Chapter 38, Problem 79PQ

(a)

To determine

The location and magnification of the final image formed by the two lenses.

Answer to Problem 79PQ

The image is formed at a distance of $17.6\text{cm}$ to the right of the diverging lens. The final magnification produced by two lenses is $-5.29$.

Explanation of Solution

Write the expression for focal length of a thin lens.

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

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

Write the expression for image distance of 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 image distance of the first lens, $f_1$ is the focal length of the first lens and $d_{0_1}$ is the object distance of the first lens.

Write the expression for magnification by first lens.

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

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

Write the expression for object distance from the second lens.

    $d_{0_2}=d-d_{{\text{i}}_1}$                                                                                              (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 for image distance for the second lens.

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

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

Write the expression for magnification by 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 final magnification.

    $M=M_1\times M_2$                                                                                             (VI)

Conclusion:

Substitute $22\text{cm}$ for $d_{0_1}$ and $15\text{cm}$ for $f_1$ in equation (I) to find $d_{{\text{i}}_1}$.

    $\begin{array}{c}{d}_{{\text{i}}_1}=\frac{\left(15\text{cm}\right)\left(\text{22}\text{cm}\right)}{22\text{cm}-15\text{cm}}\\ =47.14\text{cm}\end{array}$

Substitute $22\text{cm}$ for $d_{0_1}$ and $47.14\text{cm}$ for $d_{{\text{i}}_1}$ in equation (II) to find $M_1$.

    $\begin{array}{c}{M}_1=-\frac{47.14\text{cm}}{22\text{cm}}\\ =-2.14\end{array}$

Substitute $40\text{cm}$ for $d$ and $47.14\text{cm}$ for $d_{{\text{i}}_1}$ in equation (III) to find $d_{0_2}$.

    $\begin{array}{c}{d}_{0_2}=40\text{cm}-47.14\text{cm}\\ =-7.14\text{cm}\end{array}$

Substitute $-7.14\text{cm}$ for $d_{0_2}$ and $-12\text{cm}$ for $f_2$ in equation (IV) to find $d_{{\text{i}}_2}$    $\begin{array}{c}{d}_{{\text{i}}_2}=\frac{\left(-12\text{cm}\right)\left(-7.14\text{cm}\right)}{-7.14\text{cm}-\left(-12\text{cm}\right)}\\ =17.6\text{cm}\end{array}$

Substitute $-7.14\text{cm}$ for $d_{0_2}$ and $17.63\text{cm}$ for $d_{{\text{i}}_2}$ in equation (V) to find $M_2$.

    $\begin{array}{c}{M}_2=-\frac{17.63\text{cm}}{-7.14\text{cm}}\\ =2.47\end{array}$

Substitute $-2.14$ for $M_1$ and $2.47$ for $M_2$ in equation (VI) to find $M$.

    $\begin{array}{c}M=\left(2.47\right)\left(-2.14\right)\\ =-5.29\end{array}$

Therefore, the image is formed at a distance of $17.6\text{cm}$ to the right of the diverging lens. The final magnification produced by two lenses is $-5.29$.

(b)

To determine

The final image formed is upright or inverted.

Answer to Problem 79PQ

The final image formed will be inverted.

Explanation of Solution

The image formed will be real as the final image distance is positive and the image formed will be inverted as the final magnification produced by two lenses is negative. Therefore, the final image formed will be real and inverted.

Conclusion:

Therefore, the final image formed will be inverted.