{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

phy263test

# phy263test - Name ‘ MONTGOMERY COLLEGE'~ g 6 1‘...

This preview shows pages 1–13. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Name ‘ MONTGOMERY COLLEGE '~ g 6 ' 1‘ Rockville Campus PH 263 a 28 September 2010 1. A tube 1.0 m long is closed at one end; a stretched wire is placed near the open end. The wire is 0.3 m long with a mass of 0.010 kg, and it vibrates in its fundamental mode. The wire vibration induces vibrations in the air column at the fundamental frequency of the air column. (Ignore and effects at the open end of the tube. Sound speed in air is 330 mlsec.) a) What is the frequency of oscillation of the air coluan—r—m—ﬁ» i. Egg. \x, \ r l 2+4 m . v #:X‘k-2¥2—:L=o\E—.L (/v/ a{k ! V . r What is‘the tension in the wire? “ ‘ ,,. W3 U z ‘3‘) ‘. \ \" ‘ ‘ I R l “H 1 aﬁ ._ e I, to M \\ ' iv 7 Ar W (71% is n \ 2. A ski lift cable runs 400 m up a mountain and has a mass of 80 kg. The cable is struck a (transverse) blow at one end and a return pulse is detected 12 3 later. Determine the speed of the pulse and the tension in the cable. f F“ I P LC: (LOOM' " 1 c 1' LE— NC; folczo /U \ w: 7 N VHMWZ 61/423 J'b M541... do \T: Iii/M“ \\ ,www . “2/ A: \T‘ﬂb 7 \Q‘: E:— ) (T: duoﬁ egrblt woo _ / V‘- M LC - “WC r /%' EV ’ 4Q l \ 3 r l“, _ \‘ - 43 r M (Hm/g t l 'F l 0 PH 263 2 2d Sept 2010 3. A monochromatic light beam in carbon disullide (n = 1.64) enters a glass hemisphere (n = 1.50) at the center of curvature of the spherical surface. The beam travels through the glass and reenters the surrounding carbon disultide. a) Complete the path of the beam in the diagram. b) What is the angle of deviation between the incident ray and the ray leaving the hemispherical surface? . x _ ‘ ~47 Col/«Emu. - gig“: QX-V‘u a“ WWW NA 5 \ A to Mt ﬁl’l\n .' \MOQ'MBO _,,, A Quthtwﬁaﬁkgq I ': S‘Q’bobo / v\ i . awfww-«mm— \ “HM/fgggé g; gem a eat/ma. BL : Kit/{1 V3 4. A piano—convex lens (n = 1.60) is 3.0 cm thick at its widest point. lts curved surlace has a radius of curvature of 20 cm. An object is placed 50 cm from the plane surface in air. a) Locate the image formed -- eac efracting surlace. Mg M9 M4?“ ? 3 s’ 2’ J7 4%" - 44004 £0.03 _ 55 s' m 1, L20 2 0.0’gr—l-tB-0ll g’ S? F, l-bo .— [37%04,” % a ’37:, ~ N b) Consider the lens to be "thin" (the 3 cm thickness is negligible.) f What is the focal length? 1 : Q‘ i r "' CW [L ' '— " in l: 9,9 \ ti c) Find the image distance from the "thin" lens. Compare to the image distance from the curved surface in Part a) 3 s‘ 5] z \W ob \Sr {,0 0‘5 , ‘, :(mwa—iQ " J“ A " = ———‘ \$4 L : "" #1 LPMQ J 3 \'>r m' g \ ‘ I MMiHl-bo PH 263 r 3 28 Sept 2010 5. Two thin lenses each have a focal length of 10 cm. The first is a converging lens, the second is a diverging lens. They are placed 5 cm apart and an object is placed 20 cm to the left of the first (converging) lens, a) How far from the first lens will the final image be formed? b) is the ﬁnal image real or virtual? Erect or inverted? A-q F- g H _ ,. #4:") \f’q I, r " _} N” E _ Lo _..1 ,I ma:_ "SS, :-(‘_L6I):Q,333& / 0W3”) X 90 / M<wﬂm \’M46U'g ﬁrmdl gti<o=0.