- #1
TheSodesa
- 224
- 7
The question:
--------------------
The length of a microscope pipe is $L=160\,\rm mm$,
the transverse magnification of its objective $M_o = 40\times$
and the diameter $d_o = 5\,\rm mm$.
As for the ocular/eyepiece, its magnification is $M_e = 10\times$.
1. Find out the focal length of the objective $f_o$
2. At what distance from the objective must object be placed in order for a sharp image to form
3. What is the numerical aperture of the objective?
4. Where is the exit pupil of the microscope located,
if the near distance of an average person is $25\,\rm cm$?
An attempt at a solution:
--------------------------------------
1. Using the transverse magnification equation for a thin lens,
the focal length of the objective can be found out to be
\begin{equation}
f_o = -\frac L {M_o} = -\frac{160\,\rm mm}{-40} = 4\,\rm mm\,.
\end{equation}
2. Using the focal length $f_o$, if the distance of the image
formed by the objective is known to be $s_i = f_o + L$ we can solve for $s_o$ using the thin lens equation:
\begin{equation}
s_o
= \left(\frac 1 {f_o} - \frac 1 {f_o + L}\right)^{-1}
\approx 4.1\,\rm mm\,.
\end{equation}
3. The numerical aperture $\rm NA$ is defined as
\begin{equation}
{\rm NA} = n_i\sin\theta_{\rm max},
\end{equation}
where $n_i$ is the refractive index of the substance surrounding the object.
Here it is assumed to be air, so $n_i \approx 1$.
The maximum angle where light from a point on the lens axis can penetrate the objective lens can be found out from the diameter of the lens and the object distance:
\begin{equation}
\theta_{\rm max} = \tan^{-1}\left(\frac{d_o}{2s_o}\right) \approx 0.548,
\end{equation}
so
\begin{equation}
{\rm NA} = \sin\left(0.548\right) \approx 0.52\,.
\end{equation}
4. This is where I got stuck.
The definition of the exit pupil is the image of the objective as viewed through the eyepiece. For this I need the distance between the eyepiece and objective $s_{oe} = f_o + L + f_e$, which again requires knowldge of $f_e$,
the focal length of the eyepiece, but I can't seem to figure out a way to calculate this. What I tried was to calculate the focal length using transverse magnification:
\begin{equation}
f_e = M_ex_o = M_e(L + f_o) = 1.64\rm\,m,
\end{equation}
and using this I calculated the distance of the exit pupil to be
\begin{equation}
s_i = \left(\frac 1 {f_o} - \frac 1 {f_o + L + f_e}\right)^{-1} \approx 18\,\rm m,
\end{equation}
which is preposterous, as the eye would have to be placed this far from the microscope. I would not even be able to see the microscope itself from this distance without my glasses...
What I didn't try was to use the near distance of an average person $s_l = 25\,\rm cm$ to my advantage, but I'm not sure how to go about this. I guess the microscope could be though of as a pair of correcting eyeglasses, but which part of the microscope should this function belong to?
--------------------
The length of a microscope pipe is $L=160\,\rm mm$,
the transverse magnification of its objective $M_o = 40\times$
and the diameter $d_o = 5\,\rm mm$.
As for the ocular/eyepiece, its magnification is $M_e = 10\times$.
1. Find out the focal length of the objective $f_o$
2. At what distance from the objective must object be placed in order for a sharp image to form
3. What is the numerical aperture of the objective?
4. Where is the exit pupil of the microscope located,
if the near distance of an average person is $25\,\rm cm$?
An attempt at a solution:
--------------------------------------
1. Using the transverse magnification equation for a thin lens,
the focal length of the objective can be found out to be
\begin{equation}
f_o = -\frac L {M_o} = -\frac{160\,\rm mm}{-40} = 4\,\rm mm\,.
\end{equation}
2. Using the focal length $f_o$, if the distance of the image
formed by the objective is known to be $s_i = f_o + L$ we can solve for $s_o$ using the thin lens equation:
\begin{equation}
s_o
= \left(\frac 1 {f_o} - \frac 1 {f_o + L}\right)^{-1}
\approx 4.1\,\rm mm\,.
\end{equation}
3. The numerical aperture $\rm NA$ is defined as
\begin{equation}
{\rm NA} = n_i\sin\theta_{\rm max},
\end{equation}
where $n_i$ is the refractive index of the substance surrounding the object.
Here it is assumed to be air, so $n_i \approx 1$.
The maximum angle where light from a point on the lens axis can penetrate the objective lens can be found out from the diameter of the lens and the object distance:
\begin{equation}
\theta_{\rm max} = \tan^{-1}\left(\frac{d_o}{2s_o}\right) \approx 0.548,
\end{equation}
so
\begin{equation}
{\rm NA} = \sin\left(0.548\right) \approx 0.52\,.
\end{equation}
4. This is where I got stuck.
The definition of the exit pupil is the image of the objective as viewed through the eyepiece. For this I need the distance between the eyepiece and objective $s_{oe} = f_o + L + f_e$, which again requires knowldge of $f_e$,
the focal length of the eyepiece, but I can't seem to figure out a way to calculate this. What I tried was to calculate the focal length using transverse magnification:
\begin{equation}
f_e = M_ex_o = M_e(L + f_o) = 1.64\rm\,m,
\end{equation}
and using this I calculated the distance of the exit pupil to be
\begin{equation}
s_i = \left(\frac 1 {f_o} - \frac 1 {f_o + L + f_e}\right)^{-1} \approx 18\,\rm m,
\end{equation}
which is preposterous, as the eye would have to be placed this far from the microscope. I would not even be able to see the microscope itself from this distance without my glasses...
What I didn't try was to use the near distance of an average person $s_l = 25\,\rm cm$ to my advantage, but I'm not sure how to go about this. I guess the microscope could be though of as a pair of correcting eyeglasses, but which part of the microscope should this function belong to?
