- (a) 5cm, 35cm
Explanation:
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=7$
$\text{f}_0=7\text{f}_e$
In normal adjustment, distance between the lenses,
$\text{f}_0+\text{f}_\text{e}=40$
$7\text{f}_0+\text{f}_\text{e}=40\Rightarrow\text{f}_\text{e}=\frac{40}{8}=5\text{cm}$
$\text{f}_0=7\text{f}_\text{e}=7\times5=35\text{cm}$
- (d) 20cm
Explanation:
$\text{m}=-10;\text{L}=22\text{cm}$
As, $\text{m}=\frac{-\text{f}_0}{\text{f}_e}\Rightarrow-10=-\frac{\text{f}_0}{\text{f}_\text{e}}$
$\text{f}_0=10\text{f}_\text{e}$
As, $\text{L}=\text{f}_0+\text{f}_\text{e}$
$22=10\text{f}_\text{e}+\text{f}_e=11\text{f}_\text{e}$
or $\text{f}_\text{e}=\frac{22}{11}=2\text{cm}$
$\text{f}_0=10\text{f}_\text{e}=20\text{cm}$
- (d) Large focal length.
Explanation:
Objective lens has larger focal length than eye-piece.
- (d) Astronomical telescope.
Explanation:
Astronomial telescope is used to see stars, sun etc.
- (c) f0 << fe