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.

Step-by-step solution

For final image at infinity, magnifying power of a telescope is given by m = f o f e = 9 where, m = magnification, a n d f o = focal length of eyepiece ⇒ f o = 9 f e …. (i) Also, distance between objective and eyepiece = f o + f e = 20 (given) ⇒ 9 f e + f e = 20 ⇒ f e = 2 c m f o = 9 f e = 18 c m