(7.5 points) Interference of light Consider Young's double slit apparatus that is represented in

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1 وزارة التربية والتعلين العالي الوديرية العاهة للتربية دائرة االهتحانات اهتحانات شهادة الثانىية العاهة الفرع : العلىم العاهة مسابقة في مادة الفيزياء المدة : ثالث ساعات االسن: الرقن: الدورة اإلستثنائية للعام 9 First Exercise This exm is formed of four exercises in four pges The use of non-progrmmble clcultor is recommended (7.5 points) nterference of light Consider Young's double slit pprtus tht is represented in M the djcent figure. S nd S re seprted by distnce (E) = mm. The plnes (P) nd (E) re t distnce D = m. is the S x midpoint of S S nd O the orthogonl projection of on (P). S On the perpendiculr to O t point O nd prllel to S S, point M is defined by its bsciss OM = x. O ) S nd S, illuminted by two lmps, emit synchronous S rditions. Do we observe interference fringes on the screen? Why? ) S nd S re illuminted by point source S put on O. S emits monochromtic rdition of wvelength in vcuum (or in ir). ) s the fringe obtined t O bright or drk? Why? D Fig. (P) Give, t point M,the expression of the opticl pth difference between the two rditions emitted by S, one pssing through S nd the other through S, in terms of D, x nd. c) Derive the reltion giving the bscisss of the centers of the bright fringes nd tht giving the bscisss of the centers of the drk fringes. d) For x =.4 mm, M is t the center of the fourth bright fringe (bright fringe of order 4). Clculte. ) The source S emits now white light. ) At O, we observe white light. Why? Clculte the wvelengths of the visible rditions tht give t M, of bsciss OM = x =.4mm, drk fringes. Visible spectrum:.4 m.8 m. Second Exercise (7.5 points) Determintion of the hlf-life of Polonium Polonium nucleus ( 84 Po ) is n emitter, nd it is the only polonium isotope tht exists in nture; it ws found by Pierre Curie in 898 in n ore. t is lso obtined from the decy of bismuth nucleus ( Msses of the nuclei: m (Bi) = u ; m(po) = u mss of the electron : m e =.55 u u = 9.5 MeV / c =.66-7 kg. Here is prt of the periodic tble of elements: 8 Th ; 8 Pb ; 8 Bi ; 84 Po ; 85 At ; Rn A The polonium ) ) Write down the eqution of the decy of bismuth. dentify the emitted prticle nd specify the type of this decy. ) Clculte the energy liberted by this decy Bi ).

2 ) The decy of the bismuth nucleus is ccompnied with the emission of photon of energy E() =.96 MeV nd n ntineutrino of energy. MeV. Knowing tht the dughter nucleus is prcticlly t rest, clculte the kinetic energy of the emitted prticle. B Hlf-life of polonium ) ) Write down the eqution of the decy of polonium. dentify the dughter nucleus. ) n order to determine the rdioctive period T (hlf-life) of 84 Po, we consider smple of this isotope contining o nuclei t the instnt t o =. Let be the number of the non-decyed nuclei t n instnt t. ) Write down the expression of the lw of rdioctive decy. Determine the expression of ln ( ) s function of t. ) A counter llows to obtin the mesurements tht re tbulted in the following tble : t (dys) ln ( ).4.8. ) Complete the tble. Trce, on the grph pper, the curve giving the vrition of ln ( ) s function of time. Scle: cm on the bsciss xis corresponds to 4 dys. cm on the ordinte xis corresponds to.. c) s this curve in greement with the expression found in the question (B,? Justify. d) i) Clculte the slope of the trced curve. ii) Wht does this slope represent for the polonium nucleus? iii) Deduce the vlue of T. Third Exercise (7.5 points) Exchnged Energy We connect up the circuit formed of resistor of resistnce R =. kω, n idel genertor of e.m.f E = 8 V, coil of inductnce L =.8 H nd of negligible resistnce, resistor of djustble resistnce r nd two switches K nd K (Fig.). A (RC) Series circuit At n instnt tken s n origin of time, (t = ), we close the switch K, nd K remins open. We study the chrging of the cpcitor through the vrition of the voltge u AB = u C s function of time. ) Show tht the differentil eqution in u C is: du E = u C + RC C. dt E K A K R Fig. B C r L t ) Knowing tht u C = E (- e ) is the solution of this differentil eqution, determine the expression of the time constnt τ in terms of R nd C. ) The curve of figure shows the vrition of u C s function of time. Using this curve, determine the time constnt τ (indicting the method used). 4) Clculte the vlue of C.

