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Q-A beam of light consisting of two wavelenghts 600 nm and 450 nm is used to obtain interference in Young's Double Slit experiment (YDSE). Find the least distance from the central maximum where the bright fringes due to both wavelengths coincide. The distance between the two slits is 0.4 mm and screen is placed at a distance of 1.0 m from the slits.

Now my problem is not with the question but the solution given in my book. It says

Let at distance $x$ from central bright maxima, the bright fringes due to both the wavelengths coincide for the first time. It is possible only if within the distance $x$ there are $n$ fringes for light of 600nm and $(n+1)$ fringes for light of wavelength 450 nm.

Okay so my problem is that I am unable to figure out how have they made this argument about $n$ and $n+1$ fringes? It's isn't mention anywhere in the question. Can't it be ($n$+any number)? If there's any better solution than this I would be really thankful for that

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In this case whoever set the question has chosen wavelengths with a nice ratio and not too far apart, so the first coincidence is at $n_{600}$ = 3 and $n_{450} = 4$. However you're quite correct that this isn't a general rule. For example if the two wavelengths, $\lambda_1$ and $\lambda_2$ were both primes they would coincide when $n_1 = \lambda_2$ and $n_2$ = $\lambda_1$.

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So how to solve such questions then? Like when λ1 and λ2 are both primes? – Harsh Feb 23 '13 at 14:15

Well i was also quiet confused by the solution. Here the batter one as much i know distance for the bright from the central bright fringe y=(n(lemda)D)/d Where lemda= wavelength of light D=distance of screen d=distance between slits so let the bright fringes coincides be y

                                  so y(1)=  y(2)
                        n(1)(lemda 1)D/d = n(2)(lemda 2)D/d
                               n(1)x600  =  n(2)*450
                               n(1)/n(2) =    450/600
                                         =    3/4
                            fringe width = (lemda)D/d

                                         = [(600nm) or (450nm)]*1/4*10^(-4)
                                         = [(600nm) or (450nm)]*1*10000/4
                                         = for lemda 600= 15*10^(-4)
                                       and for lemda 450 = 112.5*10(-5)

   now multiply the corresponding fringe with the n obtained to know the distance of the coincidence from the central bright fringe

                                       y= fringe width*n
                         y=for lemda 600= 15*10^(-4)*3
                      and for lemda 450 = 112.5*10(-5)*4

   There may be some calculation mistake so if answer is not correct try it by yourself by above given method
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You have the ratio n1/n2=3/4. So how are you taking 3 for 600 nm and 4 for 450 nm? – Harsh Feb 23 '13 at 14:17
yes it is the ratio of the fringe number for the bright fringes where they will coincide next position will be 6/8 hope you understood if not you are free to ask – Dimensionless Feb 23 '13 at 17:32

Yes, the ratio is for example $4/5$ and they are asking for minimum distance $n_1=4k$; $n_2=5k$ (for some $k$), the distance will be minimum when $n$ is minimum, so $k=1$ and $n_1=4$ and $n_2=5$.

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Hello, and welcome to Stack Exchange. Your answer is difficult to understand; if you feel it is a good answer, you should spend a little time elaborating and improving its formatting. – Daniel Griscom Nov 22 '15 at 19:07

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