Chapter 26, Problem 66P

(a)

To determine

The distance between the object and the screen.

Answer to Problem 66P

The distance between the object and the screen is $67.5\text{cm}$.

Explanation of Solution

Given info: The focal length of the left and right lenses are $f_1=5.00\text{cm}$ and $f_2=10.0\text{cm}$. The object distance is $7.50\text{cm}$ to the left of the lens 1. The magnification of the image is $+1.00$.

Write the expression of thin lens equation for lens 1.

$\frac{1}{q_1}=\frac{1}{f_1}-\frac{1}{p_1}$

Here,

$p_1$ is the object distance.

$q_1$ is the image distance.

Substitute $5.00\text{cm}$ for $f_1$ and $7.50\text{cm}$ for $p_1$ in above equation.

$\begin{array}{c}\frac{1}{q_1}=\frac{1}{5.00\text{cm}}-\frac{1}{7.50\text{cm}}\\ q_1=15\text{cm}\end{array}$

Write the expression for magnification.

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

Substitute $15\text{cm}$ for $q_1$ and $7.50\text{cm}$ for $p_1$ in above equation.

$\begin{array}{c}{M}_1=-\frac{15\text{cm}}{7.50\text{cm}}\\ =-2\end{array}$

Write the expression of magnification for a combination of two lenses.

$M=M_1{M}_2$

Substitute $-2$ for $M_1$ and $1$ for $M$ in above equation.

$\begin{array}{c}1=\left(-2\right)M_2\\ M_2=-\frac{1}{2}\end{array}$

Write the expression to calculate the magnification for lens 2.

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

Substitute $-\frac{1}{2}$ for $M_2$ in above equation.

$\begin{array}{c}-\frac{1}{2}=-\frac{q_2}{p_2}\\ p_2=2q_2\end{array}$ (1)

Write the expression of thin lens equation for lens 2.

$\frac{1}{p_2}+\frac{1}{q_2}=\frac{1}{f_2}$

Here,

$p_2$ is the object distance from the lens 2.

$q_2$ is the image distance from the lens 2.

Substitute $2q_2$ for $p_2$ and $10.00\text{cm}$ for $f_2$ in above equation.

$\begin{array}{c}\frac{1}{2q_2}+\frac{1}{q_2}=\frac{1}{10.00\text{cm}}\\ q_2=15\text{cm}\end{array}$

Substitute $15\text{cm}$ for $q_2$ in equation (1).

$\begin{array}{c}{p}_2=2\times 15\text{cm}\\ \text{=30}\text{cm}\end{array}$

The distance between the object and the screen is,

$D=p_1+q_1+p_2+q_2$

Substitute $7.5\text{cm}$ for $p_1$, $15\text{cm}$ for $q_1$, $30\text{cm}$ for $p_2$ and $15\text{cm}$ for $q_2$ in above equation.

$\begin{array}{c}D=7.5\text{cm}+15\text{cm}+30\text{cm}+15\text{cm}\\ \text{=67}\text{.5}\text{cm}\end{array}$

Thus, the distance between the object and the screen is $67.5\text{cm}$.

Conclusion:

Therefore, the distance between the object and the screen is $67.5\text{cm}$.

(b)

To determine

The displacement of each lens from its initial position.

Answer to Problem 66P

The displacement of each lens from its initial position is $1.28\text{cm}$ and $17.7\text{cm}$ or $0.927\text{cm}$ and $4.44\text{cm}$.

Explanation of Solution

Given info: The focal length of the left and right lenses are $f_1=5.00\text{cm}$ and $f_2=10.0\text{cm}$. The object distance is $7.50\text{cm}$ to the left of the lens 1. The magnification of the image is $+3.00$.

Write the expression of thin lens equation for lens 1.

$\frac{1}{p_1^{\text{'}}}+\frac{1}{q_1^{\text{'}}}=\frac{1}{f_1}$

Here,

$p_1^{\text{'}}$ is the object distance from the lens 1.

$q_1^{\text{'}}$ is the image distance from the lens1.

Substitute $5.00\text{cm}$ for $f_1$ in above equation.

$\begin{array}{c}\frac{1}{p_1^{\text{'}}}+\frac{1}{q_1^{\text{'}}}=\frac{1}{5.00\text{cm}}\\ q_1^{\text{'}}=\frac{5p_1^{\text{'}}}{p_1^{\text{'}}-5}\end{array}$

Write the expression of magnification for lens 1.