3 5) ) Give, in ms, the durtion t t the end of which the voltge cross the cpcitor will no longer prcticlly vry. Clculte the chrge of the cpcitor nd the energy W stored t the end of the durtion t. B (L,C) series circuit u C (V) We give r the vlue zero. The voltge cross the cpcitor is 8 V. At n instnt tken s n origin of time ( t = ), we open the switch K nd we close the switch K. ) Derive the differentil eqution tht describes the vrition 8 of the voltge u C s function of time. ) The circuit is the set of electric oscilltions of proper period T. The solution of this differentil eqution is: π 5 u C = E cos( t). Determine the vlue of T o. T ) Trce the shpe of the curve representing the vrition of u C s function of time. 4) Specify the energy exchnges tht tke plce in the circuit. C (r, L, C) series circuit We give r certin vlue. The voltge cross the terminls of the cpcitor is 8 V. We open K nd we. close K t the instnt t =. The wveform of Fig. figure shows the vrition of the voltge u C s function of time. u C (V) ) Specify the energy exchnges tht tke plce in the 8 circuit. ) ) Referring to figure, find the pseudo-period T of the electric oscilltions. Compre T nd T. ) At the end of the durtion t n = nt ( n being whole number), the energy dissipted by Joule's effect is 98.6 % of the energy W initilly stored in the cpcitor. ) At the instnt t n = nt, the energy stored in the circuit is purely electric. Why? We denote by W nd W n the electric energy of the oscilltor t the instnts t nd t n respectively. Clculte W n. Fig. c) Determine n. O () Fourth Exercise (7.5 points) Compound Pendulum The im of this exercise is to study the vrition of the proper period T of compound pendulum s function of the distnce, of djustble vlue, seprting the xis of oscilltion from the center of mss of this pendulum, nd to show evidence of some properties ssocited to this distnce. We consider homogeneous disc (D) of mss m = g, free to rotte without friction round horizontl xis (Δ) perpendiculr to its plne through point O (Fig. ). is the moment of inerti of (D) bout the xis (Δ ) prllel to (Δ) nd pssing through its center of mss G nd its moment of inerti bout the xis (Δ), (Δ ) being t distnce = OG from (Δ), so tht: = + m G Fig O () θ G t(ms) t(ms) Z Fig

4 The grvittionl potentil energy reference is the horizontl plne pssing through the center of mss G of (D) when (D) is in the position of stble equilibrium (Fig.). (D) is mde to oscillte round (Δ) nd we mesure the vlue of the proper period corresponding to ech vlue of. Tke: g = m/s ; π = ; θ For smll ngles ( in rdin); cosθ nd sin =. A Theoreticl study (D) is shifted from its stble equilibrium position by smll ngle θ m nd is then relesed from rest t the instnt t =. (D) thus oscilltes round the xis (Δ) with proper period T. dθ At n instnt t, the ngulr bsciss of the pendulum is nd its ngulr velocity is θ = (Fig. ). dt ) Write down, t the instnt t, the expression of the mechnicl energy of the system (pendulum, Erth) in terms of, m,, g, θ nd θ. ) ) Derive the second order differentil eqution in tht describes the motion of (D). Deduce tht the expression of the period T of this pendulum is given by: T. ) T nd T re respectively the periods of the pendulum when it oscilltes round (Δ) tht psses successively through O nd O where O G = nd O G =. The oscilltions hve the sme period (T = T ). nd re respectively the moments of inerti of the pendulum round (Δ) tht psses successively through O nd O. ) i) Find reltion mong,, nd. ii) Deduce tht = m. The proper period T' of simple pendulum of length, for oscilltions of smll mplitude, is given by the expression T' = g. Show tht, when the vlue of T' is equl to tht of T, we obtin = +. B Experimentl study T(s) We mesure the vlue of the period T of the pendulum for ech vlue of. The obtined mesurements llow us to trce the curve giving the vrition of T s function of. The stright line of eqution T =.s intersects this curve in two points A nd B. (Fig. ) ) ) Referring to the curve, give the vlues of nd corresponding to the period T =. s. A Deduce the vlue of nd tht of. T=. ) According to the curve of figure, T tkes minimum vlue (T min =.5 s) for certin vlue ' of. ) Give, using the curve, the vlue of ' corresponding to T min.. Find, by clcultion, the vlue of ' nd tht of Fig. T min. B (cm) 4

