Chapter 23, Problem 51AP

The lens and the mirror in figure P23.51 are separated by 1.00 m and have focal lengths of +80.0 cm and 50.0 cm., respectively. If an object is placed 1.00 m to the left of the lens, where will the final image be located? Stale whether the image is upright or inverted, and determine the overall magnification.

Figure P23.51

To determine

The position of the final image.

Answer to Problem 51AP

The position of the final image is located $160\text{}\text{cm}$ to the left of the lens.

Explanation of Solution

Given info:

The initial object distance is $1.00\text{m}$.

The distance between the lens and mirror is $1.00\text{m}$.

The focal length of lens is $80.0\text{cm}$.

The focal length of mirror is $-50.0\text{cm}$.

Formula to calculate the first image distance is,

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

• $q$ is the image distance
• $p$ is the object distance
• $f$ is the focal length

Substitute $1.00\text{m}$ for $p$ and $80.0\text{ cm}$ for $f$ to find $q$.

$\begin{array}{c}q=\frac{1}{\left(\frac{1}{80.0\text{cm}}-\frac{1}{\left(1.00\text{m}\right)\left(\frac{100\text{cm}}{1.00\text{m}}\right)}\right)}\\ =+400\text{cm}\end{array}$

Therefore, the position of the first image is $+400\text{cm}$.

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

 $p\text{'}=d-q$

• $p\text{'}$ is the object distance for the mirror
• $d$ is the distance between the lens and the mirror

Substitute $1.00\text{m}$ for $d$ and $+400\text{ cm}$ for $q$ to find $p\text{'}$.

$\begin{array}{c}p\text{'}=\left(1.00\text{m}\right)\left(\frac{100\text{cm}}{1.00\text{m}}\right)-400\text{cm}\\ =-300\text{cm}\end{array}$

Therefore, the position of the object of mirror is $-300\text{cm}$.

Formula to calculate the image distance for the mirror 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 $q\text{'}$.

$\begin{array}{c}q\text{'}=\frac{1}{\left(\frac{1}{\left(-50.0\text{cm}\right)}-\frac{1}{\left(-300\text{cm}\right)}\right)}\\ =-60.0\text{cm}\end{array}$

Therefore, the position of the image from mirror is $-60.0\text{cm}$.

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

 $p\text{'}\text{'}=d-q\text{'}$

• $p\text{'}\text{'}$ is the object distance for the mirror

Substitute $1.00\text{m}$ for $d$ and $-60.0\text{ cm}$ for $q\text{'}$ to find $p\text{'}\text{'}$.

$\begin{array}{c}p\text{'}\text{'}=100\text{cm}-\left(-60.0\text{cm}\right)\\ =160\text{cm}\end{array}$

Therefore, the position of the object of lens is $160\text{cm}$.

Formula to calculate the final image distance is,

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

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

Substitute $160\text{m}$ for $p\text{'}\text{'}$ and $80.0\text{cm}$ for $f$ to find $q\text{'}\text{'}$.

$\begin{array}{c}q\text{'}\text{'}=\frac{1}{\left(\frac{1}{\left(80.0\text{cm}\right)}-\frac{1}{\left(160\text{cm}\right)}\right)}\\ =+160.0\text{cm}\end{array}$

Therefore, the position of the final image is located $160\text{}\text{cm}$ to the left of the lens.

Conclusion:

The position of the final image is located $160\text{}\text{cm}$ to the left of the lens.

To determine

The total magnification of the system.

Answer to Problem 51AP

The total magnification of the system is $-0.800$.

Explanation of Solution

Given info:

Formula to calculate the total magnification of the system is,

 $\begin{array}{c}M=M_1{M}_2{M}_3\\ =\left(-\frac{q}{p}\right)\left(-\frac{q\text{'}}{p\text{'}}\right)\left(-\frac{q\text{'}\text{'}}{p\text{'}\text{'}}\right)\end{array}$

• $M$ is the magnification of the total system
• $M_1$ is the magnification of first image
• $M_2$ is the magnification of second image
• $M_3$ is the magnification of third

Substitute $400\text{cm}$ for $q$, $100\text{cm}$ for $p$,$-60.0\text{cm}$ for $q\text{'}$, $-300\text{cm}$ for $p\text{'}$, $160\text{cm}$ for $q\text{'}\text{'}$ and $160\text{cm}$ for $p\text{'}\text{'}$ to find $M$.

$\begin{array}{c}M=\left(-\frac{400\text{cm}}{100\text{cm}}\right)\left(-\frac{-60.0\text{cm}}{-300\text{cm}}\right)\left(-\frac{160\text{cm}}{160\text{cm}}\right)\\ =-0.800\end{array}$

Thus, the total magnification of the system is $-0.800$.

Conclusion:

The total magnification of the system is $-0.800$.

To determine

The final image is inverted.

Answer to Problem 51AP

The final image is inverted.

Explanation of Solution

The magnification is less than zero. Thus, the final image is inverted.

Conclusion:

The final image is inverted.