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Physics 30 Workbook Unit4:EMR UNIT 4: ELECTROMAGNETIC WAVES (EMR) "The speed of Hght it is 186,000 miles per second. The speed of Ute i's only 100j000 miles per second." Einstein discovers that the speed of light is 65 MPH. p.67 Unit4:EMR Physics 30 Workbook A. TRANSVERSE WAVES Types of Waves A wave is a transfer of energy, not matter, from a source to a receiver. It is a travelling "disturbance" in a medium, where the medium is the material that the wave travels through. There are two types of waves: 1. Mechanical waves -waves that MUST travel through matter -the matter is disturbed (vibrated), but it does not move with the wave e.g. spring, water, dominoes, sound, earthquakes 2. Electromagnetic Waves -waves that can travel through a vacuum (no medium is required) e.g. light, ultraviolet, X-rays Source Vibrations The source of all waves is a vibration. e.g. Sound - vibrating guitar string or vocal chord Water - dropped rock or ocean floor The source vibration determines two qualities of a wave: • it provides the initial energy of a wave - this energy is then transferred through the medium • it determines the frequency of the wave (fsource = fwave) -once the wave is created, the frequency of the wave will never change (assuming that the source and the receiver are stationary- no Doppler effect) Frequency and Period 1. Frequency of vibration (f): -the number of cycles (oscillations, repetitions) in one second f = total # of cycles total time Units: Hertz (Hz) 2. Period of vibration (T) -the time it takes for one cycle (repetition) T= total time total # of cycles Units: Seconds (s) 3. Frequency and Period: - both are reciprocals of each other p.68 where 1Hz 1 cycle 1 s s Physics 30 Workbook Unit4:EMR Anatomy of a Transverse Wave (Ideal) - in a transverse wave, particles move in a direction perpendicular to the wave motion Crest Equilibrium Source Vibration Trough wave motion where A is the amplitude (energy) of a wave A is the wavelength (length of one cycle) -particles that are one wavelength apart will move identically (they are in phase) -due to friction, transverse waves will lose their energy (amplitude), but the frequency will never change Speed of a Wave I Universal Wave Equation If the medium doesn't change, the speed of a wave is constant. Thus, you can change the speed of a wave only by changing the medium (e.g. stretching the spring, different depths of water, etc.). The frequency and amplitude of source vibration have no effect on the speed of the wave. However, the frequency of the source vibration does affect the wavelength, as described by the universal wave equation : where f is the frequency of the source vibration (in Hz) A is the wavelength of the wave created (in m) Note: Wavelength and frequency have an inverse relationship Based on the equation A Thus, = we see that j' 1 A oc - f (assuming v remains constant) if the source frequency is high, the wavelength is short if the source frequency is low, the wavelength is long If it is a light wave, then it travels at the speed of light. In air /vacuum, light travels at 8 v = c = 3.00 x 10 m/s p.69 Unit4:EMR Physics 30 Workbook B. ELECTROMAGNETIC RADIATION (EMR) B 1. Maxwell's Laws and EMR Faraday believed that there was a relationship between electricity, magnetism , and light. This was established by James Maxwell in the 1860's. Maxwell summarized electromagnetism in 4 laws, described as follows : 1. The distribution of electric charge is related to the net electric field it produces 2. Magnetic field lines are continuous (closed loops), whereas electric field lines begin and end on electric charges. 3. Electric current can produce a magnetic field (discovered by Oersted). Also, a changing electric field can produce a magnetic field. ( E---+ B) 4. A changing magnetic field can produce an electric field. (B---+ E) And thus, it can induce a current in a conductor (based on Faraday's law) Note: An electric field can accelerate a charge from rest, and thus induce current, but a magnetic field can 't. It follows, then, that by changing the magnetic field, an electric field is created, and this induces a current. Based on Laws 3 and 4, Maxwell predicted the existence of a brand new kind of wave, which he called an electromagnetic wave (or electromagnetic radiation, EMR). • The source of all electromagnetic waves is an accelerating charge • This creates a changing magnetic field ( B),based on Law #3 • This B creates a changing electric field ( E),based on Law #4 • This E creates a B (Law #3), and the cycle continues indefinitely Illustrated as: e- I Source B E L1E B E velocity ----------- ---11>- Electromagnetic Wave The magnetic field, electric field, and the velocity are always perpendicular to each other. j p.70 Physics 30 Workbook