PHY15-B PHY47 Dt Provided: Formul sheet nd physicl constnts Dt Provided: A formul sheet nd tble of physicl constnts is ttched to this pper. DEPARTMENT OF PHYSICS & Autumn Semester 009-010 ASTRONOMY DEPARTMENT OF PHYSICS AND ASTRONOMY ADVANCED QUANTUM MECHANICS hours Autumn 016 Optics Answer question ONE (Compulsory) nd TWO other questions, one ech from section A nd section B. Instructions: All questions Answer the re compulsory mrked out question of ten. (Q1), The ndbrekdown ONE dditionl on question the right-hnd of your choice. side of the pper is ment s guide to the mrks tht cn be obtined from ech prt. 1 hour All questions re mrked out of twenty. The brekdown on the right-hnd side of the pper is ment s guide to the mrks tht cn be obtined from ech prt. Answers to different sections must be written in seprte books, the books tied together nd hnded in s one. 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. PHY15-B TURN OVER 1
PHY15-B 1. COMPULSORY () Consider light of wvelength 55 nm. i. Wht is the wvenumber? ii. Wht is the ngulr frequency? iii. Wht is the speed of the light in glss with refrctive index n g = 1.5? [3] (b) A prllel bem of light in ir mkes n ngle of 3.0 with the surfce norml of glss plte hving refrctive index of 1.55. Wht is the ngle between the refrcted bem nd the surfce norml of the glss? [] (c) Wht is the criticl ngle for light propgting in glss rod tht hs refrctive index n r = 1.60? (d) A prllel bem of unpolrised light in ir is incident on flt semiconductor surfce. If the refrctive index of the semiconductor is 3.40, wht ngle should the incident light mke to the surfce norml to result in completely linerly polrised reflected light? (e) A thin lens with focl length of 5.00 cm is used s simple mgnifier. Wht is the ngulr mgnifiction if n object is plced t the focl point of the lens? You my ssume tht the ner point is 5.0 cm from the eye nd tht the lens is very close to the eye. (f) An object 0.600 cm tll is plced 16.5 cm to the left of the vertex of concve sphericl mirror hving rdius of curvture.0 cm. Drw principl ry digrm showing the formtion of the imge. [3] (g) A diffrction grting produces third order intensity mximum t 60 from the centrl mximum for light of wvelength 650 nm. At wht ngle will the second order intensity mximum occur for light of wvelength 53 nm? [4] (h) A lens forms n imge of n object. The object is 16.0 cm from the lens. The imge is 1.0 cm from the lens on the sme side s the object. Wht is the focl length of the lens? [] [] [] [] PHY15-B CONTINUED
PHY15-B. () The imge of tree just fills the length of plne mirror 5.00 cm tll when the mirror is held 35.0 cm from the eye. The tree is 8.0 m from the mirror. Wht is the height of the tree? (b) An object to the left of lens is imged by the lens on screen 8.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to re-focus the imge. Wht is the focl length of the lens? [5] (c) The rdii of curvture of the surfces of thin, converging meniscus lens re R 1 = +8.00 cm nd R = +15.0 cm. The lens refrctive index is 1.5. An object with height of 4.00 mm mesured from the optic xis is plced 36.0 cm to the left of the lens. i. Clculte the position of the imge. [3] ii. Clculte the size of the imge. [] (d) An object 0.600 cm tll is plced 16.5 cm to the left of the vertex of convex sphericl mirror hving rdius of curvture.0 cm. i. Drw principl ry digrm showing the formtion of the imge. [3] ii. Determine the position, size, nd orienttion of the imge. Is the imge rel or virtul? [3] [4] 3. () Describe two uses of thin film cotings in opticl technologies. [4] (b) Light with wvelength 53 nm in ir is normlly incident upon glss surfce tht is coted by thin dielectric film. Wht is the thinnest film of coting with refrctive index n f = 1.35 on glss (n g = 