Question

A compound microscope consists of two converging lenses as shown to the left (not to scale.) A compound microscope has an eyepiece lens with a power of 11 D and an objective lens with a power of 40 D. The lenses are separated by 15 cm and the object is 4 cm from the objective lens. Where is the final image relative to the eyepiece lens? [-99.6 cm] What is the total magnification of the final image? [-19.9] Is the final image real or virtual, right side up or upside down, smaller or bigger than the object?

Answer 1

for objective

image distacne q1 = ?

focal length fob = 1/Po = 1/40 = 0.025 m = 2.5 cm

object distacne p1 = 4 cm

from lens equation

1/p1 + 1/q1 = 1/fob

1/4 + 1/q1 = 1/2.5

image distance q1 = 6.67 cm <<<=-==answer

magnification mo = -q1/p1 = -6.67/4 = -1.67
==================

for eyepiece

object distance p2 = 15 - 6.67 = 8.33 cm

focal length fe = 1/11 = 0.0909 m = 9.09 cm

image distance q2 = ?

1/p2 + 1/q2 = 1/fe

1/8.33 + 1/q2 = 1/9.09

q2 = -99.6 cm  

me = -q2/p2 = 99.6/8.33 = 11.9

(a)

finalimage q2 = -99.6 cm    <<<<-------------answer

(b)

total magnification M = mo*me = -1.67*11.9 = -19.9

(c)

virtual , upside down and bigger than object