Chapter 37, Problem 46PQ

(a)

To determine

The distance of the object from the mirror.

Answer to Problem 46PQ

The distance of the object from the mirror is $8.40\text{cm}$.

Explanation of Solution

Write the equation for the mirror.

  $\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_{\text{i}}}$                                                                                   (I)

Here, $f$ is the focal length of the mirror, $d_0$ is the distance of object from the mirror and $d_{\text{i}}$ is the distance of image from the mirror.

Rewrite the above expression for $d_0$.

  $\frac{1}{d_0}=\frac{1}{f}-\frac{1}{d_{\text{i}}}$                                                                                  (II)

Write the equation for magnification produced by the mirror and find $d_{\text{i}}$.

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

Here, $h_{\text{i}}$ is the height of image and $h_0$ is the height of object.

Conclusion:

Substitute $2.50h_0$ for $h_{\text{i}}$ in equation (III) to find $d_{\text{i}}$.

  $\begin{array}{c}{d}_{\text{i}}=-d_0\left(\frac{2.50h_0}{h_0}\right)\\ =-2.50d_0\end{array}$                                                                         (IV)

Substitute $\text{14}\text{.0}\text{cm}$ for $f$ and $-2.50d_0$ for $d_{\text{i}}$ in equation (II) to find $d_0$.

  $\begin{array}{c}\frac{1}{d_0}=\frac{1}{\text{14}\text{.0}\text{cm}}-\frac{1}{\left(-2.50d_0\right)}\\ \frac{1}{d_0}-\frac{1}{2.50d_0}=\frac{1}{\text{14}\text{.0}\text{cm}}\\ \frac{1}{d_0}\left(\frac{2.50-1}{2.50}\right)=\frac{1}{\text{14}\text{.0}\text{cm}}\\ d_0=8.40\text{cm}\end{array}$

Therefore, the distance of the object from the mirror is $8.40\text{cm}$.

(b)

To determine

Whether the image formed by the mirror is real or virtual.

Answer to Problem 46PQ

The image distance from the mirror is negative, so the image formed by the mirror is virtual.

Explanation of Solution

Consider equation (III) for the image distance.

Conclusion:

Substitute $8.40\text{cm}$ for $d_0$ in equation (III) to find $d_{\text{i}}$.

  $\begin{array}{c}{d}_{\text{i}}=-2.50\times 8.40\text{cm}\\ =-21.0\text{cm}\end{array}$

Therefore, the image formed at the distance of $21.0\text{cm}$ from the mirror and the negative sign indicates that the image will be virtual.

(c)

To determine

A ray diagram for showing the locations of the object and image.

Answer to Problem 46PQ

Ray diagram of the object and the image location is shown in the following figure using three rays.

Explanation of Solution

Ray 1 will pass from the top of the object through the vertex of the mirror and reflected opposite to the object position.

Ray 2 will pass from the bottom of the object and reflected back from the mirror on itself.

Ray 3 will pass from the top of the object, which is parallel to the principle axis and reflected through the mirror from the focal point.

The image formed will be virtual and upright.

Therefore, the ray diagram of the object and the image location is shown in figure 1 using three rays.

Figure (1)