A small telescope has an objective lens of focal length `140 cm` and eye piece of focal length `5.0 cm`. What is the magnifying power of telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e. when the image is at infinity)
(b) the final image is formed at the least distance of distinct vision `(25 cm)`.

Options

A. 28,33.6
B. 33.6,28
C. Both (a) and (b)
D. None of these

Answer

A

Explanation

i. When the telescope is i9n normal adjustment, the magnifying power is given by `M=(f_(o))/(|f_(e )|)=+(140)/(5)=28`
ii. When the final image is formed at the least distance of distinct vision, then M is given by
`M = (f_(o))/(|f_(e )|)(1+(f_(e ))/(D))=(140)/(5)(1+(5.0)/(25))=33.6`