f0 = 1cm, fe = 6cm, D = 24cm
For the eye piece, ve = -24cm, fe = 6cm
Now,
$\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}\Rightarrow-\Big[\frac{1}{24}+\frac{1}{6}\Big]=-\frac{5}{24}$
$\Rightarrow \text{u}_\text{e}=-4.8\text{cm}$
Now,
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{5}-\frac{1}{1}=-\frac{4}{5}$
$\Rightarrow\text{u}_0=-\frac{5}{4}=-1.25\text{cm}$
So, the magnifying power is given by,
$\text{m}=\frac{\text{v}_0}{\text{u}_\text{0}}\Big[1+\frac{\text{D}}{\text{f}}\Big]=\frac{-5}{-1.25}\Big[1+\frac{24}{6}\Big]=4\times5=20$
v0 = 11.8 - 4.8 = 7.0cm, f0 = 1cm
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{7}-\frac{1}{1}=-\frac{6}{7}$
So,
$\text{m}=-\frac{\text{v}_0}{\text{u}_\text{o}}\Big[1+\frac{\text{D}}{\text{f}}\Big]=\frac{-7}{-\Big(\frac{7}{6}\Big)}\Big[1+\frac{24}{6}\Big]=6\times5=30$So, the range of magnifying power will be 20 to 30.
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