- #1
indigojoker
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So there are two lenses. The first lens is a converging lens and the second is a diverging lens with focal length f,-f respectively, separated by a distance D. The question asks to show the combined effect of the lenses is focusing however, I am not getting them to be always focusing.
using: [tex] \frac{1}{o}+\frac{1}{i}=\frac{1}{f} [/tex] we see that an object placed at infinity gives i=f
Moving onto the second lens, the object is located at D-f so we get:
[tex] \frac{1}{D-f}+\frac{1}{i}=-\frac{1}{f} [/tex]
[tex] -\frac{1}{D-f}-\frac{1}{f}=\frac{1}{i} [/tex]
[tex] -(\frac{1}{D-f}+\frac{1}{f})=\frac{1}{i} [/tex]
[tex] -(\frac{f+D-f}{f(D-f)})=\frac{1}{i} [/tex]
[tex] -(\frac{f+D-f}{f(D-f)})=\frac{1}{i} [/tex]
[tex] \frac{D}{f^2-fD}=\frac{1}{i} [/tex]
[tex] \frac{f^2-fD}{D}=i [/tex]
we see that if D>f then the image is negative, telling use that there isn't convergence
any ideas?
using: [tex] \frac{1}{o}+\frac{1}{i}=\frac{1}{f} [/tex] we see that an object placed at infinity gives i=f
Moving onto the second lens, the object is located at D-f so we get:
[tex] \frac{1}{D-f}+\frac{1}{i}=-\frac{1}{f} [/tex]
[tex] -\frac{1}{D-f}-\frac{1}{f}=\frac{1}{i} [/tex]
[tex] -(\frac{1}{D-f}+\frac{1}{f})=\frac{1}{i} [/tex]
[tex] -(\frac{f+D-f}{f(D-f)})=\frac{1}{i} [/tex]
[tex] -(\frac{f+D-f}{f(D-f)})=\frac{1}{i} [/tex]
[tex] \frac{D}{f^2-fD}=\frac{1}{i} [/tex]
[tex] \frac{f^2-fD}{D}=i [/tex]
we see that if D>f then the image is negative, telling use that there isn't convergence
any ideas?