Chapter 17, Problem 3SP

(a)

To determine

The distance of the image.

Answer to Problem 3SP

The distance of the image is $15\text{cm}$.

Explanation of Solution

Given Info: The object distance is $10\text{ cm}$ and the focal length of the lens is $6\text{cm}$.

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

Rewrite the above equation for $i$.

$i=\frac{of}{o-f}$ (1)

Substitute $10\text{cm}$ for $o$ and $6\text{cm}$ for $f$ in equation (1) to find $i$.

$\begin{array}{c}i=\frac{\left(10\text{cm}\right)\left(6\text{cm}\right)}{10\text{cm}-6\text{cm}}\\ =\frac{60{\text{ cm}}^2}{4\text{ cm}}\\ =15\text{cm}\end{array}$

Conclusion:

Therefore, the distance of the image is $15\text{cm}$.

(b)

To determine

The magnification of the image.

Answer to Problem 3SP

The magnification of the image is $-1.5\times$.

Explanation of Solution

Given Info: The object distance is $10\text{ cm}$

Write the expression for the magnification of the image.

$m=-\frac{i}{o}$ (2)

Here,

$m$ is the magnification

Substitute $15\text{cm}$ for $i$ and $10\text{cm}$ for $o$ in equation (2) to get $m$.

$\begin{array}{c}m=-\frac{15\text{cm}}{10\text{cm}}\\ =-1.5\times \end{array}$

Conclusion:

Therefore, the magnification of the image is $-1.5\times$.

(c)

To determine

The image of tracing three rays from the top of the object.

Answer to Problem 3SP

The image formed can be drawn as,

Explanation of Solution

The ray from the object parallel to the principle axis will pass through the focus and that through the center will go un-deflected. The rays passing through the focus will go parallel to the principal axis. The diagram is shown in figure 1.

Figure 1

The object is at $10\text{cm}$ from the lens with focal length $6\text{cm}$. The image is formed $15\text{cm}$ from the lens which is beyond the focal length and on the opposite side of the lens. From the figure it is clear that the conclusions of part (a) and (b) are correct.

Conclusion:

Thus, the formation of the image is drawn and the results are confirmed.

(d)

To determine

The distance of the image and its magnification for the image formed by the second lens.

Answer to Problem 3SP

The distance of the image is $12\text{cm}$ and its magnification is $-2\times$.

Explanation of Solution

Given Info: The object distance is $\text{6 cm}$ and the focal length of the lens is $4\text{cm}$.

Substitute $6\text{cm}$ for $o$ and $4\text{cm}$ for $f$ in equation (1) to find $i$.

$\begin{array}{c}i=\frac{\left(6\text{cm}\right)\left(4\text{cm}\right)}{6\text{cm}-4\text{cm}}\\ =\frac{{\text{24 cm}}^2}{\text{2 cm}}\\ =12\text{cm}\end{array}$

Substitute $12\text{cm}$ for $i$ and $6\text{cm}$ for $o$ in equation (2) to get $m$.

$\begin{array}{c}m=-\frac{12\text{cm}}{6\text{cm}}\\ =-2\times \end{array}$

Conclusion:

Therefore, the distance of the image is $12\text{cm}$ and its magnification is $-2\times$.

(e)

To determine

The overall magnification produced.

Answer to Problem 3SP

The overall magnification produced is $3$.

Explanation of Solution

Write the expression for the overall magnification.

$M=m_1{m}_2$

Here,

$M$ is the overall magnification

$m_1$ is the magnification of the first image

$m_2$ is the magnification of the second image

Substitute $-1.5\times$ for $m_1$ and $-2\times$ for $m_2$ in the above equation to get $M$.

$\begin{array}{c}M=\left(-1.5\right)\left(-2\right)\\ =3\end{array}$

Conclusion:

Therefore, the overall magnification produced is $3$.