A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5cm and the tube length is 6.5cm. Find the focal length of the eyepiece.
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For the give compound microscope, $f_0 = 0.5cm$, tube length = 6.5cm magnifying power = 100 (normal adjustment) Since, the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece.So, $v_o + f_e = 6.5cm ...(1)$
Again, magnifying power $\text{m}=\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}$ [for normal adjustment]
$\Rightarrow\text{m}=\Big[1-\frac{\text{V}_0}{\text{f}_0}\Big]\frac{\text{D}}{\text{f}_\text{e}}$
$\Rightarrow100=-\Big[1-\frac{\text{V}_0}{0.5}\Big]\times\frac{25}{\text{f}_\text{e}}$ [Taking D = 25cm]
$\Rightarrow100\text{f}_\text{e}=-(1-2\text{v}_0)\times25$
$\Rightarrow2\text{v}_0-4\text{f}_\text{e}=1 $ ...(2)
Solving equation (1) and (2) we can get,
$V_0 = 4.5cm$ and $f_e = 2cm$
So, the focal length of the eye piece is 2cm.
Again, magnifying power $\text{m}=\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}$ [for normal adjustment]
$\Rightarrow\text{m}=\Big[1-\frac{\text{V}_0}{\text{f}_0}\Big]\frac{\text{D}}{\text{f}_\text{e}}$
$\Rightarrow100=-\Big[1-\frac{\text{V}_0}{0.5}\Big]\times\frac{25}{\text{f}_\text{e}}$ [Taking D = 25cm]
$\Rightarrow100\text{f}_\text{e}=-(1-2\text{v}_0)\times25$
$\Rightarrow2\text{v}_0-4\text{f}_\text{e}=1 $ ...(2)
Solving equation (1) and (2) we can get,
$V_0 = 4.5cm$ and $f_e = 2cm$
So, the focal length of the eye piece is 2cm.
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