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## Homework Statement

## Homework Equations

θ

_{min}=1.22λ/D

## The Attempt at a Solution

(a)θ

_{min}=1.22λ/D

(b)θ

_{s}=1.22(600nm)/30mm=24.4μm but what is spatial resolution and how to calculate it??

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- Thread starter kenok1216
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θ

(a)θ

(b)θ

- #2

SammyS

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That answer should be in units appropriate for an angle, certainly not μm .## Homework Statement

View attachment 99604

## Homework Equations

θ_{min}=1.22λ/D

## The Attempt at a Solution

(a)θ_{min}=1.22λ/D

(b)θ_{s}=1.22(600nm)/30mm=24.4μm but what is spatial resolution and how to calculate it??

- #3

blue_leaf77

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It's related to the angular resolution. You can think of the spatial resolution as the arc length of a circle subtended by the angle of angular resolution.what is spatial resolution and how to calculate it??

- #4

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sor, it is μradThat answer should be in units appropriate for an angle, certainly not μm .

- #5

SammyS

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Yes.sor, it is μrad

- #6

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s=rθ, so what is r is this question?object distance? focal length?image distance?It's related to the angular resolution. You can think of the spatial resolution as the arc length of a circle subtended by the angle of angular resolution.

- #7

blue_leaf77

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About the circle I was talking about, you can assume it to be centered at the center of the lens and since your image is located at the focal plane, the arc length of interest must touch the focal plane. What will r be?s=rθ, so what is r is this question?object distance? focal length?image distance?

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Focal lenght?About the circle I was talking about, you can assume it to be centered at the center of the lens and since your image is located at the focal plane, the arc length of interest must touch the focal plane. What will r be?

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SammyS

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Yes.Focal lenght?

- #10

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ThankYes.

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