Chapter 23, Problem 78GP

To determine

To Explain: The answers for the object distance are not the same.

Answer to Problem 78GP

The object distance has two values: ${\mathrm{d}}_{\mathrm{o}}=13.3\text{ cm; 26}\text{.7 cm}$

Explanation of Solution

Given info:

Radius of curvature of the concave mirror, $\mathrm{r}=40\text{ cm}$

Formula used:

Mirror formula is given by:

  $\frac{1}{\mathrm{f}}=\frac{1}{{\mathrm{d}}_{\mathrm{o}}}+\frac{1}{{\mathrm{d}}_{\mathrm{i}}}$

Here,

  $\mathrm{f}$ is the focal length

  ${\mathrm{d}}_{\mathrm{o}}$ is the object distance

  ${\mathrm{d}}_{\mathrm{i}}$ is the image distance

Calculation:

Firstly, the focal length can be calculated from the given data as follows:

  $\begin{array}{l}\mathrm{f}=\frac{\mathrm{r}}{2}=\frac{40\text{ cm}}{2}\\ \mathrm{f}=20\text{ cm}\end{array}$

Now, according to the given statement, the magnification can be written as:

  $\begin{array}{l}\mathrm{m}=-\frac{{\mathrm{d}}_{\mathrm{i}}}{{\mathrm{d}}_{\mathrm{o}}}=\pm 3\\ {\mathrm{d}}_{\mathrm{i}}=\mp 3{\mathrm{d}}_{\mathrm{o}}\end{array}$

So, the object distance can be calculated as:

  $\begin{array}{l}\frac{1}{\mathrm{f}}=\frac{1}{{\mathrm{d}}_{\mathrm{o}}}+\frac{1}{{\mathrm{d}}_{\mathrm{i}}}\\ \frac{1}{{\mathrm{d}}_{\mathrm{o}}}+\frac{1}{\mp 3{\mathrm{d}}_{\mathrm{o}}}=\frac{1}{\mathrm{f}}\\ {\mathrm{d}}_{\mathrm{o}}=\frac{2\mathrm{f}}{3};\frac{4\mathrm{f}}{3}\\ {\mathrm{d}}_{\mathrm{o}}=\frac{2\left(20\text{ cm}\right)}{3};\frac{4\left(20\text{ cm}\right)}{3}\\ {\mathrm{d}}_{\mathrm{o}}=13.3\text{ cm; 26}\text{.7 cm}\end{array}$

Hence, the image distance can be given as:

  $\begin{array}{l}{\mathrm{d}}_{\mathrm{i}}=\mp 3{\mathrm{d}}_{\mathrm{o}}\\ {\mathrm{d}}_{\mathrm{i}}=-\left(13.3\times 3\right)=-40\text{ cm}\\ {\mathrm{d}}_{\mathrm{i}}=\left(26.7\times 3\right)=80\text{ cm}\end{array}$

Hence, the students don’t need to repeat the lab.

Conclusion:

Thus, the object distance has two values: ${\mathrm{d}}_{\mathrm{o}}=13.3\text{ cm; 26}\text{.7 cm}$