Chapter 38, Problem 114PQ

(a)

To determine

The magnification due to each lens and the magnification of the final image.

Answer to Problem 114PQ

The magnification of the first lens is $3$, the magnification of the second lens is $0.21$ and the total magnification of the final image is $0.63$.

Explanation of Solution

Write the expression for thin lens equation.

    $\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_{\text{i}}}$

Here, $d_{\text{i}}$ is the distance of the image from the lens, $f$ is the focal length of the lens and $d_0$ is the distance of the object.

Rearrange the above equation for $d_{\text{i}}$.

    $\begin{array}{c}\frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_0}\\ =\frac{d_0-f}{f{d}_0}\\ d_{\text{i}}=\frac{f{d}_0}{d_0-f}\end{array}$

Write the expression to calculate the image distance for the first lens.

    $d_{{\text{i}}_1}=\frac{f_1{d}_{0_1}}{d_{0_1}-f_1}$                                                                                             (I)

Here, $d_{{\text{i}}_1}$ is the distance of the image from the first lens, $f_1$ is the focal length of the first lens and $d_{0_1}$ is the distance of the object from the first lens.

Write the expression for the magnification produced by the first lens.

    $M_1=\frac{d_{{\text{i}}_1}}{d_{0_1}}$                                                                                                 (II)

Here, $M_1$ is the magnification produced by the first lens.

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

    $d_{0_2}=d_{{\text{i}}_1}+d$                                                                                             (III)

Here, $d_{0_2}$ is the object distance from the second lens and $d$ is the distance between the two lenses.

Write the expression to calculate the image distance for the second lens.

    $d_{{\text{i}}_2}=\frac{f_2{d}_{0_2}}{d_{0_2}-f_2}$                                                                                          (IV)

Here, $d_{{\text{i}}_2}$ is the distance of the image from the second lens, $f_2$ is the focal length of the second lens and $d_{0_2}$ is the distance of the object from the second lens.

Write the expression for the magnification produced by the second lens.

    $M_2=\frac{-d_{{\text{i}}_2}}{d_{0_2}}$                                                                                               (V)

Here, $M_2$ is the magnification produced by the second lens.

Write the expression for the total magnification.

    $M=M_1{M}_2$                                                                                             (VI)

Here, $M$ is the total magnification.

Conclusion:

Substitute $30.0\text{cm}$ for $f_1$ and $20.0\text{cm}$ for $d_{0_1}$ in equation (I) to find $d_{{\text{i}}_1}$.

    $\begin{array}{c}{d}_{{\text{i}}_1}=\frac{\left(30.0\text{cm}\right)\left(20.0\text{cm}\right)}{20.0\text{cm}-30.0\text{cm}}\\ =-60.0\text{cm}\end{array}$

Substitute $20.0\text{cm}$ for $d_{0_1}$ and $-60.0\text{cm}$ for $d_{{\text{i}}_1}$ in equation (II) to find $M_1$.

    $\begin{array}{c}{M}_1=\frac{-\left(-60.0\text{cm}\right)}{20.0\text{cm}}\\ =3\end{array}$

Substitute $60.0\text{cm}$ for $d_{{\text{i}}_1}$ and $15.0\text{cm}$ for $d$ in equation (III) to find $d_{0_2}$.

    $\begin{array}{c}{d}_{0_2}=60.0\text{cm}+15.0\text{cm}\\ =\text{75}\text{.0}\text{cm}\end{array}$

Substitute $-20.0\text{cm}$ for $f_2$ and $75.0\text{cm}$ for $d_{0_2}$ in equation (IV) to find $d_{{\text{i}}_2}$.

    $\begin{array}{c}{d}_{{\text{i}}_2}=\frac{\left(-20.0\text{cm}\right)\left(75.0\text{cm}\right)}{75.0\text{cm}-\left(-20.0\text{cm}\right)}\\ =-15.8\text{cm}\end{array}$

Substitute $75.0\text{cm}$ for $d_{0_2}$ and $-15.8\text{cm}$ for $d_{{\text{i}}_2}$ in equation (V) to find $M_2$.

    $\begin{array}{c}{M}_2=\frac{-\left(-15.8\text{cm}\right)}{75.0\text{cm}}\\ =0.21\end{array}$

Substitute $3$ for $M_1$ and $0.21$ for $M_2$ in equation (VI) to find $M$.

    $\begin{array}{c}M=3\times 0.21\\ =0.63\end{array}$

Therefore, the magnification of the first lens is $3$, the magnification of the second lens is $0.21$ and the total magnification of the final image is $0.63$.

(b)

To determine

The final height of the image.

Answer to Problem 114PQ

The final height of image formed is $10.71\text{cm}$.

Explanation of Solution

Write the expression for the magnification produced by the first lens.

    $M_1=\frac{h_{{\text{i}}_1}}{h_{0_1}}$

Here, $h_{{\text{i}}_1}$ is the height of the image produced by first lens and $h_{0_1}$ is the height of the object in case of first lens.

Rearrange the above equation for $h_{{\text{i}}_1}$.

    $h_{{\text{i}}_1}=M_1{h}_{0_1}$                                                                                               (VII)

Write the expression for the height of the image produced by second lens.

    $h_{{\text{i}}_2}=M_2{h}_{0_2}$                                                                                            (VIII)

Here, $h_{{\text{i}}_2}$ is the height of the image produced by second lens and $h_{0_2}$ is the height of the object in case of second lens.

Conclusion:

Substitute $17.0\text{cm}$ for $h_{0_1}$ and $3$ for $M_1$ in equation (VII) to find $h_{{\text{i}}_1}$.

    $\begin{array}{c}{h}_{{\text{i}}_1}=\left(3\right)\left(17.0\text{cm}\right)\\ =51\text{cm}\end{array}$

Substitute $51\text{cm}$ for $h_{0_2}$ and $0.21$ for $M_2$ in equation (VIII) to find $h_{{\text{i}}_2}$.

    $\begin{array}{c}{h}_{{\text{i}}_2}=\left(0.21\right)\left(51\text{cm}\right)\\ =10.71\text{cm}\end{array}$

Therefore, the final height of image formed is $10.71\text{cm}$.