A microscope is having an objective of focal length 1 cm and eyepiece of focal length 6 cm. If tube length is 30 cm and the image is formed at the least distance of distinct vision, what is the magnification produced by the microscope? (Take D = 25 cm)

Options

A. 6
B. 150
C. 25
D. 125

Answer

B

Explanation

Given, length of tube, L=30cm, focal length of objective lens, `f_(v)=1cm focal length of eye piece, `f_(e)=6cm, 2D=25cm By compound microscope, `m=(L)/(f_(u))(1+(D)/(f_(e)))`
`m=(30)/(1)(1+(25)/(6))=30xx((6+25))/(6)`
`5xx31=155cm approx 150`