Answer to Question #113352 in Optics for Trudi Hagen

Question

the image of a crayon appears to be 22.8 cm behind the surface of a convex mirror and is 3.38 cm tall. The mirrors focal length is 51.2 cm (a) how far in front of the mirror is the crayon positioned?

Answer 1

As per the question,

Image position from the convex mirror $(v)=22.8cm$

Height of the image $(h_i)=3.38 cm$

The mirror of the focal length $(f)=51.2cm$

distance of the object from the convex mirror=?

Let it is at a distance u.

Now, applying the mirror formula,

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\Rightarrow \dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\Rightarrow \dfrac{1}{u}=\dfrac{1}{51.2}-\dfrac{1}{22.8}$

$\Rightarrow u=-41.10cm$

Height of the object

$\dfrac{-v}{u}=\dfrac{h_i}{h_o}$

$h_o=-\dfrac{u}{-v}\times h_i=\dfrac{41.10}{22.8}\times 3.38 cm=6.084cm$