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 $$L<< \left| v-f \right| $$

Answer 1

As the mean distance of object from mirror is $$u$$
$$\therefore$$ $${u}_{1}=u-\cfrac{L}{2}$$ and $${u}_{2}=(u+\cfrac{L}{2})$$
Let the image of the two ends of object form at distance $${v}_{1}$$ and $${v}_{2}({v}_{1}> {v}_{2})$$. So lenght of image on principle axis is $$L'({v}_{1}-{v}_{2}$$)
$$\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } $$ or $$\cfrac { 1 }{ v } +\cfrac { 1 }{ f } =\cfrac { 1 }{ u } $$
$$\Rightarrow$$ $$\cfrac { 1 }{ v } =\cfrac{u-f}{uf}\Rightarrow v=\cfrac{uf}{u-f}$$

So $$L'={v}_{1}-{v}_{2}==\cfrac { \left( u-\cfrac { L }{ 2 } \right) f }{ \left( u-\cfrac { L }{ 2 } \right) -f } -\cfrac { \left( u+\cfrac { L }{ 2 } \right) f }{ \left( u+\cfrac { L }{ 2 } \right) -f } \Rightarrow L'=f\cfrac { u-\cfrac { L }{ 2 } }{ \left( u-f-\cfrac { L }{ 2 } \right) } -\cfrac { u+\cfrac { 1 }{ 2 } }{ \left( u-f+\cfrac { L }{ 2 } \right) } $$

$$=f\left[ \cfrac { \left( u-\cfrac { L }{ 2 } \right) \left( u-f+\cfrac { L }{ 2 } \right) -\left( u+\cfrac { L }{ 2 } \right) \left( u-f-\cfrac { L }{ 2 } \right) }{ \left( u-f-\cfrac { L }{ 2 } \right) \left( u-f+\cfrac { L }{ 2 } \right) } \right]$$

$$ =\cfrac { f\left[ { u }^{ 2 }-uf+\cfrac { uL }{ 2 } -\cfrac { uL }{ 2 } +\cfrac { fL }{ 2 } -\cfrac { { L }^{ 2 } }{ 4 } -\left( { u }^{ 2 }-uf+\cfrac { uL }{ 2 } -\cfrac { uL }{ 2 } +\cfrac { fL }{ 2 } -\cfrac { { L }^{ 2 } }{ 4 } \right) \right] }{ { \left( u-f \right) }^{ 2 }-\cfrac { { L }^{ 2 } }{ 4 } } $$

$$\therefore L<<(u-f)$$
So neglecting the terms $$\cfrac { { L }^{ 2 } }{ 4 } $$
$${ L }^{ 2 }=f\left[ \cfrac { \cfrac { fL }{ 2 } +\cfrac { fL }{ 2 } }{ { (u-f) }^{ 2 } } \right] \quad \quad $$
$$L'=\cfrac { ffL }{ { (u-f) }^{ 2 } } \Rightarrow L'=\cfrac { L{ f }^{ 2 } }{ { (u-f) }^{ 2 } } $$
it is the length of image $$f$$