Unit4:EMR Verified by Heinrich Hertz (1886): inductJon coil t .. •• ' . switch e-l B E B E receiver primary coil Hertz used a step-up transfonner to induce a high, alternating voltage in the secondary coils. This high voltage created an accelerating charge across the spark gap, which according to Maxwell, would create electromagnetic waves. The EMR then travelled across a distance and induced a current in the receiver wire. The current was detected by a galvanometer. This verified the existence of electromagnetic waves. In fact, these waves were later called radio waves. B2. The Electromagnetic Spectrum -using his device, Hertz was able to show that E-M waves had the same characteristics as light waves e.g. speed= c in vacuum, reflection, refraction, diffraction, interference, and polarization - this was convincing proof that light and radio waves are both electromagnetic waves - we have now discovered that there are far more types of electromagnetic waves ----+ they differ only in frequency (determined by the source vibration) and wavelength ( A, = c I f) lo-s .5x 10,.0 10"8 10-10 Pinpoint Protozoans Molecules Atoms Wavelength (m) 10·12 About the size oL Buildings Humans Honey Bee 1o1s p.71 1Ql6 1Q18 Atomic Nuclei 1Q20 Frequency (Hz) Unit4:EMR Physics 30 Workbook HOMEWORK (EMR) For the questions below, allforms of EMR travel at the speed of light. Both A and B 1. Arrange the following forms of EMR in order of increasing wavelength : ultraviolet A. 2. 3. 4. B. 5. 6. 7. radio x-rays visible infrared microwaves A source vibration completes 2.53 x 1010 cycles in 3.00 minutes. What is the wavelength of the EMR that it produces? For the EMR wave shown, determine the frequency of its source electrical vibration. d(km) A DC transformer is set up to create radio waves. Describe the nature of the EMR when the switch is closed just once. Why? Consider the carrier waves of two radio stations: 820kHz (AM) and 96.3 MHz (FM). Compare these waves in terms of wavelength, speed, and time to travel in 45.0 km (in JlS). determine the resulting wavelength. For the source vibration shown, p C\ Q 8.40 t(JlS) v v v If a radio tower's transmitter is oriented vertically , how should the antenna on your home radio be oriented for the best reception? Why? SOLUTIONS 1. x-rays , ultraviolet, visible, infrared, microwaves, radio 2. f= 1.4056 x 108 Hz ; A,= 2.13 m 4. 3. 4 A-=4.5km; f=6.7 x 10 Hz When the switch is closed, there is a brief increase in current, which creates a brief change in magnetic field. This will result in a short burst of EMR . But if the switch remains closed, the current will become constant, which will result in a constant magnetic field (constant flux). Constant magnetic fields (currents) do not create EMR, so no EMR will be created. 5. f= 820kHz: f= 96.3 MHz: v = 3.00 x 108 m/s ; A= 366m ; t4s Ian = 150 JlS Same speed ; A= 3.12 m (shorter wavelength, since the frequency is higher) Same time to travel 45 km (150 JlS) 6. T= 2.8 JlS ; f= 3.571 x 10 Hz ; 7. The antenna should also be oriented vertically (same as transmitter) . The changing electric field is parallel to the accelerating electrons in the antenna, thus it is vertical as well. If it is to generate current in a receiver , the conductor must also be vertical. 5 A-= 840 m p.72 Physics 30 Workbook Unit4:EMR C. THE SPEED OF LIGHT (EMR) C 1. Early Methods Galileo's Lanterns The first serious attempt to determine the speed oflight was done by Galileo (1564- 1642). He and his assistant stood on two distant hilltops with shuttered lanterns . Galileo opened his lantern first, and when his assistant saw the flash of light, he was to immediately open his lantern . Galileo then used his pulse to determine the time from the initial opening of the shutter to when he saw his assistant's flash of light. Finally, knowing the distance between the hills, he would calculate the speed of light. However, it turned out that light travelled so fast that their reaction times were way too slow to accurately measure the time . (In fact, light would have taken about 70 !J.S to travel the distance, which would have been impossible even with digital timers!) Roemer and Huygens Olaus Roemer (1644- 1710), a Danish astronomer, observed the motions of the moons around Jupiter. Although other scientists had already determined their periods of revolution , Roemer wanted to determine the exact moment that one of the moons would experience an eclipse - when Jupiter would be in front of the moon. He noticed that as Earth got closer to Jupiter, the times of the eclipse came earlier and earlier than expected; as Earth moved further away from Jupiter, the times came later and later. Roemer concluded that as the Earth got further away, light had to travel further to reach the Earth and thus the eclipse was seen later. This is illustrated in the diagram below : diameter of Earth's orbit ------, ,, , I I I I I ,.... T ... , , B I I I I I I sun \ ' I '... _._ .... ' I I I I I ' \ Jupiter (and moon) , I '' ' ' '' ... ,, I I I I I He was able to calculate that when Earth was furthest from Jupiter (B), it took 22 min (or 1320 s) longer than when Earth was closest (A). He concluded that this was (approximately) the time light took to travel across the diameter of Earth's orbit around the sun. A few years later, the Dutch mathematician and scientist Christiaan Huygens calculated the 11 diameter of Earth 's orbit (3.0 x 10 m), and using Roemer 's data, calculated the speed of light as follows: d 3 O x l0 11 v= = · m = 2.3 x 108 m/s 11! 