Diffraction Grating number of slits

[Ginerva123]

[SOLVED] Diffraction Grating

Homework Statement

Find the number of slits per centimeter of a grating designed to disperse the first-order visible spectrum through an angular range of 15.0. Find also the angles at which the first-order visible spectrum begins and ends.

Homework Equations

d sin \theta = n \lambda
d = 1/N

The Attempt at a Solution

Okay, if it's first order, then n = 1, I suppose, but lambda and theta are unknown and I need to get d to find N...
For the second part, once I have d, I can input 400 and 700 nm for lambda to get the two values for theta, but obviously I need d for both answers and I have no clue how to get it. Any help would be appreciated. Thanks!

 

[Kurdt]

I think for this you will just solve for d and you can use your good assumptions (400nm-700nm) for the range of visible spectrum. If you read the question carefully again it says 'designed to disperse the first-order visible spectrum through an angular range'. That means the angle from the start of the first order spectrum to the angle at the end of the first order spectrum has to be 15 degrees. Can you proceed from there?

 

[Joza]

Bump. I had a look at this too.

If you have a RANGE of wavelengths, how can you solve for d with that equation?

 

[mda]

You might consider using the "small" angle approximation to simplify the algebra a bit.
The answer will be out by 10% or so.

 

[Ginerva123]

I'm sorry, but there just seems to be to many unknowns in the question for me... using a small angle approximation won't eliminate any of them, as far as I can see. Is there perhaps another equation I could use to find d?

 

[Kurdt]

\Delta \theta = \theta_1 - \theta_2 for a range of wavelengths. What is theta? You can rearrange the equation you gave above for theta.

 

[Ginerva123]

Theta = sin-1(n lambda / d)... but without values for lambda or d (my 400 - 700 nm are only approximations for the second part) how would I solve this? (My profuse apologies if I'm coming across as particularly obtuse here, but I just can't grasp this question!)

 

[Kurdt]

You would use your assumptions. Like I said they are good assumptions for the range of visible light. Then you can solve for d.