Physics 232 Exam II Mar. 24, 2004

Transcription

1 Soc. Sec # Physics 3 Exam II Mar. 4, 004 Name. Linseed oil is slowly added to a beaker of water until the surface turns dark when viewed with light of wavelength 550 nanometers at near normal incidence. What is the thickness of the layer of linseed oil? Take n(linseed).48 and n(water).33. The total phase difference is given by δ Total δ Path Difference + δ air,linseed + δ linseed,water The phase shift due to the path difference is given by µ t δ Path Difference π λ 0 with λ 0 λ air n linseed where t is the thickness of the linseed oil. At the air-linseed interface the light ray is being reflected from a medium of higher index so there is a phase shift of 80 0 or π δ air,linseed π At the linseed-water interface the light ray is being reflected from a medium of lower index so there is no additional phase shift δ linseed,water 0 So the toal phase shift is then given by µ t δ Total π λ 0 + π Since the surface of the linseed oil goes dark, this is destructive interfence, so the total phase shift must be ( m +)π. π µ t λ 0 + π (m +)π 4 t λ 0 m t mλ0 mλ air n linseed Minimum thickness of the linseed oil occurs when m. t meters 85.8 nm (.48)

2 Soc. Sec # Name. Two loudspeakers are separated by four feet from each other. A note of frequency 9900 Hz is played. Take the velocity of sound to be 00 ft/sec. (a) As you walk along the wall opposite to the speakers, you notice that the distance between successive minima is 4 inches. How far from the speakers is the wall? (b) A new note is played on the speakers. It is noted that the first maximum beyond the central one is at a distance of inch. What is the new frequency? The phase difference in this problem is due only to a path difference. a.) Having a minima means that the phase difference is an odd multiple of π. δ (n +)π The path difference is given by d sin θ. So the phase angle difference is given by The sin θ can be approximated by δ d sin θ π (n +)π λ sin θ ' θ x D We then have Successive minima are given by or or dx π (n +)π Dλ dx Dλ π (n +)π dx Dλ π (n +)π (x x ) πd Dλ (n n ) π D (x x ) d (n n ) λ

3 Converting units 4 inch ft and the wavelength is given by λ v f 00 /9ft 9900 So D (/3) 4 (/9) ft b.) For a maximum the phase difference has to be nπ.wethenhave or The frequency is then dx π nπ Dλ λ dx (4) (/) 4 Dn () () 44 ft f v λ 00 39, 600 Hz 4/44 3

4 . (a) Soc. Sec # Name 3. Each end of a glass rod with n glass.5, which is 0 cm in diameter and 50 cm long, has been ground and polished into convex hemispheres each of radius 5 cm. The hemisphere at the left end has been silvered. An object is placed 5 cm to the right of the right end. (a) Where is the final image? Clearly delineate your calculations!! (b) Is the final image real or virtual? (c) What is the overall magnification of the final image? a.) This is three step problem. The first step is refraction through the right hand curved surface. Since the light is going from right to left is also the same direction as going from the surface to the center of curvature of that surface, so R is positive. For this step we also have that n a and n b.5. n a S + n b S 0 n b n a R S S S 0 45cm Since S is positive, the image is real. The magnification from step is given by m n a s 0 n b s () (45) (.5) (5) The second step is a relfection through the mirror that is on the left hand side. The image from step is the object for step, so the object distance is 5 cm. The radius of curvature for the mirror is positive, since to get to the center of curvature of the mirror you get left to right and this is the same direction as the reflected light ray, R is positive. S + S 0 R 5 + S 0 5 S S 0 5cm Since S is positive, the image from the reflection is also real. The magnification for step is given by m S0 S 5 5 4

5 For step 3, we go back through the curved surface on the right. But now the radius of curvature for the surface is negative, because the ongoing light ray is going from left to right, while to get to the center of curvature you go right to left. Now also n a.5 and n b. The object distance is now 45cm. n a S + n b S 0 n b n a R S S S 0 5cm 5 Since S is positive, the final image is real. The magnificationforthisstepisgivenby m 3 n a S 0 n b S (.5) 5 () The total magnification is given by m m m m 3 ( ) ( ) ( 0.5) The final image is the same size as the original object but inverted with respect to the original orientation. 5

6 Soc. Sec # Name 4. A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.6c. The pursuit ship is traveling at a speed of 0.8c relative to Tatooine in the same direction as the cruiser. (a) What is the speed of the cruiser relative to the pursuit ship? (b) According to an observer on Tatooine the two ships were initially separated by a distance of meters. According to the pilot on the pursuit ship, how long is it before he overtakes the cruiser? a.) Attach the unprimed coordinate system to the planet Tatooine, and attach the primed system to the pursuit spacecraft. We then have that v 0.8c, while v x 0.6c.Now apply the Lorentz transformation equation for velocity in the x-direction. vx 0 v x v (v/c )v x 0.6c 0.8c (0.8) (0.6) 0.c c The cruiser appears to be coming at the pursuit craft. b.) The pursuit craft sees a "shortened" distance between the two ships. Thetimeittakestocoverthisdistanceisgivenby L L 0 p (v/c) p (0.6) meters t L/vx c (0.385) (3 0 8 ) s 6

7 3. Maybe Useful formulae and constants: v 0 x S + S 0 R ; n a S + n b S 0 S + S 0 f ; x0 v x v (v/c )v x ; n b n a R ; m y0 y S0 S x vt p ; T T 0/ p (v/c) (v/c) y 0 y; L L 0 p (v/c) vy 0 v p y (v/c) ; (v/c )v x vz 0 v p z (v/c) ; (v/c )v x t 0 z 0 z; m n a s 0 n b s y0 y t vx/c p (v/c) ; δ δ p.d. + δ inherent ; c m/ sec ; C /N m µ 0 4π 0 7 T m/a v sound 00 ft/sec 340 m/sec 7