PHY 114 Summer 2009 Final Exam Solutions

Transcription

1 PHY 4 Summer 009 Final Exam Solutions Conceptual Question : A spherical rubber balloon has a charge uniformly istribute over its surface As the balloon is inflate, how oes the electric fiel E vary (a) outsie the balloon, at some point well away from the surface? (b) at the outer surface of the balloon? (c) insie the balloon? Assume the balloon remains spherical uring inflation (a) Since the fiel outsie for a spherically symmetric charge istribution is essentially ientical to a point charge at the center of the istribution, the fiel far away stays the same (b) As the raius of the balloon increases, the fiel on the surface ecreases as the square of the raius (c) Since there are no charges insie the balloon, the flux is zero An since there is spherical symmetry, the fiel is zero as well Conceptual Question : In the following figure, a close loop moves at a constant spee parallel to a long, straight, current-carrying wire Is there a current in the loop? If so, is this current circulating clockwise or counterclockwise? The magnetic fiel create by the current in the straight wire is a function of the raial istance from the wire However, as the loop moves parallel to the wire, an oes not get closer or farther away, the flux through it oes not change over time Therefore, there is no inuce current in the loop Conceptual Question 3: The lens in an overhea projector forms an image P` of a point P on an overhea transparency If the screen is move closer to the projector, how must we ajust the lens to keep the image on the screen in focus? Since the screen is move closer, the image istance is being ecrease To compensate, the object istance must increase an therefore, the lens must be move up

2 Conceptual Question 4: Two glass slies form a narrow wege as shown in the figure Taking into account only the reflections from surfaces A an B, escribe how the interference pattern will look like when the slies are illuminate with white light an the pattern is viewe from above A B Interference occurs between waves that reflect from A an B so that the spacing between the two slies acts as the thin film The light reflecte from A oes not have a 80 phase change upon reflection while the light reflecte from B oes Therefore the conition for constructive interference is: t ( m + )λ an the conition for estructive interference is: t mλ At the point where the two slies meet the estructive interference conition is satisfie for every wavelength (t 0) so we will see a ark fringe As the istance between the slies increases, the conition for constructive interference will be satisfie for light of larger wavelengths So looking from above we will see a succession of fringes with a rainbow of colors interrupte by white or ark bans where the interference conitions are satisfie by multiple wavelengths Conceptual Question 5: ight of frequency f illuminating a long narrow slit prouces a iffraction pattern (a) If we switch to light of frequency 3f, oes the pattern expan away from the center or contract towar it? (b) Does the pattern expan or contract if, instea, we submerge the equipment in clear corn syrup? (a) If the frequency of light is increase, its wavelength will be less Since the iffraction pattern is proportional to the ratio of the wavelength to the slit with, the pattern will contract (less iffraction) (b) Since the wavelength of light is going to be shorter in another meium with higher inex of refraction, again the pattern will contract

3 Problem : In the figure two semicircular curve plastic ros, one uniformly charge with +Q, the other uniformly charge with -Q, form a circle of raius centere at P (a) What is the magnitue an irection of the electric fiel at P? (b) What is the electric potential at P? +Q -Q P This is very similar to the practice test +Q -Q question We can treat each semi-circle separately For the +Q half, iviing the arc into infinitesimal pieces with length l, charge q an subtening an angle θ, we can see that every piece above E x E x the horizontal has a symmetric P P counterpart below the horizontal for which the vertical components of the electric fiel cancel leaving only the horizontal (x) component Also, since the charge is positive, the fiel at P points away from the arc In the same way the Q half will only prouce a fiel that has an x component pointing towar the negative half Therefore the fiel contributions from the two sies will be equal in magnitue an in the same irection, so they will a ooking at the left half (+Q): (a) The ro length is: l π The charge ensity is, λ Q l Then, q λl λθ q q λrθ E ke an E cosθ cosθ x ke k e λrθ λ λ Q E x ke cosθ k cos k k e θ θ e e π Q The total fiel is: ET Ex 4ke, pointing towars the +x-irection π (b) For potential, again, each semicircle can be treate separately For the left half, π π q λθ Q V ke ke an V ke ke ke ke λ θ λ θ πλ For the right half, V π π π keλθ keλ θ keπλ k π π π Therefore the total potential at P is: V V + V 0 T e Q 3

4 Problem : A solenoi of raius r an length l has N turns an I carries a current I (a) Calculate the flux through the surface of a r/ isk of raius r/ that is positione perpenicular to an centere on the axis of the solenoi (b) If the current in the solenoi is increasing l at a rate of 05 A/s, what is the inuce current in a metallic ring positione perpenicular to an centere on the axis of the solenoi? The ring has a raius r r an resistance per unit length of r (a) The fiel prouce by the solenoi is ieally a uniform fiel, confine to the insie of the solenoi an irecte along the axis of the solenoi N B sol μ 0nI μ0 I l The flux through the isc is: Φ B isk v B sol v A B sol A isk N r μ0 Nπr μ 0 Iπ l 4l I (b) The inuce emf is equal to the rate of change of the magnetic flux through the ring For the area we only take the area of the solenoi because the fiel is confine within the solenoi Φ E t B ( BA) t π r t μ0 N l I πr μ0 N l I t N 05πr μ0 l The current is, I E N r N 05 r 0 π μ μ 0 l ( π r) 8l 4

