Solution to PHYS 1112 In-Class Exam #1B

Transcription

1 n observer O, facing a mirror, observes a light urce S. Where does O perceive the mirror age of S to be located? Physics Solution to PHYS 1112 In-Class Exam #1B Thu. Feb. 5, 2009, 2:00pm-3:15pm Conceptual Problems. The image of S cannot be seen by O in this configuration. Problem 1: An observer O, facing a mirror, observes a light source S, with the observer, source and mirror positioned as shown here: O mirror 1 S Where does O perceive the mirror image of S to be located? (A) Position 1. (B) Position 2. (C) Position 3. (D) Position 4. (E) The image of S cannot be seen by O in the configuration shown above. Answer: (D) Any rays emerging from S which strike the mirror are reflected off the mirror (acc. to Archimedes law) in such a way that, after reflection, they appear to be emerging from Position 4. This is true (a) for all rays emerging from S regardless of where the rays from S strike the mirror; and (b) regardless of how far up or down the mirror extends, i.e., even if the mirror surface does not cover the wall area immediately opposite of S. It is critically important that you draw your own big, clean picture of this set-up, tracing a few (at least two) rays from S to the mirror and then, after reflection with Θ = Θ, tracing the reflected rays backwards to see where they intersect behind the mirror: at Position 4. This is similar to the in-class quiz on buying the Shortest Possible Mirror on the Wall. You can still see your own feet (on the ground) in this mirror, even though the mirror extends downward to only half of your height above the ground. 1

2 Problem 2: Sound waves (including ultrasound) have a speed of wave propagation v Air = 346m/s in air and v Water = 1497m/s in water. Also, note that sin( o ) = 346/1497. A narrow ultrasound beam striking the flat water surface of your swimming pool (A) will undergo total internal reflection if incident from below the water surface for any angle of incidence greater than o ; (B) will undergo total internal reflection if incident from above the water surface with an angle of incidence of 8.5 o ; (C) will undergo total internal reflection if incident from below the water surface with an angle of incidence of 8.5 o ; (D) will have an angle of refraction smaller than the angle of incidence if the beam is incident from below the water surface; (E) will have an angle of refraction greater than the angle of incidence if the beam is incident from below the water surface. Answer: (D) Refraction of sound at the air-water interface differs from refraction of light at the same interface in that light travels slower in water than in air, whereas sound travels faster in water than in air. Total internal reflection (TIR) occurs if (a) the sound beam is incident from the low-speed medium (=air, for sound) and gets refracted into the high-speed medium (=water, for sound) and (b) the angle of incidence Θ 1 exceeds the critical angle, defined by sin(θ (crit) ) = v 1 /v 2, with v 1 < v 2. Hence, the sound beam can undergo TIR only if incident from air into water, but never if incident from water into air! Also, for incidence from air into water, sin(θ (crit) Air ) = 346/1497, i.e., Θ (crit) = o from the information given. Air The foregoing two TIR prerequsites (a) and (b) are not both satisfied by the conditions stated in (A), (B) or (C). For this reason, (A), (B) and (C) are wrong. [In (A) and (C) incidence is from below, i.e. from water into air, hence TIR cannot occur, regardless of angle of incidence. In (B), incidence is from above, i.e. from air into water, as required for TIR condition (a); but the angle of incidence, Θ = 8.5 o, is less than the critical angle = o, hence again TIR does not happen.] Θ (crit) Air Regardless of whether the sound beam is incident from air into water or from water into air, by Snell s law sin(θ Water ) = (v Water /v Air ) sin(θ Air ) where Θ Air and Θ Water are the angles of the sound beam in air and water, respectively, measured from the normal, and v Air = 346m/s and v Water = 1497m/s are the corresponding speeds. Since v Water > v Air this implies that sin(θ Water ) > sin(θ Air ) and hence Θ Water > Θ Air That is, as a matter of general principle, the beam always has the greater angle to the normal when traveling in the medium with the greater speed. (And this is always true for any kind of wave being refracted between any two different media!) 2

