Solution to PHYS 1112 In-Class Exam #1A

Solution to PHYS 1112 In-Class Exam #1A Tue. Feb. 8, 2011, 09:30am-10:45am Conceptual Problems Problem 1: A student runs northward at 5m/s, away from a vertical plane mirror, while the mirror, mounted on wheels, travels northward at 7 m/s (with both speeds given relative to the ground). The speed at which the student s image moves and its direction, relative to the ground, is (A) 3m/s northward (B) 3m/s southward (C) 2m/s northward (D) 9m/s northward (E) 9m/s southward Answer: (D) Solution: With North chosen as the positive and South then being the negative velocity direction, let velocities be defined as follows: v = +5m/s=runner rel. to ground; u = +7m/s=mirror rel. to ground; v =runner rel. to mirror; w =image of runner rel. to mirror; w=image of runner rel. to ground. To visualize this, make a drawing and indicate distances traveled by runner, mirror and image during a short time interval, e.g., during 1 second. Keep in mind that the runner and his/her image must be always be at equal distances from the mirror, and on opposite sides of the mirror. Then v = v u = 2m/s, i.e., as seen by an observer traveling with the mirror, the runner is moving in southward direction at speed of 2m/s (away from mirror). Then w = v = +2m/s, i.e., as seen by an observer traveling with the mirror, the image is moving in northward direction at speed of 2m/s. (away from mirror). Then w = w + u = +2 + 7m/s = +9m/s i.e., as seen by an observer on the ground, the image is moving in northward direction at speed of 9m/s. Problem 2: UGA waves (we didn t cover those in class, but they do obey Snell s law!) have a speed of wave propagation v A = 2097m/s in apple juice and v B = 522m/s in butter milk. Also, assume that sin(14.414 o ) = 522/2097. A narrow beam of UGA waves striking a flat horizontal interface between apple juice and butter milk, with the apple juice above and the butter milk below the interface (A) will not undergo total internal reflection if incident from above the interface with an angle of incidence of 29.0 o ; 1

(B) will always have an angle of refraction not exceeding 14.414 o if the beam is incident from below the interface without total internal reflection; (C) will undergo total internal reflection if incident from above the interface with an angle of incidence of 8.5 o ; (D) will not undergo total internal reflection if incident from below the interface with an angle of incidence of 29.0 o ; (E) will have an angle of refraction greater than the angle of incidence if the beam is incident from above the interface without total internal reflection. Answer: (A) Solution: Regardless of the beam s travel direction (incident from above or incident from below), let its angles measured from the normal to the interface be defined as follows: Θ A =angle above the interface (in medium A=apple juice); Θ B =angle below the interface (in medium B=butter milk). Snell s Law: sin(θ A ) v A = sin(θ B) v B (S1) or sin(θ B ) sin(θ A ) = v B v A (S2) or sin(θ B ) = v B v A sin(θ A ) (S3) Given velocities obey v B < v A, hence the ratio (v B /v A ) < 1 in Eq.(S3). So, by Eq.(S3), sin(θ B ) < sin(θ A ), and, in the absence of total internal reflection, Θ B < Θ A. Thus, the angle Θ B below the interface is always less than the angle Θ A above the interface. So, for incidence from above, the beam is refracted towards the normal; for incidence from below it s refracted away from the normal. Make a drawing of beams, normal and interface for incidence from above; and likewise, a drawing for incidence from below; each drawing showing this relation between the two angles. Which angle, Θ A or Θ B, is angle of incidence; and which is angle of refraction now depends on the beam s travel direction, as follows: If incident from above: Θ A = Θ =angle of incidence; Θ B = Θ =angle of refraction. If incident from below: Θ B = Θ =angle of incidence; Θ A = Θ =angle of refraction. Total internal reflection (TIR) occurs when the sine of the angle of refraction, sin(θ ), predicted by Snell s Law, exceeds the value of 1, while the sine of the angle of incidence, sin(θ), is still less than 1. In the present case, where sin(θ B ) < sin(θ A ), this can happen only for incidence from below, and only when sin(θ B ) exceeds the critical angle of incidence, Θ crit = 14.414 o. When Θ B reaches that value, Θ crit, then, sin(θ B ) = v B /v A ; and by Snell s Law, Eq.(S3), sin(θ A ) reaches 1 and Θ A reaches 90 o. When Θ B exceeds the value Θ crit, then, by Snell s Law, sin(θ A ) exceeds 1 and Θ A does not exist, i.e., there is no refracted beam anymore. Therefore, (A) is correct, since TIR cannot occur for incidence from above, for any angle of incidence Θ A. 2

