Question

The focal eyepiece length of a compound microscope’s objective lens is 0.400 cm and the focal length of the is 3.20 cm. The image formed by the objective lens is 20.0 cm from the objective lens.

a) Where is the object under study located? (1.5 pts)

b) If the image formed by the objective lens is located 3.00 cm from the eyepiece, what is the distance l between the two lenses? (1 pt)

c) Where is the final image located? (1.5 pts)

d) Find the total magnification of the microscope using the formula we derived in class with the standard near point distance N = 25cm. (1.5 pts)

e) Now find the total magnification using the approximate formula derived in class and compare your result with the result from part c. (1.5 pts)

f) Draw a ray diagram with a straight-edge roughly to scale and at least half a 8.5 x 11 inch page in size showing the object, the intermediate image and the final image and all of the principal rays for this two-lens microscope. (3 pts)

Answer 1

from the given data

focal length of objective, fo = 0.4 cm

focal length of the eyepiece, fe = 3.2 cm

distance between lenses = d

image distance for objective, vo = 20 cm

hence

1/vo - 1/uo = 1/fo

uo = -0.40816 cm

a. the object under study is uo = -0.40816 cm from the objective lens

b. now

u = -3 cm

now

vo - u = d

hence

d = 23 cm

c. final image distance from the eyepiece = v

then

1/v - 1/u = 1/fe

1/v + 1/3 = 1/3.2

v = -48 cm form the eyepiece

d. for near point distance

v = -25 cm

1/v - 1/u = 1/fe

u = -2.8368 cm

hence

vo = d + u = 20.16312056 cm

hence

1/vo - 1/uo = 1/fo

uo = -0.4080958 cm

hence

magnification = (vo/uo)(v/u) = -435.351053