Question
A compound microscope has a magnifying power of $100$ when the image is formed at infinity. The objective has a focal length of $0.5\ cm$ and the tube length is $6.5\ cm$. Find the focal length of the eyepiece.
✓
Answer
For the give compound microscope $, f_0 = 0.5\ cm,$ tube length $= 6.5\ cm$ 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 $, v_o + f_e = 6.5\ cm ...(1)$
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 $D = 25\ cm]$
$\Rightarrow100\text{f}_\text{e}=-(1-2\text{v}_0)\times25$
$\Rightarrow2\text{v}_0-4\text{f}_\text{e}=1 ...(2)$
Solving equation $(1)$ and $(2)$ we can get,
$V_0 = 4.5\ cm$ and $f_e = 2\ cm$
So, the focal length of the eye piece is $2\ cm.$
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 $, v_o + f_e = 6.5\ cm ...(1)$
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 $D = 25\ cm]$
$\Rightarrow100\text{f}_\text{e}=-(1-2\text{v}_0)\times25$
$\Rightarrow2\text{v}_0-4\text{f}_\text{e}=1 ...(2)$
Solving equation $(1)$ and $(2)$ we can get,
$V_0 = 4.5\ cm$ and $f_e = 2\ cm$
So, the focal length of the eye piece is $2\ cm.$
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