Diffraction grating, magnification

In summary, the sketch of the rays and intensity distribution at different distances is helpful in understanding a microscope.
  • #1
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Homework Statement



Part (a):Sketch zeroth and first order rays. Find the angle of first order rays make to normal.
Part (b): Sketch intensity distribution at 7.5mm and 10mm.
Part (c): Explain how this is useful in understanding a microscope.

293dmck.png


Homework Equations





The Attempt at a Solution



Part (a)

Using ##d sin\theta = n\lambda##, ##\theta = 2.9^o##. Thus angle to normal is 87.1o.

2ibj67a.png


Part (b)

At 7.5mm

Intensity distribution is given by:
[tex]I = I_0 \frac{sin \left(\frac{Nkd sin\theta}{2}\right)}{sin \left(\frac{kd sin\theta}{2}\right)}[/tex]

Principal Maximas occur at ##\theta_{max} = \frac{n\lambda}{d}##. Separation between a principal maxima and a minima is ##\frac{\lambda}{16d}##.

dwz7r8.png


At 10mm

There will be bright spots separated by distances ##a##. This is equivalent to a N-slit grating. To find a:

[tex]a = \theta f = f \frac{\lambda}{d}[/tex]
[tex]d' = f\frac{\lambda}{a} = d[/tex]

Thus we see that at 10mm, the image of the grating slit is simply reproduced. If we observe at distances greater than 10mm, ##d' > d## is amplified.

2h2l9gz.png


Part (c)
By varying the distances of the final plane, we can set the magnification and is given by ##\frac{v}{u}## where ##v## is distance from lens to final plane, ##u## is distance from lens to grating.
 
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  • #2
What is your question ?

By the way, the drawing for the first order maximum is very strange: do both maxima end up on the same side of the zeroth order maximum on the axis ?
 
  • #3
BvU said:
What is your question ?

By the way, the drawing for the first order maximum is very strange: do both maxima end up on the same side of the zeroth order maximum on the axis ?

You're right. It should be something like this:

2s96pmd.png


I'm trying to ascertain whether my answers are right, do you mind taking a look at them??
 
  • #4
If this the alternative for the 1st maximum, it's worse than the first attempt. Positive lenses do not diffract light rays away from the axis !

The drawing for the 0th maximum is a lot better

Perhaps a suggestion to make life easier: Start with a grating from x=-w to +w. You can erase the rays from the 0 to -w half later on.

How did you derive the intensity distribution in part b from the relevant equations you should have listed under 2. in the template ? What is N, k ? Are he intensities of the 0th and 1th maxima equal ? What about the widths ?

Part c is not about magnification but about limits to resolving power.
 
  • #5
Time to pick up on this one. Any new drawings ? Start with the 0th maximum in post 1. Picture isn't all that clear: Do I see all rays go through one point at 7.5 mm and what do I see at 10 mm ?

Then the +1st maximum. rays from all these 16 slits are parallel, but at an angle with the axis. So they go through "one off-axis point" at 7.5 mm. What about where they end up on the 10 mm plane ?

Finally the -1st maximum. Also off axis but in the other direction. So where is the peak at the 7.5 mm plane ?

On the 10 mm plane we expect a depiction of the grating, right ? If the object is at x > 0, where is the image ?
And: is it infinitely sharp ? Why / why not ?
nice picture I found here -- can't one hide a link or a :frown: in a spoiler ?
 
  • #6
BvU said:
Time to pick up on this one. Any new drawings ? Start with the 0th maximum in post 1. Picture isn't all that clear: Do I see all rays go through one point at 7.5 mm and what do I see at 10 mm ?

Then the +1st maximum. rays from all these 16 slits are parallel, but at an angle with the axis. So they go through "one off-axis point" at 7.5 mm. What about where they end up on the 10 mm plane ?

I realized my first picture at 7.5mm is correct, and my second one is wrong.

They split again, and form back the grating.

BvU said:
Finally the -1st maximum. Also off axis but in the other direction. So where is the peak at the 7.5 mm plane ?

Yeah they form below the central maxima which is halfway along the grating.

BvU said:
On the 10 mm plane we expect a depiction of the grating, right ? If the object is at x > 0, where is the image ?
And: is it infinitely sharp ? Why / why not ?
nice picture I found here -- can't one hide a link or a :frown: in a spoiler ?

If the original grating is infinitely sharp, then the image at 10mm is infinitely sharp.
 
  • #7
If the original grating is infinitely sharp, then the image at 10mm is infinitely sharp
Only if the lens aperture and the difraction pattern in the focal plane are infinitely big, otherwise higher orders get cut off, causing an unsharper image (the image is the Fourier transform of the diffraction pattern).

The f/2w is 0.5, a pretty wide aperture, that picks up a considerable number of orders, but not infinitely many.

This detail might well be within the scope of the exercise, since they give lens height as 2w. I think it is a bit much to drag in rayleigh resolution and such, but who knows. Slit height isn't given, only "narrow"(which is why you draw your 1st order same height as 0th). Your call.
 
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Related to Diffraction grating, magnification

1. What is a diffraction grating?

A diffraction grating is a surface with a regular pattern of parallel lines or grooves that are closely spaced. It is used to diffract light into its component wavelengths, making it a valuable tool for studying the properties of light.

2. How does a diffraction grating work?

When light passes through a diffraction grating, the parallel lines or grooves act as slits and cause the light to diffract, or spread out, into its component wavelengths. This results in a spectrum of colors being produced, with shorter wavelengths (such as blue) being diffracted more than longer wavelengths (such as red).

3. What is the purpose of using a diffraction grating?

A diffraction grating is commonly used in scientific experiments to separate and analyze the different wavelengths of light. It is also used in optical devices such as spectrometers and telescopes to study the properties of light and to make precise measurements of wavelengths.

4. How is the magnification of a diffraction grating determined?

The magnification of a diffraction grating is determined by the ratio of the distance between the grooves on the grating (known as the grating constant) to the distance between the grating and the viewing screen or detector. This ratio is also known as the grating's angular dispersion.

5. Can a diffraction grating be used to magnify an image?

No, a diffraction grating is not typically used for magnifying images. Its main purpose is to separate and analyze the different wavelengths of light. However, it can be used to magnify the spectrum of an object by increasing the distance between the grating and the viewing screen or detector.

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