Also, (B) is wrong, since Θ A is the angle of refraction for incidence from below. As shown above, when Θ B approaches Θ crit = 14.414 o, the angle Θ A, approaches 90 o and thereby clearly exceeds 14.414 o, as well as any other value below 90 o. That s also consistent with Θ A > Θ B, as shown above. Also, (C) is wrong for the same reason that (A) is correct: TIR cannot occur for incidence from above, for any angle of incidence Θ A. Also, (D) is wrong because TIR does occur for incidence from below when the angle of incidence, Θ B, exceeds Θ crit = 14.414 o. That is the case here, for Θ B = 29.0 o. Lastly, (E) is wrong, since Θ B < Θ A (as shown above); and Θ B is the angle of refraction, and Θ A is the angle of incidence, for incidence from above. Problem 3: Two identically shaped solid blocks, S and T, made from the two different transparent materials, are immersed in the same liquid L. A ray of light strikes each block at the same angle of incidence, as shown. According to the figure below, what is the relative magnitude of the indices of refraction of the solid blocks, n S and n T, and liquid, n L? L L S T (A) n T < n L < n S ; (B) n S < n T < n L ; (C) n T < n S < n L ; (D) n L < n S < n T ; (E) n S < n L < n T. Answer: (C) Solution: First off, before you read any further, make a large (re-)drawing of the figure above wherein the normal to the top surface of each solid block, and the corresponding angle of incidence Θ and angles of refraction, Θ S or Θ T, respectively, at the top surface, are clearly shown. If you really know your stuff, this quick n dirty argument will give you the answer, fast: 3

(1) As shown in the figure, the initial incident ray (above block S or T ) is refracted by the block away from the normal by both blocks S and T ; hence n S < n L and n T < n L. (2) As also shown in the figure, in block T, the incident ray is refracted (i.e., bent by the block) away from the normal more strongly than in block S. Hence, angle of incidence and index of refraction (IoR) n L of the liquid L being the same in both cases, we must have n T < n S. (3) So, combining the three inequalities from (1) and (2), we must have n T < n S < n L which is Answer (C). If you don t know your stuff quite yet, you are strongly advised to work through and absorb the the following more detailed and mathematically rigorous argument: At the top surfaces of both blocks, compare the relative sizes of: (1) angle of incidence Θ (same in either liquid), (2) angle of refraction Θ S inside block S, (3) angle of refraction Θ T inside block T. With normals to top surfaces carefully drawn in and angles carefully labeled, you can read off from the figure that Θ < Θ S < Θ T Hence, since the sine increases with angle (for angles between 0 o and 90 o ): sin(θ) < sin(θ S) < sin(θ T ). Thus, after dividing both inequalities by sin(θ) on both sides: 1 < sin(θ S) sin(θ) < sin(θ T ) sin(θ). But, by Snell s law, n T sin(θ T ) = n L sin(θ) and n S sin(θ S) = n L sin(θ), or equivalently: sin(θ T ) sin(θ) = n L n T and sin(θ S) sin(θ) = n L n S. Thus replacing sine-ratios in the foregoing inequalities by corresponding IoR-ratios: 1 < n L n S Dividing both these inequalities by n L then gives < n L n T 1 < 1 < 1 n L n S n T Taking the reciprocal (1/...) on both sides of both of these two inequalities thus gives Answer (C). (Recall here that, if 1/a < 1/b for positive numbers a and b, then a > b) Problem 4: If a virtual object (d < 0) is presented to a divergent lens (f < 0), at an absolute distance d less then the absolute focal length f, i.e., d < f, then the image is 4

