The 37th International Physics Olympiad Singapore. Experimental Competition. Wednesday, 12 July, Sample Solution

Transcription

1 The 37h Inernainal Physics Olypiad Singapre Experienal Cpeiin Wednesday, July, 006 Saple Sluin

2 Par a A skech f he experienal seup (n required) Receiver Raing able Gnieer Fixed ar Bea splier Gnieer Mvable ar Transier Hlder Reflecr b Daa shee (n required) Psiin (c) Meer reading (A) Psiin (c) Meer reading (A) Psiin (c) Meer reading (A) Psiin (c) Meer reading (A)

3 c Meer reading (A) c psiin (c) 3

4 Fr he graph (n required) r herwise, he psiins f he firs iu pin and h iu pin are easured a 878 c and 036 c The wavelengh is calculaed by Thus, c 8 c 8 arks Errr analysis d, d 005 x c 0 c d c < 0 0 c 0 arks 4

5 Par (a) Deducin f inerference cndiins A θ θ B n θ Assue ha he hickness f he fil is and refracive index n Le θ be he inciden angle and θ he refraced angle The difference f he pical pahs L is: Law f refracin: Thus L ( n / cs θ an θ sin θ ) sin θ n sin θ L n sin θ Cnsidering he 80 deg (π) phase shif a he air hin fil inerface fr he refleced bea, we have inerference cndiins: n sin θ in (,, 3,) fr he desrucive peak and n sin θ ( ± ) fr he cnsrucive peak ark If hickness and wave lengh are knwn, ne can deerine he refracive index f he hin fil fr I θ specru (I is he inensiy f he inerfered bea) 5

6 (b) A skech f he experienal seup Receiver Plan cnvex cylinder lens θ Gnieer Mvable ar Raing able θ Thin fil Transier Gnieer Fixed ar ark Sudens shuld use he labeling n Page (c) Daa Se X: θ / degree Y: Meer reading S/A

7 Uncerainy: angle θ ± 0 5, curren: ±000 A 05 arks Meer reading (A) θ ( ) Fr he daa, θ in and θ can be fund a 48 and 705 respecively T calculae he refracive index, he fllwing equains are used: 09 arks 06 arks 7

8 8,) 3,, ( 48 sin n () and ) ( 5 70 sin n () In his experien, 58 c, 85c (easured using her ehd) Slving he siulaneus equains () and (), we ge 5 0 ) ( 48 sin 5 70 sin Subsiuing 5 in (), we ge n 54 Subsiuing 5 in (), we als ge n 54 Errr analysis: ) ( sin n θ + ) (sin ) (sin ) ( sin 3 3 n n θ θ θ θ θ If we ake θ ±05 ±00087 rad, ±005 c, ±00 c, and θ ) sin ( n Thus, n + n 54 ±00 ark 05 arks 05 arks

9 Par 3 Saple Sluin Task Skech yur final experienal seup and ark all cpnens using he labels given a page In yur skech, wrie dwn he disance z (see Figure 3), where z is he disance fr he ip f he pris he cenral axis f he ransier Transier Lens z Pris Pris d Receiver (Sudens shuld use labels n page ) Task Tabulae yur daa Perfr he experien wice Daa Se X: d(c) X(c) Se Se S average S (A) S (A) (A) S(A) # I (A) * (I ) $ Y: ln(i (A) ) Y & # S 00 A (fr each se f curren easureens) * S prprinal he inensiy, I $ (S ) I S S & Y (lni ) (I )/I 9

10 Task 3 By pling apprpriae graphs, deerine he refracive index, n, f he pris wih errr analysis Wrie he refracive index n, and is uncerainy n, f he pris in he answer shee prvided Leas Square Fiing X d(c) X(c) Y ln(i ) Y Y XY X Y ΣX ΣY Σ Y Σ( Y) ΣXY ΣX ΣY

11 Fr I I ( γ d ) exp 0, aking naural lg n bh sides, we bain: which is f he fr y x + c ln( I ) γ d + ln( I ) 0 T calculae he gradien, he fllwing equain was used, where N is he nuber f daa pins: ( ) ( )( ) N X ( X ) N XY X Y 347 T calculae he sandard deviain σ Y f he individual Y daa values, he fllwing equain was used: σ Y ( Y ) 0064 N Hence he sandard deviain in he slpe can be calculaed: σ N σ 008 Y ( ) N X X Fr he gradien: γ 347 ± ± 008 Using: n k + γ k sin θ where θ 60, k π/ 0 (using he wavelengh deerined fr earlier par (using (85 ± 00)c), we bain: n ± n 434 ± ± 00

12 Errr Analysis fr refracive index f n ( k + γ ) d ( k + γ ) d n k + γ dk k sinθ dγ k sin θ ( k + γ ) ( k + γ ) γ ( k + γ ) n k + γ sinθ k sinθ k sin θ where: π k 005 Ne: Oher ehds f errr analysis are als acceped

13 Par 4 Task Tp view f a siple square laice a d 05 arks Figure 4: Scheaic diagra f a siple square laice wih laice cnsan a and inerplaner d f he diagnal planes indicaed Deriving Bragg's Law Cndiins necessary fr he bservain f diffracin peaks: The angle f incidence angle f scaering The pahlengh difference is equal an ineger nuber f wavelenghs 3

14 Figure 4: Scheaic diagra fr deriving Bragg's law h d sinθ () The pah lengh difference is given by, h d sinθ () Fr diffracin ccur, he pah difference us saisfy, d sinθ,,, 3 (3) a d 05 arks Figure 43 Illusrain f he laice used in he experien (his Figure is n required) 4

15 Fig 44 Acual laice used fr he experien (his Figure is n required) Fig 44 The acual laice used in he experien (n required) Task (a) Digial Mulieer D θ Laice Bx I Raing Table I n L ζ 80 θ Plan cylindrical Lens n Hlder J N A Micrwave Transier n Hlder B Micrware Receiver n Hlder 5 arks Fig 45 Skech f he experienal se up 5

16 Task (b) & (c) Daa Se θ/ ζ/ Oupu curren S (A) Inensiy IS (A) Relaive inensiy Inensiy (A) (W) Thea (degree) 7 arks Task (d) Fr eq 3 and le, d sin θ (4) Fr Fig 43, 6

17 a d (5) Cbine eqs (4) and (5), we bain, a sin θ Fr he syery f he daa, he peak psiin is deerined be: θ 3 (The hereical value is θ 3 ) a sin θ 85 c sin c (Acual value a 380 c) [The value 355 in he arking schee is derived fr: 83 c a 3 58 c sin θ sin 34 where 83 c and 34 deg are he in and allwed values fr wavelengh and peak psiin Siilarly: The value 40 is derived fr: The value 355 is derived fr: a a sin θ sin θ 87 c sin c sin c 3 58 c The value 340 is derived fr: The value 40 is derived fr: a a sin θ sin θ 83 c sin c sin c 4 8 c ] Errr analysis: Knwn uncerainies: 00 c; θ 05 deg 004 rad (uncerainy in deerining θ fr graph) 7

18 Fr: a a sin θ a ( sin θ sin θ (sin θ d (sin θ d θ ) θ ) a ( c θ θ ) ( c( 3 ) ( 0 04 )) c 85 0 c 0 ) d (sin θ d θ ) θ Hence: a ± a 393 ± 0 39 ± 0 c 08 arks 8