Chapter 25, Problem 44P

Repeat Problem 46 assuming that the final image is located 25 cm from the eyepiece (near point of a normal eye)

To determine

Part (a)To determine:

The total magnification of the given microscope.

Answer to Problem 44P

Solution:

The magnification of the given microscope is found to be 900 x.

Explanation of Solution

Microscopes are devices used to magnify tiny objects. The construction of the microscope is similar to that of the telescope. The objective produces a real and inverted image of the object, which falls between the focus and the optic centre of the eyepiece. The eyepiece produces an enlarged virtual image of the image formed by the objective. The final image formed is inverted and enlarged.

The total magnification of the microscope is the product of the magnification of the objective and the eyepiece lenses. When the image is located at the near point of the eye, the eyepiece, which behaves as a simple magnifying lens, has its magnification increased by 1, when compared to the magnification it produces for the relaxed eye.

Given:

The magnification of the eyepiece $m_e=14.0$

The magnification of the objective $m_o=60.0$

Barrel length of the microscope $l=20.0\text{ cm}$

Formula used:

$M=m_o\times \left(m_e+1\right)$

Calculation:

Use the given values of the magnification in the formula and simplify.

$\begin{array}{c}M=m_o\times \left(m_e+1\right)\\ =\left(60.0\right)\left(14.0+1\right)\\ =900\end{array}$

To determine

Part (b)To determine:

The focal length of the objective and the eyepiece lenses.

Answer to Problem 44P

Solution:

The focal length of the objective lens was found to be 0.299 cm and the focal length of the eyepiece was found to be 1.79 cm.

Explanation of Solution

Using the magnification of the eyepiece, the focal length of the eyepiece can be determined. The image of the object is formed at the near point of the eye.

Using the thin lens equation for the eyepiece, the object distance for the eyepiece $d_{oe}$ is determined.

$\frac{1}{f_e}=\frac{1}{d_{oe}}+\frac{1}{N}$

The image distance for the objective $d_{io}$ is the difference between the barrel length l and the object distance $d_{oe}$ for the eyepiece.

$d_{io}=l-d_{oe}$

Using the expression for magnification of the objective, the object distance is calculated.

$m_o=\frac{d_{io}}{d_o}$

With the object distance for the objective determined, the thin lens equation is used to calculate the focal length of the objective.

$\frac{1}{f_o}=\frac{1}{d_o}+\frac{1}{d_{io}}$

Given:

The magnification of the eyepiece $m_e=14.0$

The magnification of the objective $m_o=60.0$

Barrel length of the microscope $l=20.0\text{ cm}$

Formula used:

For a relaxed eye, the image is formed at infinity. If the values of $f_e$ and $f_o$ are very less compared to the value of l, then,

$m_e=\frac{N}{f_e}$

$\frac{1}{f_e}=\frac{1}{d_{oe}}+\frac{1}{N}$

$d_{io}=l-d_{oe}$

The magnification of the objective is given by

$m_o=\frac{d_{io}}{d_0}$

Where, the object distance is $d_0$ and the image distance is $d_i$

The thin lens equation is written as,

$\frac{1}{f_o}=\frac{1}{d_o}+\frac{1}{d_{io}}$

Calculation:

Use the given values of N and $m_e$, calculate the value of the focal length of the eyepiece $f_e$

$\begin{array}{c}{f}_e=\frac{N}{m_e}\\ =\frac{25\text{ cm}}{14.0}\\ =1.786\text{ cm}\end{array}$

Using the given values in the thin lens equation for the eyepiece, the object distance for the eyepiece $d_{oe}$ is determined.

$\begin{array}{c}\frac{1}{f_e}=\frac{1}{d_{oe}}+\frac{1}{N}\\ \frac{1}{d_{oe}}=\frac{1}{f_e}-\frac{1}{N}\\ =\frac{1}{1.79\text{ cm}}-\frac{1}{25\text{ cm}}\end{array}$

The object distance for the eyepiece is given by $d_{oe}=1.928\text{ cm}$

Calculate the image distance for the objective, using the calculated value of $d_{oe}$ and the given value of l.

$\begin{array}{c}{d}_{io}=l-d_{oe}\\ =\left(25\text{ cm}\right)-\left(1.928\text{ cm}\right)\\ =18.072\text{ cm}\end{array}$

The object distance is calculated using the expression, $m_o=\frac{d_{io}}{d_0}$

$\begin{array}{c}{d}_o=\frac{d_{io}}{m_o}\\ =\frac{18.072\text{ cm}}{60.0}\\ =0.301\text{ cm}\end{array}$

Use the thin lens equation to calculate the focal length of the objective.

$\begin{array}{c}\frac{1}{f_o}=\frac{1}{d_o}+\frac{1}{d_{io}}\\ =\frac{1}{\left(0.301\text{ cm}\right)}+\frac{1}{\left(18.072\text{ cm}\right)}\end{array}$

On solving the equation, the focal length of the objective is found to be $f=0.296\text{ cm}$.

To determine

Part (c)To determine:

The distance at which the object must be placed to see it in focus.

Answer to Problem 44P

Solution:

The object must be placed at 0.301 cm from the objective, to see it in focus.

Explanation of Solution

From the calculated value of the focal length of the eyepiece, the object distance can be determined.

Given:

The magnification of the eyepiece $m_e=14.0$

The magnification of the objective $m_o=60.0$

Barrel length of the microscope $l=20.0\text{ cm}$

Formula used:

The magnification of the objective is given by

$m_o=\frac{d_{io}}{d_0}$

Where, the object distance is $d_0$ and the image distance is $d_{io}$.

Calculation:

The object distance is calculated using the expression, $m_o=\frac{d_{io}}{d_0}$

Use the values of $d_{io}$ calculated in the previous section to calculate the object distance.

$\begin{array}{c}{d}_o=\frac{d_{io}}{m_o}\\ =\frac{18.072\text{ cm}}{60.0}\\ =0.301\text{ cm}\end{array}$