Essential Physics I E 英語で物理学の エッセンス I Lecture 14: 13-07-15 Exam Next Lecture! (2 weeks) All lecture slides on course website: http://astro3.sci.hokudai.ac.jp/~tasker/teaching/ep1 Remember your calculator! Dictionary is OK! (Phone is not) SHOW ALL WORKING! Next week’s exam 10 multiple choice questions (A) ..... (B) ..... (C) ..... (D) ..... ~6 classical mechanics ~2 oscillations, waves & fluids ~2 optics Homework Attendance / clickers Exam Total 40 % 20 % 40 % 100 % Pass > 60 % Next week’s exam This is a question. (A) ..... (B) ..... (C) ..... (D) ..... 1 2 x = x0 + ut + at 2 p u ± u2 4(a/2)(x0 x) t= a p 12 ± 144 4(9.81/2)(17) = 9.81 If this is wrong.... But parts of this are right... You will get marks! SHOW YOUR WORKING! Last week: Mirrors & Lenses Image position can be found by drawing 2 light rays from points on the object. Rays touch a real image Last week: Mirrors & Lenses Image position can be found by drawing 2 light rays from points on the object. Rays touch a real image Rays only appear to touch a virtual image h0 Magnification: M = = h Mirror/Lens equation: (thin lenses) Inside a lens: s0 s 1 1 1 + 0 = s s f n1 n2 n2 n1 + 0 = s s R f h s h’ s’ Optics Interference & Diffraction So far... We assumed: x Light travels in a straight line: a ray (Geometrical optics) << x x But, sometimes the wave nature of light is very important. Interference and diffraction Lecture 10: waves add wave 1 + wave 2 = resulting wave wave 1 + wave 2 cancel Destructive interference waves coincide Constructive interference Light waves are the same. Coherence For continuous interference, waves must be coherent. e.g. always constructive or always destructive or always …. Coherence For continuous interference, waves must be coherent. t1 t2 t3 If phase between waves changes, wave sum changes incoherent Coherent waves Length where the wave phase is constant: coherence length Lasers: long coherence length Destructive and constructive Even for lasers, hard to keep coherence x1 Split a single light source x2 Light travels different paths Recombine If x1 = (m )x2 , constructive interference. ✓ ◆ 1 x2 , If x1 = m + 2 destructive interference patterns. Double-Slit interference One method of splitting a source is to pass through two narrow slits. Produces 2 coherent sources. cylindrical wavefronts interfere Double-Slit interference illuminating screen Constructive interference bright on screen Destructive interference dark on screen interference fringes Double-Slit interference Constructive interference Difference in path: m If L > d , rays ~ parallel Difference in path: d sin ✓ Bright fringes (constructive interference): d sin ✓ = m m = 0, 1, 2, ... fringe order Double-Slit interference m=1 m=0 m=1 Bright fringes Double-Slit interference Destructive interference Difference in path: ✓ dark fringes: ✓ ◆ 1 d sin ✓ = m + 2 1 m+ 2 ◆ m = 0, 1, 2, ... Typically L ⇠ 1m , d < 1mm and ybright y Therefore: sin ✓ ' tan ✓ = L ✓ ◆ 1 L L ydark = m + =m 2 d d < 1µm Double-Slit interference Quiz Constructive interference of 2 coherent waves will occur if their path difference is... (A) (B) 3 2 (C) 5 2 (D) 2 Double-Slit interference Example 2 slits 0.075 mm apart are located 1.5 m from a screen. The 3rd-order bright fringe is 3.8 cm from the screen centre. What is ? m=3 Use: ybright L =m d ybright d = mL (0.038m)(0.075 ⇥ 10 = (3)(1.5m) = 633nm 3 m) Double-Slit interference Quiz If you increase the slit separation in a 2 slit system, how does the spacing of the interference fringes change? (A) They become further apart (B) The spacing does not change (C) They become closer together ybright L =m d ydark = ✓ 1 m+ 2 ◆ L d Double-Slit interference Quiz A double-slit experiment has a slit spacing 0.12 mm. If the bright fringes are 5 mm apart when the slits are illuminated with 633-nm wavelength laser light.... What