Chapter 36, Problem 46P

(a)

To determine

The location of the image, the nature of the image and the magnification of the image when the object distance is $40.0\text{cm}$.

Answer to Problem 46P

The location of the image, the nature of the image and the magnification of the image when the object distance is $40.0\text{cm}$ are $\underset{\_}{-13.3\text{cm}}$, virtual, upright and $\underset{\_}{+0.333}$.

Explanation of Solution

Write the expression for mirror equation.

    $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$                                                                                                                (I)

Here, $q$ is the image distance, $f$ is the focal length and $p$ is the object distance.

Write the expression for magnification.

    $M=-\frac{q}{p}$                                                                                                                  (II)

Conclusion:

Substitute $40.0\text{cm}$ for $p$ and $-20.0\text{cm}$ for $f$ in the equation (I).

    $\begin{array}{c}\frac{1}{-20.0\text{cm}}=\frac{1}{40.0\text{cm}}+\frac{1}{q}\\ q=-13.3\text{cm}\end{array}$

Substitute $40.0\text{cm}$ for $p$ and $-13.3\text{cm}$ for $q$ in the equation (II).

    $\begin{array}{c}M=-\frac{\left(-13.3\text{cm}\right)}{40.0\text{cm}}\\ =+0.333\end{array}$

The object distance is negative, so the image is virtual. The magnification is positive, so the image is upright.

Therefore, the location of the image, the nature of the image and the magnification of the image when the object distance is $40.0\text{cm}$ are $\underset{\_}{-13.3\text{cm}}$, virtual, upright and $\underset{\_}{+0.333}$.

(b)

To determine

The location of the image, the nature of the image and the magnification of the image when the object distance is $20.0\text{cm}$.

Answer to Problem 46P

The location of the image, the nature of the image and the magnification of the image when the object distance is $20.0\text{cm}$ are $\underset{\_}{-10.0\text{cm}}$, virtual, upright and $\underset{\_}{+0.500}$.

Explanation of Solution

Write the expression for mirror equation.

    $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$                                                                                                                 (I)

Here, $q$ is the image distance, $f$ is the focal length and $p$ is the object distance.

Write the expression for magnification.

    $M=-\frac{q}{p}$                                                                                                                   (II)

Conclusion:

Substitute $20.0\text{cm}$ for $p$ and $-20.0\text{cm}$ for $f$ in the equation (I).

    $\begin{array}{c}\frac{1}{-20.0\text{cm}}=\frac{1}{20.0\text{cm}}+\frac{1}{q}\\ q=-10.0\text{cm}\end{array}$

Substitute $20.0\text{cm}$ for $p$ and $-10.0\text{cm}$ for $q$ in the equation (II).

    $\begin{array}{c}M=-\frac{\left(-10.0\text{cm}\right)}{20.0\text{cm}}\\ =+0.500\end{array}$

The object distance is negative, so the image is virtual. The magnification is positive, so the image is upright.

Therefore, the location of the image, the nature of the image and the magnification of the image when the object distance is $20.0\text{cm}$ are $\underset{\_}{-10.0\text{cm}}$, virtual, upright and $\underset{\_}{+0.500}$.

(c)

To determine

The location of the image, the nature of the image and the magnification of the image when the object distance is $10.0\text{cm}$.

Answer to Problem 46P

The location of the image, the nature of the image and the magnification of the image when the object distance is $10.0\text{cm}$ are $\underset{\_}{-6.67\text{cm}}$, virtual, upright and $\underset{\_}{+0.667}$.

Explanation of Solution

Write the expression for mirror equation.

    $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$                                                                                                                (I)

Here, $q$ is the image distance, $f$ is the focal length and $p$ is the object distance.

Write the expression for magnification.

    $M=-\frac{q}{p}$                                                                                                                  (II)

Conclusion:

Substitute $10.0\text{cm}$ for $p$ and $-20.0\text{cm}$ for $f$ in the equation (I).

    $\begin{array}{c}\frac{1}{-20.0\text{cm}}=\frac{1}{10.0\text{cm}}+\frac{1}{q}\\ q=-6.67\text{cm}\end{array}$

Substitute $10.0\text{cm}$ for $p$ and $-6.67\text{cm}$ for $q$ in the equation (II).

    $\begin{array}{c}M=-\frac{\left(-6.67\text{cm}\right)}{10.0\text{cm}}\\ =+0.667\end{array}$

The object distance is negative, so the image is virtual. The magnification is positive, so the image is upright.

Therefore, the location of the image, the nature of the image and the magnification of the image when the object distance is $10.0\text{cm}$ are $\underset{\_}{-6.67\text{cm}}$, virtual, upright and $\underset{\_}{+0.667}$.