Chapter 23, Problem 52AP

The object in Figure P23.52 is mid-way between the lens and the mirror, which are separated by a distance d = 25.0 cm. The magnitude of the mirror’s radius of curvature is 20.0 cm, and the lens has a focal length of –16.7 cm. (a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. (b) Is the image real or virtual? (c) Is it upright or inverted? (d) What is the overall magnification of the image?

Figure P23.52

To determine

The position of the final image.

Answer to Problem 52AP

The position of the final image is located $25.3\text{cm}$ behind (to the right of) the mirror.

Explanation of Solution

Given info:

The distance between the lens and mirror is $25.0\text{cm}$.

The radius of curvature is $20.0\text{cm}$.

The focal length is $-16.7\text{cm}$.

Explanation:

Formula to calculate the object distance for the mirror is,

$p=\frac{d}{2}$

• $p$ is the object distance
• $d$ is distance between the lens and mirror

Substitute $25.0\text{cm}$ for $d$ to find $p$.

$\begin{array}{c}p=\frac{25.0\text{cm}}{3}\\ =12.5\text{cm}\end{array}$

Therefore, the position of the object for the mirror is $12.5\text{cm}$.

Formula to calculate the first image distance is,

$q=\frac{1}{\left(\frac{2}{R}-\frac{1}{p}\right)}$

• $q$ is the image distance
• $p$ is the object distance
• $R$ is the radius of curvature

Substitute $12.5\text{cm}$ for $p$ and $20.0\text{cm}$ for $R$ to find $q$.

$\begin{array}{c}q=\frac{1}{\left(\frac{2}{20.0\text{cm}}-\frac{1}{12.5\text{cm}}\right)}\\ =+50.0\text{cm}\end{array}$

Therefore, the position of the first image is $+50.0\text{cm}$.

This image is the object for the lens. Formula to calculate the object distance for the lens is,

$p\text{'}=d-q$

• $p\text{'}$ is the object distance

Substitute $25.0\text{cm}$ for $d$ and $+50.0\text{cm}$ for $q$ to find $p\text{'}$.

$\begin{array}{c}p\text{'}=25.0\text{cm}-50.0\text{cm}\\ =-25.0\text{cm}\end{array}$

Therefore, the position of the object is $-25.0\text{cm}$.

Formula to calculate the image distance for the lens is,

$q\text{'}=\frac{1}{\left(\frac{1}{f\text{'}}-\frac{1}{p\text{'}}\right)}$

• $q\text{'}$ is the image distance
• $p\text{'}$ is the object distance
• $f\text{'}$ is the focal length

Substitute $-300\text{m}$ for $p\text{'}$ and $-50.0\text{cm}$ for $f\text{'}$ to find $q\text{'}$.

$\begin{array}{c}q\text{'}=\frac{1}{\left(\frac{1}{\left(-16.7\text{cm}\right)}-\frac{1}{\left(-25.0\text{cm}\right)}\right)}\\ =-50.3\text{cm}\end{array}$

Therefore, the position of the image from mirror is $50.3\text{cm}$ in front of (to the right of) the lens.

Formula to calculate the final position of the image is,

$q\text{'}\text{'}=q\text{'}+d$

• $q\text{'}\text{'}$ is the final image distance

Substitute $25.0\text{cm}$ for $d$ and $-50.3\text{ cm}$ for $q\text{'}$ to find $q\text{'}\text{'}$.

$\begin{array}{l}q\text{'}\text{'}=-50.3\text{cm}+25.0\text{cm}\\ =-25.3\text{cm}\end{array}$

Thus, the position of the final image is located $25.3\text{cm}$ behind (to the right of) the mirror.

Conclusion:

The position of the final image is located $25.3\text{cm}$ behind (to the right of) the mirror.

To determine

The characteristics of the image.

Answer to Problem 52AP

The final image is virtual.

Explanation of Solution

The final image position is less than zero. Thus, the final image is virtual.

Conclusion:

The final image is virtual.

To determine

The total magnification of the system.

Answer to Problem 52AP

The total magnification of the system is $+8.05$.

Explanation of Solution

Given info:

Explanation:

Formula to calculate the total magnification of the system is,

$\begin{array}{c}M=M_1{M}_2\\ =\left(-\frac{q}{p}\right)\left(-\frac{q\text{'}}{p\text{'}}\right)\end{array}$

• $M$ is the magnification of the total system
• $M_1$ is the magnification of first image
• $M_2$ is the magnification of second image

Substitute $50.0\text{cm}$ for $q$,   $12.5\text{cm}$ for $p$, $-50.3\text{cm}$ for $q\text{'}$ and $-25.0\text{cm}$ for $p\text{'}$ to find $M$.

$\begin{array}{c}M=\left(-\frac{50.0\text{cm}}{12.5\text{cm}}\right)\left(-\frac{-50.3\text{cm}}{-25.0\text{cm}}\right)\\ =+8.05\end{array}$

Thus, the total magnification of the system is $+8.05$.

Conclusion:

The total magnification of the system is $+8.05$.

To determine

The characteristics of the image.

Answer to Problem 52AP

The final image is upright.

Explanation of Solution

The magnification is greater than zero. Thus, the final image is upright.

Conclusion:

The final image is upright.