
You are given two converging lenses of focal length \[1.25cm\] and $5cm$ to design a compound microscope. If it is desired to have a magnification of $30$, find out the separation between the objective and the eyepiece.
Answer
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Hint: An upright microscope that uses two lens sets (a compound lens system) to achieve a higher magnification than a stereomicroscope is a compound microscope. A compound microscope produces a two-dimensional view, while a three-dimensional view is given by a stereo microscope. Tiny samples not identifiable with the naked eye are seen with the compound microscopes. Usually these samples are mounted under the microscope on a slide.
Magnification of a compound microscope can be written as,
$M = {M_0} {M_e} $
Where the magnification of microscope is $M$, ${M_0} $ is the magnification due to objective and ${M_e} $ is magnifying power by the eye piece
Complete step by step solution:
Magnification of a compound microscope is given by:
$M = {M_0} {M_e} $
Where the magnification of microscope is $M$, ${M_0} $ is the magnification due to objective and ${M_e} $ is magnifying power by the eye piece
We know that,
${M_0} = \dfrac{L}{{{f_0}}}$ and ${M_e} = \dfrac{{25}}{{{f_e}}}$
Where $L$ is the distance between objective and eyepiece, ${f_0} $ is the focal length of the objective and ${f_e} $ is the focal length of the eye piece.
Therefore, $M = \dfrac{L}{{{f_0}}} \times \dfrac{{25}}{{{f_e}}}$
It is given that, ${f_0} = 1.25cm$, ${f_e} = 5cm$ and $M = 30$
We get,
$\Rightarrow 30 = \dfrac{L}{{1.25}} \times \dfrac{{25}}{5}$
$\Rightarrow L = 7.5cm$
Therefore, the distance between the objective and eyepiece is $7.5cm$.
Note: With the aid of a compound microscope, two lenses can transcend the limits on resolution (and thus magnification power) imposed by the restrictions of the simple microscope. One of them has a small focal length and is positioned near the object under investigation. It is used to create a real picture in the middle of the front, eye or eye lens. The oculus forms a broader virtual image that the viewer can see. The compound microscope magnification strength is the combination of the lens magnification and the eyepiece magnification.
Magnification of a compound microscope can be written as,
$M = {M_0} {M_e} $
Where the magnification of microscope is $M$, ${M_0} $ is the magnification due to objective and ${M_e} $ is magnifying power by the eye piece
Complete step by step solution:
Magnification of a compound microscope is given by:
$M = {M_0} {M_e} $
Where the magnification of microscope is $M$, ${M_0} $ is the magnification due to objective and ${M_e} $ is magnifying power by the eye piece
We know that,
${M_0} = \dfrac{L}{{{f_0}}}$ and ${M_e} = \dfrac{{25}}{{{f_e}}}$
Where $L$ is the distance between objective and eyepiece, ${f_0} $ is the focal length of the objective and ${f_e} $ is the focal length of the eye piece.
Therefore, $M = \dfrac{L}{{{f_0}}} \times \dfrac{{25}}{{{f_e}}}$
It is given that, ${f_0} = 1.25cm$, ${f_e} = 5cm$ and $M = 30$
We get,
$\Rightarrow 30 = \dfrac{L}{{1.25}} \times \dfrac{{25}}{5}$
$\Rightarrow L = 7.5cm$
Therefore, the distance between the objective and eyepiece is $7.5cm$.
Note: With the aid of a compound microscope, two lenses can transcend the limits on resolution (and thus magnification power) imposed by the restrictions of the simple microscope. One of them has a small focal length and is positioned near the object under investigation. It is used to create a real picture in the middle of the front, eye or eye lens. The oculus forms a broader virtual image that the viewer can see. The compound microscope magnification strength is the combination of the lens magnification and the eyepiece magnification.
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