Chapter 23, Problem 46P

An object is placed 15.0 cm from a first converging lens of focal length 10.0 cm. A second converging lens with focal length .5.00 cm is placed 10.0 cm to the right of the first converging lens. (a) Find the position q₁ of the image formed by the First converging lens. (b) How fat from the second lens is the image of the first lens? (c) What is the value of p₂, the object position for the second lens? (d) Find the position q₂ of the image formed by the second lens. (e) Calculate the magnification of the first lens (f) Calculate the magnification of the second lens. (g) What is the total magnification for the system? (h) Is the final image real or virtual? Is it upright or inverted (compared to the original object for the lens system)?

To determine

The position of the image formed by the first converging lens.

Answer to Problem 46P

The position of the image formed by the first converging lens is $+30.0\text{cm}$.

Explanation of Solution

Given info:

The object distance for first lens is $15.0\text{cm}$.

The focal length of first lens is $10.0\text{cm}$.

Explanation:

Formula to calculate the image distance is,

$q_1=\frac{p_1{f}_1}{p_1-f_1}$

• $p_1$ is the object distance
• $q_1$ is the image distance
• $f_1$ is the focal length

Substitute $15.0\text{cm}$ for $p_1$ and $10.0\text{cm}$ for $f_1$ to find $q_1$.

$\begin{array}{c}{q}_1=\frac{\left(15.0\text{cm}\right)\left(10.0\text{cm}\right)}{\left(15.0\text{cm}\right)-\left(10.0\text{cm}\right)}\\ =+30.0\text{cm}\end{array}$

Thus, the position of the image formed by the first converging lens is $+30.0\text{cm}$.

Conclusion:

The position of the image formed by the first converging lens is $+30.0\text{cm}$.

To determine

The distance between the second lens and the image formed from the

first lens.

Answer to Problem 46P

The distance between the second lens and the image formed from the first lens is $20.0\text{cm}$ beyond the second lens.

Explanation of Solution

Given info:

The object distance for first lens is $15.0\text{cm}$.

The focal length of first lens is $10.0\text{cm}$.

Explanation:

Formula to calculate the object distance is,

$\left|p_2\right|=\left|d-q_1\right|$

• $p_2$ is the object distance
• $d$ is the distance between the lenses

Substitute $30.0\text{cm}$ for $q_1$ and $10.0\text{cm}$ for $d$ to find $p_2$.

$\begin{array}{c}\left|p_2\right|=\left|10.0\text{cm}-30.0\text{cm}\right|\\ =20.0\text{cm}\end{array}$

Thus, the distance between the second lens and the image formed from the first lens is $20.0\text{cm}$ beyond the second lens.

Conclusion:

The distance between the second lens and the image formed from the first lens is $20.0\text{cm}$ beyond the second lens.

To determine

The position of object for second lens.

Answer to Problem 46P

The object distance is $-20.0\text{cm}$.

Explanation of Solution

Formula to calculate the object distance is,

$p_2=d-q_1$

Substitute $30.0\text{cm}$ for $q_1$ and $10.0\text{cm}$ for $d$ to find $p_2$.

$\begin{array}{c}{p}_2=10.0\text{cm}-30.0\text{cm}\\ =-20.0\text{cm}\end{array}$

Thus, the object distance is $-20.0\text{cm}$.

Conclusion:

The object distance is $-20.0\text{cm}$.

To determine

The position of the image formed from second lens.

Answer to Problem 46P

The image distance is $+4.00\text{cm}$.

Explanation of Solution

Given info:

The focal length of second lens is $5.00\text{cm}$.

Explanation:

Formula to calculate the image distance is,

$q_2=\frac{p_2{f}_2}{p_2-f_2}$

• $q_2$ is the image distance
• $f_2$ is the focal length

Substitute $-20.0\text{cm}$ for $p_2$ and $5.00\text{cm}$ for $f_2$ to find $q_2$.

$\begin{array}{c}{q}_2=\frac{\left(-20.0\text{cm}\right)\left(5.00\text{cm}\right)}{\left(-20.0\text{cm}\right)-\left(5.00\text{cm}\right)}\\ =+4.00\text{cm}\end{array}$

Thus, the image distance is $+4.00\text{cm}$.

Conclusion:

The image distance is $+4.00\text{cm}$.

To determine

The magnification of first lens.

Answer to Problem 46P

The magnification of first lens is $-2.00$.

Explanation of Solution

Formula to calculate the magnification of first lens is,

$M_1=-\frac{q_1}{p_1}$

• $M_1$ is the magnification of first lens

Substitute $15.0\text{cm}$ for $p_1$ and $30.0\text{cm}$ for $f_1$ to find $M_1$.

$\begin{array}{c}{M}_1=-\frac{30.0\text{cm}}{15.0\text{cm}}\\ =-2.00\end{array}$

Thus, the magnification of first lens is $-2.00$.

Conclusion:

The magnification of first lens is $-2.00$.

To determine

The magnification of second lens.

Answer to Problem 46P

The magnification of second lens is $+2.00$.

Explanation of Solution

Formula to calculate the magnification of first lens is,

$M_2=-\frac{q_2}{p_2}$

• $M_2$ is the magnification of first lens

Substitute $-20.0\text{cm}$ for $p_2$ and $5.00\text{cm}$ for $f_2$ to find $M_2$.

$\begin{array}{c}{M}_2=-\frac{4.00\text{cm}}{\left(-20.0\text{cm}\right)}\\ =+2.00\end{array}$

Thus, the magnification of second lens is $+2.00$.

Conclusion:

The magnification of second lens is $+2.00$.

To determine

The total magnification of the system.

Answer to Problem 46P

The total magnification of the system is  $-0.400$.

Explanation of Solution

Formula to calculate the total magnification is,

$M=M_1{M}_2$

• $M$ is the total magnification of the system

Substitute $-2.00$ for $M_1$ and $+2.00$ for $M_2$ to find $M$.

$\begin{array}{c}{M}_2=\left(-2.00\right)\left(+2.00\right)\\ =-4.00\end{array}$

Thus, the total magnification of the system is  $-0.400$.

Conclusion:

The total magnification of the system is  $-0.400$.

To determine

The characteristics of the final image.

Answer to Problem 46P

The final image is real and inverted relative to the original object.

Explanation of Solution

The final image distance $q_2>0$. Therefore, the final image is real.

The total magnification $M<0$. Therefore, the final image is inverted.

Thus, the final image is real and inverted relative to the original object.

Conclusion:

The final image is real and inverted relative to the original object.