Chapter 24, Problem 19P

(a)

To determine

The location at which a scree has to be placed to display an image.

Answer to Problem 19P

The location at which a screen has to be placed to display an image is $\underset{\_}{12.0\text{cm to the right of the converging lens}}$. Image formed by the diverging lens is virtual and it cannot be displayed on screen.

Explanation of Solution

Given that the focal length of the converging lens is $3.00\text{cm}$, the distance between the object and the converging lens is $4.00\text{cm}$, the focal length of the diverging lens is $-5.00\text{cm}$, and the distance between the converging lens and the diverging lens is $17.0\text{cm}$.

Write the thin lens formula for the first lens (converging lens)

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

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

Solve equation (I) for $q_1$.

$\begin{array}{c}\frac{1}{q_1}=\frac{1}{f_1}-\frac{1}{p_1}\\ q_1=\frac{f_1{p}_1}{p_1-f_1}\end{array}$ (II)

Write the thin lens formula for the second lens.

$\frac{1}{f_{\text{2}}}=\frac{1}{p_{\text{2}}}+\frac{1}{q_{\text{2}}}$ (III)

Here, $f_{\text{2}}$ is the focal length of the second lens, $p_{\text{2}}$ is the distance between the object and the second lens, and $q_{\text{2}}$ is the distance between the image and the second lens.

Solve equation (III) for $q_2$.

$\begin{array}{c}\frac{1}{q_{\text{2}}}=\frac{1}{f_{\text{2}}}-\frac{1}{p_{\text{2}}}\\ q_{\text{2}}=\frac{p_2{f}_2}{p_2-f_2}\end{array}$ (IV)

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

$p_2=s-q_1$ (V)

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

Use equation (II) in (IV).

$p_2=s-\frac{f_{\text{1}}{p}_{\text{1}}}{p_{\text{1}}-f_{\text{1}}}$ (VI)

Conclusion:

Substitute $17.0\text{cm}$ for $s$, $3.00\text{cm}$ for $q_1$, and $4.00\text{cm}$ for $p_1$ in equation (VI) to find $p_2$.

$\begin{array}{c}{p}_2=17.0\text{cm}-\frac{\left(3.00\text{cm}\right)\left(4.00\text{cm}\right)}{\left(4.00\text{cm}\right)-\left(3.00\text{cm}\right)}\\ =5.0\text{cm}\end{array}$

Substitute $5.0\text{cm}$ for $p_2$, and $-5.00\text{cm}$ for $f_2$ in equation (IV) to find $q_2$.

$\begin{array}{c}{q}_{\text{2}}=\frac{\left(5.0\text{cm}\right)\left(-5.00\text{cm}\right)}{5.0\text{cm}-\left(-5.00\text{cm}\right)}\\ =-2.5\text{cm}\end{array}$

This indicates that the image is virtual, so it cannot be displayed on the screen. On the other hand, a card can be placed in between the two lenses, at $12.0\text{cm}$ to the right of the converging lens to intercept the light and catch the real image formed by the first lens.

Therefore, the location at which a screen has to be placed to display an image is $\underset{\_}{12.0\text{cm to the right of the converging lens}}$. Image formed by the diuverging lens is virtual and it cannot be displayed on screen

(b)

To determine

The location at which a scree has to be placed to display an image when the separation of the lenses is $10.0\text{cm}$.

Answer to Problem 19P

The location at which a screen has to be placed to display an image is $\underset{\_}{3.3\text{cm to the right of the diverging lens}}$. The real image formed by the converging lens cannot be displayed on the same time since the second lens intercepts the light before this image forms.

Explanation of Solution

Given that the focal length of the converging lens is $3.00\text{cm}$, the distance between the object and the converging lens is $4.00\text{cm}$, the focal length of the diverging lens is $-5.00\text{cm}$, and the distance between the converging lens and the diverging lens is $10.0\text{cm}$.

Equation (VI) gives the distance between the diverging lens and its object.

$p_2=s-\frac{f_{\text{1}}{p}_{\text{1}}}{p_{\text{1}}-f_{\text{1}}}$

Equation (IV) gives the distance between the diverging lens and its image.

$q_{\text{2}}=\frac{p_2{f}_2}{p_2-f_2}$

Conclusion:

Substitute $10.0\text{cm}$ for $s$, $3.00\text{cm}$ for $f_1$, and $4.00\text{cm}$ for $p_1$ in equation (VI) to find $p_2$.

$\begin{array}{c}{p}_2=10.0\text{cm}-\frac{\left(3.00\text{cm}\right)\left(4.00\text{cm}\right)}{\left(4.00\text{cm}\right)-\left(3.00\text{cm}\right)}\\ =-2.0\text{cm}\end{array}$

Substitute $-2.0\text{cm}$ for $p_2$, and $-5.00\text{cm}$ for $f_2$ in equation (IV) to find $q_2$.

$\begin{array}{c}{q}_{\text{2}}=\frac{\left(-2.0\text{cm}\right)\left(-5.00\text{cm}\right)}{-2.0\text{cm}-\left(-5.00\text{cm}\right)}\\ =3.3\text{cm}\end{array}$

This indicates that the image is real, so it can be displayed on the screen. A screen can be placed at $3.3\text{cm}$ to the right of the diverging lens to display this image. The real image formed by the converging lens cannot be displayed on the same time since the second lens intercepts the light before this image forms.

Therefore, the location at which a screen has to be placed to display an image is $\underset{\_}{3.3\text{cm to the right of the diverging lens}}$. The real image formed by the converging lens cannot be displayed on the same time since the second lens intercepts the light before this image forms.