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Transcript
Problem 1 In the figure shown above, an object is placed a distance in front of a converging lens equal to twice the focal length f1 of the lens. On the other side of the lens is a concave mirror of focal length f2 separated from the lens by a distance 2∗(f1 +f2 ). Determine the location, type, orientation, and lateral magnification of the final image, as seen by an eye looking toward the mirror through the lens and just past (to one side of) the object. What is the type of the final image? Physics ideas: • Each optical element in the problem can be treated separately in sequence • A lens directs parallel beams to the focus and vice versa • Reflections maintain the same angle with respect to normal of the interface Consider the first optical element, i.e. the lens. We can draw three rays to determine the location of the image due to the lens (technically two rays should suffice): 1. Parallel to the optical axis through the focus 2. Through the focus and then parallel to the optical axis 3. Through the optical center 1 Now that we have the image of the lens, we use it as the object for the mirror: The image is upright and real. Additionally, the image is smaller than the object (you can check by using the lens maker’s formula). 2 Problem 2 A plane electromagnetic wave, with wavelength λ, travels in vacuum in the positive x direction with its electric vector E, of amplitude E0 , directed along y axis. What is the time-averaged rate of energy flow in watts per square meter associated with the wave? y E v x z Physics ideas: • The Poynting vector gives the direction of motion of an EM wave • The magnitude of the Poynting vector is the rate of flow of energy ~ ×B ~ ~= 1E (1) S µ0 Recall that for an EM wave, there is a relationship between the electric field ~ and the magnetic field (B), ~ specifically: (E) ~ = c|B| ~ |E| (2) With this relationship, we can obtain an expression for the Poynting vector: ~= 1 E ~ ·E ~ S cµ0 (3) ~ Here we note that we Next we need an expression for the electric field (E). have a periodic function, specifically a sine (or cosine) wave. Since we’re asked about the time averaged value, we can choose any phase since we average over a single period. Since it matches my diagram, I’ll choose ~ = −E0 sin(ωt)ˆĵ E (4) Lastly, recall the general definition of an average (discrete or continuous): N Vavg = 1X vi N i=1 Vavg = 3 1 X Z X vdx 0 (5) Plugging (4) into (3) and applying the definition of an average (5), we obtain: Savg = E02 cµ0 T T Z sin(ωt)2 dt (6) 0 Savg = 4 1 E02 2 cµ0 (7) Problem 3 A plane EM wave with wavelength λ, travels in vacuum in the x direction with its electric field vector E of amplitude E0 directed along the y axis. If E = E0 sin(kx − ωt), what are the values of k and ω? y E v x z Here we recall that there is a specific relationship between the wavelength (λ) and the frequency (ν) of a wave that obeys the wave equation: ~ ~ 1 δ2 E δ2 E = 2 2 2 δx c δt (8) Specifically: λ k c= (9) T ω The problem gives the value of the wavelength and we are told that the EM wave travels in vacuum so we know the speed of the wave. As a note, recall that the wavenumber (k) is related to the wavelength (λ) in the same way that the angular frequency (ω) is related to the period (T ). c = λν = k= 2π λ ω = 2πν = 5 2π T (10) Problem 4 A plane EM wave with wavelength λ, travels in vacuum in the x direction with its electric field vector E of amplitude E0 directed along the y axis. What is the amplitude of the magnetic field associated with the wave? ~ and |B| ~ in Problem 2 We’ve already addressed the relationship between |E| (see equation 2). 6 Problem 5 A concave shaving mirror has a radius of curvature of R. It is positioned so that the (upright) image of a man’s face is M times the size of the face. How far is the mirror from his face? Physics ideas: • Using a concave mirror, magnification of an object is only possible if the object is between the mirror and focal point. • Magnification of the mirror is related to the ratio of the image and object distances • Radius of curvature of the mirror is related to the focal distance of the mirror • Lens maker’s equation relates the image and object distances to the focal length Starting with the first idea, we can draw a ray diagram. C f We know the magnification is the ratio of the