Chapter 24, Problem 81P

(a)

To determine

The position of the final image.

Answer to Problem 81P

The position of the final image is $5.3\text{ cm}$ to the left of the converging lens.

Explanation of Solution

Write lens equation for converging lens.

  $\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{f_1}$

Here, $p_1$ is the object distance of for converging lens, $q_1$ is the image distance for converging lens and $f_1$ is the focal length of converging lens.

Rearrange above equation to get $q_1$.

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

The image produced by the converging lens acts as object for the diverging lens.

Write the expression for diverging lens.

  $p_2=s-q_1$                                                                                                   (II)

Here, $p_2$ is the object distance for diverging lens and $s$ is the distance between converging and diverging lens.

Write lens equation for diverging lens to get $q_2$.

  $\begin{array}{c}\frac{1}{p_2}+\frac{1}{q_2}=\frac{1}{f_2}\\ q_2={\left(\frac{1}{f_2}-\frac{1}{p_2}\right)}^{-1}\end{array}$                                                                                           (III)

Here, $q_2$ is the image distance for diverging lens and $f_2$ is the focal length of diverging lens.

Conclusion:

Given that focal length of converging lens is $5.500\text{cm}$, object distance is $9.000\text{cm}$ and focal length of diverging lens is $-4.20\text{cm}$.

Substitute $5.500\text{cm}$ for $f_1$ and $9.000\text{cm}$ for $p_1$ in equation (I) to get $q_1$.

  $\begin{array}{c}{q}_1={\left(\frac{1}{5.500\text{cm}}-\frac{1}{9.000\text{cm}}\right)}^{-1}\\ =14.14\text{cm}\end{array}$

Substitute $8.00\text{cm}$ for $s$ and $14.14\text{cm}$ for $q_1$ in equation (II) to get $p_2$.

  $\begin{array}{c}{p}_2=8.00\text{cm}-14.14\text{cm}\\ \text{=}-6.14\text{cm}\end{array}$

The negative sign for $p_2$ indicates that the image is virtual.

Substitute $-4.20\text{cm}$ for $f_2$ and $-6.14\text{cm}$ for $p_2$ in equation (III) to get $q_2$.

  $\begin{array}{c}{q}_2={\left(\frac{1}{-4.20\text{cm}}-\frac{1}{-6.14\text{cm}}\right)}^{-1}\\ =-13.3\text{cm}\end{array}$

Since both the lenses are separated by $8.00\text{cm}$, final image is formed at $13.3\text{cm}-8.00\text{cm}=\text{5}\text{.3}\text{cm}$ to the left of the converging lens.

Therefore, the final image is formed at $5.3\text{ cm}$ to the left of the converging lens.

(b)

To determine

The height of the final image.

Answer to Problem 81P

The height of the final image is $3.4\text{cm}$.

Explanation of Solution

Write the expression for the magnification of image.

  $m=-\frac{q}{p}$

Here,$m$ is the magnification, $p$ is the object distance and $q$ is the image distance.

Analogues to above equation, write magnification of the image formed by converging and diverging lens.

  $m_1=-\frac{q_1}{p_1}$                                                                                                              (IV)

  $m_2=-\frac{q_2}{p_2}$                                                                                                               (V)

Here,$m_1$ is the magnification of the image formed by converging lens and $m_2$ is the magnification of the image formed by diverging lens.

Write the expression for the total magnification.

  $m=m_1\times m_2$

Substitute (IV) and (V) in above equation.

  $m=-\frac{q_1}{p_1}\left(-\frac{q_2}{p_2}\right)$                                                                                                     (VI)

Write magnification equation in terms of height of object and image.

  $m=\frac{h^{\prime }}{h}$                                                                                                                   (VII)

Here, $h^{\prime }$ is the height of image and $h$ is the height of object.

Equate Equations (VI) and (VII) to get $h^{\prime }$.

  $h^{\prime }=\frac{q_1}{p_1}\left(\frac{q_2}{p_2}\right)h$                                                                                                     (VIII)

Conclusion:

Substitute $14.14\text{cm}$ for $q_1$, $-13.3\text{cm}$ for $q_2$, $9.000\text{cm}$ for $p_1$, $1.0\text{cm}$ for $h$ and $-6.14\text{cm}$ for $p_2$ in (VIII) to get $h^{\prime }$

  $\begin{array}{c}{h}^{\prime }=\frac{14.14\text{cm}}{9.000\text{cm}}\left(\frac{-13.3\text{cm}}{-6.14\text{cm}}\right)\left(1.0\text{cm}\right)\\ =3.4\text{cm}\end{array}$

Therefore, the height of the final image is $3.4\text{cm}$.

(c)

To determine

Whether the image is upright or inverted.

Answer to Problem 81P

The image is upright, since transverse magnification is positive.

Explanation of Solution

According to sign convention heights above principle axis are positive whereas heights below principle axis are negative.

In the question, since object is real upright $h$ is positive. Since transverse magnification is positive, height of image should be positive. Therefore, image is upright.

Conclusion:

It is found that transverse magnification of the image is positive. This clearly indicates that the image is upright.

The image is upright, since transverse magnification is positive.