Diffraction Grating, focal length of lens

[unscientific]

Homework Statement

Light from infinity strikes a diffracting grating normally with 600 lines per mm. The emitted light then passes through a lens to project visible light (400-800 nm) spectrum onto a screen that is 50mm, just enough to cover it. Find the focal length of the lens.

Homework Equations

The Attempt at a Solution

Lens maker's Equation:
$$1/u + 1/v = 1/f$$

Let u -> ∞,

Thus $v = f$, so the emitted light is focused on the paper.

$$d sin \theta = n\lambda$$

I can't find a way to solve this question unless $n$ is known...I shall let n = 1.

$$\frac{y}{\sqrt{y^2 + f^2}}$$ = \lambda [/tex]

Solving,

$$f = 22.4 m$$

This is a strange setup, imagine the photographic paper would have to be placed so far away from the lens! (Relative to its size)

 

[Simon Bridge]

Did I read that right: The screen is w=50mm wide and exactly covers a single spectrum from the grating - but only after passing through a lens focal length f. The distance from the lens to the grating or from the lens to the screen are not given?

Note: What do you mean by v=f?
Surely: The image is at the focal length only for an object very far away.
If the object is at the focus, then there is no image.
When light is "focussed on" a surface, you get a very bright spot - which maybe burns.
When an image is said to be "focussed" then it is sharp (i.e. not blurry) - this is not usually at the focal length.

I don't see your final working - what value did you use for wavelength?

 

[unscientific]

Did I read that right: The screen is w=50mm wide and exactly covers a single spectrum from the grating - but only after passing through a lens focal length f. The distance from the lens to the grating or from the lens to the screen are not given?

Note: What do you mean by v=f?
Surely: The image is at the focal length only for an object very far away.
If the object is at the focus, then there is no image.
When light is "focussed on" a surface, you get a very bright spot - which maybe burns.
When an image is said to be "focussed" then it is sharp (i.e. not blurry) - this is not usually at the focal length.

I don't see your final working - what value did you use for wavelength?

The screen is after the lens, that is right. But the question didn't specify how many spectrums did the screen cover..

v=f means that the light comes to a focus at distance f away.

I used λ = 800nm as that gives maximum angle at first maxima.

 

[Simon Bridge]

the question didn't specify how many spectrums did the screen cover

Don't you think that's important?
But I think that the question, as you wrote it in post #1, does specify how many orders the screen covers.
It only mentions one spectrum.

v=f means that the light comes to a focus at distance f away.

No it doesn't.
It is not even clear what you mean by "light comes to a focus" here - since "light coming to a focus" would just make a bright dot rather than an image.

I mean is v supposed to be the object distance or the image distance?

I'm guessing, since you have u at infinity that you mean it to be the image distance and u is the object distance.
The light from infinity is "parallel", but after it strikes the diffraction grating, the rays are no longer parallel. So why did you assume u as infinite?

I have a feeling about this - see below.

I used λ = 800nm as that gives maximum angle at first maxima.

... so what is the picture you expect to see on the screen?

I have a feeling you are expected to do something like for the setup in section 3 here:
http://physics-animations.com/Physics/English/DG10/DG.htm
... is that correct?

That would make sense of your approach since you want the central maxima, which is the image of the "slot" in the link to come to a focus on the screen - the screen will be one focal length away. Your version does not have the slot or the lens L1 but it amounts to the same thing.

Once you are clear about what is being described, it becomes easier to figure out what to do.

 

[unscientific]

Don't you think that's important?
But I think that the question, as you wrote it in post #1, does specify how many orders the screen covers.
It only mentions one spectrum.

No it doesn't.
It is not even clear what you mean by "light comes to a focus" here - since "light coming to a focus" would just make a bright dot rather than an image.

I mean is v supposed to be the object distance or the image distance?

I'm guessing, since you have u at infinity that you mean it to be the image distance and u is the object distance.
The light from infinity is "parallel", but after it strikes the diffraction grating, the rays are no longer parallel. So why did you assume u as infinite?

I have a feeling about this - see below.

... so what is the picture you expect to see on the screen?

I have a feeling you are expected to do something like for the setup in section 3 here:
http://physics-animations.com/Physics/English/DG10/DG.htm
... is that correct?

That would make sense of your approach since you want the central maxima, which is the image of the "slot" in the link to come to a focus on the screen - the screen will be one focal length away. Your version does not have the slot or the lens L1 but it amounts to the same thing.

Once you are clear about what is being described, it becomes easier to figure out what to do.

while we assume that the grating-lens distance is really far away compared to width of the slits, so rays coming out at angle θ can be assumed parallel, with adjacent rays having path difference dsinθ.

These parallel rays strike the lens, which converges at distance focal length away.

First maximum of 800nm - first maximum of 400nm = length of screen.

Using that approach, it gives a reasonable answer for focal length.

 

[Simon Bridge]

while we assume that the grating-lens distance is really far away compared to width of the slits, so rays coming out at angle θ can be assumed parallel, with adjacent rays having path difference dsinθ.

These parallel rays strike the lens, which converges at distance focal length away.

First maximum of 800nm - first maximum of 400nm = length of screen.

Using that approach, it gives a reasonable answer for focal length.

Well done - this is pretty much what I was hoping you'd figure out.
See how that description of the situation is much clearer than the one in post #1?
A very large part of doing science is just asking a question or describing a situation.

Now I've just got to work out a more effective way of approaching the issue without actually providing the answer.

 

[unscientific]

Well done - this is pretty much what I was hoping you'd figure out.
See how that description of the situation is much clearer than the one in post #1?
A very large part of doing science is just asking a question or describing a situation.

Now I've just got to work out a more effective way of approaching the issue without actually providing the answer.

Yup, this problem is done, thanks alot!