Chapter 24, Problem 5P

To determine

The locations and sizes of the images formed by the two lenses.

Answer to Problem 5P

The location of the image formed by the first lens is $\underset{\_}{12.0\text{cm}}$ and that by second lens is $\underset{\_}{-4.0\text{cm}}$, the height of the first image is $-4.00\text{mm}$, and the height of the second image is $\underset{\_}{4.0\text{mm}}$.

Explanation of Solution

Given that $f_1$ is $4.00\text{cm}$, $f_2$ is $-2.00\text{cm}$, the distance between the lenses is $8.00\text{cm}$, the object distance from the first lens is $+6.00\text{cm}$,and the height of the object is $2.00\text{mm}$.

Write the thin lens formula.

  $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$                                                                                                        (I)

Here, $p$ is the object distance from the lens, $q$ is the image distance from the lens, $f$ is the focal length of the lens.

Solve equation (I) for $q$.

  $q=\frac{1}{\left(\frac{1}{f}-\frac{1}{p}\right)}$                                                                                                      (II)

Rewrite equation (II) for the image distance of the first lens and the image distance of the second lens.

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

  $q_2=\frac{1}{\left(\frac{1}{f_2}-\frac{1}{p_2}\right)}$                                                                                                (IV)

Write the expression for the object distance for the second lens.

  $p_2=s-q_1$                                                                                                         (V)

Write the expression for the magnification of the image in terms of object distance and image distance.

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

Here, $m$ is the magnification.

Write the expression for magnification of image in terms of height of object and image.

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

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

Equate the right hand sides of equations (VI) and (VII) and solve for $h^{\prime }$.

  $\begin{array}{c}-\frac{q}{p}=\frac{h^{\prime }}{h}\\ h^{\prime }=-\frac{qh}{p}\end{array}$                                                                                                     (VIII)

Conclusion:

Substitute $4.00\text{cm}$ for $f_1$, $6.00\text{cm}$ for $p_1$ in equation (III) to find $q_1$.

    $\begin{array}{c}{q}_1=\frac{1}{\left(\frac{1}{4.00\text{cm}}-\frac{1}{6.00\text{cm}}\right)}\\ =12.0\text{cm}\end{array}$

Substitute $12.0\text{cm}$ for $q_1$, and $8.00\text{cm}$ for $s$ in equation (V) to find $p_2$.

    $\begin{array}{c}{p}_2=8.00\text{cm}-12.0\text{cm}\\ \text{=}-4.0\text{cm}\end{array}$

Substitute $-4.0\text{cm}$ for $p_2$, and $-2.00\text{cm}$ for $f_2$ in equation (IV) to find $q_2$.

  $\begin{array}{c}{q}_2=\frac{1}{\left(\frac{1}{-2.00\text{cm}}-\frac{1}{-4.0\text{cm}}\right)}\\ =-4.0\text{cm}\end{array}$

Substitute $12.0\text{cm}$ for $q$ and $2.00\text{mm}$ for $h$ and $6.00\text{cm}$ for $p$ in equation (VIII) to find the height of the image formed by the first lens ($h_1{}^{\prime }$).

    $\begin{array}{c}{h}_1{}^{\prime }=-\frac{\left(12.0\text{cm}\right)\left(\text{2}\text{.00}\text{mm}\right)}{6.00\text{cm}}\\ =-\frac{\left(12.0\text{cm}\times \frac{10\text{mm}}{1\text{m}}\right)\left(\text{2}\text{.00}\text{mm}\right)}{6.00\text{cm}\times \frac{10\text{mm}}{1\text{m}}}\\ =-4.00\text{mm}\end{array}$

Since the image of the first lens is the object for the second lens, the height of the first lens’s image has to be taken as the height of the object of the second lens.

Substitute $-4.0\text{cm}$ for $q$ and $-4.00\text{mm}$ for $h$ and $-4.00\text{cm}$ for $p$ in equation (VI) to find the height of the image formed by the first lens ($h_2{}^{\prime }$)

    $\begin{array}{c}{h}_2{}^{\prime }=-\frac{\left(-4.0\text{cm}\times \frac{10\text{mm}}{1\text{cm}}\right)\left(-4.\text{00}\text{mm}\right)}{-4.0\text{cm}\times \frac{10\text{mm}}{1\text{cm}}}\\ =4.0\text{mm}\end{array}$

Therefore, the location of the image formed by the first lens is $\underset{\_}{12.0\text{cm}}$ and that by second lens is $\underset{\_}{-4.0\text{cm}}$, the height of the first image is $-4.00\text{mm}$, and the height of the second image is $\underset{\_}{4.0\text{mm}}$.