(A) virtual, inverted and enlarged in height relative to the virtual object (B) virtual, inverted and reduced in height relative to the virtual object (C) virtual, erect and reduced in height relative to the virtual object (D) real, inverted and enlarged in height relative to the virtual object (E) real, erect and enlarged in height relative to the virtual object Answer: (E) Solution: If 0 > d = d (virtual object!) and 0 > f = f (divergent lens!) and d < f, then: 1/ d > 1/ f and thus 1/d = 1/f 1/d = 1/ f ( 1/ d ) = 1/ d 1/ f ) > 0. Hence d = (1/f 1/d) 1 = (1/ d 1/ f ) 1 = [ ] 1 [ ] ( f d ) f = d > 0; d f ( f d ) and d > 0 means we have a real image. Also, together with d < 0, this implies that m = d /d = + d / d > 0; and m > 0 means the image is erect. Also, since 0 < d < f, we have 0 < f d < f, f /( f d ) > 1, and thus and therefore [ ] [ ] f > 1, or d f = d ( f d ) ( f d ) m = d / d = Here m > 1 means that the image is enlarged. [ f ] ( d f ) > 1. > d ; Lastly, if you do not like general mathematical proofs, like the one above, you could also answer this question simply by plugging in some numbers for d and f. Take, e.g, f = 4cm and d = 3cm, chosen so that d < 0 and f < 0 and d < f. Then, you ll get d = (1/f 1/d) 1 = [ (1/4) + (1/3)] 1 cm = +12cm and from that m = d /d = (+12)/( 3) = +4. Therefore: Since d > 0, the image is real. Since m > 0, the image is erect, relative to object. Since m > 1, the image is enlarged in absolute height, relative to object. 5

Numerical Problems Problem 5: The state highway patrol radar guns send out a a microwave frequency of 8.090GHz. You have just passed, and are receding from, a radar speed trap driving 29.3m/s and the radar gun measures the frequency of the microwave reflecting from your car. How does this reflected microwave frequency, as detected by the receiver on the radar gun, differ from the original frequency sent out by the gun? (A) The reflected is 790Hz higher than the original frequency. (B) The reflected is 1580Hz higher than the original frequency. (C) The reflected is 395Hz lower than the original frequency. (D) The reflected is 1580Hz lower than the original frequency. (E) The reflected is 790Hz lower than the original frequency. Answer: (D) Solution: See assigned HWP on radar gun. Let u =speed of car C moving away from radar gun; f =frequency sent out by radar gun source S; f =frequency of wave received by moving car=frequency of reflected wave, as sent out by moving car; and f =frequency of reflected wave, as received by radar gun receiver R. Then f C S := f f = f u c, f R C := f f = f u c = f u c where the last approximate equality holds because u/c 1, hence f C S f. So, f := f f = f C S + f R C = 2f u c or f = 2 (8.090 10 9 Hz) ( 29.3m/s ) = 1580Hz. 3.00 10 8 m/s Problem 6: Visible light has a range of wavelengths from 400nm (violet) to 700nm (red) in vacuum. A beam of electromagnetic (EM) waves with a frequency of 714.3THz in air travels from air into water, with indices of refraction n Air = 1.00 and n Water = 1.333. To an under-water observer, the beam while traveling in water will (A) have a wavelength of 420.0nm, have a frequency of 535.9THz, and be visible to the human eye; (B) have a wavelength of 559.9nm, have the same frequency as in air, and be visible to the human eye; (C) have a wavelength of 315.1nm, have the same frequency as in air, and be invisible to the human eye; (D) have a wavelength of 315.1nm, have a frequency of 952.2THz, and be invisible to the human eye; 6

