Chapter 24, Problem 78P

(a)

To determine

The distances of the intermediate and final image relative to the corresponding lenses.

Answer to Problem 78P

The distances of the intermediate and final image relative to the corresponding lenses are $\underset{\_}{60.0\text{cm}}$ to the right of the first lens and $\underset{\_}{5.45\text{cm}}$ to the right of the second lens.

Explanation of Solution

Write the expression for the lens maker’s formula.

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

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

Rewrite the above equation to find the image distance of the first lens.

    $q_1=\frac{f_1{p}_1}{p_1-f_1}$                                                                                                              (I)

Write the expression to find the object distance of the second lens.

    $p_2=s-q_1$                                                                                                                (II)

Here, $s$ is the separation distance of the lenses.

Write the expression for the lens maker’s formula.

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

Here, $p_1$ is the object distance, $q_2$ is the image distance and $f_2$ is the focal length of the second lens.

Rewrite the above equation to find the image distance of the second lens.

    $q_2=\frac{f_2{p}_2}{p_2-f_2}$                                                                                                          (III)

Conclusion:

Substitute $15.0\text{cm}$ for $f_1$ and $20.0\text{cm}$ for $p_1$ in the equation (I).

    $\begin{array}{c}{q}_1=\frac{\left(15.0\text{cm}\right)\left(20.0\text{cm}\right)}{20.0\text{cm}-15.0\text{cm}}\\ =60.0\text{cm}\end{array}$

Substitute $50.0\text{cm}$ for $s$ and $60.0\text{cm}$ for $q_1$ in the equation (II).

    $\begin{array}{c}{p}_2=50.0\text{cm}-\left(60.0\text{cm}\right)\\ =-10.0\text{cm}\end{array}$

Substitute $12.0\text{cm}$ for $f_2$ and $-10.0\text{cm}$ for $p_2$ in the equation (III).

    $\begin{array}{c}{q}_2=\frac{\left(12.0\text{cm}\right)\left(-10.0\text{cm}\right)}{-10.0\text{cm}-\left(12.0\text{cm}\right)}\\ =5.45\text{cm}\end{array}$

Therefore, the distances of the intermediate and final image relative to the corresponding lenses are $\underset{\_}{60.0\text{cm}}$ to the right of the first lens and $\underset{\_}{5.45\text{cm}}$ to the right of the second lens.

(b)

To determine

The total magnification.

Answer to Problem 78P

The total magnification is $\underset{\_}{-1.64}$.

Explanation of Solution

Write the expression to find the magnification of one lens.

    $m=\frac{-q}{p}$

Here, $m$ is the magnification of the lens.

Write the expression for total magnification.

    $M=m_1\times m_2$

Here, $m_1$ is the magnification of the first lens and $m_2$ is the magnification of the magnification of the second lens.

Rewrite the above equation in terms of object and image distances.

    $m=\frac{-q_1}{p_1}\times \frac{-q_2}{p_2}$

Conclusion:

Substitute $60.0\text{cm}$ for $q_1$, $20.0\text{cm}$ for $q_2$, $5.455\text{cm}$ for $p_1$ and $-10.0\text{cm}$ for $p_2$ in the above equation.

    $\begin{array}{c}m=\frac{-60.0\text{cm}}{20.0\text{cm}}\times \frac{-\left(5.455\text{cm}\right)}{-10.0\text{cm}}\\ =-1.64\end{array}$

Therefore, the total magnification is $\underset{\_}{-1.64}$.

(c)

To determine

The height of the final image.

Answer to Problem 78P

The height of the final image is $\underset{\_}{4.91\text{cm}}$.

Explanation of Solution

Write the expression to find the height of the final image.

    $h^{\prime }=mh$

Here, $h$ is the height of the initial image.

Conclusion:

Substitute $1.636$ for $m$ and $3.00\text{cm}$ for $h$ in the above equation.

    $\begin{array}{c}{h}^{\prime }=\left(1.636\right)\left(3.00\text{cm}\right)\\ =4.91\text{cm}\end{array}$

Therefore, the height of the final image is $\underset{\_}{4.91\text{cm}}$.