Question

a double slit is illuminated by two wavelengths 450nm and 600nm. What is the lowest order at which the maxima of one wavelength coincides with minima of the other?
what i did is n*450=[n-(1/2)]*600 this gives me 2. answer is 3. whats wrong?

Answer 1

@Mashy

Answer 2

@shubhamsrg

Answer 3

well your method seems right to me.

Answer 4

it looks right to me too. i checked it on google. got a couple of google books results which have the answer 3 too. must be something wrong with my logic.

Answer 5

Leme try again.

Answer 6

wait, what was I saying,nopes the method is wrong. :P

Answer 7

hmm. a book has a similar question. so i followed that. :P

Answer 8

maxima for 450 = m(450)
minima for 450 = (m -1/2)450

maxima for 600 = n(600)
minima for 600 = (n- 1/2)600

thus,
either 450m = (n- 1/2)600
or
600n = (m- 1/2)(450)

=> m/(n -1/2) = 4/3
and
n/(m- 1/2) = 3/4

on comparison,n and m are symmetrical .
hit and trial shows n=2 and m=2 satisfy it
ans = 4

lol, my method is thus flawed, sorry about that.

Answer 9

nice try. better than mine for sure.

Answer 10

I did what you have done. But, you can not take both orders to be n, your eqn is correct with one n and other m. Order then = n+m, but but, it fails! :3

Answer 11

hmm.

Answer 12

can you tell me when is minima = (n -1/2)lambda and when is it (n +1/2)lambda ?

Answer 13

Reason I ask is (n+ 1/2) guides us to the ans i.e. 3! :P

Answer 14

i have always used n-1/2. n+1/2 is used for diffraction max. i can be wrong.

Answer 15

can't *

;)

Answer 16

haha. if that yields the answer i am obviously wrong. :P

Answer 17

http://answers.yahoo.com/question/index?qid=20071031230123AApqPpL

Answer 18

as according to him, I'll take n+1/2
and thus, 3 is the ans! :P

Answer 19

But am not too comfortable with it,

Answer 20

@Mashy
@Vincent-Lyon.Fr
@DLS (the champ)

Answer 21

i dont know. i checked in 2 books now. they have n-1/2.
and champ is still in 11th. :P

Answer 22

hmm,must be some problem in the logic then.

Answer 23

yeah. lets see.

Answer 24

its n + 1/2 lamda.. when n starts from 0,1,2,3 etc

its n -1/2 lambda when n starts from 1, 2,3 etc.. you can take anything

however.. since in this case we are considering both min and max together, for max we usually start with 0, i prefer to take the formula which allows me to start with 0 for min too.. so i would go with n+1/2

Answer 25

so when i do

450m = 600 (n +1/2)
i get m = 2 and n = 1

which means second order maxima of 450nm light coincides with 2nd order minima (remember we have chosen 0 as the first order minima) of the 600nm light.. i dunno how shubhang got 3? :O

Answer 26

well n and m are respective orders right? n=1 and m=2 , so net =3 ?

Answer 27

Well I may be wrong, I am not too comfortable with wave optics at the moment.

Answer 28

shubhang? o.O

Answer 29

so is this net thing right?

Answer 30

what net thing?? .. the question is quite ambigious :-/.. the question doesn't really say we need to add up the orders :-/

Answer 31

and besides i got.. both 2nd order .. ! :P

Answer 32

That is what I had yielded in the first place! B|

Answer 33

think about it.. you usually get the answer after some time. take your time. :P
also if we dont take n,m and just do n*600=(n+1/2)*450, we get n=2. i know its not correct but i have done this in questions involving nth fringe where we do n*a=(n+1)*b where nth fringe coincides. :P

Answer 34

where a and b are fringe widths.

Answer 35

well thats because.. it turns out the second orders DO CONINCIDE..

if the second order's didn't conincide.. then in your equation you would get n as some fraction!

Answer 36

hmmmm.. got that part.
and what do you think of n=3?

Answer 37

look at this question. why have they taken n only?

Answer 38

I will not tell you.

Answer 39

:O was that a stupid question?

Answer 40

well thats because the incident light itself is made up of two colors

you ll see a maxima on the screen, when the maximas of BOTH the yellow light conincide .. !

Answer 41

nopes, because I don't know! -_-

Answer 42

in this quesiton.. we have two seperate patterns itself..

Answer 43

hmm i see.

Answer 44

wait.. what?.. maxima co inciding with minima!!.. wait wait wait!

