Applied Physics - II

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1 F.E. Sem. II Applied Physics - II Time : Hrs.] Prelim Question Paper Solution [Marks : 60 Q.1 Attempt any Five questions of the following : [15] Q.1(a) Fringes of equal thickness are observed in a thin glass wedge of [3] R.I-1.5. The fringe spacing is 1mm and wavelength of light used is 5893A. Calculate the angle of the wedge. (A) Given data : = 1.5 = 1 mm = 10 3 m = 5893 A =? = = 103 = = radian = sec. of an arc. = sec. of an arc. [3 marks] Q.1(b) What is Rayleigh s criteria of resolution? What is resolving power [3] of diffraction grating? (A) According to Rayleigh s criterion, closely spaced point sources of light are said to be just resolved by an optical instrument only if the central maximum in the diffraction pattern of one falls over the first minimum in the diffraction pattern of the other & viceaversa. [1½ marks] The expression for the resolving power of a grating is given by [1½ marks] R.P. = m.n. d Where, wavelength of a line and d difference in the wavelength between this line & the neighbouring line such that the lines appear to be just resolved, m = order of the spectrum, N = no. of slits. Q.1(c) The core diameter of multimode step Index fibre is 50m. The numerical aperature is 0.5. Calculate the no. of guided modes at an operating wavelength of 0.75 m. [3] 1

2 Vidyalankar : F.E. Physics (A) Given data d = 50 m NA = 0.5 N =? = 0.75 m N =? formula V d = NA = 5.33 No. of guided mode V N = N = 1369 = = [3 marks] Q.1(d) What is acronym of LASER? How are they different than X- [3] rays? (A) LASER is the acronym for Light Amplification by Stimulated Emission of Radiation. [1 mark] Given below are some important differences between LASER & X-rays : [ marks] SR. LASER X-rays No. 1. Their Wavelenth () varies between 38 to 740 nm.. They are only light electromagnetic waves Their wavelength () varies between 0.01 to 10 nm. They are radioactive electromagnetic waves. 3. Can be focussed sharply. Cannot be focussed. 4. They are highly coherent. Noncoherent. 5. Highly directional possess high energy 6. Can be produced artificially using stimulated emission. They are scattered. possess low energy. These are produced naturally. Q.1(e) An electron is bound in an one dimensional potential well of width of [3] Å, but of infinite height. Find its energy values in the first excited state. (A) [3 marks] Given Required electron m = kg, a = Å E 1 First excited state n =

3 Energy of electron in one dimensional potential well is given by nh E = 8 ml E 1 = = J Prelim Question Paper Solution Q.1(f) How is phase difference between two A.C. signals measured by [3] CRO? (A) XY Method [3 marks] (i) This method makes use of lissajous pattern (ii) The time/div knob is set to X-Y EXT state. In this state the signal given to CH acts as the sweep signal. (iii) The vertical mode switch is kept in X-Y position An ellipse such as the one illustrated (pattern) obtained on the CRO screen. (iv) The ACGNDDC switch is set to GND position. (v) The vertical length. A & B of ellipse are measured = sin 1 (A/B) = sin 1 (.4/4) = A B Q.1(g) What is the vortex state of a superconductor? [3] (A) The term vortex or mixed state refers to type II super-conductors. Shown below is the magnetization characteristics for a type II superconductor. [1 mark] [1 mark] Between H C1 & H C the specimen is magnetically in the in the mixed state but electrically it is in the superconducting state. Hence this region is known as mixed or vortex state. In this state, the specimen has both superconducting & normal regions posses zero resistance and partial penetrating flux. [1 mark] Type II superconductors e.g. Nb 3 Sn, Nb 3 Si, Nb 3 Al, VaGa etc. have the vortex state. 3

