Imaging resolution for a microscope and eye

[StillAnotherDave]

Homework Statement

Change in resolution to microscope.

Relevant Equations

$NA=nsin\theta =\frac{nD}{2f}$
$x_{min}=1.22\frac{f\lambda }{D}$

Hello folks,

I have the two following questions I'm working on:

Q1
An optical microscope uses a lens with NA = 0.7 and a focal length of f = 20 mm. What is the smallest spatial distance that can be resolved if a wavelength of λ = 633nm is used? An iris is introduced in front of the lens with a diameter of 10 mm at a distance of 10 mm from the object. How does the resolution of the microscope change?

Q2
A digital camera equipped with a f = 300mm focal length lens is mounted on a survey satellite orbiting the Earth at a height of 150 km. The camera must be designed such, that photographs with 1.2m in spatial resolution can be taken. Determine the diameter of the lens. The lens casts an image onto the digital camera with a finite pixel size. Calculate the pixel size required to allow this spatial resolution. The wavelength of light can be assumed to be λ = 500 nm.The second parts of each question (in bold) are what I'm struggling with. I've worked out the smallest spatial distance in Q1 and the diameter in Q2. Any help on how to tackle (1) the introduction of the iris into Q1, and the pixel size in Q2.

For Q1:

$$NA=\frac{nD}{2f}$$

therefore, $D=28mm$ (assuming $n=1$ as given in a previous part of the question). Plugging this value into:
$x_{min}=1.22\frac{f\lambda }{D}$ gives $x_{min}=552nm$

For Q2:

$$NA=\frac{nD}{2f}$$

Plugging in the given values for $NA$, $f$ and $\lambda$ gives $D=153nm$

 

[StillAnotherDave]

I've been trying to read around. I can't find a clear connection between spatial resolution and pixel size (Q2), but it seems that there's a dependency on the height of the satellite.. is that correct?

And for Q1 is this something to do with angular resolution once you introduce the eye?

 

[Steve4Physics]

Homework Statement:: Change in resolution to microscope.
Relevant Equations:: $NA=nsin\theta =\frac{nD}{2f}$
$x_{min}=1.22\frac{f\lambda }{D}$

Q1
An optical microscope uses a lens with NA = 0.7 and a focal length of f = 20 mm. What is the smallest spatial distance that can be resolved if a wavelength of λ = 633nm is used? An iris is introduced in front of the lens with a diameter of 10 mm at a distance of 10 mm from the object. How does the resolution of the microscope change?

Q2
A digital camera equipped with a f = 300mm focal length lens is mounted on a survey satellite orbiting the Earth at a height of 150 km. The camera must be designed such, that photographs with 1.2m in spatial resolution can be taken. Determine the diameter of the lens. The lens casts an image onto the digital camera with a finite pixel size. Calculate the pixel size required to allow this spatial resolution. The wavelength of light can be assumed to be λ = 500 nm.

The second parts of each question (in bold) are what I'm struggling with. I've worked out the smallest spatial distance in Q1 and the diameter in Q2. Any help on how to tackle (1) the introduction of the iris into Q1, and the pixel size in Q2.

For Q1:

$$NA=\frac{nD}{2f}$$

therefore, $D=28mm$ (assuming $n=1$ as given in a previous part of the question). Plugging this value into:
$x_{min}=1.22\frac{f\lambda }{D}$ gives $x_{min}=552nm$

For Q2:

$$NA=\frac{nD}{2f}$$

Plugging in the given values for $NA$, $f$ and $\lambda$ gives $D=153nm$

Not a familiar area for some years, but since no one has answered, see if this helps...

Q1.$$NA=nsin\theta =\frac{nD}{2f}$$is for small values of NA; it is derived using the small angle approximation (θ ≈ sinθ ≈ tan θ). For values of NA<0.3 the error is less than 5%. But for NA = 0.7 the error is much larger and to find D you would need the full formula:$$NA=nsin[arctan{(\frac{D}{2f})}]$$(e.g. see https://en.wikipedia.org/wiki/Numerical_aperture). This gives D = 39mm, not 28mm.