image over the object. Be careful here, the negative sign ist: M =− simage |simage | = sobject sobject 7 (11) The focal length of the (spherical) mirror is given by: f =− R 2 (12) Note that since the mirror is convex, we set the radius R < 0 so the negative signs will cancel (see below). Lastly, the lens maker’s equation stats: 1 1 1 = + f sobject simage (13) Subbing in equations 11 and 12 and recalling that the image distance is negative we obtain: 1 1 2 = − |R| sobject M sobject (14) Which we can then solve for sobject sobject = R 1 (1 − ) 2 M 8 (15) Problem 6 A plane electromagnetic wave, with wavelength λ, travels in vacuum in the positive x direction with its electric vector E, of amplitude E0 , directed along y ~ associated with the wave? axis. What is the direction of the magnetic field ((B)) Physics ideas: • The Poynting vector gives the direction of motion of an EM wave Essentially we have already discussed the solution (see problem 2). We simply need to recall the definition of the Poynting vector and how to apply the right-hand-rule. ~= 1E ~ ×B ~ S µ0 (16) ~ 1. Fingers in the direction of the first vector (E) ~ 2. Palm in the direction of the second vector (B) ~ 3. Thumb gives the direction of the third vector (S) ~ (fingers) is in the ĵ direction and S ~ is in the î direction. Here we know that E ~ to be in the k̂ direction. This requires B î = ĵ × k̂ 9 (17) Problem 7 A plane electromagnetic wave, with wavelength λ, travels in vacuum in the positive x direction with its electric vector E, of amplitude E0 , directed along y axis. What is the frequency of the wave? This problem has been previously answered in Problem 3. 10 Problem 8 In the figure above, a ray of light is perpendicular to the face ab of a glass prism (nglass = 1.5). Find the largest value for the angle φ so that the ray is totally reflected at face ac if the prism is immersed in a fluid of index of refraction nf luid . a nfluid nglass φ b c Physics ideas: • Snell’s law gives the relationship between index of refraction and angle of refraction. • Total internal reflection occurs when ’exiting’ angle is 90◦ . Let’s first draw the path of the light through the prism. If the incident angle is 0◦ then there is no deflection at the ab interface according to Snell’s law (since n1 & n2 6= 0). Furthermore, at the ac interface the exit angle is 90◦ . n1 sin(θ1 ) = n2 sin(θ2 ) (18) a nfluid φ 90o-φ nglass φ b c Applying Snell’s law to the ac interface: nglass sin( π π − φ) = nf luid sin( ) 2 2 (19) Using the trig identity sin(θ ± φ) = cos(θ)sin(φ) ± cos(φ)sin(θ) 11 (20) the above equation can be rewritten as: nglass cos(φ) = nf luid φ = cos−1 ( nf luid ) nglass (21) (22) Note that nf luid < nglass in order for total internal reflection to occur. MAKE SURE YOUR CALCULATOR IS IN THE CORRECT UNITS! 12 Problem 9 Figure shows a small light bulb suspended y2 above the surface of the water in a swimming pool. The water is y1 deep, and the bottom of the pool is a large plane mirror. How far below the mirror’s surface is the image of the bulb? y2 y1 Water Mirror Physics ideas: text If there were no water, then the total distance would be 2(y1 + y2 ), i.e. the bulb will look like it is a distance of (y1 + y2 ) below the surface of the mirror. However, since the water is present, light will refract and so the total distance will actually look shorter. 13 Problem 10 Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of θ with the normal to the surface, while in the glass it makes an angle of φ with the normal. What is the index of refraction of the glass? Physics ideas: • Snell’s law relates the angles to the index of refraction φ θ nvacuum sin(θ) = nglass sin(φ) nglass = sin(θ) sin(φ) (23) (24) Sanity check tells us that the nglass > nvacuum which is the case since sin(θ) > sin(φ). 