3 So, if incident from air, the sound beam is refracted away from the normal upon entering the water; and if incident from water, the sound beam is refracted towards the normal upon entering the air (provided that refraction occurs at all, i.e., absent TIR). For these reasons answer (D) is correct and (E) is wrong. Problem 3: If a real object is placed more than two focal lengths away from a concave mirror(f > 0), then the image is (A) virtual, erect and enlarged in height relative to the object (B) virtual, erect and reduced in height relative to the object (C) real, inverted and reduced in height relative to the object (D) real, inverted and enlarged in height relative to the object (E) real, erect and reduced in height relative to the object Answer: (C) Being more than two focal lengths away means d > 2f. From that, and from f > 0, it follows that (1/d) < 1/(2f) = 1 2 (1/f) (by taking the inverse of the inequality d > 2f). Therefore (1/d ) = (1/f) (1/d) > (1/f) 1 2 (1/f) = 1 2 (1/f) and hence [by taking the inverse of the inequality (1/d ) > 1 (1/f) and using d > 2f again] 2 it follows that d < 2f < d but also (1/d ) > 0 which implies d > 0, i.e., the image is real. From 0 < d < d and m = d /d it then follows that m = d /d < 1 i.e., the image is reduced in absolute height, compared to the object; and it also follows that m = d /d < 0 i.e., the image is inverted, relative to the object. So, the answer is (C). Lastly, if you do not like general mathematical derivations, 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 = 10cm so that d > 2f = 8cm, i.e., the object is placed more than two focal lengths from mirror. Then, you ll get d = (1/f 1/d) 1 = [(1/4) (1/10)] 1 cm = +(20/3)cm and from that m = d /d = (20/3)/10 = 2/3. Therefore: Since d > 0, the image is real. 3

4 Since m < 0, the image is inverted, relative to object Since m < 1, the image is reduced in absolute height, relative to object. Problem 4: Two non-parallel light rays initially converge to a single point on a flat screen so that the normal to the screen is enclosed between the two incident rays, as shown here: A slab of glass is now placed somewhere in front of the screen, in the path of the light rays, so that the slab s two planar glass surfaces are parallel to the screen. The index of refraction (IoR) of the glass is greater than the IoR of the surrounding air. Where is the new convergence point of the rays? (A) On the screen (unchanged); (B) Toward the glass slab, in front of the screen (i.e., between slab and screen); (C) Further away from the glass slab, behind ( to the right of) the screen; (D) Inside the glass slab; (E) Any of the above [(A), (B), (C) or (D)] are possible, depending on the angles of incidence of the two rays. Answer: (C) At the 1st refraction, when entering the slab, both rays are refracted towards the normal inside the slab, since n Glass > n Air. So, Ray 1 is bent upward and Ray 2 is bent downward, inside the slab, as shown in the figure above. Then, at the 2nd refraction, when leaving the slab, both rays are refracted away from the normal (again because n Glass > n Air ) so that the 2nd refraction is exactly un-doing the change of direction the two rays suffered as a result of the 1st refraction. This happens as a result of Snell s law and as a result of the fact that, inside the slab, the angle of refraction 4

