Magnification of Microscope

Notation

\(Z_{1}= 30\)	linear magnification of the objective lens
\(γ_{2}=5\)	angular magnification of the ocular lens
\(Δ=15 \mathrm{cm}= 0{.}15 \mathrm{m}\)	optical tube length of the microscope
\(γ= ?\)	magnification of the microscope
\(f_{1}= ?\)	focal length of the objective lens
\(f_{2}= ?\)	focal length of ocular lens

Hint 1

What is the microscope composed of and how does it work?

Solution of Hint 1

A microscope consists of two converging lenses. The first is called an objective lens and the second an ocular lens or eyepiece. The focal length of the objective lens is small, the focal length of the eyepiece is larger.

Description of the picture

\(F_{1}\)	object focal point of the objective lens
\(F'_{1}\)	image focal point of the objective lens
\(F_{2}\)	object focal point of the ocular lens
\(F'_{2}\)	image focal point of the ocular lens
\(τ'\)	magnified angle of view

We put a very small object that, in case of observing with the naked eye, we would see at a very small angle of view \(τ\), near the object focal point of the lens \(F_{1}\) (before the objective focal point of the lens). A magnified, inverted and real image will appear behind the lens. The eyepiece is placed so that the image shown by the lens would appear in the object focal point of the eyepiece \(F_{2}\). The ocular lens then works as a magnifying glass.

Hint 2

What is the relation between the total magnification of the microscope \(γ\), linear magnification of the objective lens \(Z_{1}\) and the angular magnification of the ocular lens \(γ_{2}\)?

Solution of Hint 2

We obtain the total magnification of the microscope as the product of the linear magnification of the objective lens and the angular magnification of the eyepiece. Thus: \[γ=Z_{1}\cdot γ_{2}\tag{1}\]

Numerical solution

Known from the task assignment:

\(Z_{1}=30\)

\(γ_{2}=5\)

Substitution for the values:

\(γ=Z_{1} \cdot γ_{2} = 30 {\cdot} 5=150.\)

Hint 3

What is an optical tube length \( Δ\)?

Solution of Hint 3

This is the distance between the image focus of the objective lens and the object focus of the ocular lens.

Hint 4

Deduce the relation for the linear magnification of the objective lens. Then determine the focal length of the lens.

Solution of Hint 4

Linear magnification of the objective lens \( Z_{1}\) tells us how many times the image created by the objective lens is larger than the projected object, thus it applies: \[Z_{1}=\frac{y'}{y},\tag{2}\] where \(y'\) is the height of the image and \(y\) is the height of the object.

Using the following picture, we can determine the ratio of the height of the image to the height of the object.

The yellow triangle is similar to the blue triangle. From the similarity it follows that: \[\frac{y'}{Δ}=\frac{y}{f_{1}},\] where \(f_{1}\) is the focal length of the objective lens. We multiply both sides of the equation by \(\frac{Δ}{y}\) and obtain the ratio of the height of the object to the height of the image: \[\frac{y'}{y}= \frac{Δ}{f_{1}}.\] And in accordance with (2): \[Z_{1}= \frac{Δ}{f_{1}}.\] Therefrom we express the focal length of the objective lens: \[f_{1}= \frac{Δ}{Z_{1}}.\tag{3}\]

Numerical solution

According to the sign convention, the linear magnification of the objective lens should have been negative, but since we are interested only in the height of the image and not in its orientation, we will not write the negative value of the linear magnification in the numerical solution. This „neglect“ is going to affect the solution of the problem in no way.

Known from the task assignment:

\(Z_{1}=30\)

\(Δ=0{.}15 \mathrm{m}\)

Substitution for the values:

\(f_{1}= \frac{Δ}{Z_{1}}=\frac{0{.}15}{30} \mathrm{m}=5{\cdot}10^{−3} \mathrm{m}=5 \mathrm{mm}.\)

Hint 5

What is the angular magnification of the ocular lens? How do we calculate it? Determine the focal length of the ocular lens.

Solution of Hint 5

Angular magnification of the ocular lens \(γ_{2}\) tells us how many times the angle at which we observe the object by the ocular lens (magnified angle of view \(τ'\))) is greater than the angle at which we observe the object by the naked eye (angle of view \(τ\)). The microscope’s eyepiece works as a magnifying glass. It applies: \[γ_{2}=\frac{τ'}{τ}.\tag{4}\]

We draw a picture from which we can determine the value of the angle of view \(τ\) at which we observe the object of the height \(y\) from the conventional distance of distinct vision \(d\) by the naked eye.

We see that it holds true: \[\mathrm{tg}\,{τ}=\frac{y}{d}.\] Since angle \(τ\) is very small, we can assume that \(\mathrm{tg}\,{τ} \dot= τ\) and thus we can write: \[τ=\frac{y}{d}.\tag{5}\]

The angle at which we observe the object will be increased, if we observe the object through the eyepiece.

Let us imagine that we have placed the ocular lens between the object and the eye so that the object is at the object focal point of the ocular lens. Then the light beams in the image space are parallel and the image is created without the eye being accommodated.

