Question

What is a mirror's radius of curvature if cars 16.0 m away appear 0.35 their normal size? follow the sign conventions.


express your answer to two significant figures and include the appropriate units?

Answer 1

It is not specified if the image is real or virtual: I assume it is real, so $d_i$, the distance of the image from the mirror, is positive.

The ratio between the size of the image and the size of the object is equal to the ratio between the distance image-mirror and the distance object-mirror:
$\frac{h_i}{h_o}= \frac{d_i}{d_o}$
the problem says that the size of the image is 0.35 times the size of the object, so this means that the ratio between the distances is also 0.35, and so we can write 
$d_i = 0.35 d_o$

Then we can use the mirror equation to find the focal length f:
$\frac{1}{f} = \frac{1}{d_o}+ \frac{1}{d_i}$
Substituting $d_o = 16.0 m$ and $d_i = 0.35 d_o = 5.6 m$, we find
$f=4.1 m$
with positive sign, this means it is a concave mirror for the sign convention.

The radius of the mirror is the negative of half the focal length:
$r=- \frac{f}{2}=- \frac{4.1 m}{2}=2.1 m$

Answer 2

look a rear view mirror uses convex mirror for wider field of view.. so the image is always virtual and erect..
also, magnification = -v/u = 0.36

so, v = 8.28 m = 828 cm..

now applying mirror's formula: 1/v+1/u=1/f
or, 1/828-1/2300=1/f
or f=1293.75 cm= 12.9375m

clearly, R=2f=12.9375*2=25.8750 m