(E) have a wavelength of 315.1nm, have the same frequency as in air, and be visible to the human eye. Answer: (E) Solution: The frequency f = 714.3THz is unchanged as the wave travels from one medium into another. Its vaccum wavelength is: λ Vac = c f = 3.00 10 8 m/s 714.3 10 12 Hz = 420.0 10 9 m = 420.0nm. This is clearly within the visible vacuum wavelength range from λ Vac,min =400nm to λ Vac,max = 700nm, i.e., λ Vac,min λ Vac λ Vac,max. Therefore the EM wave beam will be visible to the human eye. Note: As we discussed in class, visibility of an EM wave depends only on the frequency, not on the wavlength it may have in any particular medium it travels through before it reaches the eye. That s why the vacuum wavelength is the relevant quantity here: the vacuum wavelength is determined only by the frequency. The vaccum wavelength range from λ Vac,min =400nm to λ Vac,max =700nm corresponds to a frequency range from to f min = f max = c = 3.00 108 m/s λ Vac,max 700 10 9 m = 429 10 12 Hz = 429THz c = 3.00 108 m/s λ Vac,min 400 10 9 m = 750 1012 Hz = 750THz The actual frequency f = 714.3THz falls within this visible range: f min f f max. Therefore, we again conclude that the EM wave beam is visible. So, one can determine visibility either by checking whether λ Vac falls within the range from λ Vac,min = 400nm to λ Vac,max = 700nm; or by checking whether f falls within the range from f min = 429THz to f max =750THz. The speed of wave propagation in a medium with index of refraction n is v = c/n. Hence, the wavelength in the medium is λ = v f = c/n f So, in water with n = n Water = 1.333, = 1 n λ = 420.0nm 1.333 c f = 1 n λ Vac. = 315.1nm. In summary, the EM wave beam in water has the same frequency as in air, f = 714.3THz, and a wavelength of λ = 315.1nm; and it is visible in water. Problem 7: A very thin, flat circular mirror lies flat on the floor. Hanging from the ceiling and centered 1.6m above the mirror is a small lamp. The circular spot formed on the flat ceiling 3.0m above the floor, by the reflection of the light from the lamp, has a radius of 1.29m. What is the radius of the mirror? 7

(A) 0.45m (B) 0.90m (C) 0.22m (D) 0.69m (E) 1.38m Answer: (A) Solution: See HWP01.12 for details and for a drawing. Let r = radius of the mirror, R = 1.29m= radius of circular spot on the ceiling, H = 3.0m= height of the ceiling above floor, and h = 1.6m= height of the lamp above floor. Then, by similar triangles (see drawing in HWP01.12 solution, with x = R r here): r h = R r H or H h = R r r = R r 1 or H + h h = R r hence, after solving for r: r = R h H + h = (1.29m) 1.6 3.0 + 1.6 = 0.45m. Problem 8: Two laser beams are passing from water through the glass wall of an aquarium tank into air, as shown below. The glass wall is x = 34cm thick and has two parallel planar surfaces. Both beams enter the wall at the same point, beam 1 at normal incidence, beam 2 at an angle α = 50 o to the wall. The two beams points of exit from the wall are y = 16.6cm apart. Air and water have indices of refraction n Air = 1.00 and n Water = 1.333, respectively. What is the index of refraction of the glass? x beam 2 α beam 1 beam 1 y beam 2 (A) 2.33 (B) 1.95 (C) 1.75 (D) 1.47 (E) 0.91 Answer: (B) Solution: Note first that beam 1 is always the normal to the surface, at both surfaces. Only the angles of incidence and refraction of beam 2 at the left surface are needed here. Namely: 8

The angle of incidence of beam 2, Θ, is the angle between beam 2 and normal ( beam 1), to the left of the left surface: Θ = 90 o α = 90 o 50.0 o = 40.0 o. The angle of refraction of beam 2, Θ, is the angle between beam 2 and normal ( beam 1) to the right of the left surface: tan(θ ) = y x or Θ = arctan ( y ) ( 16.6) = arctan = 26.023 o x 34.0 Since the medium to the left of the left surface is water and to the right of it it s glass, Snell s law gives n Water sin(θ) = n Glass sin(θ ), hence: n Glass = n Water sin(θ) sin(θ ) = 1.333 sin(40.0o ) sin(26.023 o ) = 1.95 Problem 9: A postage stamp placed 32.00cm to the left of a lens produces an image 4.00cm to the left of the lens. If the postage stamp is 2.48cm tall its image produced by the lens will be (A) real, inverted and 19.5cm tall in absolute size. (B) virtual, erect and 19.5cm tall in absolute size; (C) virtual, erect and 0.31cm tall in absolute size; (D) virtual, inverted and 19.5cm tall in absolute size; (E) real, inverted and 0.31cm tall in absolute size; Answer: (C) Solution: Object is to the left of lens; light originates from object and light passes through lens. Therefore, light enters lens from the left and exits lens to the right. Therefore left=incoming side of lens and right=outgoing side of lens. Draw this! Show: the object, the lens, arrows indicating incoming light entering and outgoing light rays exiting, labels for in to the left and out to the right of lens. Object is 32.00cm to left of lens =on incoming side: d = +32.00cm> 0, by sign convention. Image is to 4.00cm to left of lens =not on outgoing side: d = 4.00cm< 0, by sign convention. Lateral magnification: Image size: m = d d = ( 4.00) (+32.00) = +0.125 h = mh = (+0.125) (2.48cm) = 0.31cm 9