4 Vidyalankar : F.E. Physics Q.(a) Show that the diameter of Newton s n th ring is directly proportional [8] to square root of ring number. In a Newton's ring pattern one of the dark ring due to light of wavelength 7000A is found to coinside with the dark ring of next order due to 5000A. If the radius of curvature of the lens is 148.8cm. Find the diameter of the overlapping dark ring. (A) From fig. [ marks] In ADB AB = AD + DB R = (R t) + r n R = R Rt + t + r n t is very small as compare to R R >>> t, t will neglected. r n = Rt (1) for dark ring Optical path difference is t cos r = n normal incidence r = 0 & surrounding medium is air t = n t = n/ () from (1) & () r n = R n/ = n R D n 4 = n R D n = 4n R D n = R n AB = AO = R BC = t AD = R t OC = r n D n n [ marks] Given data 1 = 7000 A = 5000 A D n =? R = cm D n = 4n 1R D n+1 = 4(n1) R 4n 1 R = 4(n + 1) R n 1 = (n + 1) n( 1 ) = n = =.5 [ marks] 1 4

5 Prelim Question Paper Solution D n = = = m D n =.86 mm [ marks] Q.(b) What is N.A.? Consider a multimode step index fibre with [7] n 1 = 1.53 & n = 1.5. The core radius = 50 m. If the light of wavelength 1m is transmitted through the fibre, calculate (a) Normalised frequency V of the fibre & (b) The number of modes the fibre will support. (A) Numerical Aperture (N.A.) : It is the light gathering capacity of an optical fibre. It is defined as the sine of the acceptance angle a, Thus, N.A. = sin a = 1 3 where, 1 & are the refractive indices of the core & cladding materials respectively. N.A. ranges between 0.13 to 0.50 [3 marks] Given Required n 1 = 1.53, n = 1.5, r = m V = m Nm (I) V = d (N.A.) = 6 (r) n 1 n ( ) = (1.53) (1.5) = [ marks] (II) Nm = V = (94.53) = say 4468 [ marks] Q.3(a) What is the difference between holography and photography? [8] Discuss the construction and reconstruction of image in holography with neat diagram. (A) Holography is a technique of producing three dimensional images. In conventional photography, two dimensional images are formed as only the amplitude of light coming from an object is recorded on a photographic plate. In holography, the amplitude as well as the phase of light waves coming from the object are stored in the form of an interference pattern. The technique involves two steps-recording and reconstruction. Best interference patterns are produced using lasers which are the most coherent sources. [ marks] 5

6 Vidyalankar : F.E. Physics Recording : Coherent light from a laser is partly reflected from a mirror (known as reference beam) and partly reflected from the object (known as object beam). The interference pattern of these two beams is recorded on a photographic plate as shown in Fig.1. The developed photographic plate is known as hologram. It consists of alternate bright and dark fringes. [3 marks] Fig.1 Reconstruction : To see the three dimensional image of the object, the hologram is illuminated by the reference beam only or a laser beam with the orientation of the reference beam with respect to the hologram. The hologram acts as a diffraction grating. One of the first order diffracted beam produces a real three dimensional and inverted image and the other first order diffracted beam produces a virtual three dimensional image which is formed at the same place where the object was kept during recording as shown in Fig.. [3 marks] Fig. Q.3(b) Why are the fringes in Wedge shaped film straight? Drive the conditions of maxima and minima for interference in wedge shaped film. [7] 6

7 Prelim Question Paper Solution (A) For wedge shaped air film, the locus of all the points at equal thickness are straight lines parallel to the edge of the wedge so fringes appears straight & parallel. [ marks] Let a beam AB of monochromatic light wavelength be incident at an angle i' on the upper surface of the film. It is reflected along BR 1 & transmitted along BC. At C the beam suffers partial reflect & partial transmission. Finally we have DR in reflected system. We have two coherent rays BR 1 & DR is reflected system. These rays interfere constructively or distructively according to path difference between them. Optical path difference between them is = (BC + CD) BF [1 mark] Draw r to surface OX at point C. BQC = 0 1 = Draw r DE to BC from D. As is small BE = EC QBE = r = BDE Also CP = CD = (BC + CD) BF = (BE + EC + CP) BF from dia. = sin i = BF BF = BE sinr BE = (BE + EC + CP) BE = EP [ marks] Now In DPC CP = CD, CPD = r + & DPE is right angle triangle cos(r + ) = EP DP 7