We can deduce the equation for the magnified angle of view \(τ'\) from the picture: \[\mathrm{tg}\,{τ'}=\frac{y}{f_{2}},\] where \(f_{2}\) is the focal length of the ocular lens. The magnified angle of view is small and thus we can write: \[τ'=\frac{y}{f_{2}}.\tag{6}\]

We substitute for relations (5) and (6) in the relation for the angular magnification of the ocular lens (4) and adjust the relation:

\[γ_{2}=\frac{τ'}{τ}=\frac{\frac{y}{f_{2}}}{\frac{y}{d}},\] \[γ_{2}=\frac{y}{f_{2}}\frac{d}{y},\] \[γ_{2}=\frac{d}{f_{2}}.\]

Now we can express the focal length of the ocular lens: \[f_{2}=\frac{d}{γ_{2}}.\tag{7}\]

Numerical solution

Known from the task assignment:

\(γ_{2}=5\)

We look up the conventional distance of distinct vision in the Tables:

\(d= 0{.}25 \mathrm{m}\)

We substitute for the values:

\(f_{2}=\frac{d}{γ_{2}}=\frac{0{.}25}{5} \mathrm{m}=5{\cdot}10^{−2} \mathrm{m}=50 \mathrm{mm}.\)

Overall Solution

Picture of the microscope

The microscope consists of two converging lenses (objective and ocular lens).

Description of the picture

\(F_{1}\)	object focal point of the objective lens	\(y\)	height of the object
\(F'_{1}\)	image focal point of the objective lens	\(y'\)	height of the image created by the objective lens
\(F_{2}\)	object focal point of the ocular lens	\(f_{1}\)	focal length of the objective lens
\(F'_{2}\)	image focal point of the ocular lens	\(f_{2}\)	focal length of the ocular lens
\(τ'\)	zvětšený zorný úhel	\(Δ\)	optical tube length

Total magnification of the microscope

We obtain the total magnification of the microscope as the product of the linear magnification of the objective lens and the angular magnification of the eyepiece. Thus: \[γ=Z_{1}\cdot γ_{2}\tag{1}\]

Linear magnification of the objective lens and its focal length

Linear magnification of the objective lens \( Z_{1}\) tells us how many times the image created by the objective lens is larger than the projected object, thus it applies: \[Z_{1}=\frac{y'}{y},\tag{2}\] where \(y'\) is the height of the image and \(y\) is the height of the object.

We determine the ratio of the height of the image to the height of the object.

The yellow triangle is similar to the blue triangle. From the similarity it follows that: \[\frac{y'}{Δ}=\frac{y}{f_{1}},\] where \(f_{1}\) is the focal length of the objective lens. We express the ratio of the height of the object to the height of the image: \[\frac{y'}{y}= \frac{Δ}{f_{1}}.\] And in accordance with (2): \[Z_{1}= \frac{Δ}{f_{1}}.\] Therefrom we express the focal length of the objective lens: \[f_{1}= \frac{Δ}{Z_{1}}.\tag{3}\]

Angular magnification of the ocular length and its focal length

The ocular lens of the microscope works as a magnifying glass. It applies: \[γ_{2}=\frac{τ'}{τ}.\tag{4}\]

We draw a picture from which we can determine the value of the angle of view \(τ\) at which we observe the object of the height \(y\) from the conventional distance of distinct vision \(d\) with the naked eye.

Now we place the ocular lens between the object and the eye so that the object is at the object focal point of the ocular lens. Angle at which we observe the object will increase (magnified angle of view \(τ'\)).

Object observed with the naked eye and through the eyepiece

We deduce from the picture that if we observe the object with the naked eye it holds true: \[\mathrm{tg}\,{τ}=\frac{y}{d}.\] Since angle \(τ\) is very small, we can assume that \(\mathrm{tg}\,{τ} \dot= τ\) and thus we can write: \[τ=\frac{y}{d}.\tag{5}\]

If we observe the object trough the eyepiece, we can deduce the relation for the magnified angle of view \(τ'\): \[\mathrm{tg}\,{τ'}=\frac{y}{f_{2}}.\] The magnified angle of view \(τ'\) is small and thus we can write: \[τ'=\frac{y}{f_{2}}.\tag{6}\]

We substitute for relations (5) and (6) in the relation for the angular magnification of the ocular lens (4) and adjust the relation:

\[γ_{2}=\frac{τ'}{τ}=\frac{\frac{y}{f_{2}}}{\frac{y}{d}},\] \[γ_{2}=\frac{d}{f_{2}}.\]

We express the focal length of the ocular lens: \[f_{2}=\frac{d}{γ_{2}}.\tag{7}\]

Numerical solution

Known from the task assignment:

\(Z_{1}=30\)

\(γ_{2}=5\)

\(Δ=0{.}15 \mathrm{m}\)

We look up the conventional distance of distinct vision in the Tables:

\(d= 0{.}25 \mathrm{m}\)

We substitute for the values in the following relationships:

for the total magnification of the microscope (1):

\(γ=Z_{1} \cdot γ_{2} = 30 {\cdot} 5=150.\)

for the focal length of the objective lens (3):

\(f_{1}= \frac{Δ}{Z_{1}}=\frac{0{.}15}{30} \mathrm{m}=5{\cdot}10^{−3} \mathrm{m}=5 \mathrm{mm}.\)

for the focal length of the ocular lens (7):

\(f_{2}=\frac{d}{γ_{2}}=\frac{0{.}25}{5} \mathrm{m}=5{\cdot}10^{−2} \mathrm{m}=50 \mathrm{mm}.\)

Answer

The magnification of the microscope is \(150\). The focal length of the objective lens is \(5 \mathrm{mm}\) and the focal length of the ocular lens is \(50 \mathrm{mm}\).