1320 s Interestingly , the speed they determined was so fast that most scientists of their day rejected it. It was not accepted until after their death. p.73 Unit4:EMR Physics 30 Workbook C2. Michelson In 1905, the American scientist Albert Michelson used a rotating mirror to make a very accurate measurement of the speed of light. His device is as shown below: Light source Rotating 8-sided Mirror Fixed mirror telescopeD very large distance (D) < When the mirror is rotated at the minimum frequency (about 32,000 rpm) to see continuous light: Time for 1/8 revolution = Time for light to travel to the fixed mirror and back Equation: Speed of light ::::; where d 2D t Ysr D is the distance between the mirrors (in m) T is the period of rotation of rotating mirror (in s) Note: Speed of light ::::; d 2D • For ann-sided mirror, usae the formula • Today, we use lasers to calculate the speed of light through a vacuum. The currently accepted value is c = 2.997 924 562 ± 0.000 000 011 x 108 m/s. In Physics 30, we round this value to 3.00 x 108 m/s Ynr p.74 Physics 30 Workbook Unit4:EMR Homework (Speed of light) Both A and B 1. In a Michelson experiment, the 8-sided mirror is rotating at 166,000 rpm. If the distance between the mirrors is 5.8 km, then what is the measured speed oflight? A. 2. It takes a total time of83.0 ms (milliseconds) for a microwave signal to travel to a satellite and reflect back to the Earth, then how far is the satellite from the Earth (transmitter)? 3. During a lightning storm, the temperature of the air is 28° C (speed of sound is 347.8 m/s). If lightning strikes 16 km away, then compare the times for sound and light to reach you. 4. In a Michelson-like experiment, a 6-sided mirror is rotating with a period of355 J.lS. If the measured speed oflight was 3.2 x 108 rn/s, then fmd the distance between the mirrors (in km). B. 5. A radar tower sends out a signal at an angle of 41 .0° above the horizontal. The signal reflects off of a plane and returns to the transmitter in a total time of 7.50 J.lS . Determine the height of the plane. 6. In a Michelson experiment, the two mirrors are located 4.7 km apart. If the speed oflight was measured at 2.9 x 108 m/s, then at what frequency is the 8-sided mirror rotating? 7. In a Michelson-like experiment, ann-sided mirror rotates at a minimum frequency of 435.5 Hz for the light to be seen in the telescope . If the distance between the two mirrors is 31.0 km and the measured speed of light is 2.7 x 108 rn/s, then determine the number of sides n. 8. A light year (ly) is the distance that light travels in one year (365 days). If the width of our galaxy is 5.4 x 1020 m, then how many light years is this? SOLUTION 1. 2. 3. 4. 5. 6. 7. 8. T= 3.6145 X 10-4 s ; v = 2.6 X 108 rn/s Time one-way: 41.5 ms ; d= 1.25 x 107 m = 1.25 x 104 km Sound: t = 46 s ; Light: t = 5.3 X 10-5 s n=6;D=9.5x10 3 m=9.5km t1-way = 3.75 J.lS ; 1-way distance : 1125 m ; h = 1125 sin 41.0° = 738 m 4 3 T= 2.5931 X 10- s ; f= 3.9 X 10 Hz T= 2.296 X 10-3 s ; n = 10 (i.e. 10-sided mirror) 15 lly = 9.4608 x 10 m ; The galaxy is 5.7 x 104 ly wide p.75 Unit4:EMR Physics 30 Workbook Light as a Wave or Particle? Throughout history , the nature oflight has been debated. Many scientists (such as Newton) thought light was composed of particles, while other scientists (such as Huygens) were convinced that light was a wave . We will now investigate the many properties of light that were well known in the times of "classical physics", which was before the 20 1 h century. Many of these properties gave convincing evidence that light behaved like a wave. D. REFLECTION OF LIGHT (EMR) IN PLANE MIRRORS Dl. Law of Reflection (Particles and waves) normal Angle of Incidence Angle of Reflection inciden ray : !Y /'eflected ray mrrror **Angles are always measured with respect to the normal** - this law is true for both particles and waves D2. Images in a Plane Mirror Consider how we see the image of an object: Object hnage : . ..........................IN· ···········.................... . ............................t···········....