Chapter 36, Problem 77AP

(a)

To determine

The location of the final image formed when the light has gone twice through the lens.

Answer to Problem 77AP

The location of the final image formed when the light has gone twice through the lens is $\underset{\_}{160\text{cm}}$ to the left of the lens.

Explanation of Solution

Write the expression for lens equation as the light first passes through the lens.

    $\frac{1}{q_1}=\frac{1}{f_1}-\frac{1}{p_1}$                                                                                                               (I)

Write the expression for the object distance of the mirror.

    $p_2=q_1-100\text{cm}$                                                                                                       (II)

Write the expression for lens equation as the light passes through the mirror.

    $\frac{1}{q_2}=\frac{1}{f_2}-\frac{1}{p_2}$                                                                                                           (III)

Write the expression for the object distance of the right side of lens.

    $p_3=100\text{cm}+q_2$                                                                                                      (IV)

Write the expression for lens equation as the light passes through the lens the second time.

    $\frac{1}{q_3}=\frac{1}{f_1}-\frac{1}{p_3}$                                                                                                            (V)

Conclusion:

Substitute $80.0\text{cm}$ for $f_1$ and $100\text{cm}$ for $p_1$ in equation (I).

    $\begin{array}{c}\frac{1}{q_1}=\frac{1}{80.0\text{cm}}-\frac{1}{100\text{cm}}\\ q_1=400\text{cm}\end{array}$

Substitute $400\text{cm}$ for $q_1$ in equation (II).

    $\begin{array}{c}{p}_2=-\left(400\text{cm}-100\text{cm}\right)\\ =-300\text{cm}\end{array}$

Substitute $-50.0\text{cm}$ for $f_2$ and $-300\text{cm}$ for $p_2$ in equation (III).

    $\begin{array}{c}\frac{1}{q_2}=\frac{1}{-50.0\text{cm}}-\frac{1}{-300\text{cm}}\\ q_2=-60.0\text{cm}\end{array}$

Substitute $-60.0\text{cm}$ for $q_2$ in equation (IV).

    $\begin{array}{c}{p}_3=100\text{cm}-\left(-60.0\text{cm}\right)\\ =160\text{cm}\end{array}$

Substitute $80.0\text{cm}$ for $f_1$ and $160\text{cm}$ for $p_3$ in equation (V).

    $\begin{array}{c}\frac{1}{q_3}=\frac{1}{80.0\text{cm}}-\frac{1}{160\text{cm}}\\ q_3=160\text{cm}\end{array}$

Therefore, the location of the final image formed when the light has gone twice through the lens is $\underset{\_}{160\text{cm}}$ to the left of the lens.

(b)

To determine

The overall magnification of the image.

Answer to Problem 77AP

The overall magnification of the image is $\underset{\_}{-0.800}$.

Explanation of Solution

Write the expression for magnification.

    $M=\frac{-q}{p}$                                                                                                                  (VI)

Write the expression for overall magnification.

    $M=M_1{M}_2{M}_3$                                                                                                       (VII)

Conclusion:

Substitute $400\text{cm}$ for $q_1$ and $100\text{cm}$ for $p_1$ in equation (VI) to find $M_1$.

    $\begin{array}{c}{M}_1=\frac{-\left(400\text{cm}\right)}{100\text{cm}}\\ =-4.00\end{array}$

Substitute $-60.0\text{cm}$ for $q_2$ and $-300\text{cm}$ for $p_2$ in equation (VI) to find $M_2$.

    $\begin{array}{c}{M}_2=\frac{-\left(-60.0\text{cm}\right)}{-300\text{cm}}\\ =-\frac{1}{5}\end{array}$

Substitute $160.0\text{cm}$ for $q_3$ and $160.0\text{cm}$ for $p_3$ in equation (VI) to find $M_3$.

    $\begin{array}{c}{M}_2=\frac{-\left(160.0\text{cm}\right)}{160.0\text{cm}}\\ =-1\end{array}$

Substitute $-4.00$ for $M_1$, $-\frac{1}{5}$ for $M_2$ and $-1$ for $M_3$ in equation (VII).

    $\begin{array}{c}M=\left(-4.00\right)\left(-\frac{1}{5}\right)\left(-1\right)\\ =-0.800\end{array}$

Therefore, the overall magnification of the image is $\underset{\_}{-0.800}$.

(c)

To determine

Whether the image is upright or inverted.

Answer to Problem 77AP

The image is inverted.

Explanation of Solution

The overall magnification is less than $0$. This proves that the image is inverted.

Therefore, the image is inverted.