# Young's Experiment: Find Bright Fringe Distance

In summary, the fringe difference is when the brightest parts are. Then you can find the point the two cross.
In a double-slit experiment, the slit separation is 2.0mm, and two wavelengths, 750nm and 900nm, illuminate the slits. A screen is placed 2.0m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with the bright fringe from the other?

what i know:
d=2.0mm or 2.0e-3m
lambda1=750nm or 750e-9m
lambda2=900nm or 900e-9m
y1=y2=Y <-- this is what we're looking for

y=n lambda L / d
d sin@ = n lambda

by equating both Ys i get n1/n2 = lambda2/lambda1

but I end up nowhere, any suggestions?

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anyone have an idea?

still no takers?

You need to work out the fringe difference, that is when the brightest parts are. Then you can find the point the two cross. Simple as that really.

do u mean trial and error?

i don't know if this is correct, but if i use n<=(less than or equal)d/lambda
there's going to be too many maximas on both wavelengths (thousands of them)

In a double-slit experiment, the slit separation is 2.0mm, and two wavelengths, 750nm and 900nm, illuminate the slits. A screen is placed 2.0m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with the bright fringe from the other?

You need the equation: $$\frac{\lambda}{d} = \frac{x}{L}$$

So for the wavelength of 750nm the fringes will be:

$$\frac{750 \times 10 ^{-9}}{2.0 \times 10^{-3}} = \frac{x}{2.0}$$

$$\frac{750 \times 10 ^{-9}}{2.0 \times 10^{-3}} \times 2.0 = x = 7.5 \times 10^{-4} \ m$$

And for the wavelength of 900nm the fringes will be:

$$\frac{900 \times 10 ^{-9}}{2.0 \times 10^{-3}} = \frac{x}{2.0}$$

$$\frac{900 \times 10 ^{-9}}{2.0 \times 10^{-3}} \times 2.0 = x = 9.0 \times 10^{-4} \ m$$

Then $$\frac{9.0 \times 10^{-4}}{7.5 \times 10^{-4}} = 1.2$$

1.2 needs to be multiplied by 5 times to get the first integer (which will be a bright fringe) which is 6. This means there will be 5 fringes of wavelength 900nm and 6 fringes of wavelength 750nm. This means the mixing fringe will be 4500nm or 4.5 x 10-6m from the central brightest fringe.

No trial and error at all. Can you see anything wrong? (serious question. I like a second opinion)

thanks for the help, but i actually got it already from my original equation which i got n2/n1 = lambda1/lambda2

I actually factored (700nm/900nm) smaller and got n2=n1 5/6

which gives me n1=6 and n2=5

which is the same as what you got :)

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how did u type in the lambda sign? so i don't have to type it all the time next time lol

thanks for the help, but i actually got it already from my original equation which i got n2/n1 = lambda1/lambda2

I actually factored (700nm/900nm) smaller and got n2=n1 5/6

which gives me n1=6 and n2=5

which is the same as what you got :)
I was simply showing you another way of doing it because you said:

by equating both Ys i get n1/n2 = lambda2/lambda1

but I end up nowhere, any suggestions?

how did u type in the lambda sign? so i don't have to type it all the time next time lol

The typing is using a system called LaTeX mathematical typesetting. This thread has more information but you simply type [tex ] /lambda [/ tex] but without the spaces in the tex bit.

thanks for the help, but i actually got it already from my original equation which i got n2/n1 = lambda1/lambda2

I actually factored (700nm/900nm) smaller and got n2=n1 5/6

which gives me n1=6 and n2=5

which is the same as what you got :)
For my interest, how did you factor down 700/900 and get 5/6??

oh sorry its 750/900

don't get me wrong, i do appreciate (a lot) the help. it just amazes me how simple the problem is, but it took me so long to figure it out

don't get me wrong, i do appreciate (a lot) the help. it just amazes me how simple the problem is, but it took me so long to figure it out
It's alright. I thought that there must have been something else to it as well because it didn't seem that bad but I have made silly posts in the past they have led other places and I have learned other stuff from them. I guess I am saying don't worry. There are people that will mock you but we learn more from making mistakes than always getting the right answer. Questions are for the brave, cowards don't ask.

Anyway, enough of the philsophy, I was happy to help and don't worry about how simple it was. In fact you have helped my revise for my test tomorrow.

## 1. What is Young's Experiment?

Young's Experiment, also known as the double-slit experiment, is a classic experiment in physics that demonstrates the wave nature of light. It was first conducted by Thomas Young in the early 1800s and has since been replicated and expanded upon by many other scientists.

## 2. How does Young's Experiment work?

In this experiment, a beam of light is passed through two narrow slits and then projected onto a screen. The result is a pattern of light and dark fringes, known as an interference pattern. This pattern is caused by the constructive and destructive interference of the light waves passing through the two slits.

## 3. What is the purpose of finding the bright fringe distance in Young's Experiment?

The bright fringe distance, also known as the fringe spacing, is the distance between two consecutive bright fringes in the interference pattern. It is an important measurement in Young's Experiment as it allows us to determine the wavelength of the light being used, which is a fundamental property of light.

## 4. How is the bright fringe distance calculated in Young's Experiment?

The bright fringe distance can be calculated using the formula d = λL/D, where d is the fringe spacing, λ is the wavelength of the light, L is the distance from the slits to the screen, and D is the distance between the two slits. This formula is based on the principles of diffraction and interference.

## 5. What factors can affect the bright fringe distance in Young's Experiment?

The bright fringe distance can be affected by the wavelength of the light, the distance between the slits and the screen, and the distance between the two slits. Any changes in these variables can alter the interference pattern and therefore affect the measurement of the bright fringe distance.

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