light Chapter Type equation here. Important long questions

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1 Type equation here. Light Chapter 9 Important long questions Q.9.1 Describe Young s double slit experiment for the demonstration of interference of. Derive an expression for fringe spacing? Ans. Young s Double Slit Experiment Interference of can be described by young s double slit experiment. The experimental arrangement is shown in fig (a) A beam of monochromatic is incident on a screen having two narrow slits S1 and S. The portion of the wave front falls on the slits behave as a source of secondary wavelets (Huygen principle). The secondary wavelets leaving the slits are coherent. Superposition (lying on one another) of these wavelets produce a s eries of bright and dark fringes (or band) which are observed on a second screen placed at some distance parallel to the first screen. If the wavelets arrive at the screen in such a ay that at some points crests fall on crests and troughs on troughs then a bright fringe is seen on the screen due to constructive interference. There are some points on the screen where crests meet troughs then a dark fringe is seen on the screen due to destructive interference. The bright fringes are called maxima and dark fringes as minima. Prof. Muhammad Amin 1

2 Path Difference Consider a point P on the screen at which we want to see the interference of two wavelets coming from the slits A and B. The separation between the two slits is AB = d. P A y d C O D B L A screen is placed at distance L from the slits and Y is the distance of P from the centre of the screen O.The wavelet coming from B to point P on the screen covers a longer distance than the wavelet coming from A. A perpendicular AD is drawn from A to BP. Then AP = DP.The angle between CP and CO is. As the angle between two lines is equal to the angle between their perpendiculars. Therefore the angle between AD and AB is also.the path difference between the wavelets, leaving the slits and arriving at P is BP AP = BD From ABD, Or For the position of bright fringe, From equation () and (3), we get For the position of dark fringe, BD Sinθ= AB BD Sinθ = (1) d BD = d sinθ () BD = m (3) Where m = 0,1,, m = dsin (4) BD = Where m = 0,1,,3, Thus 1 BD = m+ λ 1 dsinθ = m+ λ (5) From POC, PO tanθ= CO Y tanθ= L Prof. Muhammad Amin

3 If the angle is small then Sinθ tanθ Y Sinθ = L If bright fringe is seen at P then by using Y mλ = d L Sinθ= Y L in equation (4), we have Or Lλ Y = m ( 6 ) d If dark fringe is seen at P then by using Fringe Spacing Sinθ= Y L Y 1 d = m+ λ L 1Lλ Y = m+ (7) d in equation (5), we get The distance between any two consecutive bright or dark fringes is called fringe spacing λl F.S = d Let us find the distance between two adjacent bright fringes on the screen. Consider the position of mth and (m+1)th bright fringes. For the mth bright fringe, Lλ Y m = m d For (m+1)th bright fringe Lλ Y m+1 = m+1 d If the distance between adjacent fringes is y (fringe spacing) then Δy = Y Similarly this is true for dark fringes. Y m+1 m Lλ Lλ Δy = m+1 m d d λl ΔY = d Q.9. What are Newton s rings? Describe the experimental arrangement for producing Newton s rings? Ans. Newton s Rings: Circular dark and bright interfere interference fringes produced by Newton s are called Newton s rings. Experimental Arrangement: A plan convex lens L of long focal length is placed on a plane glass plate P, a thin air film is enclosed between.upper surface of the glass plate P and the lower surface of the lens L. Prof. Muhammad Amin 3

