Diffraction Grating, focal length of lens

In summary: So why did you assume u as infinite?The diffraction grating causes the light to come to a focus at a distance of f away from the lens.
  • #1
unscientific
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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:
[tex]1/u + 1/v = 1/f[/tex]

Let u -> ∞,

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

[tex]d sin \theta = n\lambda[/tex]

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

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

Solving,

[tex] f = 22.4 m[/tex]

This is a strange setup, imagine the photographic paper would have to be placed so far away from the lens! (Relative to its size)
 
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  • #2
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?
 
  • #3
Simon Bridge said:
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.
 
  • #4
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.
 
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  • #5
Simon Bridge said:
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.
 
  • #6
unscientific said:
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.
 
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  • #7
Simon Bridge said:
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!
 

FAQ: Diffraction Grating, focal length of lens

1. What is a diffraction grating?

A diffraction grating is a device that consists of a large number of parallel, equally spaced slits or lines. It is used to separate light into its component wavelengths by causing interference and diffraction.

2. How does a diffraction grating work?

When light passes through a diffraction grating, it diffracts and interferes with itself, creating a pattern of bright and dark fringes. The spacing between the fringes is dependent on the wavelength of the light and the spacing of the slits on the grating.

3. How is the focal length of a lens determined?

The focal length of a lens is the distance between the lens and the point where parallel rays of light converge to a single point after passing through the lens. It can be determined by measuring the distance between the lens and the point where the image of an object is formed when the object is placed at a known distance from the lens.

4. What factors affect the focal length of a lens?

The focal length of a lens is primarily determined by the shape and curvature of the lens. It also depends on the refractive index of the material the lens is made of and the medium through which the light is passing. Additionally, the distance between the lens and the object or image can also affect the focal length.

5. How is a diffraction grating used to determine the focal length of a lens?

A diffraction grating can be used in conjunction with a lens to determine its focal length. By measuring the distance between the fringes on the diffraction pattern produced by the grating, the wavelength of the light can be calculated. This, along with the known distance between the lens and the object, can be used to calculate the focal length of the lens.

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