Find Sin θ and D in Diffraction Equation

[magma_saber]

Homework Statement

I have to find "sinθ" and "d" in the diffraction equation.

Homework Equations

The formula is n(lambda) = d(sinθ)
I have the n, length, width, and hypotenuse.
http://img81.imageshack.us/my.php?image=captureee1.jpg
n=1, L=335 W=42.5 hyp=337.8

The Attempt at a Solution

I forgot how to get the angle for sin. So I'm pretty much stuck.

 

[tiny-tim]

Welcome to PF!

I have to find "sinθ" and "d" in the diffraction equation.
…
The formula is n(lambda) = d(sinθ)
I have the n, length, width, and hypotenuse.
http://img81.imageshack.us/my.php?image=captureee1.jpg
n=1, L=335 W=42.5 hyp=337.8

…I forgot how to get the angle for sin. So I'm pretty much stuck.

Hi magma_saber! Welcome to PF!

Do you mean that if you have, say sinθ = x, you want to get θ as a function of x?

If so, the solution is θ = sin^-1x (also written arcsin(x)), which most electronic calculators will do … alternatively, use sine tables "backwards"!

Or am I misunderstanding the problem?

 

[magma_saber]

I got sinθ. In my book it says sinθ = opposite over hypotenuse. By doing that i got 42.5/337.8. so sinθ should equal 0.126 right?
Now how do i get "d"? The equation is (n)(lambda)=(d)(sinθ). lambda is 650 btw.
So would this be the equation?
(1)(650)=d(sin0.126)

 

[magma_saber]

anyone?

 

[tiny-tim]

I got sinθ. In my book it says sinθ = opposite over hypotenuse. By doing that i got 42.5/337.8. so sinθ should equal 0.126 right?
Now how do i get "d"? The equation is (n)(lambda)=(d)(sinθ). lambda is 650 btw.
So would this be the equation?
(1)(650)=d(sin0.126)

Hi magma_saber!

I think people have been avoiding answering because you haven't made it clear what W and d are.

If the correct equation is (n)(lambda)=(d)(sinθ), and if W and hyp are the correct opposite and hypotenuse, then 650 = d times 0.126.

(not sin(0.126) … 0.126 is the sin )

 

[magma_saber]

thanks i got it now.