14 Problem 11 Initially unpolarized light is sent through three polarizing sheets whose polarizing directions make angles of θ1 = θ2 = θ3 with the direction of y axis. What percentage of the initial intensity is transmitted by the system of three sheets? θ1 θ3 θ2 Physics ideas: • Unpolarized light → polarized light cuts the intensity by 12 . • Polarized light → polarized light decreases the intensity by a factor of cos2 (φ), where φ is the angle between the initial and final polarizations. Take one filter at a time! I1 = I0 2 (25) I2 = I1 cos2 φ = I1 cos2 (π − 2θ) (26) I3 = I2 cos2 φ = I2 cos2 (π − 2θ) (27) Putting everything together, we have: 1 If = I0 cos4 (π − 2θ) 2 (28) Applying some trig. formulas 1 If = I0 (cos(π)cos(2θ) + sin(2θ)sin(π))4 2 (29) 1 If = I0 cos4 (2θ) 2 (30) 15 Problem 12 A double-slit arrangement produces interference fringes for sodium light (λ = 589nm) that are θ apart. What is the angular fringe separation if the entire system is immersed in acetone (nacetone )? Physics ideas: • Fringes are caused by a superposition of waves • Constructive interference corresponds to an integer phase difference of 2π. • A phase difference can be cause by different path lengths, specifically integer differnces of λ. • Immersion in a different medium changes the wavelength of the light (but not the frequency). θ d ΔL We’ll start by using the given information (in air) to determine the slit separation. The path length difference (DeltaL) is ∆L = nλ (31) From the geometry of the situation and assuming that the screen is very far from the slits, we have: dsin(θ) = ∆L = nλ (32) We are told both the angle θ for the first fringe (n = 1) and the wavelength λ so we can obtain the slit separation distance d. 16 d= λ sin(θ) (33) We also need an expression for the wavelength when the apparatus is immersed in acetone. We recall that the frequency of the light does not change when entering a different medium. νair = νacetone (34) But we know that since the nacetone > 1 the speed of light in the acetone is less than that in air: nacetone = c vacetone (35) Lastly, the speed of light in a medium is given by: v = λν (36) Taken all together, we arrive at the conclusion that: c vacetone (37) λair νair λacetone νacetone (38) nacetone = nacetone = λacetone = λair nacetone (39) Now we are in a position to find θacetone using equation 33 d= λacetone sin(θacetone ) sin(θacetone ) = (40) λair sin(θair ) nacetone λair (41) sin(θair ) nacetone (42) sin(θacetone ) = 17 Problem 13 An electromagnetic wave is traveling in the negative y (−ĵ) direction. At a particular position and time, the electric field is along positive z (k̂) axis and has a magnitude of E0 . What is the magnitude and direction of the magnetic field at that position and at that time? Physics ideas: ~ and B ~ fields are related (equation 2) • For EM wave, E • The Poynting vector gives the direction of motion of an EM wave ~ and B ~ are in phase • E This is closely related to Problem 2. 18 Problem 14 he electric field of a certain plane electromagnetic wave is given by ~ = h0, 0, E0 cos(1015 π(t − x )i (43) E c All quantities are in SI units. The wave is propagating in the positive x direction. Write an expression for the components of the magnetic field of the wave. Physics ideas: ~ and B ~ fields are related (equation 2) • For EM wave, E • The Poynting vector gives the direction of motion of an EM wave ~ and B ~ are in phase • E ~ |E| (44) c Using the right hand rule and the Poynting vector, we find that the magnetic field is in the −ĵ. ~ = |B| ~ = h0, − 1 E0 cos(1015 π(t − x ), 0i B c c 19 (45) Problem 15 We see a full Moon by reflected sunlight. How much earlier did the light that enters our eye leave the Sun? The Earth-Moon and Earth-Sun distances are 3.8 × 105 km and 1.5 × 108 km. Physics ideas: • Speed of light in vacuum is constant • Total time is the distance traveled divided by the speed d (46) v Essentially the moon and Earth are the same distance from the sun, so the time difference is due to light having to travel from the moon to the Earth ∆t = ∆t = dE−M + dE−S c 20 (47) Problem 16 When monochromatic light is incident on a slit w wide, the first diffraction minimum is observed at an angle of θ from the direction of the incident light. What is the wavelength? Physics ideas: • Superposition of waves results in interference. • Destructive interference occurs when waves are π radians out of phase. Here we take the same approach as the double slit experiment. However, since the entire distance w is illuminated, consider 4 rays of light, one from each extreme of the slit (A and D) and two in the middle of the slit (B and C). A B C θ w D ΔL ΔL If rays A and B are perfectly out of phase, then rays C and D will also be perfectly out of phase. Assuming that the screen is very far away from the slit, we can write λ w sin(θ) = (48) 2 2 We proceed to enter in the width and wavelength to determine the angle. ∆L = 21 Problem 