is the slit-to-screen distance, L? (A) 0.95cm Use: (B) 95cm ybright L y=1 d L =m d L 0 = 5mm d (C) 150cm L= (D) 15cm yd (0.12mm)(5.0mm) = = 95cm 633nm Multiple-Slit interference What if we add more slits? Constructive interference All 3 waves must be in phase Difference in paths: m For evenly spaced slits: d sin ✓ = m (same as for 2 slits) Destructive interference ... more complicated All waves must add to zero. For 3 waves, each wave must out of phase with by 1/3. d sin ✓ = ✓ ✓ 1 m+ 3 m+ 2 3 ◆ ◆ general m d sin ✓ = N slit no. m is integer but not integer multiple of N Multiple-Slit interference ✓ m+ 1 3 ◆ ✓ 2 m+ 3 ◆ 2 minima between primary maxima m and (m + 1) secondary maxima not fully constructive or fully destructive interference. Diffraction grating What if we add even more slits? Diffraction grating: ~1000s slits / cm If using multi-wavelength light: since: d sin ✓ = m for m > 0 , angular position ✓ depends on maxima occur in different places for different wavelengths Many slits gives very precise locations for the wavelengths (clear separations) Diffraction grating Example Light from glowing Hydrogen contains H↵ (hydrogen-alpha) 656.3nm and H (hydrogen-beta) 486.1 nm. Find the 1st order angular separation between these wavelengths when using a grating of 6000 slits /cm. m=1 1 d= cm = 1.667µm 6000 slits / cm 6000 ✓ ◆ ✓ ◆ 0.6563µm = 23.2 1 1 ✓↵ = sin = sin d 1.667µm ✓ ◆ ✓ ◆ 0.4861µm 1 1 = 17 ✓ = sin = sin d 1.667µm ✓ = 6.2 Diffraction grating Quiz Green light at 520 nm is diffracted by a grating with 3000 lines / cm. Through what angle is the light diffracted in 5th order? (A) 0.07 (B) 9.0 maxima at: ✓ = sin 1 d= 3000cm (C) 51 ✓ = sin (D) 0.05 1 ✓ 1 1 ✓ m d ◆ = 3.33µm 5 ⇥ 520nm 3.33µm ◆ = 51 Diffraction Why does the slit cause cylindrical waves? light bending as it passes object Diffraction Really need Maxwell’s equations but.... Huygens’ Principal: (pronounce: her-genz) All points on wave front are spherical wave sources. add for resultant wave Diffraction Spherical waves blocked by the barrier; only some pass through slit. Sum of the remaining spherical waves wave bends (diffraction) Diffraction If the slit width ~ wavelength slit = single wave source But.... If slit width is larger Each point in slit is a wave source Interference between waves in slit Diffraction Consider 5 equal spaced sources: 1 Path length for ray 1 and 3 differ by a sin ✓ 2 Destructive interference if or 1 1 a sin ✓ = 2 2 a sin ✓ = But if ray 1 and 3 interfere destructively, so do 2 and 4... a All rays in lower half of slit will interfere destructively with ray 2 above it. If you look at the slit at ✓ such that a sin ✓ = , you will see no light Diffraction Similarly 1 Ray 1 and 2 differ by a sin ✓ 4 Destructive interference if or 1 1 a sin ✓ = 4 2 a sin ✓ = 2 But is ray 1 and 2 interfere destructively, so do 3 and 4... a All rays in lower 3/4 of slit will interfere destructively with ray 4 above it. If you look at the slit at ✓ such that a sin ✓ = 2 , you will see no light Diffraction Generally: (divide slit into 6, 7, 8 etc sources) a sin ✓ = m Destructive interference, single-slit diffraction for m = 1, 2, 3, ... for m = 0 no destructive interference at the central maximum. Diffraction Limit Diffraction creates a limit for how close two object can be.... ... and still be resolved. 