1.53) for which destructive interference of light cn tke plce by reflection? (c) Light is normlly incident upon sop film with refrctive index n s = 1.33. Assuming tht ir is on either side of the sop film, wht is the thinnest sop film (excluding zero thickness) tht ppers blck when illuminted by light of wvelength 53 nm in ir? (d) Monochromtic light is used in double slit experiment to produce n interference pttern on distnt screen. The width of ech slit is third of the centre-to-centre distnce between the slits. Which interference mxim re missing in the pttern on the screen? (e) Sketch the intensity distribution of the predicted pttern on the screen from prt (d). (f) Lser light of wvelength 53 nm psses through single slit 0.750 mm wide. The diffrction pttern is observed on screen tht is 3.50 m from the slit. Wht is the distnce between the two drk fringes on either side of the centrl mximum? [4] END OF EXAMINATION PAPER [4] [4] [3] [1] 3
PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE Physicl Constnts electron chrge e = 1.60 10 19 C electron mss m e = 9.11 10 31 kg = 0.511 MeV c proton mss m p = 1.673 10 7 kg = 938.3 MeV c neutron mss m n = 1.675 10 7 kg = 939.6 MeV c Plnck s constnt h = 6.63 10 34 J s Dirc s constnt ( = h/π) = 1.05 10 34 J s Boltzmnn s constnt k B = 1.38 10 3 J K 1 = 8.6 10 5 ev K 1 speed of light in free spce c = 99 79 458 m s 1 3.00 10 8 m s 1 permittivity of free spce ε 0 = 8.85 10 1 F m 1 permebility of free spce µ 0 = 4π 10 7 H m 1 Avogdro s constnt N A = 6.0 10 3 mol 1 gs constnt R = 8.314 J mol 1 K 1 idel gs volume (STP) V 0 =.4 l mol 1 grvittionl constnt G = 6.67 10 11 N m kg Rydberg constnt R = 1.10 10 7 m 1 Rydberg energy of hydrogen R H = 13.6 ev Bohr rdius 0 = 0.59 10 10 m Bohr mgneton µ B = 9.7 10 4 J T 1 fine structure constnt α 1/137 Wien displcement lw constnt b =.898 10 3 m K Stefn s constnt σ = 5.67 10 8 W m K 4 rdition density constnt = 7.55 10 16 J m 3 K 4 mss of the Sun M = 1.99 10 30 kg rdius of the Sun R = 6.96 10 8 m luminosity of the Sun L = 3.85 10 6 W mss of the Erth M = 6.0 10 4 kg rdius of the Erth R = 6.4 10 6 m Conversion Fctors 1 u (tomic mss unit) = 1.66 10 7 kg = 931.5 MeV c 1 Å (ngstrom) = 10 10 m 1 stronomicl unit = 1.50 10 11 m 1 g (grvity) = 9.81 m s 1 ev = 1.60 10 19 J 1 prsec = 3.08 10 16 m 1 tmosphere = 1.01 10 5 P 1 yer = 3.16 10 7 s
Polr Coordintes x = r cos θ y = r sin θ da = r dr dθ = 1 ( r ) + 1r r r r θ Sphericl Coordintes Clculus x = r sin θ cos φ y = r sin θ sin φ z = r cos θ dv = r sin θ dr dθ dφ = 1 ( r ) + 1 r r r r sin θ ( sin θ ) + θ θ 1 r sin θ φ f(x) f (x) f(x) f (x) x n nx n 1 tn x sec x e x e x sin ( ) ln x = log e x 1 x cos 1 ( x sin x cos x tn ( cos x sin x sinh ( ) cosh x sinh x cosh ( ) sinh x cosh x tnh ( ) ) ) 1 x 1 x +x 1 x + 1 x x cosec x cosec x cot x uv u v + uv sec x sec x tn x u/v u v uv v Definite Integrls 0 + + x n e x dx = n! (n 0 nd > 0) n+1 π e x dx = π x e x dx = 1 Integrtion by Prts: 3 b u(x) dv(x) dx dx = u(x)v(x) b b du(x) v(x) dx dx
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 + ( x < 1)! (x ) f () +! nd (x )3 f () + 3! ( ) n n! = k (n k)!k! e x = 1+x+ x! + x3 x3 +, sin x = x 3! 3! + x5 x nd cos x = 1 5!! + x4 4! ln(1 + x) = log e (1 + x) = x x + 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 = sin cos cos = cos sin = cos 1 = 1 sin sin + sin b = sin 1( + b) cos 1 ( b) sin sin b = cos 1( + b) sin 1 ( b) cos + cos b = cos 1( + b) cos 1 ( b) cos cos b = sin 1( + b) sin 1 ( b) e iθ = cos θ + i sin θ cos θ = 1 ( e iθ + e iθ) nd sin θ = 1 ( e iθ e iθ) i cosh θ = 1 ( e θ + e θ) nd sinh θ = 1 ( e θ e θ) 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
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 = φ = φ x + φ y + φ z ( φ) = 0 nd ( A) = 0 ( A) = ( A) A ( Ax î + z A ) ( z Ay ĵ + x x A ) x y ˆk