An optical instrument used for angular magnification has a 25D objective and a 20D eyepiece. The tube length is 25cm when the eye is least strained.
- Whether it is a microscope or a telescope?
- What is the angular magnification produced?
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The optical instrument has$\text{f}_0=\frac{1}{25\text{D}}=0.04\text{m}=4\text{cm}$
$\text{f}_\text{e}=\frac{1}{20\text{D}}=0.05\text{m}=5\text{cm}$
tube length = 25cm (normal adjustment)
Now, using lens formula.
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{20}-\frac{1}{4}$
$=\frac{-4}{20}=\frac{-1}{5}\Rightarrow\text{u}_0=-5\text{cm}$
So, angular magnification $\text{m}=-\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}$[Taking D = 25cm]
$=-\frac{20}{-5}\times\frac{25}{5}=20$
$\text{f}_\text{e}=\frac{1}{20\text{D}}=0.05\text{m}=5\text{cm}$
tube length = 25cm (normal adjustment)
- The instrument must be a microscope as $f_0< f_e$
- Since the final image is formed at infinity, the image produced by the objective should lie on the focal plane of the eye piece.
Now, using lens formula.
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{20}-\frac{1}{4}$
$=\frac{-4}{20}=\frac{-1}{5}\Rightarrow\text{u}_0=-5\text{cm}$
So, angular magnification $\text{m}=-\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}$[Taking D = 25cm]
$=-\frac{20}{-5}\times\frac{25}{5}=20$
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