2 sources illuminate the slit Their waves reach the slit at different angles (assume sources are incoherent, no regular interference pattern) 2 single slit diffraction patterns As the sources get closer Central maxima begins to overlap Diffraction Limit Diffraction patterns merge 2 peak structure disappears 2 peak visible if Central maxima of peak 1 = 1st minima of peak 2 Rayleigh criterion when true, 2 sources are just resolved Diffraction Limit The angular separation, ✓ , between the peaks = The angular separation, ✓ , between the sources 1st minima: sin ✓ = since << a ✓min = a small angle approximation: sin ✓ ' ✓ (Rayleigh criterion, slit) a If using a circular aperture, not slit (e.g. a camera): ✓min 1.22 = D (Rayleigh criterion, circular aperture) aperture diameter Diffraction Limit Example An astroid 20 ⇥ 106 km away appears on a collision course with the Earth! What is the minimum size for the astroid that could be resolved with the 2.4 m diameter diffraction-limited Hubble Space Telescope? diffraction is what prevents the resolved imaged (not atmosphere .... etc) circular aperture Diffraction Limit Example An astroid 20 ⇥ 106 km away appears on a collision course with the Earth! What is the minimum size for the astroid that could be resolved with the 2.4 m diameter diffraction-limited Hubble Space Telescope? ✓min 1.22 = D l 1.22 = L D 1.22 L l= = 5.6km D l ✓ L Opposite ends of the asteroid = 2 peaks l (small angle approximation) ✓' L Diffraction Limit Example An astroid 20 ⇥ 106 km away appears on a collision course with the Earth! What is the minimum size for the astroid that could be resolved with the 2.4 m diameter diffraction-limited Hubble Space Telescope? ✓min 1.22 = D l 1.22 = L D 1.22 L l= = 5.6km D big problem! l ✓ L Opposite ends of the asteroid = 2 peaks l (small angle approximation) ✓' L Diffraction Limit Quiz What is the longest wavelength you could use to resolve an object with angular diameter 0.44 mrad, using a microscope with aperture 1.2 mm in diameter? (A) 528nm (B) 220nm (C) 130nm (D) 430nm ✓min 1.22 = D ✓min D (0.44mrad)(1.2mm) = = = 430nm 1.22 1.22 Real optics movie Real optics Quiz What has been created at the VLBA (Very Large Telescope Array)? (A) The biggest mirror on a telescope (B) An artificial star (C) A new type of laser (D) A defense shield Real optics Quiz Why is this useful? (A) The laser reflects of stars to find their distance (B) We can communicate with extra-terrestrial life (C) Allows the ‘adaptive optics’ technique to be used anywhere in the sky. (D) It tracks communication satellites around the world Real optics How big is the mirror on the telescope that the laser is launched from? (A) 10 m (B) 8.2 m (C) 3 m (D) 5 m Quiz Real optics What height is the “star” created? (A) 90 km (B) 10 km (C) The same as our nearest star (D) The same as the moon Quiz Real optics Quiz How faint (weak / hard to see) is the artificial star? (A) 20 x fainter than the faintest star that can be seen with the telescopes (B) 20 x fainter than the brightest star that can be seen by eye (C) 20 x fainter than the faintest star that can be seen by eye (D) 20 x fainter than the brightest star that can be seen by eye Real optics Quiz What limits a ground-based telescope’s image sharpness? (how good the image is) (A) Diffraction (B) Atmospheric turbulence (C) Moon light (D) Sun light Real optics Quiz How does adaptive optics solve this? (A) It uses a bigger mirror (B) The laser finds the object’s location more accurately (C) A flexible mirror corrects the image (D) The laser freezes the image Real optics Quiz Adaptive optics needs reference star. Why isn’t a real star used? (A) A suitable star isn’t always in the sky area you want to observe (B) A real star’s light is too variable (it twinkles) (C) The reference star must be red (D) The moon is too bright Real optics How does the laser make the star? (A) Shoots a beam of light into the air (B) Causes sodium atoms in the atmosphere to glow (C) The laser reflects of a real star, making it brighter (D) The laser reflects off the moon Quiz Real optics Quiz What do scientists hope to observe with this new technique? (A) Black holes forming in other galaxies (B) ‘high redshift’ (young) galaxies (C) Our galactic centre (D) All the above

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