An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102cm, find the powers of the objective and the eyepiece.
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For the astronomical telescope in normal adjustment.Magnifying power = m = 50, length of the tube = L = 102cm
Let $f_0$ and fe be the focal length of objective and eye piece respectively.
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=50\Rightarrow\text{f}_0=50\text{f}_\text{e}\dots(1)$
and, $\text{L}=\text{f}_0+\text{f}_\text{e}=102\text{cm}\dots(2)$
Putting the value of f0 from equation (1) in (2), we get,
$f_0 + f_e= 102 \Rightarrow 51f_e = 102 \Rightarrow f_e= 2cm = 0.02m$
So, $f_0 = 100cm = 1m$
$\therefore$ Power of the objective lens$=\frac{1}{\text{f}_0}=1\text{D}$
And Power of the eye piece lens$=\frac{1}{\text{f}_\text{e}}=\frac{1}{0.02}=50\text{D}$
Let $f_0$ and fe be the focal length of objective and eye piece respectively.
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=50\Rightarrow\text{f}_0=50\text{f}_\text{e}\dots(1)$
and, $\text{L}=\text{f}_0+\text{f}_\text{e}=102\text{cm}\dots(2)$
Putting the value of f0 from equation (1) in (2), we get,
$f_0 + f_e= 102 \Rightarrow 51f_e = 102 \Rightarrow f_e= 2cm = 0.02m$
So, $f_0 = 100cm = 1m$
$\therefore$ Power of the objective lens$=\frac{1}{\text{f}_0}=1\text{D}$
And Power of the eye piece lens$=\frac{1}{\text{f}_\text{e}}=\frac{1}{0.02}=50\text{D}$
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