Question

A 15.0 cm object is 12.0 cm from a convex mirror that has a focal length of -6.0 cm. What is the height of the image produced by the mirror?

Answer 1

Answer: 5 cm

In convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object .

The Mirror equation is:  

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$    (1)  

Where:  

$f=-6cm$ is the focal distance  

$u=12cm$ is the distance between the object and the mirror  

$v$ is the distance between the image and the mirror

We already know the values of $f$ and $u$, let's find $v$ from (1):  

$v=\frac{u.f}{u-f}$    (2)  

$v=\frac{(12cm)(-6cm)}{12cm-(-6cm)}$

$v=-4cm$   (3)

On the other hand, the magnification $m$ of the image is given by the following equations:  

$m=-\frac{v}{u}$   (4)

$m=\frac{h_{i}}{h_{o}}$   (5)

Where:

$h_{i}$ is the image height  

$h_{o}=15cm$ is the object height

Now, if we want to find the image height, we firstlu have to find $m$ from (4), substitute it on (5) and find $h_{i}$:

Substituting  (3) in (4):

$m=-\frac{-4cm}{12cm}$  

$m=\frac{1}{3}$    (6)

Substituting  (6) in (5):

$\frac{1}{3}=\frac{h_{i}}{15cm}$

$h_{i}=\frac{15cm}{3}$

Finally we obtain the value of the height of the image produced by the mirror:

$h_{i}=5cm$

Answer 2

Answer:

The answer is D. on edgen

Explanation:

D. 5.0