Determine lens system for known magnification and available space

I'm trying to create a system of two lenses with a set magnification in a limited space for the setup, i.e. the distance from the object to the first lens $$g_1$$ has a minimum ($$152\,$$cm) and the whole lens setup is limited ($$40\,$$cm). The defined length of the setup $$L$$ is defined below. For a sketch see image: In total there are the following important parameters:

• object size $$G=4.136\,$$mm, is known
• magnification $$V\simeq0.2579$$
• focal length $$f_i$$ of each lens (should be reasonable so one can buy it)
• distance to objects $$g_i$$ (where $$g_1$$ has a minimum of $$g_{min}=152\,$$cm)
• distance to image $$b_i$$
• image size $$B_i$$ ($$B_2$$ is set due to the magnification)
• distance between lenses $$d=b_1+g_2$$

where $$i\in\{1,2\}$$ and the setup distance $$L=(g_1-g_{min})+d+b_2\overset{!}{<}40\,$$cm. Since the image is captured by a camera in position $$B_2$$ the image distance of the second lens is required as $$b_2>0$$.

My question is, whether there is an elegant way of finding the best possible fit of lenses and positions that I don't know of or if is this only solvable by calculating the whole system for varying $$f_i,g_i,d$$?

• Does the "lens setup" refer to distance (d)? Is lens (2) the lens of your camera? Aug 9 '21 at 14:21
• I updated the question to clarify, that the setup distance is given by $L$ (so not the total length of all parameters, but a reduced length due to $g_{min}$. No, the camera is in the focal plane of the second lens or in other words at the position of $B_2$. Aug 9 '21 at 14:27
• I haven't looked closely at this question, but I'll share that I've had success getting large magnifications in small distances using negative focal length lenses. Aug 9 '21 at 14:44

You can start with a single lens to achieve the magnification you want. This will give you a range of values for the focal length. If the orientation of $$B_2$$ does not matter, then you would get two intervals for the focal length: $$f = \frac{g_1}{1 \pm \frac{1}{V}}$$