Chapter 17, Problem 5SP

(a)

To determine

The location of the image formed by the objective.

Answer to Problem 5SP

The location of the image formed by the objective is at $3.6\text{cm}$.

Explanation of Solution

Given Info: The objective lens of a microscope has a focal length of $0.9\text{cm}$ and an eyepiece of focal length $2.4\text{cm}$. The object is located $1.2\text{cm}$ in front of the objective lens.

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 $1.2\text{cm}$ for $o$ and $0.9\text{cm}$ for $f$ and re-write in terms of $i$.

$\begin{array}{c}\frac{1}{1.2\text{cm}}+\frac{1}{i}=\frac{1}{0.9\text{cm}}\\ \frac{1}{i}=\frac{1}{0.9\text{cm}}-\frac{1}{1.2\text{cm}}\\ =0.278{\text{cm}}^{-1}\\ i=3.6\text{cm}\end{array}$

Conclusion:

Therefore, the location of the image formed by the objective is at $3.6\text{cm}$.

(b)

To determine

The magnification of the image.

Answer to Problem 5SP

The magnification of the image is $-3\times$.

Explanation of Solution

Given Info: The objective lens of a microscope has a focal length of $0.9\text{cm}$ and an eyepiece of focal length $2.4\text{cm}$. The object is located $1.2\text{cm}$ in front of the objective lens.

Write the expression for the magnification of the image.

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

Here,

$m$ is the magnification

Substitute $3.6\text{cm}$ for $i$ and $1.2\text{cm}$ for $o$ to get $m$.

$\begin{array}{c}m=-\frac{3.6\text{cm}}{1.2\text{cm}}\\ =-3\times \end{array}$

This means that the image is three time larger but is inverted.

Conclusion:

Therefore, the magnification of the image is $-3\times$.

(c)

To determine

The location of the image formed by the eyepiece.

Answer to Problem 5SP

The location of the image formed by the eyepiece is at $-7.2\text{cm}$.

Explanation of Solution

Given Info: The objective lens of a microscope has a focal length of $0.9\text{cm}$ and an eyepiece of focal length $2.4\text{cm}$. The object is located $1.2\text{cm}$ in front of the objective lens. The eyepiece is $1.8\text{cm}$ beyond the image formed 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 $1.8\text{cm}$ for $o$ and $2.4\text{cm}$ for $f$ and re-write in terms of $i$.

$\begin{array}{c}\frac{1}{1.8\text{cm}}+\frac{1}{i}=\frac{1}{2.4\text{cm}}\\ \frac{1}{i}=\frac{1}{2.4\text{cm}}-\frac{1}{1.8\text{cm}}\\ =-0.138{\text{cm}}^{-1}\\ i=-7.2\text{cm}\end{array}$

Conclusion:

Therefore, the location of the image formed by the eyepiece is at $-7.2\text{cm}$.

(d)

To determine

The magnification of the image.

Answer to Problem 5SP

The magnification of the image is $4\times$.

Explanation of Solution

Given Info: The objective lens of a microscope has a focal length of $0.9\text{cm}$ and an eyepiece of focal length $2.4\text{cm}$. The object is located $1.2\text{cm}$ in front of the objective lens. The eyepiece is $1.8\text{cm}$ beyond the image formed of the object.

Write the expression for the magnification of the image.

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

Substitute $-7.2\text{cm}$ for $i$ and $1.8\text{cm}$ for $o$ to get $m$.

$\begin{array}{c}m=-\frac{-7.2\text{cm}}{1.8\text{cm}}\\ =4\times \end{array}$

This means that the image is four times larger.

Conclusion:

Therefore, the magnification of the image is $4\times$.

(e)

To determine

The overall magnification produced.

Answer to Problem 5SP

The overall magnification produced is $-12\times$.

Explanation of Solution

Given Info: The objective lens of a microscope has a focal length of $0.9\text{cm}$ and an eyepiece of focal length $2.4\text{cm}$. The object is located $1.2\text{cm}$ in front of the objective lens. The eyepiece is $1.8\text{cm}$ beyond the image formed 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 objective

$m_2$ is the magnification of the eyepiece

Substitute $-3\times$ for $m_1$ and $4\times$ for $m_2$ to get $M$.

$\begin{array}{c}M=\left(-3\right)\left(4\right)\times \\ =-12\times \end{array}$

Conclusion:

Therefore, the overall magnification produced is $-12\times$.