2) (Refer to steps 2.1-2.6) An experimenter sets up a microscope with a 5.67 mm

Question

2) (Refer to steps 2.1-2.6) An experimenter sets up a microscope with a 5.67 mm objective lens. The object distance between the object and the center of the obyective lens, do 5.86 mm. Virtual rays seen by observer What is The image distance units required) between the objective lens and the image created bY the objective, di- The tube length (units required) L The lateral magnification of the objective Mo- If an eyepiece of focal length 31.3 mm is used, what is the magnifying power of this microscope? Calculated microscope magnifying power:

Explanation / Answer

2)

Part A)

1/f = 1/do + 1/di

1/5.67 = 1/5.86 + 1/di

di = 252.96 mm

Part B)

From the picture, L is given as di - fo

L = 252.96- 5.67 = 247,299230769 mm

Part C)

M = di/do

M = 252.96/5.86 = 43,16881071147282

Part D)

M = L(250)/(fo)(fe)

M = (247,2992)(250)/(5.67)(fe)

M =

3) Use the fact that a light ray passes undeviated thru the center of the lens.
The half angle of such a ray is 44 = 21.5 deg.
tan 21.5 = 18 / f
f = 45.69 mm
So 35 mm is the longest focal length from the list that can be used.

4) Telescope length:

angular magnification = -fo/fe

so f)eyepiece = - 100 mm

if the input rays are to be maintained parallel by eyepiece giving image at infinity

LENGTH = fo+fe = 350 - 100

= 250 mm

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