5 at the left surface equals angle of incidence at the right surface (... and these two angles are equal because the left and right slab surfaces are parallel planes). As a consequence, (a) the direction of Ray 1 exiting the slab (to the right of slab) is parallel to the direction of Ray 1 entering the slab, but shifted upward relative to the entering Ray 1; and (b) likewise the direction of Ray 2 exiting the slab (to the right of slab) is parallel to the direction of Ray 2 entering the slab, but shifted downward relative to the entering Ray 2. In the absence of the slab, without refraction, both rays would converge to their point of intersection located exactly on the screen, as indicated by dotted lines in the figure. Hence, as a result of the upward shift of Ray 1 and of the downward shift of Ray 2 (both caused by the refractions at the slab), this point of convergence is shifted to the right, ı.e., behind the screen, as shown in figure. Numerical Problems Problem 5: A beam of monochromatic light with a frequency of Hz in air travels from air into water, with indices of refraction n Air = 1.00 and n Water = 4/3, in air and water, respectively. To an under-water observer, the light beam while traveling in water will (A) have a wavelength of 840nm, have the same frequency as in air, and be invisible to the human eye; (B) have a wavelength of 630nm, have the same frequency as in air, and be invisible to the human eye; (C) have a wavelength of 840nm, have a frequency of Hz, and be invisible to the human eye; (D) have a wavelength of 840nm, have a frequency of Hz, and be visible to the human eye. (E) have a wavelength of 630nm, have a frequency of Hz, and be visible to the human eye; Answer: (B) The frequency f does not change when the light wave travels from air into water. However, the speed of light does change, from v Air = c/n Air = c = m/s to v Water = c/n Water. Hence, the wavelength also changes from λ Air = v Air /f to λ Water = v Water /f = (v Air /f) (n Air /n Water ) = λ Air (n Air /n Water ). Therefore, f = Hz in both air and water and λ Air = ( m/s)/( Hz) = 840nm, but λ Water = (840nm) [1.00/(4/3)] = 630nm. Here, λ Air is (approximately) the same as the vacuum wavelength (λ Vacuum c/f). Visibility of light to the human eye is determined only by its frequency f, or equivalently, by its vacuum wavelength: light is visible to the human eye if λ Vacuum c/f is between 700nm and 5

6 400nm. The light ray in this problem is not visible, since its 840nm vacuum wavelength does not fall within the nm visible range. The 630nm value of the actual wavelength in water is irrelevant for visibility to the human eye: it s only the frequency in water that matters! Problem 6: The state highway patrol radar guns send out a frequency of 9.34GHz. You re approaching a radar speed trap and the radar gun detects the frequency of the radar wave reflecting from your car. The reflected frequency is measured to be % larger than the original frequency sent out by the radar gun. Does this percentage of frequency change depend on the frequency sent out by the radar gun? How fast were you driving? (A) driving 20.94m/s, percentage does not depend on frequency; (B) driving 10.47m/s, percentage does depend on frequency; (C) driving 10.47m/s, percentage does not depend on frequency; (D) driving 5.24m/s, percentage does depend on frequency; (E) driving 5.24m/s, percentage does not depend on frequency. Answer: (C) Please read and work carefully through the posted solution of, e.g., HW Problem P01.09, for details: the frequency shift between original wave sent out by radar gun (f) and reflected wave (f ) after reflection off your car is: f f f = 2f(u/c). Therefore the percentage change in frequency is: p ( f/f) 100 % = (2u/c) 100 %. Clearly, this percentage does not depend on f; it depends only on your speed u. Solving for u, you get u = (c/2) (p/100%) and its value is u = [( m/s)/2] [( %)/(100%)] = 10.47m/s. Note that it is very difficult (if not impossible!) to evaluate u from f and p accurately if you first try to calculate the reflected wave frequency f = f + pf and then solve for u, using f = f(1 + 2u/c): f and f are so very close in value, that their difference, pf, is likely smaller than the rounding error of your calculator. So, you really need to do the algebra first in this problem [to derive p = 2(u/c)] before blindly plugging numbers into your calculator. That was one of the lessons HW P01.09 was meant to remind you of. Problem 7: A flat circular mirror of radius 0.190m lies flat on the floor. Centered above the mirror at a height of 0.590m is a light source. The circular spot formed on the ceiling by the reflection of the light has a diameter of 2.20m. How high is the ceiling, measured from the floor? (A) 1.41m (B) 2.83m (C) 3.00m (D) 5.65m (E) 6.00m 6

