JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 22)
The magnifying power of a telescope with tube
60 cm is 5. What is the focal length of its eye
piece ?
40 cm
10 cm
30 cm
20 cm
Explicació
For telescope
Tube length (L) = f0 + fe
and magnification (m) = $${{{f_0}} \over {{f_e}}}$$ = 5
where fo and fe are focal length of objective and eyepiece.
$$ \therefore $$ f0 + fe = 60
$$ \therefore $$ fo = 50 cm
fe = 10 cm
Tube length (L) = f0 + fe
and magnification (m) = $${{{f_0}} \over {{f_e}}}$$ = 5
where fo and fe are focal length of objective and eyepiece.
$$ \therefore $$ f0 + fe = 60
$$ \therefore $$ fo = 50 cm
fe = 10 cm