4 The thickness of air film is zero at the point of contact o between the len s L ad plate P. the thickness of air film increases from the centre to the edge of the lens as shown in fig (a). Light rays from a monochromatic source S becomes parallel after passing through the convex lens L1. This beam of falls on the plate G (beam splitter) which reflects some rays normally towards the air film. Light rays are reflected by the top and bottom surfaces of the enclosed air film. These rays interfere each other and circular fringes are produced as shown in fig (b). As the air gap increases in width from central contact point outward to the edges, the extra path length for the lower ray varies λ 3λ i.e., λ, λ and so on. These values of path length correspond to constructive and destructive interference. Thus alternate dark and bright circular sings are observed. Why is the central spot black? At the point of contact of the lens and the glass plate, the thickness of the air film is zero but due to reflection at the lower surface of air film from denser medium, an additional path difference of λ is produced. Therefore, the centre of Newton rings is dark due to destructive interference. Q.9.3 What is Michelson s interferometer? Describe its principle, construction and working. How can you find the wave length used? Ans. Michelson s Interferometer: It is a device which divides the into parts and then recombines them to produce interference. Principle:-It is based on the interference of rays. Construction: Prof. Muhammad Amin 4

5 It consists of two plane mirrors M1 and M the mirror M1 is movable and the mirror M is fixed. A monochromatic source of S which emits waves of the same wave length, ray. G1 is a half silvered glass plate which is placed at angle of 45 o to the incident beam coming from the source S. G is another glass plate which is equal in thickness to G1 but not silvered. This glass plate G is placed in the path of the horizontal ray in order to equalize the path length of the two beams and s called compensator plate. This arrangement each ray will the pass through the equal thickness of plate. Working When a monochromatic beam of falls on a half silvered glass plate G 1 that partially reflects it and partially transmits it. The reflected part I travels a distance L 1 to mirror M1, which reflects the beam back towards G 1. The half silvered plate G1 partially transmits this portion that finally arrives at the observer s eye. The other transmitted part II travels a distance L to the mirror M which reflects the beam back towards G1. The part II partially reflected by G 1 also arrives the observer s eye finally. The two beams having their different paths are coherent. When they arrive at observer s eye, produce interference effects. The observer then sees a series of parallel interference fringes as shown in fig. Let initially a bright fringe is visible. Now if M1 is moved backward through a distance of λ 4, then the beam I will cover a distance λ λ λ or more than the other ray no II. 4 4 Therefore, dark fringe will be seen due to destructive interference. Thus by moving M1, bright and dark fringes can be seen if mirror is further moved by λ then again bright 4 fringe is produced. If L is the distance moved by the mirror M1 and m be the number of fringes passes through a reference point then m L λ 1 or L = mλ By this formula, we can calculate the wavelength of.the motion of mirror M 1 by λ 4 produces a clear difference between brightness and darkness. Michelson measured the length of standard metre in terms of the wavelength of red cadmium and showed that the standard metre was equivalent to 1,553, wavelengths of this. Prof. Muhammad Amin 5

6 Q.9.4 What is diffraction grating? Derive a grating equation to find the wave length of used? Ans. Diffraction Grating: (Grating Plate) It is a rectangular glass plate having a large number of equally spaced non transparent parallel lines while the spacing between them is transparent. Grading element Gradient element is the adjacent distance b/w the two lines on the diffraction grating. i.e d = length of grating = L no.of rulings N Grating Equation c Fig (a) c Fig (b) Consider a parallel beam of monochromatic fails on the grating plate and diffracted through it as shown in fig. The distance between two adjacent slits is d, called grating element. Its value is obtained by dividing the length L of the grating plate by the total number N of the lines ruled on it. The parts of wave front that pass through the slits behaves as sources of secondary wavelets according to Huygen s principle. Consider the parallel rays which after diffraction through the grating plate make an angle with AB, the normal to grating. They are brought to focus convex lens. If the path difference between rays no. 1 and is one wave length, they will reinforce each other at P and show constructive interference. Path difference = ab For constructive interference condition, the path difference betwee n two consecutive rays should be equal to, i.e. From right angled triangle acb, ab = (1) ab = sinθ ac Prof. Muhammad Amin 6

