contestada

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?

Respuesta :

Answer:

The focal length of the eyepiece is 4.012 cm.

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

[tex]m=-(\dfrac{i-f_{e}}{f_{0}})\dfrac{N}{f_{e}}[/tex]

Where,

[tex]i[/tex] = distance between the lenses in a compound  microscope

[tex]f_{e}[/tex]=focal length of eyepiece

[tex]f_{0}[/tex]=focal length of the object

N = normal point

Put the value into the formula

[tex]-58=-(\dfrac{18-f_{e}\times25}{1.5\timesf_{e}})[/tex]

[tex]87f_{e}=450-25f_{e}[/tex]

[tex]f_{e}=\dfrac{450}{112}[/tex]

[tex]f_{e}=4.012\ cm[/tex]

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

Otras preguntas