Question

An object that is 25 cm in front of a convex mirror has an image located 17 cm behind the mirror. How far behind the mirror is the image located when the object is 19 cm in front of the mirror?

Answer 1

Step 1: Assuming the quantities

Let, the focal length of the convex mirror be f

Step 2: Explanation of the quantities

For convex mirror,

Object distance, ${\mathrm{u}}_1=-25\mathrm{cm}$

Image distance, ${\mathrm{v}}_1=17\mathrm{cm}$

Step 3: Calculation of the focal length

By using the mirror formula

$$\begin{gathered} \frac{1}{{\mathrm{v}}_1}+\frac{1}{{\mathrm{u}}_1}=\frac{1}{\mathrm{f}} \\ \frac{1}{17\mathrm{cm}}+\frac{1}{-25\mathrm{cm}}=\frac{1}{\mathrm{f}} \\ \frac{1}{17}-\frac{1}{25}=\frac{1}{\mathrm{f}} \\ \frac{25-17}{425}=\frac{1}{\mathrm{f}} \\ \frac{8}{425}=\frac{1}{\mathrm{f}} \\ \mathrm{f}=\frac{425}{8} \end{gathered}$$

Step 4: Assuming the quantities

Let, the image distance convex mirror be ${\mathrm{v}}_2$

Step 5: Explanation of the quantities

For convex mirror

Object distance, ${\mathrm{u}}_2=-19\mathrm{cm}$

focal length of the convex mirror, $\mathrm{f}=\frac{425}{8}\mathrm{cm}$

Step 6: Calculation of the distance of image,

By using the mirror formula

$$\begin{gathered} \frac{1}{{\mathrm{v}}_2}+\frac{1}{{\mathrm{u}}_2}=\frac{1}{\mathrm{f}} \\ \frac{1}{{\mathrm{v}}_2}+\frac{1}{-19\mathrm{cm}}=\frac{1}{{\displaystyle \frac{425}{8}}\mathrm{cm}} \\ \frac{1}{{\mathrm{v}}_2}-\frac{1}{19}=\frac{8}{425} \\ \frac{1}{{\mathrm{v}}_2}=\frac{1}{19}+\frac{8}{425} \\ \frac{1}{{\mathrm{v}}_2}=\frac{425+\left(8\times 19\right)}{8075} \\ \frac{1}{{\mathrm{v}}_2}=\frac{425+152}{8075} \\ \frac{1}{{\mathrm{v}}_2}=\frac{577}{8075} \\ {\mathrm{v}}_2=\frac{8075}{577} \\ {\mathrm{v}}_2=13.9\mathrm{cm} \\ {\mathrm{v}}_2\simeq 14\mathrm{cm} \end{gathered}$$

Thus, the image distance of from the mirror is 14 cm.