Hi there,
There are several different equations for this. It's also dependent on the wavelength and, as previously mentioned by Upasana Kachroo
on the NA of the codenser. You'll also have to take into account the refractive index of the medium between the sample and the objective.Please find attached several links related to your questions:
https://www.olympus-lifescience.com/es/microscope-resource/primer/anatomy/numaperture/#:~:text=Numerical%20aperture%20determines%20the%20resolving,system%2C%20the%20better%20the%20resolution.
https://www.microscopyu.com/tutorials/imageformation-airyna
https://micro.magnet.fsu.edu/primer/anatomy/numaperture.html
https://www.microscopyu.com/microscopy-basics/depth-of-field-and-depth-of-focus
Cheers,
J
Hi Mohit,
The numerical aperture is a measure of the amount of light absorbed by the objective lens, please compare the image attached.
According to Abbé, it is defined as the product: n x sinus α
Aperture => (latin => opening); aperire (latin => open, to open)
n is the refractive index of the medium in which the rays run from the object to the objective lens (e.g. in air or in oil).
α is the angle that forms with the optical axis of the lens that the outermost beams of the lens still has detected
Example: magnification 63 x objective: => aperture n = 0.85, α = 58.21 °
Example: magnification 4 x aperture n = 0.12, α = 6.89 °
The resolution power is the performance of the lens, to be able to depict points separately in the intermediate image.
The larger the numerical aperture, the higher the resolution!
The smaller the numerical aperture, the lower the resolution!
Best regards,
Guenter Grundmann
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