In an astronomical telescope in normal adjustment a straight black line of the length L is drawn on inside…

1. $\frac{\mathrm{L}}{l}-1$
2. $\frac{\mathrm{L}+l}{\mathrm{L}-l}$
3. $\frac{\mathrm{L}}{l}$
4. $\frac{\mathrm{L}}{l}+1$

Solution

Magnification by eyepiece
$\mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}+\mathrm{u}}$
$-\frac{\mathrm{l}}{\mathrm{L}}=\mathrm{ }\frac{{\mathrm{f}}_{\mathrm{e}}}{{\mathrm{f}}_{\mathrm{e}}+(-({\mathrm{f}}_0+{\mathrm{f}}_{\mathrm{e}})}$
$\Rightarrow \mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\frac{\mathrm{l}}{\mathrm{L}}=\frac{{\mathrm{f}}_{\mathrm{e}}}{{\mathrm{f}}_0}$
$\mathrm{m}.\mathrm{p}.\mathrm{ }=\frac{{\mathrm{f}}_0}{{\mathrm{f}}_{\mathrm{e}}}=\frac{\mathrm{L}}{\mathrm{l}}$

Asked in: NEET 2015 (Phase 2)

Practice more Optics questions on Aicharya