Chapter 24, Problem 3P

(a)

To determine

The focal length of the second lens.

Answer to Problem 3P

The focal length of the second lens is $\underset{\_}{11.8\text{cm}}$.

Explanation of Solution

Given that the distance between the two converging lenses is $88.0\text{cm}$, the distance between the first lens and the object is $1.100\text{m}$, the focal length of the first lens is $25.0\text{cm}$, the distance between the second lens and the final image is $15.0\text{cm}$.

Write the thin lens formula for the first lens.

  $\frac{1}{f_{\text{1}}}=\frac{1}{p_{\text{1}}}+\frac{1}{q_{\text{1}}}$                                                                                                               (I)

Here, $f_{\text{1}}$ is the focal length of the first lens, $p_{\text{1}}$ is the distance between the object and the first lens, and $q_{\text{1}}$ is the distance between the image and the first lens.

Solve equation (I) for $q_{\text{1}}$.

  $\begin{array}{c}\frac{1}{q_{\text{1}}}=\frac{1}{f_{\text{1}}}-\frac{1}{p_{\text{1}}}\\ q_{\text{1}}=\frac{f_{\text{1}}{p}_{\text{1}}}{p_{\text{1}}-f_{\text{1}}}\end{array}$                                                                                                              (II)

Write the thin lens formula for the second lens.

  $\frac{1}{f_{\text{2}}}=\frac{1}{p_{\text{2}}}+\frac{1}{q_{\text{2}}}$                                                                                                           (III)

Here, $f_{\text{2}}$ is the focal length of the second lens, $p_{\text{2}}$ is the distance between the object and the second lens, and $q_{\text{2}}$ is the distance between the image and the second lens.

Solve equation (III) for $f_2$.

  $\begin{array}{c}\frac{1}{f_{\text{2}}}=\frac{p_{\text{2}}+q_{\text{2}}}{p_{\text{2}}{q}_{\text{2}}}\\ f_2=\frac{p_{\text{2}}{q}_{\text{2}}}{p_{\text{2}}+q_{\text{2}}}\end{array}$                                                                                                            (IV)

Write the expression for the object distance for the second lens.

  $p_2=s-q_1$                                                                                                                 (V)

Here, $s$ is the distance of separation between the two lenses.

Use equation (II) in (IV).

  $p_2=s-\frac{f_{\text{1}}{p}_{\text{1}}}{p_{\text{1}}-f_{\text{1}}}$                                                                                                       (VI)

Conclusion:

Substitute $88.0\text{cm}$ for $s$, $25.0\text{cm}$ for $f_1$, and $1.100\text{m}$ for $p_1$ in equation (VI) to find $p_2$.

  $\begin{array}{c}{p}_2=88.0\text{cm}-\frac{\left(25.0\text{cm}\right)\left(1.100\text{m}\right)}{1.100\text{m}-25.0\text{cm}}\\ =88.0\text{cm}-\frac{\left(25.0\text{cm}\right)\left(1.100\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)}{\left(1.100\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)-25.0\text{cm}}\\ =55.6\text{cm}\end{array}$

Substitute $55.6\text{cm}$ for $p_2$, and $15.0\text{cm}$ for $q_2$ in equation (IV) to find $f_2$.

  $\begin{array}{c}{f}_2=\frac{\left(55.6\text{cm}\right)\left(15.0\text{cm}\right)}{\left(55.6\text{cm}\right)+\left(15.0\text{cm}\right)}\\ =11.8\text{cm}\end{array}$

Therefore, the focal length of the second lens is $\underset{\_}{11.8\text{cm}}$.

(b)

To determine

The total magnification of the system of the two lenses.

Answer to Problem 3P

The total magnification of the system of the two lenses is $\underset{\_}{0.0793}$.

Explanation of Solution

Given that the distance between the first lens and the object is $1.100\text{m}$, the distance between the second lens and the final image is $15.0\text{cm}$. It is obtained that the distance between the second lens and the its object is $55.6\text{cm}$.

Write the expression for the magnification of the two lens.

  $m_1=-\frac{q_1}{p_1}$

  $m_2=-\frac{q_2}{p_2}$

Here, $m_1$ is the magnification of the first lens, and $m_2$ is the magnification of the second lens.

Write the expression for the magnification of the system of two lenses.

  $m=m_1{m}_2$                                                                                                               (VII)

Here, $m$ is the total magnification.

Use the expressions for $m_1$ and $m_2$ in equation (VII).

  $\begin{array}{c}m=\left(-\frac{q_1}{p_1}\right)\left(-\frac{q_2}{p_2}\right)\\ =\frac{q_1{q}_2}{p_1{p}_2}\end{array}$                                                                                               (VIII)

The distance between the first lens and its image is given by the equation (II).

  $q_{\text{1}}=\frac{f_{\text{1}}{p}_{\text{1}}}{p_{\text{1}}-f_{\text{1}}}$

Conclusion:

Substitute $25.0\text{cm}$ for $f_1$, and $1.100\text{m}$ for $p_1$ in equation (II) to find $q_1$.

  $\begin{array}{c}{q}_{\text{1}}=\frac{\left(25.0\text{cm}\right)\left(1.100\text{m}\right)}{\left(1.100\text{m}\right)-\left(25.0\text{cm}\right)}\\ =\frac{\left(25.0\text{cm}\right)\left(1.100\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)}{\left(1.100\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)-\left(25.0\text{cm}\right)}\\ =32.35\text{cm}\end{array}$

Substitute $1.100\text{m}$ for $p_1$, $32.35\text{cm}$ for $q_1$, $55.6\text{cm}$ for $p_2$, and $15.0\text{cm}$ for $q_2$ in equation (VIII) to find $m$.

  $\begin{array}{c}m=\frac{\left(32.35\text{cm}\right)\left(15.0\text{cm}\right)}{\left(1.100\text{m}\right)\left(55.6\text{cm}\right)}\\ =\frac{\left(32.35\text{cm}\right)\left(15.0\text{cm}\right)}{\left(1.100\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)\left(55.6\text{cm}\right)}\\ =0.0793\end{array}$

Therefore, the total magnification of the system of the two lenses is $\underset{\_}{0.0793}$.