An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
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Focal length of the objective lens, $f_0 = 1.25$ cm
Focal length of the eyepiece, $f_e = 5$ cm
Least distance of distinct vision, $d = 25$ cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
$\text{m}_\text{e}=\Big(1+\frac{\text{d}}{\text{f}_\text{e}}\Big)$
$=\Big(1+\frac{25}{5}\Big)=6$
The angular magnification of the objective lens (mo) is related to meas:
$m_e m_e= m$
$\text{m}_0=\frac{\text{m}}{\text{m}_\text{e}}$
$=\frac{30}{6}=5$
We also have the relation:
$\text{m}_0=\frac{\text{image distance for the objective lenc(v}_0)}{\text{image distance for the objective lenc(-u}_0)}$
$5=\frac{\text{v}_0}{\text{-u}_0}$
$\therefore \ \text{v}_0=-5\text{u}_0$
Applying the lens formula for the objective lens.
$\frac{1}{\text{f}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}$
$\frac{1}{1.25}=\frac{1}{-5\text{u}_0}-\frac{1}{\text{u}_0}=\frac{-6}{5\text{u}_0}$
$\therefore \ \text{u}_0=\frac{-6}{5}\times1.25=-1.5 \ \text{cm}$
And, $\text{v}_0=-5\text{u}_0$
$= -5 × (-1.5) = 7.5$ cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
$\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
Where, $v_e =$ Image distance for the eyepiece $= -d= -25$ cm
$u_e =$ Object distance for the eyepiece
$\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$=-\frac{1}{25}-\frac{1}{5}=\frac{-6}{25}$
$\therefore \ \text{u}_\text{e}=-04.17 \ \text{cm}$
Separation between the objective lens and the eyepiece $=|\text{u}_\text{e}|+|\text{v}_0|$
$= 4.17 + 7.5$
$ = 11.67$ cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
Focal length of the eyepiece, $f_e = 5$ cm
Least distance of distinct vision, $d = 25$ cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
$\text{m}_\text{e}=\Big(1+\frac{\text{d}}{\text{f}_\text{e}}\Big)$
$=\Big(1+\frac{25}{5}\Big)=6$
The angular magnification of the objective lens (mo) is related to meas:
$m_e m_e= m$
$\text{m}_0=\frac{\text{m}}{\text{m}_\text{e}}$
$=\frac{30}{6}=5$
We also have the relation:
$\text{m}_0=\frac{\text{image distance for the objective lenc(v}_0)}{\text{image distance for the objective lenc(-u}_0)}$
$5=\frac{\text{v}_0}{\text{-u}_0}$
$\therefore \ \text{v}_0=-5\text{u}_0$
Applying the lens formula for the objective lens.
$\frac{1}{\text{f}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}$
$\frac{1}{1.25}=\frac{1}{-5\text{u}_0}-\frac{1}{\text{u}_0}=\frac{-6}{5\text{u}_0}$
$\therefore \ \text{u}_0=\frac{-6}{5}\times1.25=-1.5 \ \text{cm}$
And, $\text{v}_0=-5\text{u}_0$
$= -5 × (-1.5) = 7.5$ cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
$\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
Where, $v_e =$ Image distance for the eyepiece $= -d= -25$ cm
$u_e =$ Object distance for the eyepiece
$\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$=-\frac{1}{25}-\frac{1}{5}=\frac{-6}{25}$
$\therefore \ \text{u}_\text{e}=-04.17 \ \text{cm}$
Separation between the objective lens and the eyepiece $=|\text{u}_\text{e}|+|\text{v}_0|$
$= 4.17 + 7.5$
$ = 11.67$ cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
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