Question
Prove that for a converging lens, for the case $d_{i}=d_{0}$, that $d_{i}=d_{0}=2 f$.
Step 1
We know that for a converging lens, the lens formula is given by: $\frac{1}{f} = \frac{1}{d_{0}} + \frac{1}{d_{i}}$ where $f$ is the focal length, $d_{0}$ is the object distance, and $d_{i}$ is the image distance. Show more…
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Show analytically that the image formed by a converging lens is real and inverted if the object is beyond the focal point $\left(d_{0}>f\right),$ and is virtual and upright if the object is within the focal point $\left(d_{0}<f\right) .$ Describe the image if the object is itself an image, formed by another lens, so its position is beyond the lens, for which $-d_{0}>f,$ and for which $0<-d_{0}<f$
Going back to Section $5.2 .3,$ prove that for a thin lens immersed in a medium of index $n_{m}$ $$\frac{1}{f}=\frac{\left(n_{l}-n_{m}\right)}{n_{m}}\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$ That done, imagine a double-concave air lens surrounded by water; determine if it's converging or diverging.
(a) Use the thin - lens equation to derive an expression for q in terms of f and p. (b) Prove that for a real object and a diverging lens, the image must always be virtual. Hint: Set $f=-|f|$ and show that $q$ must be less than zero under the given conditions. (c) For a real object and converging lens, what inequality involving p and f must hold if the image is to be real?
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