QJEEM22Phd23a: JEE Main 2022 Physics - Class 12 JEEM22Phd23a Solution: In normal adujstment, for a refracting telescope, the distance…

$\because$ $m = {{{f_o}} \over {{f_e}}}$

$\Rightarrow 2 = {{{f_o}} \over {{f_e}}}$ ...... (i)

and, $l = {f_o} + {f_e}$

$\Rightarrow 30 = {f_o} + {f_e}$ ..... (ii)

$\Rightarrow 30 = {f_o} + {{{f_o}} \over 2}$

$\Rightarrow 30 \times {2 \over 3} = {f_o}$

$\Rightarrow {f_o} = 20$ cm