0
\$\begingroup\$

I have inverted the lens on a webcam to create a webcam microscope. For illumination , I am using an led array. For the project that I am doing I need to know what the NA of the lens, but I am confused on how to calculate or experimentally find the NA. Does anyone know how I would do this?

Thanks,

Dawn.

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

Procure a flashlight or lamp that outputs a parallel beam. Such a lamp will project a circular spot of light. Hold the lens, centered in the beam. Place a piece of paper in direct contact at the rear of the lens. You should see a circle of light on the paper. Measure the diameter of this illuminated circle, call this the exit pupil. Now determine the focal length of the lens. A good way to do this is; use the lens as a burning glass. Hold the lens so that it projects an image of the sun on a piece of paper. Adjust the distance, lens to paper to achieve the tiniest illuminated circle of sunlight (this is an image of the sun). Measure distance from the center of the lens barrel to focused image. This will be the focal length. Use a millimeter measure for both values. Divide the focal length by the exit pupil. You have calculated the focal ratio. We call this value the f-number.

\$\endgroup\$
3
  • \$\begingroup\$ Ha! Perfect answer. To the OP: The word "aperture" literally means "opening". It refers to the width of the opening of the lens when the shutter opens for an exposure. In your case, there is no shutter - the lens is always, and fully, open, hence Alan's first step to measure the width of that opening. But in photography, aperture is not measured absolutely (otherwise how can a tiny phone camera have f/4 and an SLR camera also have f/4?) In photography, it's measured as a fraction of the focual length. "f/" means "focal length divided by" hence Alan's second step then division at the end. \$\endgroup\$ Dec 8, 2016 at 22:09
  • \$\begingroup\$ PS. This is also why f/4 is smaller than f/2. One quarter is smaller than one half. \$\endgroup\$ Dec 8, 2016 at 22:13
  • \$\begingroup\$ @Alan Marcus. Thank you for your response. This is very helpful. \$\endgroup\$
    – Dawn Abbot
    Dec 9, 2016 at 6:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.