\\,q_vmn§£ (SW C) Show the object, both lenses and both images on 6. Where must a real inverted object be placed in front of a concave mirror in‘order to produce an erect image the same size as the object? ‘ . l Illustrate with a ray diagram * "He ggeiet'x “WA ‘9" MW “'2‘ M7 III.- III-- III-- III-- III-- III-- III-- III-- III-- III-.- III-III.- III-III.- III-III.- III-III.- III-III.- III-III.- III-III.- IIIIIIIII IIIIII.IIIIIIIIIIEIIIIIIIIIIIIIIIIIIIIIIIIIIIII III-III- III-III...IIIIIIII-III..-IIIISIIIIIIIIIIIIIIIIIIIIIII-IIIIIIIIIIIIIIII III-III..-IIIIII-IllIII-IIIIIISSIIIIIIIIIIIIIIIIIIIII-IIIIIIIIIIIIII-I IIIIIIII-IIIIIIIIIIIIIII-III-IIIEIIIIIIIIII-III..-III-IIIIIIIIIIIIIIII PH 263 4 28 Sept 2010 XXX. The following data was obtained by a student studying standing waves in an air column. On a separate sheet of graph paper plot 1 vs T. Choose appropriate scales for each axis and include appropriate labels. With a ruler, draw the straight line that best represents the data. From the slope of the line determine the speed of transverse waves in air. ﬁ*k*****i***k*****‘k***ﬁ**ﬁ**********************‘k**************t******* C na sinQa = nb sineb n = — v na sinBc =nb sin90° T u = fl = — I" EMPEE: nb‘na s s’ R k 27: in]— ' l l, .1..=(n—1)[i——1—) (lens in air) s s f R1 R2 2713 7 f R (mirror) 1-2-2293. "ii—ii 27: l I _ y — S MONTGOMERY COLLEGE Rockville Campus “L. l. A thin ﬁlm of oil clings to the left side of a very ﬂat and smooth glass plate, as shown. Under the inﬂuence of gravity, the oil ﬁlm takes on a wedge shape, thicker at the bottom. The thickness of the ﬁlm is essentially zero at the top and 975 nm at the bottom The indices of refraction of the oil and the glass are 1.4 and 1.6, respectively. PH 263 A parallel beam of red light (7» = 650 nm) is incident from the left. The ﬁlm is viewed from the left, and the reﬂected light forms interference fringes. How many phase changes occur at the reﬂections? \ RQVL did. a “4, Chat/1 695 ‘/ t '6 «a 414 CW ‘ 5’» l M \ ‘ What is the least thickness of th oil for which a bright band is formed? 1 0 %*~ W\ =/‘€—_ Y“??? X133- : “(U-xéwm: >< M - t : Wm Aga-lwamlnm- How many bright bands are seen? - ﬁr A- (on q“ 6H: M)70 m’ T “ ego/PAL) W‘ t *3... his“ Ed“ 5/ . d4 2. An FM transmitter operating at a frequency of 120 Mhz is at the top of a 100 m tower. An airplane is in ﬂight over the ocean at an altitude of 2000 m. Radio waves reach the plane directly from the transmitter and by reﬂection from the ocean surface. When the plane is 80 km from the tower the pilot observes that reception has faded due to destructive interference of the waves that follow the two paths. Assume the wave reﬂected from the ocean surface has undergone a half wave phase shift. The reﬂected wave travels a