A compound microscope has an objective of focal length 4 mm and an eye-piece of focal length 25 mm. The objective produces a real image at a distance of 180 mm. If the eye-piece is in normal adjustment the magnification is

Options

A. 45
B. 90
C. 225
D. 440

Answer

D

Explanation

For objective lens `1/f_(0)=1/v_(0) -1/u_(0)`
`implies1/4 =1/180 -1/u_(0)`
`implies 1/u_(0)=1/180 – ¼= (1-45)/180=(-44)/180`
`therefore u_(0)= (-180)/44`
The magnification power of a compound microscope is given by,
`m=v_(0)/u_(0) D/f_(e)=180/(180/44) xx 250/25`
( because D=250 mm “least distance of distinct vision”)
`therefore m = 44xx10=440`