Question

A microscope has an eyepiece with a 1.8 cm focal length and a 0.8 cm focal length objective lens. Assuming a relaxed eye, calculate a) the position of the object if the distance between the lenses is 16 cm, and b) the total magnification.

Answer 1

`0.848\ \text{cm}`

`232.66`

Step-by-step explanation:

N = Near point of eye = 25 cm

`f_o` = Focal length of objective = 0.8 cm

`f_e` = Focal length of eyepiece = 1.8 cm

l = Distance between the lenses = 16 cm

Object distance is given by

`v_o=l-f_e\\\Rightarrow v_o=16-1.8\\\Rightarrow v_o=14.2\ \text{cm}`

`u_o` = Object distance for objective

From lens equation we have

`(1)/(f_o)=(1)/(u_o)+(1)/(v_o)\\\Rightarrow u_o=(f_ov_o)/(v_o-f_o)\\\Rightarrow u_o=(0.8* 14.2)/(14.2-0.8)\\\Rightarrow u_o=0.848\ \text{cm}`

The position of the object is
`0.848\ \text{cm}`.

Magnification of eyepiece is

`M_e=(N)/(f_e)\\\Rightarrow M_e=(25)/(1.8)\\\Rightarrow M_e=13.89`

Magnification of objective is

`M_o=(v_o)/(u_o)\\\Rightarrow M_o=(14.2)/(0.848)\\\Rightarrow M_o=16.75`

Total magnification is given by

`m=M_eM_o\\\Rightarrow m=13.89* 16.75\\\Rightarrow m=232.66`

The total magnification is
`232.66`.