Question

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 A 4.8 cm tall image is located 3 cm in front of a lens whose focal length is -6 cm.

                   

a) Solve for the distance in which the object is placed in front of the lens.

                          

  

b) Solve for the magnification

                  

c) Solve for the height of the object.

                

d) Is the image upright or inverted?

Answer 1

Focal length of lens, $f=-6\space cm$

Image distance, $v=-3\space cm$

Height of image, $h'=+4.8\space cm$

(a) Object distance $=u$

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

$\dfrac{1}{u}+\dfrac{1}{-3}=\dfrac{1}{-6}$

$u=+6\space cm$

(b) Magnification, $m=\dfrac{v}{u}$

$m=\dfrac{-3}{6}=-0.5$

(c) Height of object $=h$

$m=\dfrac{h'}{h}$

$h=\dfrac{h'}{m}=\dfrac{4.8}{-0.5}$

$h=-9.6\space cm$

(d) The image is upright