5 Problem 3: In the arrangement at the right, light is initially in material an is incient on material After refraction, it travels through material an is incient on material 3 at the critical angle The inexes of refraction are, n 6, n 4, n 3 (a) Fin θ (b) If θ is increase, will the light ray escape into material 3? n 3 n n θ θ θ n3 (a) The critical angle between an 3 is: θ c arcsin arcsin 59 an θ θ c n 4 n 4 Then θ 90 θ 3 an θ arcsin sinθ arcsin sin3 6 8 n 6 (b) If θ is increase, θ increases an θ will ecrease Being incient at less than the critical angle, the light will escape in to material 3 See the ash-ot line as an example Problem 4: Two parallel slits are illuminate with monochromatic light of wavelength 500 nm An interference pattern is forme on a screen some istance from the slits, an the fourth ark ban is locate 68 cm from the central bright ban on the screen (a) What is the path length ifference corresponing to the fourth ark ban? (b) What is the istance on the screen between the central bright ban an the first bright ban? Assume that the angles involve are small enough thatsinθ tanθ 7 m (a) The fourth ark ban occurs at m 3 δ ( + ) λ ( nm) nm λ m + (b) For ark bans, ( ) y m For the fourth ark ban, Then, 9600 ( 500nm)( ) 7 7 y3 λ ( 750nm) 68cm y λ For the first bright ban, ( nm)( ) mm 5

6 Problem 5: In a spectrometer, a grating that has 000 lines/mm is use to sen an unknown spectrum of light to a 0 cm long etector array that is place on a screen The screen is 50 cm away an the etector can etect the full extent of the first orer spectrum when its lower en is 35 cm away from the center of the slit (a) Fin the minimum an maximum wavelengths of the spectrum (b) Does any part of the spectrum iffract into the secon orer? θ θ 50 cm λ λ 0 cm 35 cm The grating s slit spacing is, N ( 000) mm μm 35cm (a) For the shortest wavelength, θ arctan 35 50cm Then, θ λ ( μm) sin nm sin For the longest wavelength, θ ( ) cm arctan 4 50cm Then, θ λ ( μm) sin 4 669nm sin (b) In the secon orer, for the short wavelength sie of the spectrum, sinθ λ ( 5735nm) λ sinθ 47 > then no iffraction occurs in the secon orer μm 6

7 Multiple Choice Questions: A battery is use to charge a parallel-plate capacitor, after which it is isconnecte Then the plates are pulle apart to twice their original separation This process will ouble the: a) capacitance b) surface charge ensity on each plate c) store energy ) electric fiel between the two places e) charge on each plate Since the battery is isconnecte, the charge remains constant Doubling the plate separation U Q C halves the capacitance an oubles the store energy ( ) The rate at which electrical energy is use may be measure in: a) watt/secon b) watt secon c) watt ) joule secon e) kilowatt hour The rate of energy change is power which is in units of watts 3 ines of the magnetic fiel prouce by a long straight wire carrying a current: a) are in the irection of the current b) are opposite to the irection of the current c) leave the wire raially ) are circles concentric with the wire e) are lines similar to those prouce by a bar magnet The magnetic fiel will loop on itself an be perpenicular to the current that prouces it 4 In a purely resistive circuit the current: a) leas the voltage by /4 cycle b) leas the voltage by / cycle c) lags the voltage by /4 cycle ) lags the voltage by / cycle e) is in phase with the voltage Since a resistor oes not store energy, the current an voltage are in phase 7

8 5 Maxwell's equations preict that the spee of light in free space is: a) an increasing function of frequency b) a ecreasing function of frequency c) inepenent of frequency ) a function of the istance from the source e) a function of the size of the source The spee of light in free space is constant 6 Where must an object be place in front of a converging lens in orer to obtain a virtual image? a) At the focal point b) At twice the focal length c) Greater than the focal length ) Between the focal point an the lens e) Between the focal length an twice the focal length When the object istance is less than the focal length, the image istance is negative an the image is virtual 7 In a Young's ouble-slit experiment, the slit separation is ouble This results in: a) an increase in fringe intensity b) a ecrease in fringe intensity c) a halving of the wavelength ) a halving of the fringe spacing e) a oubling of the fringe spacing If the slit separation is ouble, the angle shoul be halve for a given interference conition 8 Monochromatic light, at normal incience, strikes a thin film in air If λ enotes the wavelength in the film, what is the thinnest film in which the reflecte light will be a maximum? a) much less than λ b) λ/4 c) λ/ ) 3λ/4 ε) λ In this case there is a 80 phase ifference between the light reflecte from the top an the bottom of the film so the constructive interference conition is nt ( m + )λ For minimum thickness, m 0 an t λ n λ 4 4 n 8

9 9 Two stars that are close together are photographe through a telescope The black an white film is equally sensitive to all colors Which situation woul result in the most clearly separate images of the stars? a) Small lens, re stars b) Small lens, blue stars c) arge lens, re stars ) arge lens, blue stars e) arge lens, one star re an the other blue The ayleigh criterion for circular apertures is inversely proportional to lens iameter an irectly proportional to wavelength 0 The iagrams show four pairs of polarizing sheets, with the polarizing irections inicate by ashe lines The two sheets of each pair are place one behin the other an the front sheet is illuminate by unpolarize light The incient intensity is the same for all pairs of sheets ank the pairs accoring to the intensity of transmitte light, least to greatest a),, 3, 4 b) 4,,, 3 c), 4, 3, ), 4,, 3 e) 3,, 4, Since intensity is proportional to the square of the cosine of the angle between the two polarizers, the orer shoul be from the largest angle to the smallest angle (: 60, : 90, 3: 30, 4: 75 ) 9