Question

The eyepiece of a compound microscope has a focal length of 2.70 cm , and the objective lens has f = 0.750 cm . If an object is placed 0.810 cm from the objective lens, calculate the distance between the lenses when the microscope is adjusted for a relaxed eye.
Calculate the total magnification.

Answer 1

Answer:

Explanation:

Image formation by objective lens ,

f = focal length of objective = .75 cm, ( positive )

object distance u = .81 cm ( negative )

v = image distance  

1/v - 1/u = 1/f

1/v + 1/.81 = 1 / .75

1/v = 1/ .75 - 1/.81 = .098765

v = 10.125.cm

Image formation by eye piece,

v = infinity ( for relaxed eye )

f ( eye piece ) = 2.70cm

u = ?

1/v - 1/u = 1/f

0 -1/u = 1/2.7

u = 2.7 cm

Total length  between lenses

= 2.7 + 10.125

= 12.825 cm

Total magnification = m₁ x m₂

m₁ is magnification by objective and m₂ is magnification by eye piece

m₁ = v/u = 10.125 / .810 = 12.5

m₂ =  1 + D / f = 1 + 25 / 2.7 = 10.25

Total magnification

= 12.5 x 10.25 = 128.125