An amoeba is 0.305 cm away from the objective lens of a microscope. a) if the ob

Question

An amoeba is 0.305 cm away from the objective lens of a microscope. a) if the objective lens has a focal lenght of 0.300cm, where is the image formed by the objective? B) what is the magnification of this image? C) the eyepiece of the microscope has a focal length of 2.00 cm. and 20.0 cm away from the objective. where is the final image? D) what is the magnification produced by the eyepiece? E) What is the overall magnification?

Suppose amoeba is not at 0.305 cm but is at the focal point of the objective lens then what is the answer for the part E) in unit of time (t)?

Explanation / Answer

for concave lens

1/f = 1/u + 1/v

u is object ditance

v is image distance

1/v = 1/0.3 - 1/0.305

v = 18.3 cm

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magnification m = -V/u

m = -18.3/0.3 = -61 (inverted image)
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C:

for eyepiece

object distance = 20 - 18.3 - 1.7 cm

new image 1/Vi = 1/20 - 1/1.7

Vnew = -1.86 cm

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magnifiction M = s/f = 1.7/20 = 0.085

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over all magnification m = 61* 0.085 = 5.2

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