7 ab = ac Sinθ ac = d ab = d sinθ () Putting the value of ab in equation (1) d Sin θ=λ (3) If = o o, then path difference = o So we will get a bright fringe. This is known as zero order image formed by the grating. If we increase on either side of the direction, a value of will be arrived at which d Sin =, and we will again get a bright image. In this way if we continue increasing, we will get the second, third, etc. images on either side of the zero order image according to d sin becoming equal to, 3, etc. Thus equation no (3) can be written in general form as d Sin = n where n = o, 1,,.. Q.9.5 Explain the diffraction of x rays by crystal and derive Bragg s law or Bragg s equation. Ans. Diffraction of x-rays by Crystal. X rays are similar to but having much shorter wavelength, of the order of m. Their frequency is very high. Therefore, x-rays are not diffracted by a typical diffraction grating. In 1914 W.H Bragg & W.L Bragg discovered that the diffraction of X rays is possible with crystal. consider a beam of X rays falls on a crystal in which the distance between two consecutive layers of atoms is d. d for aluminum is 4.04 A o. Consider two rays I and II. The ray I is reflected from the atom A of the top layer and ray II is reflected from the atom A of the top layer and ray II is reflected from th atom c of second layer as shown in fig. the ray II covers a longer distance than ray I. To find path difference between the rays I and II, draw perpendiculars AB and AB. Path difference = BC + CB (1) According to constructive interference condition, the path difference between two rays should be equal to m. i.e. BC + CB = m () where m = 1, 3, 4, and is the wave length of. From triangle BC ABC, =Sinθ AC Prof. Muhammad Amin 7

8 or or BC = AC Sinθ BC = d Sinθ (3) Where is the angle which the incident beam makes with crystal surface. It is called glancing angle. From triangle CB AB C, =Sinθ AC or or BC = AC Sinθ BC = d Sinθ (4) Using equation (3) and (4) in equation () dsinθ + dsinθ= mλ Or.dSinθ = mλ (5) This is called Bragg s equation or Bragg s law for X-rays diffraction Prof. Muhammad Amin 8

9 Important short questions 9.1 Under what conditions two or more sources of behave as coherent sources? Ans. Condition for coherent sources Two sources of said to be coherent sources if:- The sources emits the waves of same wavelength i.e monochromatic The waves emitted by the sources must have constant phase difference. 9. How is the distance between interference fringes affected by the separation between the slits of Young s experiment? Can fringes disappear? Ans. The fringe spacing is given by λl Δy = d This relations shows that the fringe spacing varies inversely with the separation d of the slits. If d decreases the fringes get wider and when d increases the fringes become closer. When the slit spacing is large the fringes will be so closed that these cannot be distinguished. Also if slit so is opened extremely wide the fringes disappear. 9.3 Can visible (white ) produce interference fringes Explain? Ans. yes! The interference of visible (white ) is possible Explanation:- White consists of seven colors. Each spectral color will produce its own interference pattern. Hence pattern on the screen will not be observable since colors overlap each others. 9.4 In the young s experiment one of the slits is covered with blue filter and other with red filter. What would be the pattern of intensity on the screen? Ans. No, interference will be observed on the screen Explanation For interference the two sources should emit of same wavelength i.e. monochromatic otherwise will be distributed non-uniformly on the screen. The interference pattern is not well defined. 9.5 Explain whether the young s experiment is an experiment for studying interference or diffraction effects of. Ans. Basically young s double slit experiment is designed to study interference of. Explanation:= Young s experiment is actually designed to study the interference of. But this is combination of interference and diffraction. Light is diffracted from a single slits aga in diffracted from the two slits and produce interference. 9.6 An oil film spreading over a wet footpath shows colours. Explain how does it happen? Ans. The oil film behaves as a thin film. When white falls on this film. At a place the angle of incidence and thickness of film is such that the condition for destructive Prof. Muhammad Amin 9

