Data Provided: A formula sheet and table of physical constants are attached to this paper.

Transcription

1 PHY472 Dt Provided: Formul sheet nd physicl constnts Dt Provided: A formul sheet nd tble of physicl constnts re ttched to this pper. DEPARTMENT OF PHYSICS & Autumn Semester ASTRONOMY DEPARTMENT OF PHYSICS AND ASTRONOMY ADVANCED QUANTUM MECHANICS 2 hours Spring 2015 Optics 2 hours Answer question ONE (Compulsory) nd TWO other questions, one ech from section A nd section B. Instructions: All questions Answer THREE re mrked questions. out All of questions ten. The rebrekdown mrked out of on twenty. the right-hnd The brekdown side onof the the right-hnd pper side is ment of the pper s guide is ment to sthe guide mrks to the tht mrks cn tht be obtined cn be obtined from from ech ech prt. prt. Plese clerly indicte the question numbers on which you would like to be exmined on the front cover of your nswer book. Cross through ny work tht you do not wish to be exmined. PHY227 TURN OVER 1

2 1. () Explin the role of the cldding of opticl fibres. [2] (b) Explin the dvntges for telecommunictions of single mode fibres compred with multi-mode fibres. (c) A fibre with the inner mteril hving refrctive index n f nd the cldding mteril hving refrctive index n c = 1.5 opertes t wvelength 1.55 microns. (i) The fibre core hs dimeter of 4 microns. Find the mximum n f for which the fibre only supports propgtion of single mode. [4] (ii) For the vlue of n f found in prt (i) find the number of supported modes for this fibre if the cldding is removed nd the fibre is plced in ir. [4] (iii) Wht is the mximum externl cceptnce ngle for the fibre in prt (ii) if it is plced in ir? (d) An opticl fibre is mde from n inner mteril with refrctive index n f = The fibre core hs dimeter of 6 microns. The fibre opertes t wvelength of 1.55 microns. Find the mximum ngle (mesured from the fibre xis) t which light cn propgte inside this fibre if it does not hve the cldding nd is immersed in wter. The refrctive index of wter is n w =1.33. [4] (e) Sketch grph of the intensity of light trnsmitted through silic fibre s function of the wvelength of the light. Indicte the trnsprency windows nd stte their pproximte wvelengths. [2] [2] [2] PHY227 CONTINUED 2

3 2. () A bem of unpolrised light is incident on detector mesuring its intensity. Two liner polrisers nd two qurter-wve pltes re provided. The orienttions of the fst nd slow xes of the wve-pltes re not mrked. Suggest nd explin method for lignment of the fst xes of the two wve-pltes. [8] (b) Consider light incident on n interfce between two trnsprent dielectrics hving refrctive indices n 1 nd n 2. (i) Derive the expression for Brewster s lw for this sitution. Stte the polristion of light for which Brewster s lw is pplicble. [5] (ii) Consider light incident on the surfce of wter (n w =1.33) from ir. Sketch the intensity of trnsmitted light s function of the ngle of incidence for light polrised in nd out of the plne of incidence. Add tick lbels to the horizontl xis showing the ngle of incidence. [5] (c) Give two exmples of pplictions of Brewster s lw. [2] PHY227 TURN OVER 3

4 3. () Consider Fbry-Perot interferometer. Monochromtic light is incident on the interferometer nd the trnsmitted light intensity is mesured. Clculte the reflectivity of the mirrors if the minimum trnsmitted intensity obtined by tuning the wvelength of the incident light corresponds to 20% of the mximum trnsmitted intensity. (b) Light with wvelength 500 nm is incident normlly on Fbry-Perot interferometer consisting of two mirrors with n ir gp between them. The interferometer hs coefficient of finesse of (i) The seprtion of the mirrors is tuned until mximum in the trnsmitted intensity is observed. Find by how much the gp should be chnged from this position so tht the intensity of the trnsmitted light is reduced by fctor of 2. [5] (ii) The size of the gp is set to 0.1 mm. Will this correspond to mximum or minimum in the trnsmitted intensity? [2] (iii) For the sitution in prt (ii) find the resolving power of the interferometer. [2] (c) White light is incident t n ngle θ = 30 0 on thin dielectric film hving refrctive index n=1.5. The film s thickness is 1 micron. Light is linerly polrised normlly to the plne of incidence. Find the wvelengths in the visible rnge, for which light will be reflected most strongly from the film. [6] [5] PHY227 CONTINUED 4