··-··· ·:;:;, • The image is located where the reflected rays appear to converge ft - '""... " ...---------:::- irtual Image • The eye cannot tell that the rays have been reflected - assumes they travel in straight lines _.· Characteristics of the hnage • Same size as object • Vertically upright (but laterally inverted) • Virtual (not real) • The image is located the same distance from the mirror as the object p.76 -formed by dotted lines Physics 30 Workbook Unit4:EMR Homework (Plane Mirrors) A. 1. If the angle of the incident ray is 3 go with respect to the mirror surface, then what is the angle of reflection? 2. For the mirror shown, predict the angle of reflection off of mirror B. 97° B Both A and B 3. The objects below are placed in front of plane mirrors. Roughly sketch the image. a) b) Loving the Physics plane mirror B. 4. plane mirror If light bounces off all three mirrors, find the angle of reflection off of mirror C. B 5. For the diagram shown, fmd the angle of incidence on mirror A. B SOLUTIONS 3. (Plane Mirrors) a) b) Loving the Physics p.77 ii Q). mw J ffi 2 ffFt Unit4:EMR Physics 30 Workbook E. REFRACTION Refraction the change in direction of light (or any other wave) as it passes at an angle from one medium to another El. Index of Refraction (n) - if the medium changes, the speed of light changes -when light travels slowly through a medium, we say that the medium is optically dense - to determine the optical density of a material, we use the formula where c is the speed oflight in a vacuum (3.00 x 108 m/s) v is the speed of light in the medium Note: • There is an inverse relationship between n and v -thus, the higher then (the more optically dense), the slower the speed oflight • For air I vacuum, n = 1 - this is the smallest value for n possible, and thus the highest possible speed of light E2. Patterns of Refraction (Waves only- not particles) 1. Less to More Dense (Low n to High n) normal Partial Reflection ei = Low n (fast) 8r Partial Refraction - the light ray bends towards the normal High n (slow) 2. More to Less Dense (High n to Low n) Partial Reflection ei = 8r High n (slow) Partial Refraction - the light ray bends away from the normal Low n (fast) p.78 Physics 30 Workbook Unit4:EMR Homework (Refraction) 1. If the index of refraction for a medium is 1.78, then find the time it takes for light to travel through 250 krn of this medium. Answer in milliseconds. 2. If light takes 76.3 !J.S to travel through 9.0 km of a medium, then what is the index of refraction for the medium? 3. If the index of refraction for a medium is 2.40, then find the distance that light can travel through it in 7.35 nanoseconds. 4. Predict the direction of the refracted ray out of the glass. b) a) 5. Predict the direction of the incident ray, if the refracted ray is given. b) a) SOLUTIONS (Refraction) 1. v = 1.6854 x 108 m/s ; t = 1.48 x 10-3 s = 1.48 ms 2. v= 1.1796 x 108 m/s; n=2.54 3. v= 1.25 x 108 m/s; d=0.919m 4. a) b) R 5. a) b) p.79 Unit4:EMR Physics 30 Workbook E3. Snell's Law or, in general sin 81 = = = .?::!_ sin 82 n 1 v 2 'A 2 Note: Frequency does not change (i.e. f 1 = 6) E4. Total Internal Reflection (TIR) -recall, when light travels from a slow to fast medium (high n to low n), there are two rays: • a reflected ray (where 8i = 8r) • a refracted ray, which bends away from the normal (8R > 8i) -but what happens when the angle of incidence is increased? Critical Angle (8c) - if we increase the angle of incidence enough, eventually the angle of refraction is 90 o If 8R = 90°, then 8i is called a critical angle (8i = Elc) Total Internal Reflection (TIR) -if the angle of incidence is greater than the critical angle (i.e. 8i > 8c), then there is no refracted ray (there is only reflection) R Critical angle Total internal reflection Note: TIR cannot occur when the ray goes from a fast medium to a slow medium Why? -the ray bends towards the normal (8R < 8i) - so, even if 8i is 89°, 8R will still exist p.80 Physics 30 Workbook Unit4:EMR Homework (Snell's Law I Critical Angle) A. 1. Light travels into a medium X with an angle of 25° with respect to the surface. If the angle of refraction is 17°, then find the index of refraction for medium X. 2. Light travels from water (n = 1.34) into air, and the angle of refraction is 72.0°. If the wavelength of the light in water is 510 nm, then find: a) the angle of incidence b) the wavelength of light in air c) the frequency of light in water 3. For the diagram shown, the ray of light will reflect off the plane mirror and then enter into the medium (n = 1.77). Plane muror What would be the angle of refraction when it enters into the medium? 