10 interference for one of the seven colours is satisfied. This colour becomes absent and remaining colours can be seen. 9.7 Could you obtain Newton s rings with transmitted? If yes would the pattern de different from that obtained with reflected? Ans. yes! The Newton s rings can be obtained with transmitted Pattern In case of transmitted the fringe pattern is just opposite to the reflected fringe pattern. With the transmitted the central spot in Newton s rings will appear bright. 9.8 In the white spectrum obtained with a diffraction grating the third order image of wavelength coincides the fourth order mage of a second wavelength. Calculate the ration of the two wavelengths. Ans. For one wave length m = 3 = 1 By grating equation dsinθ = mλ dsinθ =3λ (1) For the second wavelength m = 4 = Dividing equation (1) by () 1 So dsinθ =4λ () dsinθ = dsinθ 3λ1 Or 1= 4λ Or λ 1 = 3 1 λ 4 λ1 4 Or = λ 3 3λ 4λ = Prof. Muhammad Amin 10

11 9.9 How would you manage to get more orders of spectra using a diffraction grating? Ans. Since the equation for the diffraction grating Where d = grating element i.e d = d Sin θ=λ length of grating = L no.of rulings N In order to get more orders of spectra we should increase the grating element That can be done by Decreasing the number of lines on the grating. increasing the length of grating decreasing wavelength of used 9.10 Why the polaroid sunglasses are better than ordinary sunglasses? Ans. The Polaroid sunglasses produces plane polarized. The plane polarized is better than un-polarized. The other advantages of polarized glass are it reduced glare it protect eyes from bright rays of sun it blocks harmful glares only How would you distinguish between un-polarized and plane-polarized s? Ans. Un-polarized Light A which have all possible planes of vibration such a is called un-polarized. Plane Polarized Light The beam of in which all vibrations are forced to vibrate along one plane is called plane polarized. 9.1 What are conditions of detectable Interference? Ans. Conditions for detectable Interference are (i) The interfering beams must be monochromatic. i.e having single wavelength (ii) The interfering beams must be coherent i.e having constant phase diffrence (iii) The interfering beams must be close to each other Why the centre of Newton s rings is dark? Ans. At the point of the contact of the lens and the glass plate, the thickness of the film is effective zero but due to reflection at the lower surface of air film from the denser medium an additional path difference Newton s rings is dark. is introduced. Consequently, the centre of Prof. Muhammad Amin 11

12 9.14 What are different methods for obtaining the polarized? Ans. Polarized can be obtained by following methods a. Selective absorption b. Reflection from different surfaces. c. Refraction through crystals. d. Scattering by small particles 9.15 What is optical rotation? Ans. Certain crystals and liquids when placed between Polaroids rotate the plane of polarization of these are known as optical active substance e.g Sugar Quartz Tartaric acid Sodium chlorate Sugar solution rotates the plane of polarization of incident so that it is no longer horizontal but at an angle as shown in fig. This property of optical rotation can be used to determine concentration in solutions What is grating element? Ans. The distance between two adjacent slits d is called grating element. Its value is obtained by dividing the length L of the grating by the total number of lines ruled on it. Grating element = d = d = d = d = Length of grating No. of lines ruled onit L N 1cm No. of lines / cm 1 N 9.17 What is Bragg s equation? Describe its applications? Ans. The Bragg s equation is expressed as d Sin = n n=1,,3 is called order of bright fringes. Applications It can be used to find the interplaner spacing d between parallel planes of crystals. X-rays diffraction is used to find the structure of biologically molecules such as hemoglobin. It is also used to find the structure of solids 9.18 What is the diffraction of? Ans. The property of bending of around obstacles and spreading of waves into the geometrical shadow of an obstacle is called diffraction. Prof. Muhammad Amin 1