But in fact you don’t need to find D!

Use $$x_{min} = 1.22\frac{\lambda}{2NA}$$ or $$x_{min} = 1.22\frac{\lambda}{NA}$$ depending on how you are defining resolution. E.g. see here: https://web.mit.edu/2.710/Fall06/2.710-wk12-b-sl.pdf

For the second part of Q1, when you add the stop, you are changing the value of NA because you are limiting the half-angle of the outer rays, basically stopping the light from reaching the outer parts of the lens. If that doesn’t make sense, you will need to read up about how NA is defined.

From a simple diagram and geometry/trig’ you should be able to work out the new value of θ with the stop present. Then (since n=1), NA = sinθ and you can find the new resolution as above.

Q2.

You have found the lens diameter to be 153nm. Don’t you think that might be a tad small?!

For the first part you will need the angle (in radians) subtended by points 1.2m apart at ground level. Then use the standard formula for angular resolution.

There are various ways to think about the second part. One is to remember that we really have a pinhole camera with a large hole. The only function of the lens is to make the image less fuzzy. So a suitable ray diagram (with a small hole rather than the lens) and use of similar triangles should let you find the pixel size corresponding to an object size of 1.2m. This is equivalent to saying a pixel subtends the same angle at the lens as 1.2m does from the ground.

 

[StillAnotherDave]

Thanks for the really useful help!

Can I just ask about the second part of Q1 where the iris acts as a stop limiting the angle. Is there any simple example question/notes you can point me to for this? I get the basic idea but am not exactly sure of how to proceed.

 

[Steve4Physics]

Thanks for the really useful help!

Can I just ask about the second part of Q1 where the iris acts as a stop limiting the angle. Is there any simple example question/notes you can point me to for this? I get the basic idea but am not exactly sure of how to proceed.

Follow these steps carefully.

Look at this diagram. https://www.newport.com/medias/sys_master/images/images/h92/h9e/8798397923358/LS-015a-600w.gif

Draw this diagram for yourself but using D=39mm (see my previous answer) and f=20mm. You don’t actually need these values but a good diagram will make it easier for you to see what’s going on. If you draw it correctlty you should see θ is about 45º.

The iris acts as a stop - an opaque sheet with a hole light can pass through,

Add the stop to your diagram, between O and the lens. The top (T) of the stop’s opening is 5mm above the axis and the bottom (B) of the stop’s opening is 5mm below the axis. The stop is 10mm from O.

Draw a line from O through T till it reaches the lens. Same for a line from O through B.

The effective size of the lens has been reduced because only light which passes through the stop's hole (between T and B) can reach the lens.

The new limiting angle θ is the angle between OT (or OB) and the axis. Can you proceed from there?

 

[StillAnotherDave]

Follow these steps carefully.

Look at this diagram. https://www.newport.com/medias/sys_master/images/images/h92/h9e/8798397923358/LS-015a-600w.gif

Draw this diagram for yourself but using D=39mm (see my previous answer) and f=20mm. You don’t actually need these values but a good diagram will make it easier for you to see what’s going on. If you draw it correctlty you should see θ is about 45º.

The iris acts as a stop - an opaque sheet with a hole light can pass through,

Add the stop to your diagram, between O and the lens. The top (T) of the stop’s opening is 5mm above the axis and the bottom (B) of the stop’s opening is 5mm below the axis. The stop is 10mm from O.

Draw a line from O through T till it reaches the lens. Same for a line from O through B.

The effective size of the lens has been reduced because only light which passes through the stop's hole (between T and B) can reach the lens.

The new limiting angle θ is the angle between OT (or OB) and the axis. Can you proceed from there?

This makes perfect sense! Thoroughly appreciate the careful explanation.