Question

An object is 12.0 cm from aconcave mirror with f = 18.0 cm.Find the magnification.(Mind your minus signs.)(Unit = cm)

Answer 1

Answer:

Magnification = 3

Explanation:

Given:

Object distance = 12.0 cm

Focal length, f = 18.0 cm

The image distance, v = ?

$\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \\ \frac{1}{18}=\frac{1}{12}+\frac{1}{v} \\ \\ \frac{1}{v}=\frac{1}{18}-\frac{1}{12} \\ \\ \frac{1}{v}=-0.02777777777 \\ \\ v=-\frac{1}{0.02777777777} \\ \\ v=-36 \\ \\ \\ \end{gathered}$

The magnification, M, is calculated as:

$\begin{gathered} M=|\frac{v}{u}| \\ \\ M=|-\frac{36}{12}| \\ \\ M=|-3| \\ \\ M=3 \end{gathered}$