5 4. () Light of wvelength λ is incident normlly on screen with periodic rry of N slits seprted by centre-to-centre distnce nd hving width b. Another screen is plced t distnce l wy from the slits nd is used for imging the diffrction pttern produced by the slits. Clculte the full width of diffrction fringe in the diffrction pttern observed on the screen. Assume tht b nd λ. [7] (b) Formulte the Ryleigh criterion for resolution in line spectr obtined by periodic rry of slits nd derive the expression for the resolving power of the screen with N slits from prt (). (c) Sketch schemtic of grting spectrometer nd explin the principle of its opertion bsed on your drwing. Explin how spectrum is mesured. [5] (d) Wht re the dvntges of grting spectrometer compred with Fbry-Perot interferometer? Justify your nswer. [2] [6] PHY227 TURN OVER 5

6 5. () Light from HeNe lser (λ = 633 nm) with bem dimeter of 1 mm is pssing through n djustble perture. (i) Wht is the size of the lser spot on the screen 10 m wy from the perture if the perture dimeter is reduced to 100 microns. [4] (ii) A detector with dimeter 5 mm is plced 5 m wy from the perture. How should the size of the perture be djusted so tht the whole re of the detector is illuminted by the lser? (b) Formulte the Ryleigh criterion describing ngulr resolution for n imge obtined by lens. Write mthemticl expression describing this criterion nd giving the best theoreticlly chievble ngulr resolution with lens of dimeter for light with wvelength λ. [4] (c) In blu-ry plyer the informtion is red by lser operting t wvelength of 405 nm focused on the disk using 0.5 mm focl length lens hving dimeter of 2 mm. Estimte the highest rel density of fetures tht cn be detected by the blu-ry plyer. (d) Find the mximum distnce t which two ircrft flying 1000 m prt cn be resolved by n ir trffic control rdr operting t frequency of 1 GHz. Assume tht the rdr ntenn hs circulr shpe with dimeter of 5 metres. [4] [4] [4] END OF EXAMINATION PAPER 6

7 PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE Physicl Constnts electron chrge e = C electron mss m e = kg = MeV c 2 proton mss m p = kg = MeV c 2 neutron mss m n = kg = MeV c 2 Plnck s constnt h = J s Dirc s constnt ( = h/2π) = J s Boltzmnn s constnt k B = J K 1 = ev K 1 speed of light in free spce c = m s m s 1 permittivity of free spce ε 0 = F m 1 permebility of free spce µ 0 = 4π 10 7 H m 1 Avogdro s constnt N A = mol 1 gs constnt R = J mol 1 K 1 idel gs volume (STP) V 0 = 22.4 l mol 1 grvittionl constnt G = N m 2 kg 2 Rydberg constnt R = m 1 Rydberg energy of hydrogen R H = 13.6 ev Bohr rdius 0 = m Bohr mgneton µ B = J T 1 fine structure constnt α 1/137 Wien displcement lw constnt b = m K Stefn s constnt σ = W m 2 K 4 rdition density constnt = J m 3 K 4 mss of the Sun M = kg rdius of the Sun R = m luminosity of the Sun L = W mss of the Erth M = kg rdius of the Erth R = m Conversion Fctors 1 u (tomic mss unit) = kg = MeV c 2 1 Å (ngstrom) = m 1 stronomicl unit = m 1 g (grvity) = 9.81 m s 2 1 ev = J 1 prsec = m 1 tmosphere = P 1 yer = s

