arnoke
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Hi all,
This is probably a silly questions, but I want to be sure :).
I'm wondering if a real image, created by a convex lens, can appear larger than the lens diameter itself.
As an example, I'm thinking about the following:
- I have a an object of height h_0=7.76" (display size of my IPAD :p)
- Two convex fresnel lenses, each width diameter of 8" and focal length 12.5"
- The lenses will be put back-to-back, so the combined focal length is f=6.25"
- I will place the IPAD at distance d_0=1.2f=7.5" from the lens
Now I can calculate at which distance the real image will appear to float in front of the lens:
\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_1}
\Rightarrow d_i = \frac{1}{1/f - 1/d_0} = \frac{1}{(1/6.25 - 1/7.5} = 37.5"
Also I can calculate the magnification:
\frac{h_i}{h_0}=-\frac{d_i}{d_0}
\Rightarrow h_i = -h_0\frac{d_i}{d_0} = -7.76\frac{37.5}{12.5} = -23.28"
So the real object would be at 37.5" from the lens, inverted, and would be 23.28" high.
Now my question is; would I indeed see an IPAD of height 23.28" if my (fresnel) lens has the same size as the original Ipad? Or should my lens be at least of diameter 23.28", the same size as the magnified, real image?
Thanks a lot for your input!
This is probably a silly questions, but I want to be sure :).
I'm wondering if a real image, created by a convex lens, can appear larger than the lens diameter itself.
As an example, I'm thinking about the following:
- I have a an object of height h_0=7.76" (display size of my IPAD :p)
- Two convex fresnel lenses, each width diameter of 8" and focal length 12.5"
- The lenses will be put back-to-back, so the combined focal length is f=6.25"
- I will place the IPAD at distance d_0=1.2f=7.5" from the lens
Now I can calculate at which distance the real image will appear to float in front of the lens:
\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_1}
\Rightarrow d_i = \frac{1}{1/f - 1/d_0} = \frac{1}{(1/6.25 - 1/7.5} = 37.5"
Also I can calculate the magnification:
\frac{h_i}{h_0}=-\frac{d_i}{d_0}
\Rightarrow h_i = -h_0\frac{d_i}{d_0} = -7.76\frac{37.5}{12.5} = -23.28"
So the real object would be at 37.5" from the lens, inverted, and would be 23.28" high.
Now my question is; would I indeed see an IPAD of height 23.28" if my (fresnel) lens has the same size as the original Ipad? Or should my lens be at least of diameter 23.28", the same size as the magnified, real image?
Thanks a lot for your input!