A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take $L < < |v - f|.$
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Thin mirror formula: $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
u = object distance and v = image distance.
Since, the object distance is u. Let us consider the two ends of the object be at distance $u_1$ and $u_2$ respectively,
so that du $= |u_1 - u_2| = L$. Hence size of image can be written as dv.
By differentiating Both sides $\Big(-\frac{\text{dv}}{\text{v}^2}\Big)+\Big(-\frac{\text{du}}{\text{u}^2}\Big)=0\Rightarrow\ \frac{\text{dv}}{\text{v}^2}=-\frac{\text{du}}{\text{u}^2}$
or $\text{dv}=-\Big(\frac{\text{v}^2}{\text{u}^2}\Big)\text{du}\ .....(\text{i})$
As $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}\Rightarrow\ \text{v}=\frac{\text{fu}}{\text{u}-\text{f}}\Rightarrow\ \text{v}=\frac{\text{fu}}{\text{u}-\text{f}}$
Or $\frac{\text{v}}{\text{u}}=\frac{\text{f}}{\text{u}-\text{f}}\ .....(\text{ii})$
From (i) and (ii) $\text{dv}=-\Big(\frac{\text{f}}{\text{u}-\text{f}}\Big)^2\text{du}$
But du = L, hence the length of the image is $\frac{\text{f}^2}{(\text{u}-\text{f})^2}\text{L}$
This is the required expression for lenght of image.
u = object distance and v = image distance.
Since, the object distance is u. Let us consider the two ends of the object be at distance $u_1$ and $u_2$ respectively,
so that du $= |u_1 - u_2| = L$. Hence size of image can be written as dv.
By differentiating Both sides $\Big(-\frac{\text{dv}}{\text{v}^2}\Big)+\Big(-\frac{\text{du}}{\text{u}^2}\Big)=0\Rightarrow\ \frac{\text{dv}}{\text{v}^2}=-\frac{\text{du}}{\text{u}^2}$
or $\text{dv}=-\Big(\frac{\text{v}^2}{\text{u}^2}\Big)\text{du}\ .....(\text{i})$
As $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}\Rightarrow\ \text{v}=\frac{\text{fu}}{\text{u}-\text{f}}\Rightarrow\ \text{v}=\frac{\text{fu}}{\text{u}-\text{f}}$
Or $\frac{\text{v}}{\text{u}}=\frac{\text{f}}{\text{u}-\text{f}}\ .....(\text{ii})$
From (i) and (ii) $\text{dv}=-\Big(\frac{\text{f}}{\text{u}-\text{f}}\Big)^2\text{du}$
But du = L, hence the length of the image is $\frac{\text{f}^2}{(\text{u}-\text{f})^2}\text{L}$
This is the required expression for lenght of image.
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