# Light: Fringes and Interference

• Soaring Crane
In summary, at the point where the bright fringes are observed, the wavelength of coherent light with a frequency of 597 nm passes through two very narrow slits. 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.
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.

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.

What value is y in the problem? 4.84 mm?

Soaring Crane said:
What value is y in the problem? 4.84 mm?
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.

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

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.

Well, for the second, is it this?

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

lambda_2 = (2*y*d)/L

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

It's twice the value of lambda_1.

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

## What is the concept of interference in relation to light?

Interference in relation to light refers to the phenomenon where two or more light waves overlap and interact with each other, resulting in either constructive or destructive interference. This can lead to the formation of bright or dark fringes, respectively.

## How do fringes form in interference patterns?

When two coherent light waves overlap, they create regions of constructive and destructive interference. The bright and dark fringes are formed where these two regions meet, resulting in a distinct pattern of light and dark bands.

## What factors affect the spacing between fringes in an interference pattern?

The spacing between fringes in an interference pattern is affected by the wavelength of the light used, the distance between the light source and the screen, and the distance between the two slits or sources of light. These factors can be adjusted to produce different interference patterns.

## What is the difference between Young's double slit experiment and the Michelson interferometer?

In Young's double slit experiment, a single light source is split into two coherent sources, creating an interference pattern. In contrast, the Michelson interferometer uses two separate light sources that interfere with each other, creating an interference pattern. Additionally, the Michelson interferometer can be used to measure small changes in the wavelength of light, while Young's double slit experiment is used to demonstrate the wave-like nature of light.

## Can interference patterns only be observed with light waves?

No, interference patterns can also be observed with other types of waves, such as sound waves or water waves. However, the principles of interference and the formation of fringes are the same regardless of the type of wave being used.

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