- Magnifying power is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.
Expression
$\text{m} = \beta\big/\alpha = \text{f}_{o}\big/\text{f}_{e}$
or $\text{m} = \text{f}_{o} /_{ \text{f}_{e}}\bigg(1 + \frac{\text{f}_{e}}{\text{D}}\bigg)$
- Using, the lens equation for objective lens,:
$\frac{1}{\text{f}_{o}} =\frac{1}{\text{v}_{o}}-\frac{1}{\text{u}_{o}}$
$ => \frac{1}{150} = \frac{1}{\text{v}_{o}} - \frac{1}{-3\times10^{5}}$
$ => \frac{1}{\text{v}_{o}} =\frac{1}{150} -\frac{1}{-3\times10^{5}} = \frac{2000-1}{3\times10^{5}}$
$ =>\text{v}_{o} = \frac{3\times10^{5}}{1999}\text{cm}$
$\approx 150 \text{cm}$
Hence, magnification due to the objective lens
$\text{m}_{o} =\frac{\text{v}_{o}}{\text{u}_{o}} = \frac{150\times10^{-2}\text{m}}{3000\text{m}}$
$\approx\frac{10^{-2}}{20} = .05\times10^{-2}$
Using lens formula for eyepiece
$\frac{1}{\text{f}_{e}} =\frac{1}{\text{v}_{e}} - \frac{1}{\text{u}_{e}}$
$ = > \frac{1}{5} = \frac{1}{-25} -\frac{1}{\text{u}_{e}}$
$ = > \frac{1}{\text{u}_{e}} =\frac{1}{-25} - \frac{1}{5} = \frac{-1-5}{25}$
$ = >\text{u}_{e} =\frac{-25}{6}\text{cm}$
$\therefore\text{ Magnification due to eyepiece} \text{ m}_{e} =\frac{-25}{-\frac{25}{6}}= 6 $
Hence, total magnification $ = > \text{m} = \text{m}_{e}\times\text{m}_{o}$
$\text{m} = 6 \times5\times10^{-4} = 30 \times10^{-4}$
Hence, size of final image
$ = 30 \times10^{-4}\times100 \text{m}$
$ = 30 \text{ cm}.$
