Chapter 17, Problem 3SP

(a)

To determine

The distance of the image.

Answer to Problem 3SP

The distance of the image is $24\text{cm}$.

Explanation of Solution

Given Info: An object of height $3.6\text{cm}$ lies in front of a lens with focal length $8\text{cm}$ at a distance of $12\text{cm}$.

Write the expression for the relation between the image distance and the focal length.

$\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$

Here,

$o$ is the object distance

$i$ is the image distance

$f$ is the focal length

Substitute $12\text{cm}$ for $o$ and $8\text{cm}$ for $f$ and re-write in terms of $i$.

$\begin{array}{c}\frac{1}{12\text{cm}}+\frac{1}{i}=\frac{1}{8\text{cm}}\\ \frac{1}{i}=\frac{1}{8\text{cm}}-\frac{1}{12\text{cm}}\\ =0.042{\text{cm}}^{-1}\\ i=24\text{cm}\end{array}$

Conclusion:

Therefore, the distance of the image is $24\text{cm}$.

(b)

To determine

The magnification of the image.

Answer to Problem 3SP

The magnification of the image is $-2\times$.

Explanation of Solution

Given Info: An object of height $3.6\text{cm}$ lies in front of a lens with focal length $8\text{cm}$ at a distance of $12\text{cm}$.

Write the expression for the magnification of the image.

$m=-\frac{i}{o}$

Here,

$m$ is the magnification

Substitute $24\text{cm}$ for $i$ and $12\text{cm}$ for $o$ to get $m$.

$\begin{array}{c}m=-\frac{24\text{cm}}{12\text{cm}}\\ =-2\times \end{array}$

Conclusion:

Therefore, the magnification of the image is $-2\times$.

(c)

To determine

To draw the image formed by tracing three rays from the top of the object.

Answer to Problem 3SP

The image formed can be drawn as,

Explanation of Solution

The object is at $12\text{cm}$ from the lens with focal length $8\text{cm}$. The image is formed $24\text{cm}$ from the lens which is beyond the focal length and on the opposite side of the lens. The image formed can be drawn as,

Figure 1

Conclusion:

Therefore, the formation of the image is drawn and the result is confirmed.

(d)

To determine

The distance of the image and its magnification.

Answer to Problem 3SP

The distance of the image is $24\text{cm}$ and its magnification is $-2\times$.

Explanation of Solution

Given Info: An object of height $3.6\text{cm}$ lies in front of a lens with focal length $8\text{cm}$ at a distance of $12\text{cm}$. A second lens of focal length $+6\text{cm}$ is placed $9\text{cm}$ beyond the image of the object.

Write the expression for the relation between the image distance and the focal length.

$\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$

Substitute $9\text{cm}$ for $o$ and $+6\text{cm}$ for $f$ and re-write in terms of $i$.

$\begin{array}{c}\frac{1}{9\text{cm}}+\frac{1}{i}=\frac{1}{+6\text{cm}}\\ \frac{1}{i}=\frac{1}{6\text{cm}}-\frac{1}{9\text{cm}}\\ =0.055{\text{cm}}^{-1}\\ i=18\text{cm}\end{array}$

Write the expression for the magnification of the image.

$m=-\frac{i}{o}$

Substitute $18\text{cm}$ for $i$ and $9\text{cm}$ for $o$ to get $m$.

$\begin{array}{c}m=-\frac{18\text{cm}}{9\text{cm}}\\ =-2\times \end{array}$

Conclusion:

Therefore, the distance of the image is $24\text{cm}$ and its magnification is $-2\times$.

(e)

To determine

The overall magnification produced.

Answer to Problem 3SP

The overall magnification produced is $4\times$.

Explanation of Solution

Given Info: An object of height $3.6\text{cm}$ lies in front of a lens with focal length $8\text{cm}$ at a distance of $12\text{cm}$. A second lens of focal length $+6\text{cm}$ is placed $9\text{cm}$ beyond the image of the object.

Write the expression for the overall magnification.

$M=m_1{m}_2$

Here,

$M$ is the overall magnification

$m_1$ is the magnification due to the first lens

$m_2$ is the magnification of the second lens

Substitute $-2\times$ for $m_1$ and $-2\times$ for $m_2$ to get $M$.

$\begin{array}{c}M=\left(-2\right)\left(-2\right)\times \\ =4\times \end{array}$

Conclusion:

Therefore, the overall magnification produced is $4\times$.