Question

The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscope is to provide an angular magnification of -58 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece?

Answer 1

The focal length of the eyepiece is 4.012 cm.

Step-by-step explanation:

Given that,

Focal length = 1.5 cm

angular magnification = -58

Distance of image = 18 cm

We need to calculate the focal length of the eyepiece

Using formula of angular magnification

`m=-((i-f_(e))/(f_(0)))(N)/(f_(e))`

Where,

`i` = distance between the lenses in a compound microscope

`f_(e)`=focal length of eyepiece

`f_(0)`=focal length of the object

N = normal point

Put the value into the formula

`-58=-((18-f_(e)*25)/(1.5*f_(e)))`

`87f_(e)=450-25f_(e)`

`f_(e)=(450)/(112)`

`f_(e)=4.012\ cm`

Hence, The focal length of the eyepiece is 4.012 cm.