In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.
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According to prinicipal of reversibility, the position of object and image ate interchangeable. According to lens maker's formula, We have $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}\ .....(1)$ Hence, there are two positions for which there shall be an image. Let us suppose the first position be when the lens is at O. Finding u and v ans substituting in lens formula. According to the question, we have -u + v = D ⇒ u = -(D - v) Substituting u = -(D - v) in (1), we get $\frac{1}{\text{v}}+\frac{1}{\text{D}-\text{v}}=\frac{1}{\text{f}}$ $\Rightarrow\ \frac{\text{D}-\text{v}+\text{v}}{(\text{D}-\text{v})\text{v}}=\frac{1}{\text{f}}$ $\Rightarrow\ \text{v}^2-\text{Dv}+\text{Df}=0$ $\Rightarrow\ \text{v}=\frac{\text{D}}{2}\pm\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$ $\therefore\ \text{u}=-(\text{D}-\text{v})=-\Big(=\frac{\text{D}}{2}\pm\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}\Big)$
When, the object distance is $\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$, then the image forms at $=\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$ Similarly, when the object distance is $=\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$, then the image forms at $=\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$ The distance between the poles for these two object distance is given by v - u $=\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}-\bigg(\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}\bigg)$ $=\sqrt{\text{D}^2-4\text{Df}}$ Let us suppose $\text{d}=\sqrt{\text{D}^2-4\text{Df}}$ if $\text{u}=\frac{\text{D}}{2}+\frac{\text{d}}{2}$, then the image is at $\text{v}=\frac{\text{D}}{2}-\frac{\text{d}}{2}$,then the magnification is given by $\text{m}_1=\frac{\text{v}}{\text{u}}=\frac{\text{D}-\text{d}}{\text{D}+\text{d}}$ If $\text{u}=\frac{\text{D}-\text{d}}{2}$, then $\text{v}=\frac{\text{D}+\text{d}}{2}$, then the magnification is given by $\text{m}_2=\frac{\text{v}}{\text{u}}=\frac{\text{D}+\text{d}}{\text{D}-\text{d}}$ Thus, the ration of the images sizes is given by $\frac{\text{m}_2}{\text{m}_1}=\frac{\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}{\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}=\Big({\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}\Big)^2$
When, the object distance is $\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$, then the image forms at $=\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$ Similarly, when the object distance is $=\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$, then the image forms at $=\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$ The distance between the poles for these two object distance is given by v - u $=\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}-\bigg(\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}\bigg)$ $=\sqrt{\text{D}^2-4\text{Df}}$ Let us suppose $\text{d}=\sqrt{\text{D}^2-4\text{Df}}$ if $\text{u}=\frac{\text{D}}{2}+\frac{\text{d}}{2}$, then the image is at $\text{v}=\frac{\text{D}}{2}-\frac{\text{d}}{2}$,then the magnification is given by $\text{m}_1=\frac{\text{v}}{\text{u}}=\frac{\text{D}-\text{d}}{\text{D}+\text{d}}$ If $\text{u}=\frac{\text{D}-\text{d}}{2}$, then $\text{v}=\frac{\text{D}+\text{d}}{2}$, then the magnification is given by $\text{m}_2=\frac{\text{v}}{\text{u}}=\frac{\text{D}+\text{d}}{\text{D}-\text{d}}$ Thus, the ration of the images sizes is given by $\frac{\text{m}_2}{\text{m}_1}=\frac{\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}{\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}=\Big({\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}\Big)^2$
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