Question

The objective lens in a microscope with a 17.0 cm long tube has a magnification of -50.0 and the eyepiece has a magnification of 20.0. (a) What is the focal length of the objective? cm (b) What is the focal length of the eyepiece? cm (c) What is the overall magnification of the microscope?

Answer 1

Step-by-step explanation:

It is given that,

Length of microscope, l = 17 cm = 0.17 m

Magnification of objective lens, m₀ = -50

Magnification of eyepiece,
`m_e=20`

(a) Magnification in terms of length is given by :

`m_o=-(l)/(f_o)`

`f_o=-(l)/(m)`

`f_o=-(0.17)/(-50)`

`f_o=0.0034\ m`

`f_o=3.4\ mm`

(b) Focal length of eyepiece is given by :

`m_e=(25)/(f_e)`

`f_e=(25)/(m_e)`

`f_e=(25)/(20)`

`f_e=1.25\ m`

(c) Total magnification of the microscope,

`M=m_o* m_e`

`M=-50* 20`

M = -100

Hence, this is the required solution.