13 Example Consider a small and smooth steel ball of about 3mm in dia meter is illuminated by a point source of S.The shadow of the ball is received on a screen as shown in fig. S The figure shows that the shadow of the ball is not completely dark but its central part is bright due to diffraction What is difference between interference and diffraction? Ans. Interference Diffraction Interference of is defined as a phenomenon of super position of two Diffraction of is defined as the phenomenon of bending of a coherent waves of same around the edges of a opening or amplitude moving in a medium at same time and in same direction. Interference fringes are of the same intensity. obstacle placed in its path Diffraction fringes are not of the same intensity. Diffraction fringes are not of same Interference fringes are of same width. width. Diffraction fringes are not equally Interference fringes are equally spaced. spaced. The points of minimum intensity are The points of minimum intensity are not perfectly dark. perfectly dark. Diffraction is the result of Interference is a result of super position of only a few secondary wavelets from two coherent sources. superposition of a very large number of secondary wave lets coming from single source. (0) Define state Huygens principle? Ans. Knowing the shape and location of a wavefront at any instant t. Huygen s principle enables us to determine the shape and location of the new wavefront at a later time t t. This principle consists of two parts (i) Every point of a wavefront may be considered as a source of secondary wavelets which spread out in forward direction with a speed equal to the speed of propagation of the wave. (ii) The new position of the wavefront after a certain interval of time can be found by constructing a surface touches all the secondary wavelets. Prof. Muhammad Amin 13

14 Numerical Problems 9.1 In a double slit experiment the second order maximum occurs at is 650 nm. Determine the slit separation. Solution: Data: Wavelength of 650nm Angle = For second order m = To find: Slit separation = d =? Calculation: Using the formula or dsinθ=mλ mλ d= sinθ 0.5 o Putting the values, we get d= sin 0.5 o = or -6-4 = = m or d =3 10 = d = 0.3mm -4-3 Ans 9 m 0.5 o. The wavelength 9. A monochromatic of 588nm is allowed to fall on the half silvered glass plate G1, in the Michelson interferometer. If mirror M1 is moved through 0.33 mm, how many fringes will be observed to shift? Solution: Data: Wavelength of = 588 nm = 588 x 10 m Distance through which mirror is moved = L = 0.33 mm = 0.33 x 10-3 m To find: Number of fringes shifted = m =? Calculation: mλ L= Or L m= λ Putting the values we get m= = = = m = Ans 5 Prof. Muhammad Amin 14

15 9.3 A second order spectrum is formed at one angle of 38.0 o when falls normally on a diffraction grating having 5400 lines per centimeter. Determine the wavelength of used. Solution: Data: For second order spectrum = n = Angle = = 38.0 o Number of lines on grating = N = 5400 lines per cm d = = = lines/cm N = lines/m To find: Wavelength of the = λ =? Calculation: As we know that dsin θ = nλ dsinθ λ = n sin38.0 λ = -6 o λ = λ = λ = m λ = m λ = 570nm Ans 9.4 A is incident normally on a grating which has 500 lines per centimeter. Compute the wavelength of a spectral line for which the deviation in second order is 15.0 o. Solution: Data: To find: Calculation: Number of lines on grating =N = 500 lines / cm d= = =4 10 lines/cm N 500 d = 4 x 10-6 lines / m For second order=n= o Angleof deviation=θ=15 Wavelength of =λ=? dsinθ=nλ dsinθ λ= n 4 10 sin 15 λ= -6 o λ= Prof. Muhammad Amin 15

16 λ= m λ = 518nm Ans 9.5 Sodium 589nm is incident normally on a grating having 3000 lines per centimeter. What is the highest order of the spectrum obtained with this grating? Solution: Wavelength of sodium = λ=589nm o Angle = θ = 90 for highest order = nm Nunber of lines on the grating = N=3000lines/cm To find: Order of spectrum = n =? Calculation: 1 1 N d= = lines /cm -6 d = lines /m dsinθ = nλ dsinθ n= λ -6 o sin(90 ) n= n = n = 5th Ans 9.6 X-rays of wavelength nm are observed to undergo a first order reflection at a Bragg angle of 13.3 o from a quartz. (SiO) crystal. What is the interplaner spacing of the reflecting planes in the crystal? Solution: Data: Wavelength of X-rays =λ=0.150nm To Find: λ= m For first order reflection=m=1 o Braggs Angle =θ=13.3 Interplaner spacing=d=? Calculation: As we know that the Bragg s law is; dsinθ=mλ mλ d= sinθ d= sin13.3 o d= d = m d = 0.36nm Ans Prof. Muhammad Amin 16