How Do Diffraction Gratings Aid in Microscope Magnification?

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SUMMARY

This discussion focuses on the role of diffraction gratings in enhancing microscope magnification. The angle of the first-order rays to the normal is calculated as 2.9 degrees, leading to an angle of 87.1 degrees. The intensity distribution is derived using the equation I = I_0 sin(Nkd sinθ/2) / sin(kd sinθ/2), with principal maxima occurring at θ_max = nλ/d. The analysis concludes that varying the distance from the lens to the final plane allows for setting magnification, described by the formula v/u, where v is the distance from the lens to the final plane and u is the distance from the lens to the grating.

PREREQUISITES
  • Understanding of diffraction principles and equations, specifically d sinθ = nλ.
  • Familiarity with intensity distribution calculations for diffraction gratings.
  • Knowledge of microscope optics, including magnification formulas.
  • Basic grasp of the relationship between aperture size and image sharpness in optical systems.
NEXT STEPS
  • Study the derivation of intensity distribution in diffraction gratings using the equation I = I_0 sin(Nkd sinθ/2) / sin(kd sinθ/2).
  • Learn about the Rayleigh criterion for resolution in optical systems.
  • Explore the effects of aperture size on image sharpness in microscopy.
  • Investigate advanced topics in diffraction theory, including Fourier optics and its applications in microscopy.
USEFUL FOR

This discussion is beneficial for optics students, microscopy researchers, and professionals in fields requiring precise imaging techniques, such as biomedical engineering and materials science.

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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:
I = I_0 \frac{sin \left(\frac{Nkd sin\theta}{2}\right)}{sin \left(\frac{kd sin\theta}{2}\right)}

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:

a = \theta f = f \frac{\lambda}{d}
d' = f\frac{\lambda}{a} = d

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.
 
Last edited:
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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 ?
 
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??
 
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.
 
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 ?
 
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.
 
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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