the central fringe are in a straight line. We end this section by quoting from the Nobel lecture of Dennis Gabor* The wave nature of light was demonstrated convincingly for the first time in 1801 by Thomas Young by a wonderfully simple experiment.

color{blue} ✍️He let a ray of sunlight into a dark room, placed a dark screen in front of it, pierced with two small pinholes, and beyond this, at some distance, a white screen.

color{blue} ✍️He then saw two darkish lines at both sides of a bright line, which gave him sufficient encouragement to repeat the experiment, this time with spirit flame as light source, with a little salt in it to produce the bright yellow sodium light.

color{blue} ✍️This time he saw a number of dark lines, regularly spaced; the first clear proof that light added to light can produce darkness. This phenomenon is called We should mention here that the fringes are straight lines although S_1 and S_2 are point sources. If we had slits instead of the point sources (Fig. 10.14), each pair of points would have produced straight line fringes resulting in straight line fringes with increased intensities.

Q 3138578402

Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation when bluegreen light of wavelength 500 nm is used?
Class 12 Chapter 10 Example 3
Solution:

Fringe spacing = (Dλ)/d = (1xx5xx10^(-7))/(2xx10^(-3)

= 5 × 10–4 m = 0.5 mm
Q 3148578403

What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations:

(a) the screen is moved away from the plane of the slits;
(b) the (monochromatic) source is replaced by another
(monochromatic) source of shorter wavelength;
(c) the separation between the two slits is increased;
(d) the source slit is moved closer to the double-slit plane;
(e) the width of the source slit is increased;
(f ) the monochromatic source is replaced by a source of white light?
(In each operation, take all parameters, other than the one specified, to remain unchanged.)
Class 12 Chapter 10 Example 4
Solution:

(a) Angular separation of the fringes remains constant
(= λ//d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the two slits.
(b) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below.
(c) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below.
(d) Let s be the size of the source and S its distance from the plane of the two slits. For interference fringes to be seen, the condition s//S < λ//d should be satisfied; otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as S decreases (i.e., the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed.
(e) Same as in (d). As the source slit width increases, fringe pattern gets less and less sharp. When the source slit is so wide that the condition s//S ≤ λ//d is not satisfied, the interference pattern disappears.
(f ) The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is white. For a point P for which S_2P –S_1P = λb//2, where λb (≈ 4000 Å) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away where S_2Q–S_1Q = λb = λr//2 where λr (≈ 8000 Å) is the wavelength for the red colour, the fringe will be predominantly blue.
Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

### DIFFRACTION

color{blue} ✍️If we look clearly at the shadow cast by an opaque object, close to the region of geometrical shadow, there are alternate dark and bright regions just like in interference.

color{blue} ✍️This happens due to the phenomenon of diffraction. Diffraction is a general characteristic exhibited by all types of waves, be it sound waves, light waves, water waves or matter waves.

color{blue} ✍️Since the wavelength of light is much smaller than the dimensions of most obstacles; we do not encounter diffraction effects of light in everyday observations.

color{blue} ✍️However, the finite resolution of our eye or of optical instruments such as telescopes or microscopes is limited due to the phenomenon of diffraction. Indeed the colours that you see when a CD is viewed is due to diffraction effects. We will now discuss the phenomenon of diffraction.

color{brown} {bbul{"The single slit"}}
color{blue} ✍️In the discussion of Young’s experiment, we stated that a single narrow slit acts as a new source from which light spreads out. Even before Young, early experimenters – including Newton – had noticed that light spreads out from narrow holes and slits.

color{blue} ✍️It seems to turn around corners and enter regions where we would expect a shadow. These effects, known as diffraction, can only be properly understood using wave ideas. After all, you are hardly surprised to hear sound waves from someone talking around a corner.

color{blue} ✍️When the double slit in Young’s experiment is replaced by a single narrow slit (illuminated by a monochromatic source), a broad pattern with a central bright region is seen. On both sides, there are alternate dark and bright regions, the intensity becoming weaker away from the centre (Fig. 10.16).

color{blue} ✍️To understand this, go to Fig. 10.15, which shows a parallel beam of light falling normally on a single slit LN of width a. The diffracted light goes on to meet a screen. The midpoint of the slit is M. A straight line through M perpendicular to the slit plane meets the screen at C.

color{blue} ✍️We want the intensity at any point P on the screen. As before, straight lines joining P to the different points L,M,N, etc., can be treated as parallel, making an angle θ with the normal MC. The basic idea is to divide the slit into much smaller parts, and add their contributions at P with the proper phase differences.

color{blue} ✍️We are treating different parts of the wavefront at the slit as secondary sources. Because the incoming wavefront is parallel to the plane of the slit, these sources are in phase. The path difference NP – LP between the two edges of the slit can be calculated exactly as for Young’s experiment. From Fig. 10.15,

color{purple}{NP – LP = NQ}

color{purple}{= a sin θ}

color {blue}{≈ aθ}

.............(10.21)
color{blue} ✍️Similarly, if two points M_1 and M_2 in the slit plane are separated by y, the path difference M_2P – M_1P ≈ yθ.

color{blue} ✍️We now have to sum up equal, coherent contributions from a large number of sources, each with a different phase. This calculation was made by Fresnel using integral calculus, so we omit it here. The main features of the diffraction pattern can be understood by simple arguments.