From m > 0, by sign convention: image is erect. From d < 0, by sign convention: image is virtual. From h = 0.31cm: image is 0.31cm tall. Problem 10: What is the focal length of the lens used in Problem 9? (A) 28.00cm (B) 4.57cm (C) 3.56cm (D) +4.57cm (E) +3.56cm Answer: (B) Solution: Object is to the left of lens; light originates from object and light passes through lens. Therefore, light enters lens from the left and exits lens to the right. Therefore left=incoming side of lens and right=outgoing side of lens. Draw this! Show: the object, the lens, arrows indicating incoming light entering and outgoing light rays exiting, labels for in to the left and out to the right of lens. Object is 32.00cm to left of lens =on incoming side: d = +32.00cm> 0, by sign convention. Image is to 4.00cm to left of lens =not on outgoing side: d = 4.00cm< 0, by sign convention. Use (1/f) = (1/d) + (1/d ), to find focal length f: f = ( 1 d + 1 ) 1 ( 1 = d (+32.00) + 1 ) 1cm = 4.571cm ( 4.00) Problem 11: A microsope produces a final image of a microbial cell (as seen through the eyepiece) which appears to be 2.4cm in diameter and located 40.0cm from the eyepiece on the incoming side of the eyepiece lens. If the cell has an actual diameter of 2.8 µm, what is the angular magnification achieved with the microscope, compared to viewing the cell from a 20.0cm nearpoint (reference) distance without instrument? (A) 830 (B) 1990 (C) 4290 (D) 6430 (E) 9710 Answer: (C) 10

Solution: Angle Θ e subtended at eye by final image, of size h e = 2.40cm, as seen through optical instrument, at distance d e = 40.0cm: Θ e = tan(θe ) = h e /d e = (2.40cm)/(40.0cm) = 0.060rad Angle subtended at eye by original objects, size h o = 2.8 10 4 cm, as seen without optical instrument, from reference distance d ref = d near = 20cm: Θ ref = tan(θref ) = h o /d ref = (2.80 10 4 cm)/(20.0cm) = 1.40 10 5 rad Angular magnification: M = Θ e /Θ ref = (0.060)/(1.40 10 5 ) = 4290. Problem 12: A convergent lens (Lens1), placed to the right of a small grain of sand produces an image of the grain to the right of Lens1. If a divergent lens (Lens2) of focal length f 2 = 6.42cm is now placed somewhere to the right of Lens1, the final image produced by Lens2 appears approximately 32.0cm to the left of Lens2. The object of Lens2 is (A) real and located 5.35cm to the right of Lens2 (B) virtual and located 5.35cm to the left of Lens2 (C) real and located 8.03cm to the right of Lens2 (D) virtual and located 8.03cm to the right of Lens2 (E) virtual and located 8.03cm to the left of Lens2. Answer: (D) Solution: Original object=object1 is to the left of Lens1; Lens2 is to the right of Lens1. Light originates from Object1 and light passes through Lens1, then through Lens2. Therefore, light enters Lens1 from the left and exits Lens1 to the right. Then, light enters Lens2 from the left and exits Lens2 to the right. Therefore left=incoming side of Lens2 and right=outgoing side of Lens2. Draw this! Show: Object1, Lens1, Lens2, arrows indicating incoming light entering and outgoing light rays exiting each lens, labels for in to the left and out to the right of Lens2. Since both f 2 and d 2 are given and d 2 is asked for, only image formation at Lens2 needs to be considered, as follows: Image2 is 32.0cm to the left=not on outgoing side of Lens2, hence d 2 convention: d 2 = 32.0cm< 0. Use (1/f 2 ) = (1/d 2 ) + (1/d 2), to find object distance of Object2, d 2 : d 2 = ( 1 1 ) 1 ( 1 = f 2 d 2 ( 6.42) + 1 ) 1cm = 8.03cm ( 32.0) < 0, by sign Negative object distance, d 2 < 0, means: Object2 is virtual and it is not on incoming side, by sign convention. 11

Since the incoming side is to the left of Lens2, being not on incoming side means: to the right of Lens2. So, Object2 is to the right=not on incoming side of Lens2, d 2 = 8.03cm from Lens2. 12