The focal lengths of the eye piece and objective of a compound microscope are `5 cm and 1 cm` respectively, and the length of the tube is `20 cm`. Calculate magnifying power of microscope when the final image is formed at infinity. The least distance of distinct vision is `25 cm`.

Answer

`70`

Explanation

Here, `f_(e) = 5 cm, f_(0) = 1 cm, L = 20 cm , m = ?`
`v_(e) = oo, d = 25 cm`
As final image is at infinity, image formed by objective lies at the focus of eye piece.
`:. v_(0) = L - f_(e) = 20 - 5 = 15 cm`
From `(1)/(u_(0)) = (1)/(v_(0)) - (1)/(f_(0)) = (1)/(15) - (1)/(1) = -(14)/(15)`
`u_(0) = -(15)/(14)cm`
As final image is at infinity.
`:. m = (v_(0))/(u_(0))((d)/(f_(e))) = (15)/(15//14) xx (25)/(5) = 70`