8 Vidyalankar : F.E. Physics EP = DP cos (r + ) = t cos (r + ) where DP = DI = t = EP = t cos (r + ) Due to reflection of B. additional path differ is /. Hence for maxima. = t cos(r + ) + / = n or t cos (r + ) = (n 1) /. n = 1,, 3 for minima = t cos (r + ) = (n 1) / Or t cos(r + ) = n n = 0, 1,, 3 [ marks] Q.4(a) Monochromatic light of wavelength 6560 Å falls normally on a grating cm wide. The first order spectrum is produced at an angle of 16 17' from the normal. Calculate total no. of lines on the grating. (A) Given Required = cm, W = cm, m = 1 1 (ab) = 1617 = 16.8 [5] (i) (a + b) sin = m (ii) (a + b) sin(168) = (a + b) = cm [ marks] 1 (iii) = = (N/W) [ marks] (a b) total no. of lines = N = 473 = 8546 [1 mark] Q.4(b) Derive Schrodingers time-dependent wave equation. [5] (A) Using de Broglie hypothesis = h/p wave function can be written as = A e it A = amplitude = Angular frequency Classical wave equation dy dx = 1 dy v dt d 1 d = dx v dt [1½ mark] 8

9 Prelim Question Paper Solution solution for above equation i(et px)/ (x 1 t) = A e Taking differentiation i = Ae (EtPx)/ t t x i = E = P [1½ mark] We have energy of frequency particle. E = P /m Total energy E = P V m P Or m = E V E i t P = E V m P x Then, = V m x i t Vi [ marks] m x t this is time dependent Schrodinger wave equation. Q.4(c) Derive condition for maximum diffraction at diffraction grating. [5] (a) Consider a parallel beam of monochromatic light of wavelength incident on a slit of width a diffracted at an angle as shown the Fig. The path difference between extreme rays diffracted from the slit is = BC = AB. Sin = a. sin and the phase difference between these rays is given as follows :. asin [1 mark] The slit is now divided into N parts of equal width x as shown in Fig. Since width x is very small, each part behaves like a point source. As all parts are of equal width, the amplitude E of the waves transmitted by them will be the same. 9

10 Vidyalankar : F.E. Physics The path difference between waves transmitted by two adjacent parts is = x. sin the phase difference is = x sin. [1 mark] Fig. below represents the phasor diagram using polygon law [1 mark] (b) To find the resultant of all E. It is = length of the arc AB with C. [ marks] as the centre & radius R = AC = BC (radius of the circle). Let = a sin. In ADC, sin = AD E. But AD = DB = AC E / sin = R E Em R = Now arc length AB = R. = Em = R. or R = sin. E Em R = sin Em. sin E Clearly E will be maximum if is minimum = 0 = a.sin = 0 i.e. = 0 the principal maximum is formed along incident direction it is also called the central maximum. This is the required condition. Q.5(a) Derive the condition for absent spectra in grating. [5] (A) (i) Consider a plane transmission grating as shown in the following fig. It has N number of parallel equidistant slits each of width a & separated by opaque space b. [ marks] 10

11 Prelim Question Paper Solution (ii) The intensity of light due to the diffraction grating in a direction making an angle with the normal to the grating & at a point P is given by sin sin N I = Im. [1 mark] sin where, = a sin & = (a b) sin sin (iii) The factor Im gives the intensity distribution due to a single slit sin N while the factor gives that due to the combined effect of N slits. sin [½ mark] (iv) The principle maxima in the case of a grating are obtained in the directions given by (a + b) sin = m (1) where, m = order of the spectrum. [½ mark] (v) The minima in the case of a single slit are obtain in the directions given by [½ mark] a sin = n () n = 1,, (vi) If both the conditions (1) & () are satisfied simultaneously, a particular maximum order m will be missing in the grating spectrum. [½ mark] dividing equation (1) by (), we get a b = m a n which is the condition for the absent spectra. 11