4. B. 5. Light travels into an air-glass interface, where the index of refraction for the glass is 1.51. If the wavelength of light is the glass is 720 nm, fmd: a) the critical angle b) the frequency of the light in air Light with a wavelength of 660 nm travels into glass with an angle of 30.0° with respect to the surface. If the index of refraction for glass is 1.57, then fmd: a) the angle of refraction inside the glass b) the wavelength of light inside the glass c) the frequency of the light inside the glass 6. Light in medium A (n = 1.81) has a frequency of2.50 x 1014 Hz. When it enters into medium B (n = 1.27), the angle of refraction is 56.0°. Find: a) the wavelength of the light in medium B b) the angle of incidence 7. For the diagram shown, the ray oflight will reflect off the plane mirror and then enter into the medium (n = 2.06). If the angle of refraction is 12.0°, then fmd e. p.81 mirror Unit4:EMR Physics 30 Workbook 8. If the critical angle for a liquid-air interface is 37°, then fmd the time it takes for light to travel through 2.1 km of the liquid. 9. The index of refraction for medium A is 1.92, while the speed oflight in medium B is 2.50 x 108 m/s. Find the critical angle for the A-B interface. 10. A horizontal ray oflight enters a piece of glass (n = 1.52) shaped as an equilateral triangle. What is the angle of refraction when it leaves the glass on the other side? SOLUTIONS (Snell's Law) 1. B1=65o ; n2=3 .1 2. a) b) C) 3. 45.2° 683 nm Vj = 2.2388 X 108 rn/s ; ji = 4.39 X 14 10 Hz The angle of incidence as it enters the medium is B1 = 33o Then, B2 = 17.9° ; 14 b) A- 2 = 1087.2 nm ; f2 = 2.76 x 10 Hz 4. a) 41.5° 5. a) b) c) 33.5° 420 nm 14 14 Method 1: Ji = 4.55 x 10 Hz ; f2 = Ji = 4.55 x 10 Hz 14 Method 2: v2 = 1.91 x 108 rn/s ; f2 = 4.55 x 10 Hz 6. a) fs = 2.5 X 10 Hz ; b) 35.6° 7. The angle of incidence as it enters the medium is B1 = 25.36° 8. n1 = 1.6616 ; 9. ns = 1.20 ; 14 VJ VB= 2.3622 X = 1.8054 X 108 rn/s ; 108 m/s ; As t= 1.2 X = 10-S S Be= 38.7° 10. The angle of refraction as it enters the glass is 19.205° The angle of incidence as it leaves the glass is 40.7951 o The angle of refraction leaving the glass is 83.3° p.82 9.45 X 10-7 m Then, B= 17.4° Physics 30 Workbook Unit4:EMR F. CURVED MIRRORS I LENSES Fl. Types of Curved Mirrors I Lenses There are two types of curved mirrors I lenses: Converging Lenses (convex) 1. Converging Mirrors (concave) < PA > f+ Diverging Lenses (Concave) 2. Diverging Mirrors (Convex) ' ......................... . ........ ..................................................... PA ·-----·· V _,.--_,/p < !- .... C ) < Note : Spherical Aberration if the mirror is spherical, the parallel incident rays will not converge on the same focus the further the rays are away from the principal axis, the further they go away from the principal focus a better focus is obtained using parabolic mirrors - similar with lenses p.83 f- Unit4:EMR Physics 30 Workbook F2. Images in Curved Mirrors I Lenses Step 1: Sketch the following reflected I refracted rays: • Ray parallel to the PA reflects I refracts through F (or diverges from F) - c ----- F : ;< F • Ray through centre reflects back I refracts on same path ..............----- --- --- --- ......................... ··················· ·········•···························· ·· . ... . ................... C F ...... ........................................................ F • Ray through F reflects I refracts parallel to PA .......... C ----··· .. . .............................. ............................................... .. F Step 2: Locate the image • image is always located where the reflected I refracted rays converge (intersect) • sketch the image from the PA to the intersection Step 3: Describe the image • Size - larger I smaller I same size as the object • Attitude -inverted (flipped vertically) or upright • Type- real (formed by real rays) or virtual p.84 Physics 30 Workbook Unit4:EMR SUMMARY FOR CURVED MIRRORS I LENSES 1. Converging Mirrors (for lenses, C = 2F) Beyond C OnC Between C and F OnF Between F and V Image Characteristics Location of Image Smaller, Inverted, Real Same size, Inverted, Real Larger, Inverted, Real No image Larger, Upright , Virtual Btw F and C OnC Beyond C ---------Other side of mirror 2. Diverging Mirrors I Lenses All locations In general: • • Other side of mirror Smaller, Upright, Virtual Real images are always inverted. Virtual images are always upright. <------->· -------------> do : d; F3. Equations for Curved Mirrors I Lenses 1 1. 