Fresnel Lens - Solve for Lithography & Power

Thread starter Schreiber__
Start date Apr 29, 2024
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Fresnel diffraction Optics optics problem

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The discussion focuses on calculations involving a Fresnel lens for lithography and power. The initial formula used, M = (lambda*f)/r_n, yielded a magnification of 29.48 nm, which was verified as correct. However, the subsequent calculation for focal length using f = r^2/(n*lambda) resulted in a focal length of 390.625, leading to a power in diopters of 2.56x10^03, which the user feels is too small. The user expresses uncertainty about the calculations and seeks clarification on potential errors or incorrect expressions used. The thread highlights the complexities of optical calculations in lithography applications.

Apr 29, 2024

Schreiber__

Homework Statement: Fresnel lens Part 1: You would like to design a 25 mm diameter blazed Fresnel zone plate with a first-order power of +1.5 diopters. What is the lithography requirement (resolution required) for making this lens that is designed for 550 nm? Express your answer in units of 𝜇𝑚 to one decimal point.

Fresnel lens Part 2: What would the power of the first diffracted order of this lens be at wavelength of 400 nm? Express your answer in diopters to one decimal point.

Relevant Equations: M = lambda/NA = (lambda*f)/r_n -> Part 1
f = r^2/(n*lambda) -> Part 2

For the first part, I got correct:
M = (lambda*f)/r_n
Converting units to meters (m) then plugging them in:
(550x10^-9 m) * (0.67 m)/(0.0125 m) = 2.948x10^-5 or 29.48x10^-6 m or 29.48 nm
This checked out.

For the second part, using the information from the first part:
f = r^2/(n*lambda) = (0.0125 m)^2/(1*400x10^-9m) = 390.625

To calculate the power in diopters we take 1/f = 1/390.625 = 2.56x10^03 ! This seems too small of a value.

I am missing something here or have used an incorrect expression.

Thanks for the help!

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