Chapter 17, Problem 5SP

(a)

To determine

The distance between the objective lens and the image formed by the objective lens.

Answer to Problem 5SP

The distance between the objective lens and the image formed by the objective lens is $4.0\text{cm}$.

Explanation of Solution

Given info: The focal length of the objective lens is $0.8\text{cm}$ and the distance between objective lens and the object is $1.0\text{cm}$.

Write the object-image equation of a lens.

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

Here,

$o$ is the distance between the object and lens

$i$ is the distance between the image and the lens

$f$ is the focal length of the lens

Solve for $i$.

$\begin{array}{c}\frac{1}{i}=\frac{1}{f}-\frac{1}{o}\\ i=\frac{fo}{o-f}\end{array}$

Substitute $0.8\text{cm}$ for $f$ and $1.0\text{cm}$ for $o$ to find the distance between the image and the lens.

$\begin{array}{c}i=\frac{\left(0.8\text{cm}\right)\left(1.0\text{cm}\right)}{1.0\text{cm}-0.8\text{cm}}\\ =4.0\text{cm}\end{array}$

Conclusion:

Therefore, the distance between the objective lens and the image formed by the objective lens is $4.0\text{cm}$.

(b)

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 distance between the lens and the object is $1.0\text{cm}$ and the distance between the image and the lens is $4.0\text{cm}$.

Write the expression for magnification of a lens.

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

Here,

$m$ is the magnification

Substitute $1.0\text{cm}$ for $o$ and $4.0\text{cm}$ for $i$ to find the magnification $m$.

$\begin{array}{c}m=-\left(\frac{4.0\text{cm}}{1.0\text{cm}}\right)\\ =-4\end{array}$

The negative sign of the magnification indicates the image is inverted and hence the magnification of the image is $-4\times$.

Conclusion:

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

(c)

To determine

The distance between the eyepiece lens and the image formed by the eyepiece lens.

Answer to Problem 5SP

The distance between the eyepiece lens and the image formed by the eyepiece lens is $-10\text{cm}$.

Explanation of Solution

Given info: The focal length of the eyepiece lens is $2.5\text{cm}$ and the distance between eyepiece lens and the image formed by the objective lens (which is the object of the eyepiece) is $2.0\text{cm}$.

Write the object-image equation of a lens.

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

Solve for $i$.

$\begin{array}{c}\frac{1}{i}=\frac{1}{f}-\frac{1}{o}\\ i=\frac{fo}{o-f}\end{array}$

Substitute $2.5\text{cm}$ for $f$ and $2.0\text{cm}$ for $o$ to find the distance between the image and the lens.

$\begin{array}{c}i=\frac{\left(2.5\text{cm}\right)\left(2.0\text{cm}\right)}{2.0\text{cm}-2.5\text{cm}}\\ =-10\text{cm}\end{array}$

The negative sign shows that the image is virtual.

Conclusion:

Therefore, the distance between the objective lens and the image formed by the objective lens is $-10\text{cm}$.

(d)

To determine

The magnification of the image.

Answer to Problem 5SP

The magnification of the image is $5\times$.

Explanation of Solution

Given info: The distance between the lens and the object is $2.0\text{cm}$ and the distance between the image and the lens is $-10\text{cm}$.

Write the expression for magnification of a lens.

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

Substitute $2.0\text{cm}$ for $o$ and $-10\text{cm}$ for $i$ to find the magnification $m$.

$\begin{array}{c}m=-\left(\frac{-10\text{cm}}{2.0\text{cm}}\right)\\ =5\end{array}$

The negative sign of the magnification indicates the image is inverted and hence the magnification of the image is $5\times$.

Conclusion:

Therefore, the magnification of the image is $5\times$.

(e)

To determine

The total magnification of the two-lens system.

Answer to Problem 5SP

The total magnification of the two-lens system is $-20\times$.

Explanation of Solution

Given info: The magnification of the objective lens is $-4\times$ and that of the eyepiece lens is $5\times$.

The total magnification of a combination of two lenses is the product of their individual magnification.

Write the expression for the total magnification of the combined lens system of objective lens and eyepiece lens.

$M=m_o\times m_e$

Here,

$M$ is the total magnification

$m_o$ is the magnification of the objective lens

$m_e$ is the magnification of the eyepiece lens

Substitute $-4$ for $m_o$ and $5$ for $m_e$ to find the total magnification $M$.

$\begin{array}{c}M=-4\times 5\\ =-20\end{array}$

Hence, the total magnification produced by the two-lens system is $-20\times$.

Conclusion:

Therefore, The total magnification of the two-lens system is $-20\times$.