Double Slits - Measuring Wavelength from the number of bright fringes

[Glorzifen]

Homework Statement

A double-slit interference pattern is observed on a screen 1.0m behind two slits placed on 0.3mm apart. Ten bright fringes span a distance of 1.65cm. What is the wavelength of the light?

Homework Equations

Since we're looking at bright fringes:
(m+1/2)$$\lambda$$ = dsin$$\theta$$ = d(x/L)

The Attempt at a Solution

d = 3E^-4m
m = 4 (I feel like I'm supposed to use just one half of the bands, excluding the middle band. Is this right? Why do I do this?)
L = 1.0m
y = 8.25E^-3m (this is the spread of the 10 fringes divided by 2 since I'm using half of them)

(m+1/2)$$\lambda$$ = d(x/L)
(4+1/2)$$\lambda$$ = 3E^-4(8.25E^-3/1.0)
$$\lambda$$ = 5.5E^-7m

On my formula sheet there is an 'x'...the variable for the spread of the fringes is traditionally 'y'...I figured they were just the same thing so used y for x in the above example. If that's wrong then could someone explain where I get 'x' from? Any explanation would be appreciated.

Thanks!

 

[Glorzifen]

Just realized that (m+1/2) is for DARK fringes. Would that be correct for dark fringes at least?

 

[flyers]

If you use m=4 that is 4+4+1 which only gives you 9 fringes. The question doesn't say these are all the possible fringes, maybe only a part of all fringes. What I would do is use m= 9 (that is 10 because we include m=0) and use distance y as 1.65cm. And yes you are supposed to use (m)LaTeX Code: \\lambda