Seperation between central bright frindge and third dark frindge

[crazyog]

Homework Statement

Two slits are separated by 2.00x10^-5 m. They are illuminated by a wavelength 5.60x10^-7m. If the distance from the slits to the screen is 6.00 m, what is the separation between the central bright fringe and the third dark fringe?
Answer: 0.421 m

Homework Equations

I thought I would use
d=(m+1/2)(lamba)*L/y for dark and then d=m(lamba)*L/y for bright

The third dark fringe would be m= 2 if it starts at 0, right?
so (2.5)(5.6x10^-7)(6)/(2.0x10^-5) = 0.42
buttt. what about the central bright fringe?
I plug in m = 0 and get 0 so 0.42-0 = 0.42

is this correct?

The Attempt at a Solution

 

[alphysicist]

Homework Statement

Two slits are separated by 2.00x10^-5 m. They are illuminated by a wavelength 5.60x10^-7m. If the distance from the slits to the screen is 6.00 m, what is the separation between the central bright fringe and the third dark fringe?
Answer: 0.421 m

Homework Equations

I thought I would use
d=(m+1/2)(lamba)*L/y for dark and then d=m(lamba)*L/y for bright

The third dark fringe would be m= 2 if it starts at 0, right?
so (2.5)(5.6x10^-7)(6)/(2.0x10^-5) = 0.42
buttt. what about the central bright fringe?
I plug in m = 0 and get 0 so 0.42-0 = 0.42

is this correct?

That looks right to me. The difference 0.001m in your result and the given result is because you used the small angle approximation. If you had used the set of original formulas:

$$\begin{align}<br /> d\sin(\theta)&=\left(m+\frac{1}{2}\right)\lambda\nonumber\\<br /> \tan(\theta)&=\frac{y}{L}\nonumber<br /> \end{align}$$

to find $y$, there would have been no discrepancy. (Here $d$ is the slit separation and $y$ is the distance on the screen, which I believe is the way most books have it. My guess is that you got those two confused with each other when you did the algebra of your solution; in your equation it did not matter, but in some others it would.)

 

[crazyog]

and L is the distance from the screen to the slits right?
what did I mix up then?

 

[alphysicist]

and L is the distance from the screen to the slits right?
what did I mix up then?

I was just saying that in your original post, you set $y=2\times 10^{-5}$ and you solved for $d$.

Now, I don't know what book you are using, so there's no way for me to know for sure what variables they use for what, but as far as I can remember, $d$ always stands for the slit separation and $y$ was always the position along the screen. Then your original post should have had $d=2\times 10^{-5}$ and you would be solving for $y$.

In the small angle approximation, it turned out not to matter because of the form of the equation; but if they had asked for the angle, for example, you would use

$$d \sin\theta = m\lambda$$

for the bright fringe angles, and there it definitely would matter; in that equation, $d$ has to be the slit separation!

So I was just suggesting you check the definitions of $d$ and $y$ to make sure you are following your book's definition.

 

[crazyog]

oh ok, i understand now
thank you so much for your help! =)

 

[alphysicist]

Sure, glad to help!