7 Answer: (B) This problem is a slight variation on HW Problem P Please read and work carefully through the posted solution of P01.12, for a detailed drawing, for details of the solution, and for the notation used here: In this version of the problem, r =radius of mirror on the floor, h =height of light above floor and D = 2(r + x) =diameter of bright circle on ceiling are given; and we need to solve for H =height of ceiling above floor. So, solving for x first, x = D/2 r = (2.20/ )m = 0.910m. Then, using H/x = h/r, we get H = xh/r = (0.910m) (0.590/0.190) = 2.83m. Problem 8: In flint glass, red light has an index of refraction (IoR) n R = and violet light has an IoR n V = 1.796, while both have an IoR n Air = in air. A beam of white light enters a triangular prism of flint glass from air at normal incidence at the front surface and then stikes the prism s back surface at an angle of incidence φ = 25.0 o as shown here: φ Red Violet The prism is surrounded by air. What is the angle of divergence, enclosed between the red and the violet beam after leaving the prism through the back surface? (A) o (B) o (C) o (D) o (E) o Answer: (D) This problem is similar to HW Problem P01.19(a). Please read and work through the posted solution of that problem carefully. At the front (lefthand) prism surface, the incident (red+violet) beam has a 0 o angle of incidence, and therefore by Snell s law also a 0 o angle of refraction. Therefore, φ = 25.0 o is the given angle of incidence of both read and violet rays at the back (righthand) surface of the prism. Let φ R and φ V denote the angle of refraction of the red and violet light rays after leaving the prism, respectively. 7

8 Then, by Snell s law: φ R = arcsin[n R sin(φ)] = arcsin[(1.765 sin(25.0 o )] = o ; and likewise, φ V = arcsin[(n V sin(φ)] = arcsin[1.796 sin(25.0 o )] = o. Hence, the angle of divergence between red and violet is: φ φ R φ V = o o = o Problem 9: A candle placed 40.00cm to the left of a curved mirror produces an image 60.00cm to the left of the mirror. What is the spherical radius of the mirror and where is the center C of that sphere? (A) 24.00cm, C to the left of the mirror; (B) cm, C to the left of the mirror; (C) 48.00cm, C to the right of the mirror; (D) cm, C to the left of the mirror; (E) Cannot be determined from infomation given. Answer: (D) The lefthand side of the mirror is both incoming side and outgoing side because the object (candle) is placed to the left of the mirror and the image-forming rays travel to the left after reflection at the mirror. Thus d = cm > 0 (object on incoming side)and d = cm > 0 (image on outgoing side). Then R = 2f = 2(1/d + 1/d ) 1 or R = 2 [(1/40.00) + (1/60.00)] 1 cm = +48cm. Since R > 0, the center of the sphere C must be on the outgoing side, i.e., to the left of the mirror. So, the answer is (D). Problem 10: If the candle in Problem 9 is 15.0cm tall its image produced by the mirror will be (A) real, inverted and 10.0cm tall; (B) virtual, erect and 10.0cm tall; (C) virtual, inverted and 22.5cm tall; (D) virtual, erect and 22.5cm tall; (E) real, inverted and 22.5cm tall. Answer: (E) See Problem 9: d = cm > 0 and d = cm > 0 are given. So, m = d /d = 60.00/40.00 = 1.50 and h = mh = ( 1.50) (15.0cm) = 22.5cm. Since d > 0, the image is real. Since m < 0, the image is inverted. 8

9 Problem 11: A convergent lens (no.1), placed to the right of a small gem stone produces an image of the gem 22.65cm to the right of lens no.1. If a divergent lens (no.2) of focal length f 2 = 6.42cm is now placed somewhere to the right of lens no.1, the final image produced by lens no.2 appears approximately 35.0cm to the right of lens no.2. Approximately, how far apart are the two lenses? Is the object of lens no. 2 virtual or real? (A) lenses are 17.2cm apart, lens no.2 object is real (B) lenses are 17.2cm apart, lens no.2 object is virtual (C) lenses are 14.8cm apart, lens no.2 object is real (D) lenses are 14.8cm apart, lens no.2 object is virtual (E) lenses are 57.7cm apart, lens no.2 object is real Answer: (B) This problem is similar to (but a lot easier than!) HW Problems P02.08 and P Please review the posted solutions of these two problems carefully before embarking on this one. Also, it is critically important that you draw a big, clean picture of this set-up, showing the relative positioning of lenses, original object, image by 1st lens; and also showing the arrows for relevant distances such as d 1 and d 2 and L. Then you will see that: d 1 = cm and d 2 = +35.0cm. Then d 2 = (1/f 2 1/d 2) 1 = [1/( 6.42) 1/(+35.0)] 1 cm = 5.425cm So, lens no.2 has a virtual object, since d 2 < 0. Also, the lens-to-lens spacing is L = d 1 + d 2 = ( )cm = 17.2cm. Problem 12: A telescope produces a final image of a skyscraper (as seen through the eyepiece) which appears to be 3.6cm tall and located 40.0cm from the eyepiece on the incoming side of the eyepiece lens. Assume the skyscraper is at a distance of 20km; and the telescope has achieved a 9.0-fold angular magnification, compared to viewing the building from the actual 20.0km distance without instrument. What is the approximate height of the actual skyscraper? (A) 500m (B) 450m (C) 400m (D) 300m (E) 200m Answer: (E) Again, it is critically important that you draw a big, clean picture of this set-up. Given: h 2 = 3.6cm and d 2 = 40.cm. Also, d ref = 20.0km and M =

