Connection Microscope Objective - Pinhole

[Kiffdino]

thanks alot guys, that was very helpful

[Johnfp]

Well, actually I didn't have a 5x that was clean enough to use. So i took the tip lens out of a 10 or 20x, cant remember which, and it then had the throw of about a 4x (a little less then a 5x). Since it was only the single internal lens it was beautifully clean. But as it was in a 10 or 20x objective casing it would not fit in the SF. So I flipped it around and it worked like a charm.

[holorefugee]

Does anyone have the angular spread calculation for a microscope objective? I know I have seen it...

[Dinesh]

Does anyone have the angular spread calculation for a microscope objective? I know I have seen it...

Lets say i want a more diverged beam at a shorter range, what objective would i need to get? I guess i will also need a different pinhole.
What is the connection between those two anyway ?

Very simple. On the side of the objective is a number marked NA (Sometimes the letters NA are not printed, simply a decimal number). This stands for Numerical Aperture. The numerical aperture is, strictly speaking, the sin of half angle of entry into an aperture. That is, if you want to get a beam of light into a fibre, say, then the maximum divergence of the incoming light is given by the NA, actually half the maximum divergence. You can work this backwards. Light exiting the aperture expands by a half angle whose sin is the NA. So, let's say you're beam diverges by 20 degrees. Then the half angle would be 10 degrees, sin(10) = NA = 0.173 and this is the number printed on the side of the objective. How wide will the diameter of the expanded beam be? Well, at a distance d from the objective, the (half) beam divergence is given as arcsin(NA), but the tan of the half angle is the ratio of beam diameter (d) to the distance from objective to the point of measurement (h). So, if the half angle of the beam divergence is theta, then:
theta/2 = (arcsin)(NA) = (arctan)(h/d)

If you want more divergence at a shorter range, use a higher magnification. The reason the beam diverges at all is that the beam enters the objective lens, focuses in, then expands past, like an "X" at the focus of the objective. You can now see the the nearer the X is to the objective, the faster will be the expansion and the greater the beam divergence. Also, the higher the magnification (with 40X being a higher magnification than 10X), the shorter the focal length and so the nearer the X.

The pinhole size is given by various formulae such as 1.6 (lambda)/d, 1.22(lambda)/a etc etc. Remember that the objective is designed as an objective lens for a microscope. Hence, the design criteria is that the image formed by the objective lens be magnified and "focused" in such a position that the eyepiece of the microscope then "picks up" that image and magnifies it even further. Thus, the objective is designed to work with both an eyepiece and a standard microscope tube. A standard microscope tube is about 160 mm (for various historical and physical reasons). Thus the figure of 1.6 in the pinhole formula. In the end, the ideal pinhole size is the radius of the first Airy ring caused by the lens system in the objective. This for a simple lens is (1.22(lambda))/a, where a is the diameter of the lens. Hence the second formula

However, for holography, the beam divergence and the beam diameter of the incoming light are also important, and those are a function of the laser. Specifically, the beam waist of the laser. So, no formula can exactly give the right diameter. If you attempt to try to calculate the exact diameter, you probably won't find a commercial pinhole of this size anyway. An old holographers "rule of thumb" pertaining to HeNe was: 20X -> 25u, 30X -> 15u, 40X -> 5u. I've usually found this to be about right even when operating a large frame Argon.

[jsfisher]

These are the best sort of posts, Dinesh. You provided a very nice explanation of NA and its effect, some theory, then ended with the practical "rule of thumb". It works at so many levels for me. Thanks.

[djm]

How wide will the diameter of the expanded beam be? Well, at a distance d from the objective, the (half) beam divergence is given as arcsin(NA), but the tan of the half angle is the ratio of beam diameter (d) to the distance from objective to the point of measurement (h). So, if the half angle of the beam divergence is theta, then:
theta/2 = (arcsin)(NA) = (arctan)(h/d)

I am sorry, but I think this is a little confusing. You use 'd' for both distance and beam diameter, and write that theta is the half angle of the beam divergence, but then use theta/2, which to me accordingly should mean a quarter of the beam divergence, but following the equation theta is the whole beam divergence. Also, shouldn´t the half angle of beam divergence equal arctan of radius over distance (instead of arctan of distance over diameter), or arccot of distance over radius

theta/2 = (arctan)(r/d) = (arccot) (d/r)

where theta is the angle of beam divergence, r is radius and d is distance?

It has been a long time since I studied math, so please, if I misunderstood something, enlighten me.

[Dinesh]

Also, shouldn´t the half angle of beam divergence equal arctan of radius over distance (instead of arctan of distance over diameter), or arccot of distance over radius

theta/2 = (arctan)(r/d) = (arccot) (d/r)

where theta is the angle of beam divergence, r is radius and d is distance?

It has been a long time since I studied math, so please, if I misunderstood something, enlighten me.

Sorry. You're quite right. I wrote this too quickly. I'm calling the beam divergence - the angle between the the extremal rays of the expanded beam - theta. Thus the half angle of the cone is theta/2 and, as you say,

theta/2 = arctan(r/d) = arccot(d/r)

The funny thing is that I'd actually put this, then thought that it was wrong and changed it. I was right the first time! Anyway, the sin of the half angle is the NA, that is:

sin(theta/2) = NA.

Also apologies for the delay. I'd lost my password.