1 -+-=do J di h; R Also, = 2f Sign Convention d + f + h , Mag + Virtual Real } Measured from V I optical centre Converging Diverging Upright Inverted } Measured from PA Note: If Mag < 1, then the image is smaller than the object If Mag= 1, then the image is the same size as the object If Mag > 1, then the image is larger than the object p.85 Unit4:EMR Physics 30 Workbook Homework (Curved Mirrors I Lenses) Both A and B 1. For each, sketch the image, describe its characteristics , and show where the eye is located. a) Converging mirror / lens: Object between C and F (between F and 2F) I -- - ----- ---------------------- - --- ------------------------------- . I Image characteristics: b) Diverging mirror / lens ----------------- .l ------------------------------ - --------- -I Image characteristics: c) ------ Converging mirror / lens : Object beyond C (2F) I --- ------------------------------ ----------------------------- ------ --- ------- --------------- - ---------------- ,---- -------I I I ------ . ----------------------------------------- 1-------- ---- -------- ---------- --- ----- -- ----- ---- --- ------- -- ,--- - -------I I I Image characteristics : d) Converging mirror / lens: Object inside focal length ----- -------- - ----------- - -· - ·--- + ---------- --- - --------- -1 I --------- - ----------------- ---- "I --------- --- - ------------- ---- ---1 I -------------- - ------ -- ------- -- T ------------------ ----------- -- - ·· ·· I ---------------------------------- ·t · ---- - ----------- ----- - -- Image characteristics : I _ p.86 Physics 30 Workbook 2. Unit4:EMR Based on the descriptions below, identify the type of mirror I lens and the approximate location of the object, if possible. a) c) e) b) image is virtual and smaller d) image has the same size image has a magnification of -0.25 image is bigger and inverted image is virtual and the magnification is 3 f) there is no image For the rest of the questions, be certain to interpret all signs. A. 3. A 19 em high object is located 24 em in front of a concave (converging) mirror. If the mirror's radius of curvature is 36 em, then fmd: a) the distance to the image b) the height ofthe image 4. An 8.0 em high object is placed in front of a convex (diverging) mirror. If the virtual image is 3.0 em high and is located 5.0 em from the mirror, then find: a) the distance to the object b) the mirror's radius of curvature 5. A convex (diverging) mirror has a focal length of 17 em. If an object is placed 30 em in front of the mirror, then what is the magnification of the image? 6. An 11 em high object is placed 7.0 em in front of a concave (converging) mirror. If the virtual image has a height of 55 em, then what is the mirror's radius of curvature? 7. A convex (diverging) mirror has a radius of curvature of 42 em. If a 72 em object is placed in front ofthe mirror and the resulting image is located 16 em from the mirror, then what is the height of the image? B. 8. 9. An object is placed 21 em in front of a concave (converging) mirror. If the image is real and is 4.0 times larger than the object, then find: a) the distance to the image b) the mirror's focal length A converging (convex) lens has a radius of curvature of 12 em. If an object is placed 15 em in front of the lens, then what is the magnification of the image? 10. An object is placed 17 em in front of a concave (diverging) lens. If the image is 20% of the size of the object, then what is the lens's radius of curvature? 11. A converging (convex) lens has a radius of curvature of 40 em. If the virtual image produced is 18 em tall and 50 em from the lens, than fmd the height ofthe object. 12. A 30 em high object is located in front of a concave (diverging) lens. Ifthe height ofthe image is 11 em and it is located 5.0 em from the lens, then what is the lens's focal length? *13. A very large compound microscope is composed of two converging lenses that are 1.65 m apart. The focal length of the first lens is 20 em, while the focal length for the second lens is 15 em. If the 10 em object is placed 23 em in front ofthe first lens, then determine the total magnification. Hint: The image for the first lens becomes the object for the second lens p.87 Unit4:EMR Physics 30 Workbook SOLUTIONS (Curved Mirrors) 1. Image characteristics: Larger, real, inverted Image characteristics: Smaller, virtual, upright c) Image characteristics : Smaller, Real, Inverted p.88 Physics 30 Workbook d) Unit4:EMR : - ;;;:-;:_:_: =- ' ... -=- .... -- ").(_:_:: ----- ' . . . . Image characteristics: Larger, Virtual, Upright 2. a) Diverging mirror I lens ; Object can be anywhere b) Converging mirror I lens ; Object is between C and F (between 2F and F) c) Converging mirror I lens ; Object is on C (on 2F) d) Converging mirror I lens ; Object is between V and F (between 0 and F) e) Converging mirror I lens ; Object is beyond C (beyond 2F) f) Converging mirror I lens ; Object is on F 3. f= 18 em (real) 4. d; = -5.0 em (virtual) 5. f= -17 em (virtual) ; 6. d; = -35 em (virtual) a) d; = 72 em (real) b) a) do= 13 em (real) b) f= -8.0 em ; R = -16 em (virtual) d; = -10.85 em (virtual) ; f= 8.75 em , R = 18 em h; =-57 em (inverted) Mag=0.36 (upright, smaller) (real) 7. f=-2lcm (virtual); d;= -16cm (virtual); do = 67 .2cm (real); h;=l7cm (upright) 8. Mag= -4.0 (inverted) a) d; = 84 em (real) b) f= 