Explain why, for a fixed separation between the object and the screen(i.e., d + d' is fixed) there are two positions of the lens at which the images formed on the screen are in focus (clear images or sharp images). Please provide me with a proof with diagrams, I am unable to find it anywhere.
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Let the following figurewhere the screen is at $x=0$,the object at $x=a$ and the convex lense(with focus $f$) is at variable point $x=b$. Then for sharpe image on the screen we have : $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$ $$\frac{1}{f}=\frac{1}{b}-\frac{1}{b-a}$$ $$\frac{1}{f}=\frac{b-a-b}{(b-a)(b)}$$ $$\frac{1}{f}=\frac{-a}{b(b-a)}$$ $$b^2-ab+af=0$$ It is quadratic equation with two real roots iff $$a^2-4af\geq0$$or$$a\geq4f$$.
