General Tps on How to Do Well n Physcs Exams 1. Establsh a good habt n keepng track o your steps. For example when you use the equaton 1 1 1 + = d d to solve or d o you should rst rewrte t as 1 1 1 = d d (1) and then begn to put n numbers such as: 1 1 1 1. d = 5m 6m = 30m Then you'll know that what you have just got s not yet 1/ d. o you must stll do one more step to nd that o d (2) but s only d = 30m (3) as the nal answer. Now you don't have a good habt you would wrte only 1 1 = 1 5 6 30 wthout sayng what t s really equal to. You would then orget that you have one more step to do and smply gve your answer as d = 1 m. 30 Comparng the correct answer (3) wth the wrong answer (5) I hope you can now see where you are lkely to make mstakes. It all bols down to the bad habt o omttng the step gven as Eq. (1) and also beng lazy n not payng attenton to unts n Eq. (4) and not explctly notng that t s actually 1/ d you have calculated there not yet d. That s you should have wrtten Eqs. (1) and (2) nstead o Eq. (4). Wrtng Eq. (4) nstead o Eqs. (1) and (2) means that you have a bad habt and you are lazy. You can not blame others you get wrong answers ths way. (4) (5)
(2) I you are to draw a gure to help you gure out the answer avod puttng nto the gure normaton that s not gven by the problem. For example an angle s nvolved avod makng t a rght angle or 45 or 30 or 60 unless the problem says so. These are specal angles wth specal propertes whch are not shared by a general angle. I the problem dd not say that you have such a specal angle then you should not draw such a specal angle. That s n your drawng you should avod drawng any such specal angles. Another example s: I a rectangle s nvolved n the problem you should avod drawng t as a square (unless t s a square). It should clearly be a rectangle. In act you should avod drawng t n such a way that one sde appears to be twce or three tmes etc. as long as the other sde. Ther rato should not be a smple nteger. It can mslead you nto thnkng that you have ths extra normaton n the problem that you really don't have. Now these are just examples. I can not pont out all thngs you are lkely to do. The general gudelne here s that you should avod dong thngs to ool yoursel nto thnkng that you have thngs you really don't have. (3) Do not mprovse. Physcs laws do not allow you to mody them. For example n optcs there s the equaton or double-slt ntererence: = (*) d sn θ mλ. I you are gven one m and one d but two derent λ s then usng the rst λ you can nd the rst θ and usng the second λ you can nd the second θ but you can not put n the derence n λ and hope to drectly compute the derence n θ. That would be a msuse o the ormula. That s what I mean by mprovse! A ormula does not allow you to use t n a way that t really doesn t allow. On the other hand n the example ormula gven when d s much larger than λ so you know that both θ s are very small then you can approxmate sn θ by θ n radans. We then have dθ1= mλ1 and dθ 2 = mλ 2 so takng the derence o the two equatons does gve you d( θ θ ) = m( λ λ ) or d θ = m λ. 1 2 1 2 o n ths case you can ndeed put n the derence o λ and obtan drectly the derence n θ. But ths s an excepton. It can be done
n ths case only because both λ and θ enter the equaton lnearly. Ths s not true wth the orgnal equaton (*)! I the change n λ and θ are very small then you can derentate both sdes o the orgnal e- quaton and obtan d cosθ dθ = m d λ. From ths equaton you can ndeed put n the very small quantty dλ and obtan drectly the very small quantty dθ and vce versa. But notce that t has cos θ n t whereas the orgnal equaton has sn θ. Another example s the equaton or Doppler sht o a sound wave requency when the source s movng wth the speed v and the lstener s movng wth the speed v :. - = v v v -v where v s the speed o sound s the actual requency o the sound emtted by the source and s the apparent requency o the sound heard by the lstener. When solvng such problems you can not work wth a relatve speed as you are movng wth the source so only the lstener s movng wth the speed v - v or movng wth the lstener (who s not you) so only the source s movng wth the speed v -v. I you do that you must also change v because v s dened wth respect to statonary carrer o the wave and you are movng wth respect to the statonary carrer o the wave then the sound speed s no longer v. I you don t take care o ths pont and stll work wth a relatve speed then you are mprovsng snce you have changed the content o the equaton. I the problem s about the Doppler sht o a lght wave then the ormula changes to: R = c -v c + v where c s the speed o lght and v v -v s the relatve speed. That s only the relatve speed matters n ths equaton. Ths pont
