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What approximate fraction of the low-power field AREA would you see if you were to change to the high-power objective

4. What approximate fraction of the low-power field AREA would you see if you were to change to the high-power objective, using the microscope from Question 3? Keep it simple! And remember: A = (pi/(r2). Show calculations. 



  

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Answer #1

Field of view (diameter that can be seen at given magnification) is inversely proportional to magnification. Higher the FOV, lower the magnification and vice versa.

(FOV)1 (M)1 = (FOV)2 (M)2

FOV1 = 1600 um

M1 = 10x

FOV2 = ?

M2 = 40x

(1600)(10x) = (FOV)2(40x)

FOV2 = 400 um

At 40x, you will see 400 um of the area.

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