longer distance than the direct wave. w man ﬂa s onger. g £2 \wmuazmws Hg, )« = W \A a 99% m- A to ~< w I ol= 8W WK —, we a». x423?» C/g V X & 6, Name kgSgio l. Efggtgggn ‘ <17 2 November 2010 :. PH 263 2 NOV 10 3. Monochromatic light falls normally on a screen containing several evenly spaced identical slits. l The slit separation is three times the slit width: (1 = 3 a . . A . . . . _ . The slit width is four times the wavelength of the incident light: _ i 7 _ _ g a = 4 9» y ( M v j a) Find the angle from the center line to :@c iffraction pattern minimum. : :a: : i : i : r v w\ - rm n amgznAsumgs ad ,szm .J [7 . t 4 ‘ b) How many interference maxima will be observed insi ' 'on maximum? ' a ANNE 54 WW _ same _ \QWQ- O/flMQ-L Vle 2‘3 w _ T :4/ x , 0V“ 4. A 10 kg satellite circles the earth once every 2 hours in an orbit having a radius of 8000 km. Assume that Bohr's angular momentum postulate [L = = n (h/21t)] applies to an earth satellite just as it does to an electron in the hydrogen atom. Find the quantum number (n) of the orbit of the satellite. If the satellite changed quantum orbits from n to n-l would the difference between the radii of the two orbits be measurable? ‘ e! {l W‘N L9, Wuer . 2 : am I” M=I09¢ -3- 2Nov10 PH 263 5. Light of wave length 7. = 525 nm passes through two slits separde by 0.500 mm and produces an interference pattern on a screen 7.80 m away. Let the intensity at the central maximum be 10. I What is the distance on the screen from the center of this central maximum to the point where the intensity has fallen to 3° ? 3.5 mm. Light of wavelength 500 nm is used to view two point sources 6. Treat your eye as a circular aperture of diameter that are 712 m distant from you. How far apart must the two point sources be if they are to be just resolved by your eye? (Assume the resolution is diﬂ‘raction limited and use the Rayleigh criterion.) -'3 l: ver :rHAf'” , )g .; S‘an: SYO‘UUB m 7, 6‘ :THAM 5 3- WWW ‘ ’E'J‘X Ms" 8 '; WWWM PH 263 2 Nov 10 EXTRA... Two polarizing sheets have their transmission axes crossed so that no light gets through. A third sheet is inserted between the ﬁrst two so that its transmission axis makes an angle 6 with that of the ﬁrst sheet. Unpolarized light of intensity Io falls on the ﬁrst sheet. Express the intensity of light transmitted through all three sheets in terms of lo and 9. m“ 2?. ﬂ” “ \ x ‘ 1 ‘ '1 ‘\ L0 K g) L3: \_\WC£: \$06ch 2” \ 33 : 1; Wo£<§ 41\$ i> \lg :. it; gﬁ‘ﬁd aWEﬂW stuff: - - - - - - - - - - - - - - - - - - - — - - - - - - - - - - — - - - - - - - - - - - - - - - - - - - - - - -- 2t/N=In 2t/N=m+1/2 dsin9=nl N=Mn v=f7t asin6=m7t n=c/v 1= 10 00526 cos(90-9)=sin9 0:1.22—D- I=Icoszﬁ 0 _ 27: A ﬁ_7rdsm0 “ A a), 9““ MONTGOMERY COLLEGE Rockville Campus 3 Name \' @[H M]. 30 PH 263 30 November 2010 1. The energy level diagram represents a hypothetical one-electron atom. From its ground state the ionization energy is 17.50 ev. When the atom is excited to the n = 2 state a photon is emitted n in the transition to the n = 1 state; its wavelength is 94.54 nm. What is the energy of the electron E in the n = 1 and the n = 2 states? 