What Is the Required Distance for Young's Double-Slit Experiment?

[Kris1120]

Homework Statement

Light of wavelength 473 nm falls on two
slits spaced 0.389 mm apart.
What is the required distance from the slit
to a screen if the spacing between the first and
second dark fringes is to be 6.71 mm?
Answer in units of m.

Homework Equations

sin\theta= m\lambda/d

x=tan\theta*L

The Attempt at a Solution

\theta = sin^-1{2*473e-9m / .389e-3 m} = .139336 degrees

L = 6.71e-3 m / tan .139336 = 2.75919

 

[alphysicist]

Hi Kris1120,

Homework Statement

Light of wavelength 473 nm falls on two
slits spaced 0.389 mm apart.
What is the required distance from the slit
to a screen if the spacing between the first and
second dark fringes is to be 6.71 mm?
Answer in units of m.

Homework Equations

sin\theta= m\lambda/d

I don't believe this is the formula you want. They are referring to the dark fringes in this problem.


Also, when you solve for the distance remember they want the distance between two fringes, not the distance from one fringe to the center point.

 

[Kris1120]

Ok so I am using the equation sin(theta) = (m +.5) lambda / d ... Does that mean I do not use the equation to solve for x?

 

[alphysicist]

Ok so I am using the equation sin(theta) = (m +.5) lambda / d ... Does that mean I do not use the equation to solve for x?

You do still have to solve for x.

However, I think here you should use the small angle approximation that combines both equations (because for small enough angles sin(theta)=tan(theta)). Does your book have that formula?

 

[Kris1120]

No. I can't find it

 

[alphysicist]

No. I can't find it

For constructive interference, you have the two equations that you had in your original post:

<br /> \sin\theta= \frac{m\lambda}{d}<br />

<br /> \tan\theta= \frac{x}{L}<br />

For small angles,

<br /> \frac{x}{L} =\frac{m\lambda}{d}<br />
so you can find the distance along the screen x directly for each line order (again, when the angles are small). You can do the same thing with the destructive interference case, and get the same equation with m \to (m+1/2).

 

[Kris1120]

If I am not misunderstanding... I set the sin equation equal to the tan equation and solve for L... I am still not getting it correct though. I am trying to look the equation up on google.

 

[Kris1120]

Maybe I am using the wrong m... m=2?

 

[alphysicist]

If I am not misunderstanding... I set the sin equation equal to the tan equation and solve for L... I am still not getting it correct though. I am trying to look the equation up on google.

The formula that I mentioned for the destructive interference:

<br /> x = L \frac{(m+\frac{1}{2} )\lambda}{d}<br />

gives the x value for each fringe. What they tell you in the problem is that the difference in the x values is 6.71mm. Do you see what to do?

I just saw your last post. The first order dark fringe would have m=0, and the second order dark fringe would have m=1, based on how you wrote your destructive interference condition.

 

[Kris1120]

6.71 mm = [L((1+.5) * 473 nm)/ .389 mm] - [L((0+.5)*473nm)/ .389 mm] and solve for L? I get L = 5.51809 m is that correct?

 

[alphysicist]

6.71 mm = [L((1+.5) * 473 nm)/ .389 mm] - [L((0+.5)*473nm)/ .389 mm] and solve for L? I get L = 5.51809 m is that correct?

That looks right to me.

 

[Kris1120]

Great! Thank you for being so patient with me.

 

[alphysicist]

I'm glad to help!