# What type of LED to use with a 100 micrometer pinhole for in-line digital holography?

I want to set up an in-line digital holographic microscope that is similar to the one describe in the article, A Study of Digital In-Line Holographic Microscopy for Malaria Detection, which is they place a blue LED close to a 50 micrometer pinhole. In my set up, I have a 100 micrometer pinhole and have used several LEDs with no luck. I was able to get a good Fresnel diffraction image with my iphone flashlight but cannot get it to work with any other LEDs. Could someone explain the process as to how to get the coherent light needed with an LED for the in-line digital holographic microscope?

• can you explain more : have used several LEDs with no luck. ? – scrx2 Jun 9 '17 at 22:17
• @scrx2, what I mean by no luck is that I was expecting to see a Fresnel diffraction pattern on the detector when using the super bright LEDS like I did with a green laser that I have as well as with my iphone flashlight. Is this right? For in-line holography, should I expect to see a Fresnel diffraction pattern when there is no point source? Or do I see the diffraction pattern when there is a point source in front of the detector? – Don Abbot Jun 10 '17 at 5:27
• I'd say you expect to see diffraction fringes on the edges of an illuminated object. Because of coherence of the source, the fringes should be more visible with the laser, followed by the LED, followed by a white light. – scrx2 Jun 10 '17 at 21:37
• I have taken several images with the 100 micrometer pinhole with a lensless webcam using a needle as my object, drive.google.com/drive/folders/0B8jDaYaV0JgqRWlhanRDNTE1c1U From other examples I have seen, it seems like I should have a lot more diffraction fringes around the needle. Is that a correct assumption? And if so how could I still use an LED for my illumination source and get better results? – Don Abbot Jun 11 '17 at 21:34
• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf file. – Qmechanic Jun 16 '17 at 8:49

The further away the LED is from the camera sensor, the more coherent the light becomes, like a point source of a star. In the object plane, the coherent diameter is proportional to (λ0*z1)/dp. From the article, Compact, light-weight and cost-eﬀective microscope based on lensless incoherent holography for telemedicine applications, "it holds that as long as each cell is smaller than the coherence diameter of the illuminating light in the object plane, the individual cells will be illuminated by coherent light and contribute to the hologram"

So by increasing z1 in length and by using an object with a diameter smaller than the coherence diameter I was able to obtain an image with more diffraction fringes.

Look at tiff drill002 in folder https://drive.google.com/drive/folders/0B8jDaYaV0JgqRWlhanRDNTE1c1U

• +1 you've worked it out yourself. You might like to look at my answer for some theoretical background. – Selene Routley Jun 16 '17 at 0:38

In a phrases: a high power, small diameter, narrow linewidth one. Otherwise, as well as the LED, you need a very narrowband filter.

To do holography, you need collimated light (or well focussed light), i.e. light with very low étendue(see footnote [1]), which is the entropy of the light. You can get this with a laser, or, as in your method, you can get this from something like an LED if you're willing to throw most of the light away - which is what you're doing with the pinhole. The smaller the LED chip's angular subtense at the pinhole, the more of its light will get through the pinhole. This is the reason for your answer: you can move the LED away from the pinhole, but this throws away more light. If you could find an LED with the same power but a smaller chip, then you would get a higher efficiency.

Also in holography, you need the path difference between the reference and image-encoding beams to be small enough over the whole image for interference to happen. This means that you need a coherence length at least equal to twice the range of imaging depths in the imaged object. For example, if my nose is two centimeters long, then light reflected from it will have a range of optical path lengths equal to four centimeters. To image microscopic objects, however, you won't need a huge coherence length, which is why an LED will work in the first place. If you need more coherence length, you need to throw more of the light away by passing it through a narrowband filter.

For a Lorentzian spectral line with a Full Width Half Maximum linewidth (expressed as a wavelength) of $\Delta\lambda$, the coherence length is:

$$L_c = \frac{\lambda_0^2}{\pi\,\Delta\,\lambda}$$

Here $L_c$ is a $1/e$ value: this means that interference fringes produced with a path difference of $L$ will have a visibility of $1/e$. For a Lorentzian spectrum, the visibility varies according to:

$$V = \exp\left(-\frac{L}{L_c}\right)$$

where $L$ is the path difference.

Footnote [1]

See my answer here and also the Randall Munroe "Fire From Moonlight" article. The Wikipedia Étendue article gives a "non-abstract" definition: the simplest way to think of étendue is as the volume of the state of a light field in optical phase space, but you need a bit of background to work out how to calculate that; the Wiki article gives you the latter.