Chapter 23, Problem 75P

(a)

To determine

If the image is real or virtual.

Answer to Problem 75P

The image is virtual.

Explanation of Solution

Write the magnification formula.

    $m=-\frac{q}{p}$                                                                                            (I)

Here, $m$ is the magnification, $q$ is the image distance and $p$ is the object distance.

Conclusion:

As the image is large, the magnification is positive. This implies $q<0$. The image distance is negative.

Thus, the image is virtual.

(b)

To determine

The focal length of the mirror and if the mirror is concave or convex.

Answer to Problem 75P

The focal length of the mirror is $2.4\text{cm}$ and the mirror is concave.

Explanation of Solution

Write the focal length from the mirror formula and substitute $-mp$ for $q$.

    $\begin{array}{c}\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\\ \frac{1}{p}+\frac{1}{\left(-mp\right)}=\frac{1}{f}\\ f=\frac{1}{\frac{1}{p}-\frac{1}{mp}}\end{array}$                                                                  (II)

Here, $f$ is the focal length.

Conclusion:

Substitute $1.9\text{cm}$ for $p$ and $5.0$ for $m$ in equation (II) to find $f$.

    $\begin{array}{c}f=\frac{1}{\frac{1}{1.9\text{cm}}-\frac{1}{\left(5.0\right)\left(1.9\text{cm}\right)}}\\ =\frac{1}{\frac{1}{1.9\text{cm}}-\frac{1}{9.5\text{cm}}}\\ =2.4\text{cm}\end{array}$

As the focal length is positive, the mirror is a concave mirror.

Thus, the focal length of the mirror is $2.4\text{cm}$ and the mirror is concave.

(c)

To determine

If the image gets larger or smaller if the mirror is moved closer to the tooth.

Answer to Problem 75P

If the mirror is moved closer to the tooth, the image gets smaller.

Explanation of Solution

Write the mirror equation.

    $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

Divide the equation throughout with $-p$.

    $-1-\frac{p}{q}=-\frac{p}{f}$

Substitute $\frac{1}{m}$ for $-\frac{p}{q}$ and $2.4\text{cm}$ for $f$.

    $\begin{array}{c}-1+\frac{1}{m}=-\frac{p}{f}\\ \frac{1}{m}=1-\frac{p}{f}\\ m=\frac{1}{1-\frac{p}{2.4\text{cm}}}\end{array}$                                                                             (III)

Conclusion:

If the mirror is moved closer to the tooth, the value of $p$ gets larger. Thus, from equation (III), $m$ gets smaller.

Thus, if the mirror is moved closer to the tooth, the image gets smaller.

(d)

To determine

The range of object distance where the image is upright.

Answer to Problem 75P

The range of object distance where the image is upright is $p<f$.

Explanation of Solution

Write the equation for magnification in terms of $f$.

    $m=\frac{1}{1-\frac{p}{f}}$                                                                                        (IV)

Conclusion:

For the image to be upright, the magnification must be greater than one, $m>0$. From equation (IV), for $m>0$, the object distance must be greater than the focal length, $p>f$.

Thus, the range of object distance where the image is upright is $p<f$.