00 3 6'. \Q‘EOQN : J3 60 2 E2 ._ \ r -\3‘GOQV 7»:94.54nm r.-. “max 1 !——— E1— 3 , A3 . to M l i E. - m - _ - w G; 0 M V\ ~ 1 a a : /_____ , Z/ q. 6; t, - g ' \l' W l / 2. What is the peak wavelength emitted by the human body? (Assume a body temperature of 98.6°F.) In what part of the electromagnetic spectrum does this wavelength lie? .. '3 ‘ , "~ _ -— ’ . - J )(M‘ __ g4ng MK. T/ RWY-B it”. RN 4514?: M X - ~ g u r w“ ‘x ’ \ km ‘ 9x 301 W w j l y " ~r-u-wwv-«w-m..m...«.n.; a 07 0 K M W w «.mmwmmqhw M )(YMKY (1" / l‘“‘¥’fa_~r29, W PH 263 30 Nov 10 3. A particle in a one dimensional box is represented by the wave function L L 4 4 L L ‘1’ = 0 < —— nd > -— (x) _ for x 4 a x 4 If its position is measured, what is the probability that the particle will be found between x = 0 and x = £1? Li? 1 A L‘ t ‘ "‘ (l. 1 3‘ ’— 93 5; , p : 5 NW dk. iv E EU” 0 d Oil» Q o L L '\ 7 ‘LMXL Z V " a} , - LWC’JMW Y‘d twa’m‘: V-“ L 3(1‘JW 9)) Q " W23W 3 ~+ -(0‘ it W ‘ 1,7095" at, a“ ‘“j /{ dzc— LA Q :2W 8 W _ (a? ’6. M ‘1: s t _ W»- * htto} :Q a“ ‘6 -% 7g \ i “I: ﬁlly/3L“ / \:%‘(a it \ 6W6 W A L " L D 311:1}- ?TWELY}- V12 ‘ % aw I Mr: U09 4. During a photoelectric effect experiment the photocurrent is stopped by a retarding potential of 0.54 v for 750 nm incident radiation. ‘X _ 1:30 W W‘ x \I ,1 Q). S q \l " I What is the work function of the cathode material? at: tint W on M41) 5. via—1“:— :46 r ’— N” ii ’7‘ 7\ d W WM” .6 V f/ ,— _ We 7h 6‘5 1 W __ ’0 4 Q, l) K; Q X” "W 1/? o W : ‘W’ﬂ’ “HEB : «M'sng aw” (“MM ) / {b 1 \'|‘ Q\/ PH 263 30 Nov 10 5. (a) Deﬁne “Proper Length” . » .+ 3N0“! \BM 6% x”: Mal bus—Hmvmﬁw/ (0(aNﬁ’V’WWWI; WW6 gm M (b) A space ship of proper length 300 m takes 0756?]( 0'6 séconds to pass an Earth observer. Determine its speed as observed by the Earth observer. (Consider this as a length contraction problem.) 6. Imagine playing baseball in a universe where Planck’s constant is 0.6 J's. What would be the uncertainty in position of a 0.50 kg baseball moving at 20 m/s along an axis if the uncertainty in speed is 1.0 m/s ? kit Aw : .. QCVT x (ifs "4 ‘ Am A > E: it / aw ( (j PH 263 30 Nov 10 EXTRA . . . - List all the quantum numbers that are necessary to specify the state of an electron in the hydrogen atom. - Indicate the range of possible values of each number. ? “1';9\>3;""€/vﬁ‘ / § MIL 2 Q t\‘ TR' 1 Q 1 l ' s e + 4 stuff: - - - - - - - - - - - - - - - - - - - - - — - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- . 10 1 At = y Ato fl=— = 2 7 1 v_ c2 5 TC =3(TF —32) 6 ' 20 L Icoszed9=—— 5m ' ’ 2 4 T=TC+273.15 he = 1240 evnm 21:1.055 x10‘34 Js h = 6.626075x10'34 1-5 = 4.136x10'15 ev-s TC 1:014 o=5.6703x10’8 W/m2K4 AmT=290 x 10‘3 mK AxAth/Zn e = 1.602197x10'19c0ul me = 9.11 x 10'31 kg ...
View Full Document

{[ snackBarMessage ]}