Chapter 26, Problem 57P

(a)

To determine

Position of the image.

Answer to Problem 57P

The image is $\underset{\_}{160\text{cm}}$ to the left of the lens.

Explanation of Solution

Write the mirror equation for the first pass through the lens.

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

Here $p_1$ is the object distance, $q_1$ is the image distance and, $f_1$ is the focal length.

Rewrite (I) in terms of $q_1$.

  $q_1=\frac{p_1{f}_1}{p_1-f_1}$        (II)

The object of the mirror would be at distance,

  $p_2=q_1-1\text{m}$        (III)

Therefore, the image distance for this case would be

    $q_2=\frac{p_2{f}_2}{p_2-f_2}$        (IV)

The image formed in this mirror would be

    $p_3=1\text{m}+\left|q_2\right|$        (V)

Write the equation for image position after the second pass through the lens.

    $q_3=\frac{p_3{f}_3}{p_3-f_3}$        (VI)

Conclusion:

Substitute $80\text{cm}$ for $f_1$ and $100\text{cm}$ for $p_1$ in (II)

    $\begin{array}{c}{q}_1=\frac{\left(100\text{cm}\right)\left(80\text{cm}\right)}{\left(100\text{cm}\right)-\left(80\text{cm}\right)}\\ =400\text{cm}\\ =\text{4}\text{m}\end{array}$

Substitute $4\text{m}$ for $q_1$ in (III)

    $\begin{array}{c}{p}_2=4\text{m}-1\text{m}\\ =3\text{m}\\ =300\text{cm}\end{array}$

Substitute $\text{300}\text{cm}$ for $p_2$ and $-50\text{cm}$ for $f_2$ in (IV)

    $\begin{array}{c}{q}_2=\frac{\left(\text{300}\text{cm}\right)\left(-50\text{cm}\right)}{\left(\text{300}\text{cm}\right)-\left(-50\text{cm}\right)}\\ =-60\text{cm}\\ \text{=}-0.6\text{m}\end{array}$

Substitute $-0.6\text{m}$ for $q_2$ in (V)

    $\begin{array}{c}{p}_3=1\text{m}+\left|-0.6\text{m}\right|\\ =1.6\text{m}\\ \text{=160}\text{cm}\end{array}$

Substitute $160\text{cm}$ for $p_3$ and $80\text{cm}$ for $f_3$ in (VI)

    $\begin{array}{c}{q}_2=\frac{\left(160\text{cm}\right)\left(80\text{cm}\right)}{\left(160\text{cm}\right)-\left(80\text{cm}\right)}\\ =160\text{cm}\end{array}$

The image is $\underset{\_}{160\text{cm}}$ to the left of the lens.

(b)

To determine

Overall magnification of the image.

Answer to Problem 57P

Overall magnification is $\underset{\_}{-0.800}$.

Explanation of Solution

Write the equation for magnification of three cases

    $\begin{array}{c}{M}_1=\frac{-q_1}{p_1}\\ M_2=\frac{-q_2}{p_2}\\ M_3=\frac{-q_3}{p_3}\end{array}$        (VII)

Write the equation for overall magnification

    $M=M_1{M}_2{M}_3$        (VIII)

Substitute (VII) in (VIII)

    $M=\frac{-q_1{q}_2{q}_3}{p_1{p}_2{p}_3}$        (IX)

Conclusion:

Substitute $400\text{cm}$ for $q_1$, $-60\text{cm}$ for $q_2$, $160\text{cm}$ for $q_3$, $100\text{cm}$ for $p_1$, $-300\text{cm}$ for $p_2$ and $160\text{cm}$ for $p_3$.

    $\begin{array}{c}M=\frac{-\left(400\text{cm}\right)\left(-60\text{cm}\right)\left(160\text{cm}\right)}{\left(100\text{cm}\right)\left(-300\text{cm}\right)\left(160\text{cm}\right)}\\ =-0.800\end{array}$

Overall magnification is $\underset{\_}{-0.800}$.

(c)

To determine

Whether the final image is upright or inverted.

Answer to Problem 57P

The final image is inverted.

Explanation of Solution

The sign of magnification determines whether the image is upright or inverted.

If magnification is positive, the image is upright. The negative sign of magnification indicates that the image is inverted.

Conclusion:

The final image is inverted.