Laser beam spreader optics problem

[ice109]

Homework Statement

Two positive lens are to be used as a laserbeam expander. An axial 1mm diameter beam enteres a short focal length positive lens, which is followed by a somewhat longer focal length positive lens from which it emerges with a diameter of 8mm. Given that that first lens has a 50mm focal length determine the focal length of the second lens and the separation between them.

Homework Equations

$$\frac{1}{s_o_1}+\frac{1}{s_i_1}=\frac{1}{f}$$
$$M=\frac{-s_i_1}{s_o_1}(\frac{-s_i_2}{s_o_2})$$

The Attempt at a Solution

well if the object is at infinity, which a collimated beam, a laser, would look like, then it converges to the focus on the other side of the thins lens. But this yields a zero magnification. I can draw a ray diagram after that rays converge to the focus and continue past it they diverge again and if the point they diverge from is the focus of the second lens they will come out collimated again. The problem is that neither of those two equations really describe anything about that. So i can make the ray diagram look right but not the math.

the problem is that there's a bunch of infinities floating around and i don't know how to work with them. in another problem in the book they actually cancled an infinity over an infinity which of course makes no sense to me.

 

[anathema]

Thank you for your post. It seems like you are struggling with the concept of dealing with infinities in optics calculations. Let me try to explain it in a simpler way.

First, let's define some variables:
s_o_1 = object distance from the first lens
s_i_1 = image distance from the first lens
f_1 = focal length of the first lens
s_o_2 = object distance from the second lens
s_i_2 = image distance from the second lens
f_2 = focal length of the second lens

Now, let's look at the first lens. As you correctly stated, for a collimated beam (object at infinity), the image distance is equal to the focal length:
s_i_1 = f_1
Using this in the first equation, we get:
\frac{1}{s_o_1}+\frac{1}{f_1}=\frac{1}{f_2}
We also know that the magnification of the first lens is given by:
M_1=\frac{-s_i_1}{s_o_1}=\frac{-f_1}{s_o_1}
Now, let's move on to the second lens. We want the beam to emerge from the second lens with a diameter of 8mm. This means that the image distance from the second lens must be equal to half of the diameter, which is 4mm:
s_i_2 = 4mm
Using this in the second equation, we get:
M_2=\frac{-s_i_2}{s_o_2}=\frac{-4mm}{s_o_2}
We also know that the magnification of the entire system (both lenses) is given by:
M=M_1*M_2=\frac{-f_1}{s_o_1}*\frac{-4mm}{s_o_2}=\frac{4f_1}{s_o_1*s_o_2}
Since we want the final magnification to be 1 (meaning the beam emerges with the same diameter as it entered), we can set M=1 and solve for the object distance from the second lens:
1=\frac{4f_1}{s_o_1*s_o_2}
s_o_2=\frac{4f_1*s_o_1}{f_2}
Now, we can substitute this into our first equation and

 

[m2003]

Thank you for your question. It seems like you are on the right track in understanding the concept of beam expansion using two lenses. However, the equations you have listed are not applicable to this problem. Instead, you can use the thin lens formula, which states that 1/f = 1/s_o + 1/s_i, where f is the focal length, s_o is the object distance, and s_i is the image distance.

In this problem, the first lens has a focal length of 50mm and the second lens has an unknown focal length. The object distance for the first lens is infinity, since the beam is collimated, and the image distance is equal to the focal length of the first lens, which is 50mm. Using the thin lens formula, we can solve for the image distance of the second lens, which will also be the object distance of the second lens. This will give us the separation between the two lenses.

To find the focal length of the second lens, we can use the magnification equation, M = -s_i/s_o, where M is the magnification, s_i is the image distance, and s_o is the object distance. The magnification for a beam expander is equal to the ratio of the output beam diameter to the input beam diameter. In this case, the magnification is 8mm/1mm = 8. We know the image distance of the second lens from solving for the separation between the lenses, and we can use the known magnification to solve for the focal length of the second lens.

I hope this helps you to better understand the problem and how to approach it mathematically. Remember to always use the appropriate equations for the given situation. Best of luck with your homework!