Question

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 $L<<|v-f|$

Answer 1

Step 1: Draw diagram for the given situation.




Step 2: Find the length of the image.

Formula used: $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$


Given, the length of the object =𝐿

Let the two ends of the object be at distance $u_1=u-\frac{L}{2}$ and $u_2=u-\frac{L}{2}$ respectively,



So, the length of the object will be mean distance of $u_1$ and $u_2$



$|u_1-u_2|=L$


Let the image of the two ends be formed at $v_1$ and $v_2$

so that the image length would be

$L^{\prime }=|v_1-v_2|$

Since $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}orv=\frac{fu}{u-f}$

the image of the two ends will be at.

$v_1=\frac{f(u-L/2)}{u-f-L/2},v_2=\frac{f(u+L/2)}{u-f+L/2}$

Hence

$L^{\prime }=|v_1-v_2|=\frac{f^{2}L}{(u-f{)}^2-L^2/4}$




Since the object is short and kept away from focus, we have
$\frac{L^2}{4}<<(u-f{)}^2$


Hence finally

$L^{\prime }=\frac{f^2}{(u-f{)}^2}L$


Final answer: $\frac{f^2}{(u-f{)}^2}L$