Chapter 24, Problem 29E

(a)

To determine

Where the image is formed by the objective.

Answer to Problem 29E

The image formed by the objective is $0.133\text{ m}$ from the objective.

Explanation of Solution

The image formed by the objective is real and inside the focal length of the ocular.

Write the thin lens formula

    $\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f}$        (I)

Here, $s$ is the distance of the object, $s\text{'}$ is the distance of the image and $f$ is the focal length.

To find the image distance from ocular lens,

Substitute $0.03\text{ m}$ for $f_2$, $0.25$ for $s_2^{\text{'}}$ in equation (I) to find the value of $s_2$

$\begin{array}{c}\frac{1}{s_2}+\frac{1}{\left(-s_2^{\text{'}}\right)}=\frac{1}{f_2}\\ \frac{1}{s_2}=\frac{1}{f_2}+\frac{1}{s_2^{\text{'}}}\\ =\frac{1}{0.03\text{ m}}+\frac{1}{0.25\text{ m}}\\ =\left(33.33+4\right){\text{ m}}^{-1}\end{array}$

Solving further,

$s_2=0.027\text{ m}$

From Figure 24.19, the distance between the lenses,

$d=s_1^{\text{'}}+s_2$

Here, $s_1^{\text{'}}$ is the image distance from the objective.

$s_1^{\text{'}}=d-s_2$

Substitute $0.027\text{ m}$ for $s_2$ and $0.16\text{ m}$ for $d$ in the above equation to find the value of $s_1^{\text{'}}$

$\begin{array}{c}{s}_1^{\text{'}}=\left(0.16\text{ m}\right)-\left(0.027\text{ m}\right)\\ =0.133\text{ m}\end{array}$

Conclusion:

The image formed by the objective is $0.133\text{ m}$ from the objective.

(b)

To determine

The distance between the specimen and the objective lens.

Answer to Problem 29E

The distance between the specimen and the objective lens is $0.00412\text{ m}$.

Explanation of Solution

To solve for $s_1$:

Substitute $0.133\text{ m}$ for $s_1^{\text{'}}$ and $0.004\text{ m}$ for $f_1$ in equation (I) to find the value of $s_1$

$\begin{array}{c}\frac{1}{s_1}+\frac{1}{s_1^{\text{'}}}=\frac{1}{f_1}\\ \frac{1}{s_1}=\frac{1}{f_1}+\frac{1}{s_1^{\text{'}}}\\ =\frac{1}{0.004\text{ m}}+\frac{1}{0.133\text{ m}}\\ =\left(250-7.518\right){\text{ m}}^{-1}\end{array}$

Solving further,

$s_1=0.00412\text{ m}$

Conclusion:

Thus, the distance between the specimen and the objective lens is $0.00412\text{ m}$.

(c)

To determine

The magnification of the microscope.

Answer to Problem 29E

The magnification of the microscope is $-277.83$.

Explanation of Solution

Write the expression to find the magnification of the microscope

    $M=\frac{-s_1^{\text{'}}\times 0.25\text{ m}}{f_1{f}_2}$

Here, $M$ is the magnification, $f_1$ is the focal length of the objective and $f_2$ is the focal length of the ocular.

Substitute $0.133\text{ m}$ for $s_1^{\text{'}}$, $0.004\text{ m}$ for $f_1$ and $0.03\text{ m}$ for $f_2$ in the above equation to find the value of $M$

$\begin{array}{c}M=\frac{-\left(0.133\text{ m}\right)\times 0.25\text{ m}}{\left(0.004\text{ m}\right)\left(0.03\text{ m}\right)}\\ =-277.83\end{array}$

The negative sign indicates the image is inverted.

Conclusion:

Thus, the magnification of the microscope is $-277.83$.