# Light: Fringes and Interference

1. Apr 23, 2005

### Soaring Crane

Coherent light with wavelength 597 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe.

For what wavelength of light in micrometers will the first-order dark fringe be observed at this same point on the screen?

To solve this, I must use the formula for destructive interference:
d*sin(theta) = (m + 1/2)*lambda, where m= 0 or -1 since it is the first dark fringe,

but I am stuck on using small-angle approximations where sin(theta) = tan(theta) to find d.

Could someone please explain this method to me clearly?

After finding d, what formula must I use for the 2nd beam's interference pattern to find the 2nd wavelength?

Thanks.

2. Apr 23, 2005

### Staff: Mentor

Bright fringes are found where $d \sin\theta = m\lambda$; dark fringes where $d \sin\theta = (m + 1/2)\lambda$. m = 0, 1, 2, etc.

The position (y) of the fringes is given by $\tan\theta = y/L$. For small angles $\tan\theta = \sin\theta = \theta$. So the bright fringes are found at $y = m\lambda L /d$; the dark fringes at $y = (m + 1/2)\lambda L /d$.

Note that d, L, and y stay the same for the two wavelengths. Hint: Solve for the new wavelength symbolically before plugging in any numbers.

3. Apr 23, 2005

### Soaring Crane

What value is y in the problem? 4.84 mm?

4. Apr 23, 2005

### Staff: Mentor

That's correct. Note that the same value of y is used for both wavelengths:
For the first, y = 4.84mm is the postion of the first bright fringe
For the second, y = 4.84mm is the postion of the first dark fringe​

Hint: You don't really need to plug in that y value! What you need to do is find the second wavelength in terms of the first.

5. Apr 23, 2005

### Soaring Crane

How do I do this? Do I use the formula for bright fringes for the second beam as a starting point?

6. Apr 24, 2005

### Staff: Mentor

Here's what I suggest: Call the first wavelength $\lambda_1$ and the second $\lambda_2$. Now write the appropriate equation for each and compare. I'll write the equation for the first wavelength. Since it gives the location of the first bright fringe, I'll set m = 1:
$y = \lambda_1 L /d$
Rearranging that gives you: $\lambda_1 = yd/L$

Now you write the equation that applies to the second wavelength and compare.

7. Apr 24, 2005

### Soaring Crane

Well, for the second, is it this?

y = [(m + 1/2)*lambda_2*L]/[d], where m = 0

lambda_2 = (2*y*d)/L

8. Apr 24, 2005

### Staff: Mentor

Exactly. So... how does $\lambda_2$ compare to $\lambda_1$?

9. Apr 24, 2005

### Soaring Crane

It's twice the value of lambda_1.

10. Apr 24, 2005

### Staff: Mentor

Correct. Now express the answer in micrometers and you're done.