17 em (real) 9. f= 6 em (real) ; d; = 10 em (real); Mag= -0.67 (smaller, inverted) 10. d; = -3.4 em (virtual) ; f= -4.25 em (virtual) ; R = -8.5 em (virtual) 11. f= 20 em (real) ; do= 14.2857 em ; h0 = 5.1 em (upright) 12. d; = -5.0 em (virtual) ; do= 13.636 em (real) ; f= -7.9 em (virtual) 13. Lens 1: d; = 153.33 em ; h; = -66.667 em Lens 2: d; = -52.5 em ; h; = 300 em Mag : 30x p.89 Unit4:EMR Physics 30 Workbook G. Diffraction , Interference, and Polarization of Light (EMR) Gl. Diffraction (Waves only- not particles) When waves go through an opening or around a comer, they bend . This bending is called diffraction. The longer the wavelength of the incident waves , the more bending takes place . short A Around comers: long A ---;1less bending Through openings: more bending The smaller the opening, the greater the diffraction Wide opening: Narrow opening: Narrow opening, shorter A Significant Diffraction Nearly straight Less Diffraction G2. Interference (Waves only- not particles) Waves can pass right through each other. However , when they occupy the same position (superposed), they interfere with each other. Principle of Superposition - the resultant displacement of a particle is equal to the sum of its separate displacements There are two types of interference: 1. Constructive Interference (both+, both -) 2. Destructive Interference (one+, one-) - add to a higher amplitude (brighter) light wave - produce a smaller amplitude - if equal, they cancel each other u p.90 (\ + no wave Physics 30 Workbook Unit4:EMR G3. Double-Slit Interference (Young's Double-Slit Experiment) (Waves only- not particles) When a monochromatic wave (i.e. a wave undergoing SHM; a wave of one frequency) passes through two adjacent, narrow slits, there is diffraction and interference on the opposite side. When a screen is placed beyond the barrier, we observe bright dots I lines on the screen. Constructive 2 crests Interference 2 troughs I " \ ' ' Destructive Interference (dim) crest and trough Jr---------------------------· ent front ---------------------------- ---------------------------· trough crest Maxima Minima - This is a bright dot I line. The difference in the distances from the two slits is a multiple of the wavelength, so the waves arrive in phase and undergo constructive interference. This is a dark region. The difference in the distances from the two slits is a multiple of0.5 'A, so the waves arrive out of phase and undergo destructive interference. Mathematics ofYoung's Double-Slit L -------·-···· T d 1 Min 1 (n = 0.5) Max 1 (n = 1) X where tan () = - Equations: L A= dx nL if () < 10° p.91 Min 2 (n = 1.5) Max 2 (n = 2) Unit4:EMR Physics 30 Workbook G4. Dispersion of Light (EMR) using Prisms and Diffraction Grating When light is shone through a triangular prism or diffraction grating, the different wavelengths of light bend by different amounts. This causes the light to separate into its different wavelengths, which is called dispersion. Dispersion using a Triangular Prism (Waves only) When light passes through a triangular prism, the different wavelengths of light refract by different amounts. The shorter the wavelength, the greater the angle of refraction. R 0 y When the light disperses into its component wavelengths, all wavelengths are observed with no gaps in between them. This is called a continuous spectrum. G B I v Dispersion using Diffraction Grating (Waves only) Diffraction grating is a transparent glass or plastic that has many fme wires within it, all parallel to each other and a very small distance (in j..Lm) apart from each other. This results in many small slits, all side by side. When light passes through the diffraction grating, the light behaves much like a double-slit experiment. Each wavelength of light will have maxima and minima shown on the screen. The location of the maxima are given by the equations A.= d sine n and A.= dx nL Based on the first equation, it follows that: The shorter the wavelength, the smaller the angle of deviation The dispersion pattern (spectrum) for diffraction grating is as shown below: llight diffraction In reality, these would overlap ROYGBIV ROYGBIV ROYGBIV Note: VIBGYOR VIBGYOR VIBGYOR If the light is monochromatic (i.e. one wavelength of light), then you would see only that wavelength on the screen. This would give a pattern like Young's double-slit experiment. p.92 Physics 30 Workbook Unit4:EMR G5. Polarization of Light (EMR) (Waves only) When a wave travels through a filter that allows vibrations only in one plane, then the resulting wave that is produced will vibrate in that plane only. We call this light polarized. For example, imagine a rope that is moved up and down and then it is moved side-to-side. If this rope passes through a vertical slit, then only the vertical waves will pass through; the side-to-side