The focal length of a compound microscope\'s objective lens is 0.400 cm and the

Question

The focal length of a compound microscope's objective lens is 0.400 cm and the focal length of the eyepiece 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 /between the two lenses? (1 pt) Where is the final image located? (1.5 pts) c. d. Find the total magnification of the microscope using the formula we derived in class: I-TON with the standard near point distance N 25cm. (1.5 pts) 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) e. NL fefo 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)

Explanation / Answer

part a:

for the image formed by the objective lens:

object distance=u=-d (let)

where d>0

focal length=fo=0.4 cm

image distance=v=20 cm

using lens equation,

(1/v)-(1/u)=1/f

(1/20)+(1/d)=1/0.4

object distance, d = 0.41 cm

part b:

distance between two lenses = 20 cm + 3 cm

distance between two lenses = 23 cm

part c:

for the eye piece:

object distance =u=-3 cm

focal length=fe=3.2 cm

if image distance is v,

then (1/v)-(1/u)=1/fe

(1/v)=(1/3.2)-(1/3)

=-0.0208

v = - 48 cm

so final image is formed at 48 cm to the same side of the object from the eye piece.

part d:

total magnification=(l - fe)*N/(do*fe)

= (23 - 3.2)*25/(0.41*3.2)

M1 = 377.29

Part e:

M2 = [N*l] / [fo*fe] = [25*23] / [0.4*3.2]

M2 = 449.22

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