Question 204: Ray Optics

Question

The magnifying power of a telescope with tube
60 cm is 5. What is the focal length of its eye
piece ?

Accepted Answer

For telescope
Tube length (L) = f
0
+ f
e
and magnification (m) = $${{{f_0}} \over {{f_e}}}$$ = 5
where f
o
and f
e
are focal length of objective
and eyepiece.
$$ 	herefore $$ f
0
+ f
e
= 60
$$ 	herefore $$ f
o
= 50 cm
f
e
= 10 cm

Answer

40 cm

Answer

10 cm

Answer

30 cm

Answer

20 cm

Ray Optics

The magnifying power of a telescope with tube 60 cm is 5. What is the focal length of its eye piece ?

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