5 First exercise ) We don t observe interference fringes since the two sources re not coherent. (½) ) The two vibrtions rech O in phse, the opticl pth difference t O is equl to zero : = (SS + S O) (SS +S O) = (¾) = D x (½) x λd c) Bright fringes : = k = k x = k D λ x λ Drk fringes: = (k + ) = (k + ) D x = (k + ) d) OM = x = 4i = 4 λd λd (½) x.4 = =.56 - mm = 56 nm (¼) = 4D 4 ) ) Every rdition gives t O bright fringe nd becuse of superposition of ll colors white light. () x λ x,4 = (k + ) = = = D (k )D (k ) mm = m (k ) (k ) 4,48.4m V.8m (k + ). (k + ) (k ) 7, 9, أسس التصحيح لشهادة الثانىية العامة فرع العلىم العامة. دورة 9 االستثنائية Second exercise A ) ) 8 Bi 84 Po + z X Coservtion lws : = + = 8 = 84 + z z = - (½) The emitted prticle is electron e ; 8 Bi is - emitter (¼) ) E libe = m c where m is the mss defect m = m before m fter = m (Bi) m (Po) m (e) m = u u.55 u =.47 - u E lib = MeV c =.6 MeV. () c ) E lib = E(Po) + E( ) + E(e) + E( ) E c = E(e) = =.8 MeV (¾) B ) ) 84 Po A Z X + 4 He z ) ) = o e λt = e λt Conservtion lws : = A + 4 A = 6 84 = Z + Z = 8 (½) 6 X is 8 Pb (¼) where is the rdioctive constnt (½) ln ( ) = t ln ( ) ) missing vlues in the tble :. ;.6 ; (½) (¼) Ser ies ) = t (½) (k + ) 7 9 m () c) The curve is stright line pssing through the origin is n greement with ln ( )= t. (¼). 6 d) i) = iii) T = = 5 - dy - (½) ii) The slope of this line is the rdioctive constnt of polonium. (¼) n λ = 8.6 dys. (½)

6 Third exrcise A ) E = u C + Ri, i = C ) u C = E (- t e ) du C du C E= uc +RC dt dt du C E = dt t e E= E (- (½) t t E e ) + RC e RC RC. (¾) ) At instnt t =, u c =.6E =.6 8 = 5.4 V, From grph : t = =. ms for u c = 5.4 V. (½) 4) = RC =. =. C = -6 F = F. (½) 5) ) During t = 5 = ms. (¼) Q = CE = 8 6 C ; W = ½ CE = 6 J. () di dq duc B ) u C = L, i = - C, u C = -LC u C dt dt dt π π ) u C = Ecos( t) u T C = - E( )sin( t) T u C = - E( T By replcing u C nd LC(/T ) = π ) cos( t) T ; T u C + LC u C = (½) u C in the differentil eqution : Ecos(/T )t- LCE(/T ) cos(/t )t = 6 T = LC =.8 = 5.6 ms. (¾) ) (¼) 4) The electric energy W of the cpcitor psses to the coil tht stores it in the form of mgnetic energy nd vice vers. (¼) C ) The electric energy W of the cpcitor psses (prtilly) to the coil tht stores it in the form of mgnetic energy nd the rest of this energy is dissipted in the form of the therml energy. (¼) ) ) T = 7 ms T = 5.67 ms (¼) T slightly greter then T. (¼) ) ) becuse u C is mx. i = W mg. = energy of the circuit = electricl energy (¼) Wn W n =.4 W.4 W W n =.4 6 = c) W n = ½ Cu n u n =.95 V. Curve gives t = 7 ms = nt = n(5.6) n = (¾) J. (½) Fourth exercise A ) ) ) Em ( cos ) (¾) de m ; (½) dt = ; T = T. (½) ) ) i) ii) T' = T nd. T = ; T = T = (½) m m ( ) = m ( ) = m. (¾) T' g T' = m m m = +. () g B ) ) = cm nd = cm (½) = m = 4 - kgm ; = + = cm. () ) ) From grph : T min for ' = 4 cm. (½) m T Cte g mg T is minimum when thus T is min. when Or π mg g = m = T ( m) m = = m g is min. but 4, cm m = 4. cm T min =.5 s (½) π mg m dt = cte = d

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