Answer 45

take your time! :P

Answer 46

wait.. see its given better in this .. i dunno what your book meant :-/

http://books.google.co.in/books?id=8UphV6dA7ucC&pg=SA5-PA183&lpg=SA5-PA183&dq=upto+what+order+can+the+fringes+be+seen&source=bl&ots=UVj2ajtzID&sig=8Vv-G4J0pVql2BO9JMbhfOJECjM&hl=en&sa=X&ei=2hYuUdvvCsuwqwGQzICoDA&ved=0CDEQ6AEwAA#v=onepage&q=upto%20what%20order%20can%20the%20fringes%20be%20seen&f=false

Answer 47

they have mentioned.. that until the maxima of one conincides with minima of the other, you ll see pattern.. after that you won't.. quite frankly i dunno why? but atleast it makes more sense then what your book says!

Answer 48

oh wait.. they both say the same thing.. nv :D

Answer 49

lol. have some water sir.

Answer 50

I really got to revise my wave optics! hmm.. :(

Answer 51

this IIT physics is so stupid!.. its basically puzzles.. who wants to solve puzzles.. :-/

Answer 52

omg! look what I found
http://inst.eecs.berkeley.edu/~ee119/sp10/Homework/HW11SOL.pdf

Answer 53

see "4. Young's double slit experiment" part d)

Answer 54

and leme bring to everyone;s notice, the CHAMP has arrived! \m/

Answer 55

im getting 3 :/

Answer 56

I am not surprised !

Answer 57

but m2 = 1 is second order minima :( :( .. not first order..

Answer 58

wait a sec im posting my solution

Answer 59

NOW THATS WHY YOU ARE A CHAMP. explain, NOW!

Answer 60

@Mashy :(

Answer 61

Enjoy,goodbye.

Answer 62

http://whitsblog.com/wp-content/uploads/2012/01/okay-meme-face.jpg

Answer 63

WTF. :/

Answer 64

maybe i can help next year..:P

Answer 65

answer is 2.

Answer 66

you have a misprint

Answer 67

http://books.google.co.in/books?id=dBd4dbJX40QC&pg=PA111&lpg=PA111&dq=What+is+the+lowest+order+at+which+the+maxima+of+one+wavelength+coincides+with+minima+of+the+other?&source=bl&ots=jzkLy4rGJk&sig=uZfzece77MgaLnuPXLRozVI6YV4&hl=en&sa=X&ei=SRouUbXJKcvJrAfH8IGIAQ&ved=0CDsQ6AEwAg#v=onepage&q=What%20is%20the%20lowest%20order%20at%20which%20the%20maxima%20of%20one%20wavelength%20coincides%20with%20minima%20of%20the%20other%3F&f=false

check answer 20

Answer 68

we have a new answer then! B|

Answer 69

hmm. i guess its a misprint then.
so the link shubham posted was also wrong?

Answer 70

are you talking about yahoo answers?

Answer 71

MUHAHAHAHHAA


see i was right all along!!

Answer 72

mashy mashy he is the man
if he can't do it nobody can :D

Answer 73

what if they say first order minima? whats n?

Answer 74

see first order minima means first minima.. thats when the path difference is exactly equal to lambda /2

so you see.. that can happen if you plug in n=1 in the equation n-1/2 lambda.. or plug in n=0 in equation n+1/2 lamda..

so it really depends on what equation you choose.. but ultimately you see its the same damn thing :P

Answer 75

i got it, i guess.
you are telling me to see that if i use n-1/2, then i see, first order minima is n=1. and if, i see, we use n+1/2, then first order minima is n=0. i see its the same damn thing.

Answer 76

i see that you are trying to be funny :P.. !! now lemme take bath !

Answer 77

not really. :P

Answer 78

http://books.google.co.in/books?id=AotsqgIr000C&pg=SA3-PA274&lpg=SA3-PA274&dq=What+is+the+lowest+order+at+which+the+maxima+of+one+wavelength+coincides+with+minima+of+the+other?&source=bl&ots=2hqmpdDG_N&sig=RX00-VIiMtAdd6T8mkpsI9kg-ZU&hl=en&sa=X&ei=OBwuUaU1x46uB-aZgJgM&ved=0CDQQ6AEwAA#v=onepage&q=What%20is%20the%20lowest%20order%20at%20which%20the%20maxima%20of%20one%20wavelength%20coincides%20with%20minima%20of%20the%20other%3F&f=false

Answer 79

question no.55
answer is c)

Answer 80

LAWL

Answer 81

oh dang. whatever now. must be a misprint. :P

Answer 82

well 2 of the books say 3, one says 2 .. so you all can calculate the respective probabilities!
;)

Answer 83

i'll go with the logic. :P
and if the same question comes in the exam, i'll probably leave it. :P

Answer 84

hey, how did "LAWL" come down? o.O

Answer 85

hmm. good question.

Answer 86

I'll revise my wave optics and re-visit the question in the near future.

Answer 87

Though I don't guarantee I'll be able to resolve the conflict! -_-

Answer 88

haha. bye.