Chapter 26, Problem 70P

(a)

To determine

Position of the image.

Answer to Problem 70P

The image distance is $\underset{\_}{20\text{cm}}$.

Explanation of Solution

Write the mirror equation.

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

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

Rearrange (I) in terms of $q_1$.

    $\frac{1}{q_1}=\frac{1}{f}-\frac{1}{p_1}$        (II)

The real image is the object of second lens.

  $p_2=d-q_1$        (III)

Here $d$ is the distance between 2 lenses.

Write the mirror equation for lens 2.

    $\frac{1}{f_2}=\frac{1}{p_2}+\frac{1}{q_2}$        (IV)

Here $f_2$ is the focal length of the lens, $p_2$ is the object distance and $q_2$ is the image distance.

Rewrite (IV) in terms of $q_2$.

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

Conclusion:

Substitute $40\text{cm}$ for $p_1$ and $30\text{cm}$ for $f$ in (II)

    $\begin{array}{l}\frac{1}{q_1}=\frac{1}{30\text{cm}}-\frac{1}{40\text{cm}}\\ q_1=120\text{cm}\end{array}$

Substitute $110\text{cm}$ for $d$ and $120\text{cm}$ for $q_1$ in (III)

    $\begin{array}{c}{p}_2=110\text{cm}-120\text{cm}\\ \text{=}-\text{10}\text{cm}\end{array}$

Substitute $-10\text{cm}$ for $p_2$ and $-20\text{cm}$ for $f_2$ in (V)

    $\begin{array}{c}\frac{1}{q_2}=\frac{1}{-20\text{cm}}-\frac{1}{-10\text{cm}}\\ q_2=20\text{cm}\end{array}$

The image distance is $\underset{\_}{20\text{cm}}$.

(b)

To determine

Magnification of the final image.

Answer to Problem 70P

Magnification is $\underset{\_}{-6}$.

Explanation of Solution

Write the equation for magnification for two lenses.

    $M_1=\frac{-q_1}{p_1}$        (VI)

    $M_2=\frac{-q_2}{p_2}$        (VII)

Write the equation for overall magnification

    $M=M_1{M}_2$        (VIII)

Conclusion:

Substitute $120\text{cm}$ for $q_1$ and $40\text{cm}$ for $p_1$ in (VI)

    $\begin{array}{c}{M}_1=-\frac{120\text{cm}}{40\text{cm}}\\ =-3\end{array}$

Substitute $20\text{cm}$ for $q_2$ and $-10\text{cm}$ for $p_2$ in (VII)

    $\begin{array}{c}{M}_2=\frac{-20\text{cm}}{-10\text{cm}}\\ =2\end{array}$

Substitute $-3$ for $M_1$ and $2$ for $M_2$ in (VIII)

    $\begin{array}{c}M=\left(-3\right)\left(2\right)\\ =-6\end{array}$

Magnification is $\underset{\_}{-6}$.

(c)

To determine

Whether the image is upright or inverted.

Answer to Problem 70P

The image is inverted.

Explanation of Solution

Sign of magnification decides whether the image is upright or inverted.

If magnification is positive, image is upright. If magnification is negative the image s inverted

Conclusion:

As the magnification is $-6$ , the image is inverted.

(d)

To determine

Nature of the image for a given condition.

Answer to Problem 70P

The image would be inverted at a distance $\underset{\_}{6.67\text{cm}}$

Overall magnification is $\underset{\_}{-2}$.

Explanation of Solution

The ray diagram is as shown

Refer sub part (a) and (b) and to determine the image position and magnification

    $\frac{1}{f_2}=\frac{1}{p_2}+\frac{1}{q_2}$

    $M_2=\frac{-q_2}{p_2}$

Conclusion:

Substitute $-10\text{cm}$ for $p_2$ and $20\text{cm}$ for $f_2$ in (V)

    $\begin{array}{c}\frac{1}{q_2}=\frac{1}{20\text{cm}}-\frac{1}{-10\text{cm}}\\ q_2=6.67\text{cm}\end{array}$

Substitute $6.67\text{cm}$ for $q_2$ and $-10\text{cm}$ for $p_2$ in (VII)

    $\begin{array}{c}{M}_2=-\frac{6.67\text{cm}}{-10\text{cm}}\\ =0.667\end{array}$

Substitute $-3$ for $M_1$ and $0.667$ for $M_2$ in (VIII)

    $\begin{array}{c}M=\left(-3\right)\left(0.667\right)\\ =-2\end{array}$

As the overall magnification is negative, the image is inverted.

The image would be inverted at a distance $\underset{\_}{6.67\text{cm}}$

Overall magnification is $\underset{\_}{-2}$.