color{blue} ✍️At the central point C on the screen, the angle θ is zero. All path differences are zero and hence all the parts of the slit contribute in phase. This gives maximum intensity at C. Experimental observation shown in Fig. 10.15 indicates that the intensity has a central maximum at θ = 0 and other secondary maxima at θ l (n+1/2) λ//a, and has minima (zero intensity) at θ approx nλ//a, n = ±1, ±2, ±3, .... It is easy to see why it has minima at these values of angle. Consider first the angle θ where the path difference aθ is λ. Then,

color {blue}{θ ≈ λ /a}

............(10.22)

color{blue} ✍️, divide the slit into two equal halves LM and MN each of size a/2. For every point M_1 in LM, there is a point M_2 in MN such that M_1M_2 = a//2. The path difference between M_1 and M_2 at P = M_2P – M_1P = θ a//2 = λ//2 for the angle chosen.

color{blue} ✍️This means that the contributions from M_1 and M_2 are 180º out of phase and cancel in the direction θ = λ/a. Contributions from the two halves of the slit LM and MN, therefore, cancel each other. Equation (10.22) gives the angle at which the intensity falls to zero.

color{blue} ✍️One can similarly show that the intensity is zero for θ = nλ/a, with n being any integer (except zero!). Notice that the angular size of the central maximum increases when the slit width a decreases.

color{blue} ✍️It is also easy to see why there are maxima at θ = (n + 1/2) λ//a and why they go on becoming weaker and weaker with increasing n. Consider an angle θ = 3λ//2a which is midway between two of the dark fringes. Divide the slit into three equal parts. If we take the first two thirds of the slit, the path difference between the two ends would be

color {blue}{2/3 a xx theta = (2a)/3 xx (3λ)/(2a)=λ}

............(10.23)

color{blue} ✍️The first two-thirds of the slit can therefore be divided into two halves which have a λ/2 path difference.
The contributions of these two halves cancel in the same manner as described earlier. Only the remaining one-third of the slit contributes to the intensity at a point between the two minima.

color{blue} ✍️Clearly, this will be much weaker than the central maximum (where the entire slit contributes in phase). One can similarly show that there are maxima at color{purple}{(n + 1/2) θ//a} with color{purple}{n = 2, 3,} etc.

color{blue} ✍️These become weaker with increasing n, since only one-fifth, one-seventh, etc., of the slit contributes in these cases. The photograph and intensity pattern corresponding to it is shown in Fig. 10.16.

color{blue} ✍️There has been prolonged discussion about difference between intereference and diffraction among scientists since the discovery of these phenomena. In this context, it is interesting to note what Richard Feynman* has said in his famous Feynman Lectures on Physics:

color{blue} ✍️No one has ever been able to define the difference between interference and diffraction satisfactorily. It is just a question of usage, and there is no specific, important physical difference between them.

color{blue} ✍️The best we can do is, roughly speaking, is to say that when there are only a few sources, say two interfering sources, then the result is usually called interference, but if there is a large number of them, it seems that the word diffraction is more often used.

color{blue} ✍️In the double-slit experiment, we must note that the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole, and the double-slit interference pattern. This is shown in Fig. 10.17.

color{blue} ✍️ It shows a broader diffraction peak in which there appear several fringes of smaller width due to double-slit interference. The number of interference fringes occuring in the broad diffraction peak depends on the ratio d/a, that is the ratio of the distance between the two slits to the width of a slit. In the limit of a becoming very small, the diffraction pattern will become very flat and we will obsrve the two-slit interference pattern [see Fig. 10.13(b)].

color{blue} ✍️In the double-slit interference experiment of Fig. 10.12, what happens if we close one slit? You will see that it now amounts to a single slit. But you will have to take care of some shift in the pattern.

color{blue} ✍️We now have a source at S, and only one hole (or slit) S_1 or S_2. This will produce a single- slit diffraction pattern on the screen.

color{blue} ✍️The centre of the central bright fringe will appear at a point which lies on the straight line SS_1 or SS_2, as the case may be. We now compare and contrast the interference pattern with that seen for a coherently illuminated single slit (usually called the single slit diffraction pattern).

color{blue} {(i)} The interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima. The intensity falls as we go to successive maxima away from the centre, on either side.

color{blue} {(ii)} We calculate the interference pattern by superposing two waves originating from the two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.

color{blue} {(iii)} For a single slit of width a, the first null of the interference pattern occurs at an angle of λ/a. At the same angle of λ/a, we get a maximum (not a null) for two narrow slits separated by a distance a.

color{blue} ✍️One must understand that both d and a have to be quite small, to be able to observe good interference and diffraction patterns. For example, the separation d between the two slits must be of the order of a milimetre or so.

color{blue} ✍️The width a of each slit must be even smaller, of the order of 0.1 or 0.2 mm. In our discussion of Young’s experiment and the single-slit diffraction, we have assumed that the screen on which the fringes are formed is at a large distance.

color{blue} ✍️The two or more paths from the slits to the screen were treated as parallel. This situation also occurs when we place a converging lens after the slits and place the screen at the focus. Parallel paths from the slit are combined at a single point on the screen.

color{brown} {"Note"} that the lens does not introduce any extra path differences in a parallel beam. This arrangement is often used since it gives more intensity than placing the screen far away. If f is the focal length of the lens, then we can easily work out the size of the central bright maximum. In terms of angles, the separation of the central maximum from the first null of the diffraction pattern is λ//a. Hence, the size on the screen will be fλ/a.
Q 3158578404

In Example 10.3, what should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern?
Class 12 Chapter 10 Example 5
Solution:

We want atheta = λ,theta= λ/a

10 λ/d = 2 λ/a a= d/5 0.2 mm`

Notice that the wavelength of light and distance of the screen do not enter in the calculation of a.