Young's Experiment: Find Bright Fringe Distance

[admajoremdeigloriam]

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?

 

[admajoremdeigloriam]

anyone have an idea?

 

[admajoremdeigloriam]

still no takers?

 

[The Bob]

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.

The Bob (2004 ©)

 

[admajoremdeigloriam]

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)

 

[The Bob]

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)

The Bob (2004 ©)

 

[admajoremdeigloriam]

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 :)

 

[admajoremdeigloriam]

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

 

[The Bob]

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.

The Bob (2004 ©)

 

[The Bob]

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??

The Bob (2004 ©)

 

[admajoremdeigloriam]

oh sorry its 750/900

 

[admajoremdeigloriam]

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

 

[The Bob]

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.

The Bob (2004 ©)