Physics - Geometrical Optics Question with Solution | TestHub

${\mathrm{f}}_0=1.5\mathrm{ }\mathrm{c}\mathrm{m},\mathrm{ }{\mathrm{f}}_{\mathrm{e}}=6.25\mathrm{ }\mathrm{c}\mathrm{m},\mathrm{ }{\mathrm{u}}_0=-2\mathrm{ }\mathrm{c}\mathrm{m},\mathrm{ }{\mathrm{v}}_{\mathrm{e}}=-\mathrm{D}=-25\mathrm{ }\mathrm{c}\mathrm{m}$

By objective lens $\frac{1}{{\mathrm{f}}_0}=\frac{1}{{\mathrm{v}}_0}-\frac{1}{{\mathrm{u}}_0}$

$\frac{1}{1.5}=\frac{1}{{\mathrm{v}}_0}-\frac{1}{-2}$

$\Rightarrow$ $\frac{1}{{\mathrm{v}}_0}=\frac{1}{1.5}-\frac{1}{2}$ or ${\mathrm{v}}_0=6\mathrm{ }\mathrm{c}\mathrm{m}$

By eyepiece $\frac{1}{{\mathrm{f}}_{\mathrm{e}}}=\frac{1}{{\mathrm{v}}_{\mathrm{e}}}-\frac{1}{{\mathrm{u}}_{\mathrm{e}}}$

$\frac{1}{6.25}=\frac{1}{-25}-\frac{1}{-{\mathrm{u}}_{\mathrm{e}}}$

$\Rightarrow$ $\frac{1}{{\mathrm{u}}_{\mathrm{e}}}=\frac{1}{6.25}+\frac{1}{25}=\frac{4}{25}+\frac{1}{25}=\frac{1}{5}$

${\mathrm{u}}_{\mathrm{e}}=5\mathrm{ }\mathrm{c}\mathrm{m}$ 

Length of tube $=\mathrm{L}={\mathrm{v}}_0+{\mathrm{u}}_{\mathrm{e}}=6.0\mathrm{cm}+5.0\mathrm{cm}$

$\mathrm{L}=11\mathrm{ }\mathrm{cm}$