8 Polr Coordintes x = r cos θ y = r sin θ da = r dr dθ 2 = 1 ( r ) + 1r 2 r r r 2 θ 2 Sphericl Coordintes Clculus x = r sin θ cos φ y = r sin θ sin φ z = r cos θ dv = r 2 sin θ dr dθ dφ 2 = 1 ( r 2 ) + 1 r 2 r r r 2 sin θ ( sin θ ) + θ θ 1 r 2 sin 2 θ 2 φ 2 f(x) f (x) f(x) f (x) x n nx n 1 tn x sec 2 x e x e x sin ( ) 1 x ln x = log e x 1 x cos 1 ( x sin x cos x tn ( 1 x cos x sin x sinh ( ) 1 x cosh x sinh x cosh ( ) 1 x sinh x cosh x tnh ( ) 1 x ) ) 1 2 x x 2 2 +x 2 1 x x x 2 cosec x cosec x cot x uv u v + uv sec x sec x tn x u/v u v uv v 2 Definite Integrls x n e x dx = n! (n 0 nd > 0) n+1 π e x2 dx = π x 2 e x2 dx = 1 2 Integrtion by Prts: 3 b u(x) dv(x) dx dx = u(x)v(x) b b du(x) v(x) dx dx

9 Series Expnsions (x ) Tylor series: f(x) = f() + f () + 1! n Binomil expnsion: (x + y) n = (1 + x) n = 1 + nx + k=0 ( ) n x n k y k k n(n 1) x 2 + ( x < 1) 2! (x )2 f () + 2! nd (x )3 f () + 3! ( ) n n! = k (n k)!k! e x = 1+x+ x2 2! + x3 x3 +, sin x = x 3! 3! + x5 x2 nd cos x = 1 5! 2! + x4 4! ln(1 + x) = log e (1 + x) = x x2 2 + x3 3 n Geometric series: r k = 1 rn+1 1 r k=0 ( x < 1) Stirling s formul: log e N! = N log e N N or ln N! = N ln N N Trigonometry sin( ± b) = sin cos b ± cos sin b cos( ± b) = cos cos b sin sin b tn ± tn b tn( ± b) = 1 tn tn b sin 2 = 2 sin cos cos 2 = cos 2 sin 2 = 2 cos 2 1 = 1 2 sin 2 sin + sin b = 2 sin 1( + b) cos 1 ( b) 2 2 sin sin b = 2 cos 1( + b) sin 1 ( b) 2 2 cos + cos b = 2 cos 1( + b) cos 1 ( b) 2 2 cos cos b = 2 sin 1( + b) sin 1 ( b) 2 2 e iθ = cos θ + i sin θ cos θ = 1 ( e iθ + e iθ) 2 nd sin θ = 1 ( e iθ e iθ) 2i cosh θ = 1 ( e θ + e θ) 2 nd sinh θ = 1 ( e θ e θ) 2 Sphericl geometry: sin sin A = sin b sin B = sin c sin C nd cos = cos b cos c+sin b sin c cos A

10 Vector Clculus A B = A x B x + A y B y + A z B z = A j B j A B = (A y B z A z B y ) î + (A zb x A x B z ) ĵ + (A xb y A y B x ) ˆk = ɛ ijk A j B k A (B C) = (A C)B (A B)C A (B C) = B (C A) = C (A B) grd φ = φ = j φ = φ x î + φ y ĵ + φ z ˆk div A = A = j A j = A x x + A y y + A z z ) curl A = A = ɛ ijk j A k = ( Az y A y z φ = 2 φ = 2 φ x + 2 φ 2 y + 2 φ 2 z 2 ( φ) = 0 nd ( A) = 0 ( A) = ( A) 2 A ( Ax î + z A ) ( z Ay ĵ + x x A ) x y ˆk