actually has very sophstcated reasons behnd t and s the great dscovery o Ensten. Namely lght s movng at the same speed o lght no matter how ast you are movng. It also has to do wth the act that lght waves have no carrer so you can t move wth respect to that non-exstent carrer to change ts speed. (4) Use dmensonal analyss to help you nd some o your mstakes. For example n wave theory we have λ = c. That s the wavelength o the wave tmes the requency o the wave s equal to the speed o the wave. The dmenson or unt o λ s meter or m. The dmenson or unt o s Hz or 1/s. Hence the product on the let hand sde o the equaton has the dmenson or unt o m/s whch s precsely the unt o the wave speed c. Now suppose you care- λ = c lessly put t as. You should be able to nd ths mstake easly by lookng at the dmenson or unt on the two sdes o the equaton. The dmenson or unt o λ s m. But the dmenson or unt o c s (m/s) (1/s) = m/s 2 whch s not what you have on the let hand sde o the equaton Thereore t must be wrong. As a matter o act ths dmensonal analyss mmedately suggests to you that the correct ormula should be λ = c and not λ = c. Every ormula n physcs allows you to analyzng t ths way and the dmenson or unt must be the same on ts two sdes. (5) Many ormulas allow you to have some eelng on why certan quanttes are n the numerator and why certan other quanttes are n the denomnator. Take the ormula or the speed o transverse wave on a strng. F v= µ where F s the tenson n the strng and µ s the lnear mass densty o the strng. In ths ormula t s not dcult to see why F s n the numerator and µ s n the denomnator and not the other way a- round. It s because ncreasng F can clearly make v larger because neghborng molecules n the strng can nluence each other more easly when F s larger and ncreasng µ wll clearly make v smaller because now neghborng molecules n the strng wll have a harder tme to nluence each other when µ s larger because larger µ
means larger nerta. o you can now see that you should never wrte t as v = µ / F. (As to why there s a square root sgn dmensonal analyss can help you see t.) (6) I possble you should use a gure to help you get thngs straght. For example take the ormula I = I 2 0 cos ( φ / 2). You should remember the phasor dagram assocated wth t: φ / 2 φ Ths phasor dagram not only let you see why t s φ / 2 and not just φ as the argument o the cosne uncton t also let you see that ths ormula s assocated wth the ntererence o two narrow lght beams wth a phase derence o φ between them so t can t be used or the dracton o a sngle wde slt nor the dracton o N narrow slts wth N > 2. You wll then not use t n the wrong place. On the other hand you wll also be able to see that you can also use ths ormula or the ntererence o two sound or rado waves generated at two pont sources A and B when the recever s at dstance d 1 orm source A and d 2 rom source B. Then all you need s to also see that n ths case φ = 2π(d 1 - d 1 ) / λ. On the other hand two lght beams are ntererng n phase to begn wth and a thn glass plate wth ndex o reracton n and thckness t s nserted nto one beam perpendcular to the beam leavng the other beam undsturbed then the above ormula can also be used except that now t t φ = 2π = 2 π( n 1) t λ / n λ λ because t s the change o phase o one beam (that s a new phase 2 π[ t/( λ / n)] mnus an old phase 2 π[ t / λ ]) wth the phase o the other beam unchanged. o t s also the derence between the phases o the two beams ater the glass plate has been nserted they are n phase beore the glass plate s nserted. Note that n all these cases when the ormula can be used there are always two narrow beams o some wave ntererng whle there s a phase derence between them.
Many ormulas allow you to understand them usng some sort o gures. o you should learn such ormulas ths way and not try to just memorze them. (7) I the problem gves you a quantty and you dd not use t to get your answer then you should ask yoursel the ollowng queston: hould the answer be ndependent o ths quantty? Assumng that t s true then you should be able to change the value o ths quantty to say zero or nnty ( allowed). But the correct answer or these cases mght be very easy to see. I t dsagrees wth your answer then you know that your answer must be wrong and the correct answer should depend on ths quantty. For example take the problem o nsertng a glass plate nto one beam and you are asked the ntensty change when t s ntererng wth another beam. I your answer does not depend on the thckness t o the glass plate then you should see that t has to be wrong snce or t equal to zero we have no glass plate and the phase sht must be zero! o how can you nd a nonzero answer that s ndependent o t? (8) Even when a quantty does enter your soluton you can stll ask whether t has entered correctly. et us agan take the glass-platenserton problem as an example. I you gve the phase sht as t t φ = 2π = 2πn λ / n λ then you can ask whether n has entered correctly. Well you can look at the case when n =1. The glass plate has been replaced by a layer o ar. That s nothng s now nserted nto the beam path. The phase sht should clearly be zero n that case. But s your answer zero n that case? It s not φ s gven by the above equaton. o you know t has to be wrong! It mmedately remnds you that you orgot to subtract the phase due to a layer o ar o the same thckness. That s you dd not compute the change. O course you know nothng about what s a phase all these remarks are useless to you. You must learn the general concepts rst.