17 A generator supplies V to the primary coil of a transformer of Nprimary turns. If the secondary coil has Nsecondary turns, what is the secondary voltage? The transformer equation is Vprimary Nprimary = Vsecondary Nsecondary 22 (49) Problem 18 At what frequency does the maximum current flow through a series RLC circuit containing a resistance of R, an inductance of L, and a capacitance of C? Physics ideas: • Need to consider the impedance of an RLC circuit with AC current. • Maximum current corrsponds to a minimum impedance. • The impedance is minimized at resonance of the circuit. • Resonance of a RLC circuit (in series) requires the reactance of the inductor to equal that of the capacitor. The impedance of the circuit is: Z = ZR + ZL + ZC (50) 1 iωC (51) Z = R + iωL + r 1 2 ) ) (52) ωC To minimize the impedance we set the reactance of the inductor equal to that of the capacitor Z= R2 + (ωL − 1 ωC r 1 ω= LC ωL = 23 (53) (54) Problem 19 An RLC series circuit is connected to an AC power supply with amplitude V and a frequency of ν. What is the average power dissipated in terms of R, L, &C? Physics ideas: • The impedance of the circuit can be used to determine the current through the circuit. • The energy is lost through the resistor in the circuit. • We can obtain the angular frequency from the frequency. We first find the expression of the impedence (see equation 52). Once we have this, we can use Ohm’s law to determine the current flowing through the circuit. V (t) = I(t)Ztotal (55) V sin(2πνt) I(t) = q 1 R2 + (2πνL − 2πνC )2 ) (56) Now the power dissipated in a circuit is given by P = I 2R (57) Note that we are only concerned here with the resistor, the capacitor and inductor do not cause energy to be lost. Taking the time average is 1 hP i = T hP i = Z 0 T V 2 Rsin2 (2πνt) 1 R2 + (2πνL − 2πνC )2 ) 2(R2 V 2R + (2πνL − 24 1 2 2πνC ) ) (58) (59) Problem 20 For d = na where how many bright interference fringes lie in the central diffraction envelope? a d =na θ ΔL Physics ideas: • Single slit diffraction will create a central peak. • Double slit interference will create multiple peaks within the central diffraction peak. The angular width of the central , single slit diffraction peak can be found using equation 48 and the angle of the double slit interference is given by equation 33. If we set the angles to be the same, then we can find the value of m, the number of fringes due to the double slit interference. λ (2m − 1)λ = a 2(na) (60) Where m = 1, 2, 3.... I’ve used the expression 2m − 1 to express only the odd integers. 2n + 1 (61) 2 First, we take the floor of the number (round down to the nearest integer) since we are concerned about the total number of contained peaks. Also, the equation for the double slit interference gives the number of maximum to one side of the optical axis (either above or below) so we’ll need to double the given number. Finally, the double slit eqaution ignores the central maximum. m= 25 Problem 21 A capacitor and a resistor are connected in parallel across an AC source. The reactance of the capacitor is equal to the resistance of the resistor. Assuming that iC (t) = Isin(ωt), sketch graphs of iC (t) and iR (t) on the same axes. Physics ideas: • VR is proportional to the current flowing through the resistor (Ohm’s law). • VC is proportional to the current on the capacitor. The current through the capacitor is given and two of the four graphs for this current are not sine functions and therefore can immediately be eliminated. Furthermore, the current through the resistor should be π2 radians out of phase with that of the capacitor. This is because the voltage drop across the capacitor is proportional to the total charge whereas that of the resistor is proportional to the current through the resistor. 26 Problem 22 A television set draws an rms current of Irms from a 60-Hz power line. Find (a) the average current, (b) the average of the square of the current, and (c) the amplitude of the current. Since the current is sinusoidal, the average of the current is 0. T Z sin(ωt) = 0 (62) 0 The definition of Irms is: s Irms = 1 T Z T I2 (63) 0 So the average of the square of the current is simply 1 T Z T 2 I 2 = Irms (64) 0 Lastly the average value of sine2 over a full period is places in this document. Therefore r I02 Irms = 2 √ I0 = 2Irms 27 T 2 as used several other (65) (66)