waves would be absorbed or reflected. If these vertically polarized waves were then passed through a second slit that is horizontal, then the waves cannot pass through at all. They would be completely absorbed or reflected and the transmission would almost be entirely stopped. Vertical slit Horizontal slit Light waves radiate in all directions, and this light is referred to as unpolarized. Much like the rope, when it goes through a horizontal polarizing filter, it is plane-polarized (specifically, in this case, it is horizontally polarized), which means it oscillates entirely on one plane. If it then goes through a vertical polarizing filter ("cross-polarized "), then the light energy is almost completely absorbed. plane polarized light unpolarized light horizontal polarizer near zero intensity vertical polarizer Light can also be partially plane-polarized when it reflects off of a surface or refracts into a new medium. Applications of Polarization • Polarizing filters are used in cameras and sunglasses to remove some of the light that reflects off of surfaces or refracts (scatters) through the atmosphere, which has been partially plane-polarized. This reduces the glare. • LCD displays (in digital watches and calculators) have polarizing film within them. Millions of microscopic crystals float between electrodes. When the electrodes are charged, the electric field created causes the crystals to line up "cross-polarized" with the film. This results in dark areas (since the light cannot exit), which is seen as numbers and text. • Many insects (such as bees) use polarized sky light for navigation, since it is always perpendicular to the direction of the sun. • Radio waves transmitted from towers are typically polarized (defined by the orientation of the electric field). AM and FM are vertically polarized, while television is horizontally polarized. p.93 Unit4:EMR Physics 30 Workbook Homework (Young's Double Slit Experiment) A. 1. Monochromatic EMR is shone through diffraction grating (distance between the slits is 2.50 !liD). If the angle to the 2nd-order maximum is 11.0°, then determine the EMR frequency. 2. 620 nm EMR is shone through diffraction grating (distance between the slits is 9.9 !liD). If the distance from the grating 80 em, determine the distance between maxima. Assume B < 10°. 3. 470 nm EMR is shone through diffraction grating that is 36 em from the screen and the distance between the maxima is 11 em. Determine the distance between the slits. Both A andB 4. Which type of EMR (AM or FM) would have better reception on the other side of a hill? 5. Two microwave transmitters are placed side by side, each emitting EMR in phase with a frequency of 7.5 GHz. If a receiver is placed 12 em from one transmitter and 22 em from the other transmitter, would there be a strong signal? Explain. 6. Unpolarized light is sent through a vertical polarizing filter and it then travels to a second polarizing filter. Describe the light you would see as the second filter is rotated . B. 7. Monochromatic, 5.90 x 1014 Hz EMR is shone through diffraction grating which is rated at 200,000 slits/m. Determine the angle to the third order maximum. 8. For a Young's double-slit experiment, the diffraction grating has 800 slits per mm. When monochromatic light with a wavelength of 600 nm is shone through the grating, the distance between the maxima is 7.00 em. What is the distance (in em) from the grating to the screen? 9. A Physics 30 student manipulates the distance between the slits and measures the resulting distance between maxima . For this relationship , determine the equation for the straightened curve, determine the significance of the slope, and provide the units for the slope. SOLUTIONS IL = 2.3851 x 10-7 m ; f= 1.26 x 1015 Hz n = 1 ; x = 5.0 em 6 3. B = 16.99° ; n = 1 ; d= 1.6 x 10- m = 1.6!-lm 4. In order to have good reception, the waves must be able to diffract significantly around the hill. Since AM has a longer wavelength, it refracts more. So, AM would have better reception . 5. A= 4.0 em ; The difference in distances between the receivers is 10 em. Since this distance is a multiple of 0.5 A , the waves will arrive out of phase . This results in destructive interference, which means there will be no signal. 6. The light that leaves the first filter will be vertically polarized . Thus, when the second filter is oriented vertically , the light will go through and be bright. However, as it is rotated towards the horizontal , less light will make it and it will go dimmer. When the second filter is horizontal, the light is cross-polarized and no light will make it through . 7 6 7. d = 5.0 x 10- m ; A= 5.0847 x 10- m ; B= 17.8° 6 x 8. d=l.25 x 10- m; n=1 ; 0=28.7° ; L = 12.8cm slope 2 9. Equation: x = -- ; Slope = IL L ; Units for slope: m d 1. 2. 1 d p.94