$M_1^{\text{'}}=-\frac{q_1^{\text{'}}}{p_1^{\text{'}}}$

Substitute $\frac{5p_1^{\text{'}}}{p_1^{\text{'}}-5}$ for $q_1^{\text{'}}$ in above equation.

$\begin{array}{c}{M}_1^{\text{'}}=-\frac{\frac{5p_1^{\text{'}}}{p_1^{\text{'}}-5}}{p_1^{\text{'}}}\\ =\frac{5}{p_1^{\text{'}}-5}\end{array}$

Write the expression of magnification for the combination of lenses.

$\begin{array}{c}M\text{'}=M_1^{\text{'}}{M}_2^{\text{'}}\\ M_2^{\text{'}}=\frac{M\text{'}}{M_1^{\text{'}}}\end{array}$

Substitute $3$ for $M\text{'}$ and $\frac{5}{p_1^{\text{'}}-5}$ for $M_1^{\text{'}}$ in above equation.

$\begin{array}{c}{M}_2^{\text{'}}=\frac{3}{\frac{5}{p_1^{\text{'}}-5}}\\ =-\frac{3}{5}\left(p_1^{\text{'}}-5\right)\end{array}$

Write the expression to calculate the magnification of lens 2.

$M_2^{\text{'}}=-\frac{q_2^{\text{'}}}{p_2^{\text{'}}}$

Substitute $-\frac{3}{5}\left(p_1^{\text{'}}-5\right)$ for $M_2^{\text{'}}$ in above equation.

$\begin{array}{c}-\frac{3}{5}\left(p_1^{\text{'}}-5\right)=-\frac{q_2^{\text{'}}}{p_2^{\text{'}}}\\ q_2^{\text{'}}=\frac{3}{5}{p}_2^{\text{'}}\left(p_1^{\text{'}}-5\right)\end{array}$ (2)

Write the expression of lens equation for lens 2.

$\frac{1}{p_2^{\text{'}}}+\frac{1}{q_2^{\text{'}}}=\frac{1}{f_2}$

Substitute $\frac{3}{5}{p}_2^{\text{'}}\left(p_1^{\text{'}}-5\right)$ for $q_2^{\text{'}}$ and $10\text{cm}$ for $f_2$ in above equation.

$\begin{array}{c}\frac{1}{p_2^{\text{'}}}+\frac{1}{\frac{3}{5}{p}_2^{\text{'}}\left(p_1^{\text{'}}-5\right)}=\frac{1}{10\text{cm}}\\ p_2^{\text{'}}=\frac{10\left(3p_1^{\text{'}}-10\right)}{3\left(p_1^{\text{'}}-5\right)}\end{array}$

Substitute $\frac{10\left(3p_1^{\text{'}}-10\right)}{3\left(p_1^{\text{'}}-5\right)}$ for $p_2^{\text{'}}$ in equation (2).

$\begin{array}{c}{q}_2^{\text{'}}=\frac{3}{5}\left[\frac{10\left(3p_1^{\text{'}}-10\right)}{3\left(p_1^{\text{'}}-5\right)}\right]\left(p_1^{\text{'}}-5\right)\\ =2\left(3p_1^{\text{'}}-10\right)\end{array}$ (3)

The distance between the object and the screen is,

$D=p_1^{\text{'}}+q_1^{\text{'}}+p_2^{\text{'}}+q_2^{\text{'}}$

Substitute $\frac{5p_1^{\text{'}}}{p_1^{\text{'}}-5}$ for $q_1^{\text{'}}$, $\frac{10\left(3p_1^{\text{'}}-10\right)}{3\left(p_1^{\text{'}}-5\right)}$ for $p_2^{\text{'}}$, $2\left(3p_1^{\text{'}}-10\right)$ for $q_2^{\text{'}}$ and $67.5\text{cm}$ for $D$ in above equation.

