Question

The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal length of objective and eyepiece are respectively.

पर्याय

• 10 cm, 10 cm

• 18 cm, 2 cm

• 15 cm, 5 cm

• 11 cm, 9 cm

Answer 1

18 cm, 2 cm

Explanation:

For final image at infinity, magnifying power of a telescope is given by

m = `"f"_"o"/"f"_"e"` = 9

where, m = magnification,

f_o = focal length of objective

and f_e = focal length of eyepiece

⇒ f_o = 9f_e     ....(i)

Also, distance between objective and eyepiece

= f_o + f_e = 20 (given)

⇒ 9f_e + f_e = 20 ⇒ f_e = 2 cm

f_o = 9f_e = 18 cm