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example #2: ocular power = 10x power objective = 10x high power objecti…

Question

example #2: ocular power = 10x power objective = 10x high power objective = 40x the diagram shows the edge of a millimeter ruler viewed under the microscope with the lenses listed above. the field shown is the low power field of view. a) what is the approximate width of the field of view in micrometers? b) what would be the width of the field of view under high power? c) if 5 cells fit across the high power field of view, what is the approximate size of

Explanation:

Step1: Recall the relationship for field - of - view and magnification

The relationship between the field of view (FOV) and magnification (M) is inversely proportional. That is, $FOV_1\times M_1 = FOV_2\times M_2$. First, we need to know that for a typical light - microscope, when using a low - power objective, we can estimate the field of view by counting the number of millimeter intervals visible under the low - power lens. Assume that the low - power field of view shows 4 mm intervals.

Step2: Calculate the low - power field of view in micrometers

Since 1 mm = 1000 micrometers ($\mu$m), if the low - power field of view shows 4 mm, then the low - power field of view ($FOV_{low}$) in micrometers is $4\times1000 = 4000\ \mu$m. The low - power magnification ($M_{low}$) is the product of the ocular power and the low - power objective power. $M_{low}=10\times10 = 100\times$.

Step3: Calculate the high - power magnification

The high - power magnification ($M_{high}$) is the product of the ocular power and the high - power objective power. $M_{high}=10\times40 = 400\times$.

Step4: Calculate the high - power field of view

Using the formula $FOV_1\times M_1 = FOV_2\times M_2$, we have $FOV_{low}\times M_{low}=FOV_{high}\times M_{high}$. Substituting the known values: $4000\times100 = FOV_{high}\times400$. Solving for $FOV_{high}$, we get $FOV_{high}=\frac{4000\times100}{400}=1000\ \mu$m.

Step5: Calculate the size of a cell

If 5 cells fit across the high - power field of view, and the high - power field of view is 1000 $\mu$m, then the size of one cell is $\frac{1000}{5}=200\ \mu$m.

Answer:

a) 4000 $\mu$m
b) 1000 $\mu$m
c) 200 $\mu$m