Chapter 37, Problem 31PQ

To determine

The object distance.

Answer to Problem 31PQ

The object distance is $1.20\text{m}$.

Explanation of Solution

Write the expression for magnification produced by the mirror.

    $\begin{array}{c}m=\frac{-d_{\text{i}}}{d_{\text{0}}}\\ =\frac{h_{\text{i}}}{h_0}\end{array}$

Here,$m$ is the magnification produced by the mirror, $d_{\text{0}}$ is the object distance, $d_{\text{i}}$ is the image distance from the mirror, $h_{\text{i}}$ is the size of image and $h_{\text{0}}$ for size of the object.

Rewrite the above equation to determine $d_{\text{i}}$.

    $\begin{array}{c}\frac{h_{\text{i}}}{h_{\text{0}}}=\frac{-d_{\text{i}}}{d_0}\\ d_{\text{i}}=-d_0\left(\frac{h_{\text{i}}}{h_0}\right)\end{array}$                    (I)

Write the mirror equation for a convex mirror.

    $\frac{1}{f}=\frac{1}{d_{\text{0}}}+\frac{1}{d_{\text{i}}}$               (II)

Here, $f$ is the focal length of the mirror.

Rewrite the equation (II) to determine $f$.

    $\begin{array}{l}\frac{1}{f}=\frac{d_{\text{i}}+d_{\text{0}}}{d_{\text{0}}{d}_{\text{i}}}\\ f=\frac{d_{\text{0}}{d}_{\text{i}}}{d_{\text{i}}+d_{\text{0}}}\end{array}$                      (III)

Conclusion:

Substitute $60.0\text{cm}$ for $d_{\text{0}}$, $\frac{h_{\text{0}}}{2}$ for $h_i$ in equation (I) to calculate $d_{\text{i}}$.

    $\begin{array}{c}{d}_{\text{i}}=-\left(60.0\text{cm}\right)\left[\frac{\frac{h_0}{2}}{h_{\text{0}}}\right]\\ =-\left(60.0\text{cm}\right)\left[\frac{1}{2}\right]\\ =-30.0\text{cm}\end{array}$

Substitute $-30.0\text{cm}$ for $d_{\text{i}}$ and $60.0\text{cm}$ for $d_0$ in equation (III) to calculate $f$.

    $\begin{array}{c}f=\frac{\left(60.0\text{cm}\right)\left(-30.0\text{cm}\right)}{-30.0\text{cm}+60.0\text{cm}}\\ =-60.0\text{cm}\end{array}$

Substitute $\frac{h_0}{3}$ for $h_i$ in equation (I) to calculate $d_{\text{i}}$.

    $\begin{array}{c}{d}_{\text{i}}=-d_{\text{0}}\left[\frac{\frac{h_0}{3}}{h_0}\right]\\ =-d_{\text{0}}\left[\frac{1}{3}\right]\\ =-\frac{d_{\text{0}}}{3}\end{array}$

Substitute $-60.0\text{cm}$ for $f$ and $-\frac{d_{\text{0}}}{3}$ for $d_{\text{i}}$ in equation (II) to calculate $d_{\text{o}}$.

    $\begin{array}{l}\frac{1}{-60.0\text{cm}}=\frac{1}{d_{\text{0}}}+\frac{1}{\left(-\frac{d_{\text{0}}}{3}\right)}\\ \frac{1}{-60.0\text{cm}}=\frac{1}{d_{\text{0}}}-\frac{3}{d_{\text{0}}}\\ \frac{1}{-60.0\text{cm}}=\frac{1}{d_{\text{0}}}\left(1-3\right)\\ -60.0\text{cm}=-\frac{d_{\text{0}}}{2}\end{array}$

Simplify the above equation.

    $\begin{array}{c}60.0\text{cm}=\frac{d_{\text{0}}}{2}\\ d_{\text{0}}=2\times 60.0\text{cm}\\ =120.0\text{cm}\left(\frac{{10}^{-2}\text{cm}}{1.0\text{cm}}\right)\\ =1.20\text{m}\end{array}$

Therefore, the object distance is $1.20\text{m}$.