CBSE 2022 · Region 5 · Set 3 · Q6 · 3 marks
With the help of a ray diagram, show how a compound microscope forms a magnified image of a tiny object, at least distance of distinct vision. Hence derive an expression for the magnification produced by it.
Marking-scheme solution
Ray diagram of Compound Microscope
Derivation of magnification
$\displaystyle 1 \frac{1}{2}$
The linear magnification due to objective lens\mathrm{m}_{0}=\frac{h^{\prime}}{h}=\frac{L}{f_{0}} \longrightarrow \cdots \cdots(\mathrm{i})h= size of object
h'= size of first image
As $\displaystyle \tan \beta=\frac{h}{f_{0}}=\frac{h^{\prime}}{L}$
magnification due to eye piece\mathrm{m}_{\mathrm{e}}=1+\frac{D}{f_{e}} \longrightarrow \ldots \ldots-\ldots-\ldots-\ldots(\mathrm{ii})Total Magnification $\displaystyle \mathrm{m}=\mathrm{m}_{\mathrm{o}} \times \mathrm{m}_{\mathrm{e}}$\mathrm{m} .=\frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)
$$
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.