12 Vidyalankar : F.E. Physics Q.5(b) Show that the energy of an electron in a box varies as the square [5] of the natural number. (A) Outside the box V = & particle cannot have infinite energy therefore it cannot exist outside the box. Schrodinger's equation d 8 m (E ) = 0 dx h Inside the box V = 0 Schrodinger's equation d 8 m E = 0 dx h d k = 0 dx v = v = 0 v = k 8 m ME v = E = h x = 0 x = a [ marks] x Solution for above equation. = A cos kx + B sin kx When x = 0 = 0 we get 0 = A cos + B sin 0 A = 0 [1 mark] When x = a, (x) = 0 0 = A cos ka + B sin ka But A = 0 B sin ka = 0 sin ka = 0 ka = ME a = n n ME = a n n E n = = Ma 8Ma E n n [ marks] 1

13 Prelim Question Paper Solution Q.5(c) Draw the schematic diagram of SEM and explain its construction and [5] working. (A) (i) Construction : [3 marks] (ii) Working [ marks] The electron beam is obtained from an electron gun. It is made to pass through condenser lens. The scanning coil in the next stage is used for focusing the electron beam on a small spot on the specimen surface & also to scan the surface. On reaching the specimen surface, the electrons get scattered & form the image signals. During the scanning, the scattered electron intensities are measured by detector & then displayed on the screen. For high scattering, the respective point will be bright & for low scattering, the corresponding point on the screen will be dark. Due to this, required contrast for a clear image of the specimen is developed. Q.6(a) Find the energy of the neutron in units of electron volts where Debroglie [5] wavelength is 1Å mass of neutron = kg Planck s constant = J.sec. (A) [5 marks] Given Required m n = kg, = 1Å E h = J-sec 13

14 Vidyalankar : F.E. Physics = h me E 34 h = = 7 10 m (1 10 ) = J E = ev ( 1 ev = J) E = ev Q.6(b) Explain the principle, construction and working of CRT with neat diagram. (A) [5] [ marks] electron gun Grid Anode depletion system floroscent screen In the CRT the electron gun generate an electron beam focuses it and accelerates it towards a florescent screen located at the further end of the tube. The screen emits a small round glow at the point where energetic electrons strike it the electron beam may be move dot spot on the screen with the help of depletion system electron gun, grid Accelerating Anodes deflection system floroscent screen. [3 marks] Q.6 (c) What are carbon tubes and what are their properties? [5] (A) Carbon Nano Tubes (i) Introduction : [1½ marks] In graphite, each carbon atom is linked to three other carbon atoms through covalent bonds forming hexagonal rings connected to each other to form plane sheets known as grapheme. The structure of CNT is a cylinder formed by rolling a grapheme sheet & then closing it on both the ends by fullerene hemispheres. 14

15 Prelim Question Paper Solution (ii) Structure : [1½ mark] there are mainly two types of CNT s depending upon their structures. : 1. Single Walled Carbon Nanotubes (SWCNT) : It consists of a single seamless cylinder having a diameter of a nm. & a length of range of mm.. Multi Walled Carbon Nanotubes (MWCNT) : They consists of multilayered concentric cylinders of single grapheme sheets with the outer tube having diameter of the order of 10 to 30 nm. The properties of the CNT s depend upon their diameters. (iii) Properties : [ marks] 1. Electrical : These vary between semiconducting to metallic depending upon their diameter & length.. Mechanical : CNT s are stiffest & strongest in terms of elastic modulus & tensile strength. 3. Thermal : Very high thermal conductivity. Also they have high thermal stability. 15

Dept. of Physics, MIT Manipal 1

Dept. of Physics, MIT Manipal 1 Chapter 1: Optics 1. In the phenomenon of interference, there is A Annihilation of light energy B Addition of energy C Redistribution energy D Creation of energy 2. Interference fringes are obtained using