Chapter 37, Problem 19PQ

To determine

The distance between the image of your roommate’s face and you when she stands behind you in such a way that her face is $18\text{in}$ behind your face.

Answer to Problem 19PQ

The distance between the image of your roommate’s face and you is $8.5\text{ft}$.

Explanation of Solution

Write the expression for magnification of the plane mirror.

    $M=-\frac{d_{\text{i}}}{d_0}$                                                                                                           (I)

Here, $d_0$ is the object distance, and $d_{\text{i}}$ is the image distance.

The magnification of the plane mirror will be constant for both person and his roommate.

    $M_{\text{p}}=M_{\text{f}}$

Here, $M_{\text{p}}$ is the magnification of the plane mirror for the person, and $M_{\text{f}}$ is the magnification of the mirror for his roommate.

Write the mirror equation for the person and his roommate.

    $\frac{1}{{\left(d_0\right)}_{\text{p}}}+\frac{1}{{\left(d_{\text{i}}\right)}_{\text{p}}}=\frac{1}{{\left(d_0\right)}_{\text{f}}}+\frac{1}{{\left(d_{\text{i}}\right)}_{\text{f}}}$                                                                      (II)

Write the expression for the distance between the images of the person roommate from the person.

    $d={\left(d_0\right)}_{\text{p}}-{\left(d_{\text{i}}\right)}_{\text{f}}$                                                                                    (III)

Here, ${\left(d_0\right)}_{\text{p}}$ is the object distance for person, and ${\left(d_{\text{i}}\right)}_{\text{f}}$ is the image distance for his roommate, and $d$ is the image distance between roommate and person.

Substitute $-\frac{{\left(d_{\text{i}}\right)}_{\text{p}}}{{\left(d_0\right)}_{\text{p}}}$ for $M_{\text{p}}$, and $-\frac{{\left(d_{\text{i}}\right)}_{\text{f}}}{{\left(d_0\right)}_{\text{f}}}$ for $M_{\text{f}}$ in the equation (I).

    $\begin{array}{c}-\frac{{\left(d_{\text{i}}\right)}_{\text{p}}}{{\left(d_0\right)}_{\text{p}}}=-\frac{{\left(d_{\text{i}}\right)}_{\text{f}}}{{\left(d_0\right)}_{\text{f}}}\\ \frac{{\left(d_{\text{i}}\right)}_{\text{p}}}{{\left(d_0\right)}_{\text{p}}}=\frac{{\left(d_{\text{i}}\right)}_{\text{f}}}{{\left(d_0\right)}_{\text{f}}}\end{array}$

Conclusion:

Substitute $3.5\text{ft}$ for ${\left(d_0\right)}_{\text{p}}$ and $3.5\text{ft}+1.8\text{in}$ for ${\left(d_0\right)}_{\text{f}}$ in the above equation to calculate ${\left(d_{\text{i}}\right)}_{\text{p}}$.

    $\begin{array}{c}\frac{{\left(d_{\text{i}}\right)}_{\text{p}}}{3.5\text{ft}}=\frac{{\left(d_{\text{i}}\right)}_{\text{f}}}{3.5\text{ft}+1.8\text{in}\times \left(\frac{1\text{ft}}{12\text{in}}\right)}\\ {\left(d_{\text{i}}\right)}_{\text{p}}=\frac{3.5\text{ft}}{\left(3.5\text{ft}+1.5\text{ft}\right)}{\left(d_{\text{i}}\right)}_{\text{f}}\\ =\left(\frac{3.5\text{ft}}{5}\right){\left(d_{\text{i}}\right)}_{\text{f}}\end{array}$

Substitute $3.5\text{ft}$ for ${\left(d_0\right)}_{\text{p}}$, $\left(\frac{3.5\text{ft}}{5}\right){\left(d_{\text{i}}\right)}_{\text{f}}$ for ${\left(d_{\text{i}}\right)}_{\text{p}}$, and $5\text{ft}$ for ${\left(d_0\right)}_{\text{f}}$ in the equation (II) to calculate ${\left(d_{\text{i}}\right)}_{\text{f}}$.

    $\begin{array}{c}\frac{1}{3.5\text{ft}}+\frac{1}{\left(\frac{3.5\text{ft}}{5}\right){\left(d_{\text{i}}\right)}_{\text{f}}}=\frac{1}{5\text{ft}}+\frac{1}{{\left(d_{\text{i}}\right)}_{\text{f}}}\\ \frac{1}{3.5\text{ft}}+\frac{5}{3.5\text{ft}\times {\left(d_{\text{i}}\right)}_{\text{f}}}=\frac{1}{5\text{ft}}+\frac{1}{{\left(d_{\text{i}}\right)}_{\text{f}}}\\ {\left(d_{\text{i}}\right)}_{\text{f}}=-5\text{ft}\end{array}$

Here, the negative sign indicate the distance of the image from the mirror.

Substitute $3.5\text{ft}$ for ${\left(d_0\right)}_{\text{p}}$, and $-5\text{ft}$ for ${\left(d_{\text{i}}\right)}_{\text{f}}$ in the equation (III) to calculate $d$.

    $\begin{array}{c}d=3.5\text{ft}-\left(-5\text{ft}\right)\\ =8.5\text{ft}\end{array}$

Therefore, the image of the roommate from the person is $8.5\text{ft}$.