$\begin{array}{c}{p}_1^{\text{'}}+\frac{5p_1^{\text{'}}}{p_1^{\text{'}}-5}+\frac{10\left(3p_1^{\text{'}}-10\right)}{3\left(p_1^{\text{'}}-5\right)}+2\left(3p_1^{\text{'}}-10\right)=67.5\text{cm}\\ \left[\begin{array}{l}\left[3\left(p_1^{\text{'}}-5\right)\right]p_1^{\text{'}}+15p_1^{\text{'}}+10\left(3p_1^{\text{'}}-10\right)\\ +2\left(3p_1^{\text{'}}-10\right)\left[3\left(p_1^{\text{'}}-5\right)\right]\end{array}\right]=67.5\left[3\left(p_1^{\text{'}}-5\right)\right]\\ \left[\begin{array}{l}3{\left(p_1^{\text{'}}\right)}^2+30p_1^{\text{'}}-100+18{\left(p_1^{\text{'}}\right)}^2\\ -150p_1^{\text{'}}+300-202.5p_1^{\text{'}}+1012.5\end{array}\right]=0\\ 21{\left(p_1^{\text{'}}\right)}^2-322.5p_1^{\text{'}}+1212.5=0\end{array}$

Solve the above quadratic equation to find the value of $p_1^{\text{'}}$.

$\begin{array}{c}21{\left(p_1^{\text{'}}\right)}^2-322.5p_1^{\text{'}}+1212.5=0\\ \left(p_1^{\text{'}}-8.784\right)\left(p_1^{\text{'}}-6.573\right)=0\\ p_1^{\text{'}}=8.784\text{cm}\\ p_1^{\text{'}}=6.573\text{cm}\end{array}$

When the value of $p_1^{\text{'}}$ is, $8.784\text{cm}$.

The displacement of object is,

$\begin{array}{c}{p}_1^{\text{'}}-p_1=8.784\text{cm}-\text{7}\text{.50}\text{cm}\\ \text{=1}\text{.28}\text{cm}\end{array}$

Substitute $8.784\text{cm}$ for $p_1^{\text{'}}$ in equation (3).

$\begin{array}{c}{q}_2^{\text{'}}=2\left(3\times 8.784\text{cm}-10\right)\\ =32.7\text{cm}\end{array}$

The displacement of the image is,

$\begin{array}{c}{q}_2^{\text{'}}-q_2=32.7\text{cm-15}\text{.0}\text{cm}\\ \text{=17}\text{.7}\text{cm}\end{array}$

When the value of $p_1^{\text{'}}$ is, $6.573\text{cm}$.

The displacement of object is,

$\begin{array}{c}{p}_1^{\text{'}}-p_1=6.573\text{cm}-\text{7}\text{.50}\text{cm}\\ \text{=}-\text{0}\text{.927}\text{cm}\end{array}$

Substitute $6.573\text{cm}$ for $p_1^{\text{'}}$ in equation (3).

$\begin{array}{c}{q}_2^{\text{'}}=2\left(3\times 6.573\text{cm}-10\right)\\ =19.44\text{cm}\end{array}$

The displacement of the image is,

$\begin{array}{c}{q}_2^{\text{'}}-q_2=19.44\text{cm-15}\text{.0}\text{cm}\\ \text{=4}\text{.44}\text{cm}\end{array}$

Thus, the displacement of each lens from its initial position is $1.28\text{cm}$ and $17.7\text{cm}$ or $0.927\text{cm}$ and $4.44\text{cm}$.

Conclusion:

Therefore, the displacement of each lens from its initial position is $1.28\text{cm}$ and $17.7\text{cm}$ or $0.927\text{cm}$ and $4.44\text{cm}$.

(c)

To determine

Whether the lens can be displaced by more than one way.

Answer to Problem 66P

It is possible to displace the lens in more than one way.

Explanation of Solution

Given info: The focal length of the left and right lenses are $f_1=5.00\text{cm}$ and $f_2=10.0\text{cm}$. The object distance is $7.50\text{cm}$ to the left of the lens 1. The magnification of the image is $+1.00$.

Yes the lens can be displaced in more than one way.

The first lens can be displaced $1.28\text{cm}$ far from the object and the second lens can be moved by $17.7\text{cm}$ distance toward the object.

Another way is, the first lens can be moved $0.927\text{cm}$ toward the object and the second lens can be moved by $4.44\text{cm}$ distance toward the object.

Thus, it is possible to displace the lens in more than one way.

Conclusion:

Therefore, it is possible to displace the lens in more than one way.