10 Then: Θ e = tan(θe ) = h 2/d 2 = 3.6/40. = 0.090rad. Use M = Θ e /Θ ref to solve for the reference angle, subtended at the eye by original object, when viewed from reference distance d ref without instrument: Θ ref = Θ e / M = 0.090rad/9.0. = 0.01rad. Then, use Θ ref = tan(θref ) = h 1 /d ref to solve for the height of the original object: h 1 = Θ ref d ref = (0.01) (20000.m) = 200m. So, (E) is the answer. 10

11 Formula Sheet Wave Propagation and Wave Nature of Light Periodic Wave Condition: v = λf = λ τ Index of Refraction (IoR) for electromagnetic waves, definition: n = c v Doppler Effect: f = f ( 1 ± u v ) Reflection and Refraction of a Single Ray For angle of incidence (Θ 1 ), angle of reflection ( Θ 1 ) and angle of refraction (Θ 2 ): Archimedes Law of Reflection: Θ 1 = Θ 1 Snell s Law of Refraction: sin(θ 1 )/v 1 = sin(θ 2 )/v 2 where v 1 and v 2 are the speeds of wave propagation in the respective media. For electromagnetic waves, this is also written as Snell s Law in IoR Form: n 1 sin(θ 1 ) = n 2 sin(θ 2 ) where n 1 c/v 1 and n 2 c/v 2 are the corresponding IoR. 11

12 Image Formation and Optical Instruments Image-Object-Relations for a Single Device (Lens or Mirror): Mirror / Thin Lens Equations: 1 d + 1 d = 1 f, m h h = d d where d = object distance, d = image distance, h = object height, h = image height, f = focal length, m = lateral magnification. Focal length for reflection at a curved surface (mirror): f = R/2 Sign Conventions for Single Device (Lens or Mirror): d > 0 (real object) if object on incoming side; else d < 0 (virtual object). d > 0 (real image) if image on outgoing side; else d < 0 (virtual image). If m > 0 then image erect (upright) rel. to object; else, if m < 0 then image inverted (upside-down) rel. to object. If f > 0 then F on incoming and F on outgoing side; else, if f < 0 then F not on incoming and F not on outgoing side. R > 0 if center of spherical surface on outgoing side; else R < 0. (For reflection at a curved surface only.) Compound Instrument: If image by device 1 serves as object for device 2 and L =separation of devices then d 1 + d 2 = L, h 2 = h 1, m 12 = m 1 m 2, where m 12 is total lateral magnification of image by 2 relative to object of 1. Angular Magnification: Definition: M = Θ e /Θ ref where Θ e =angle subtended at eye by optical instrument s final image; and Θ ref =reference angle subtended at eye by original object viewed without optical instrument at reference distance d ref 12

13 Algebra and Trigonometry az 2 + bz + c = 0 z = b ± b 2 4ac 2a sin θ = opp hyp, adj cos θ = hyp, opp tan θ = adj = sin θ cos θ For very small angles θ (with θ 90 o ): sin 2 θ + cos 2 θ = 1 sin θ = tan θ = θ (in radians) Numerical Data Speed of light in vacuum: c = m/s Range of vacuum wavelengths visible to the human eye: from 700nm to 400nm. Index of refraction of vacuum: n V = 1.0 (exact). air: n Air = 1.0. water: n Water = 4/3 = Other numerical inputs (IoR, angles, lengths, etc.) statement. will be provided with each problem 13