Question

example #3: ocular = 10x low power objective = 20x high power objective = 40x. the picture shows the low power field of view for the microscope with the lenses listed above. a) what is the approximate size of the cell in micrometers? b) what would be the high power field of view? c) how many cells like the one in the picture could fit in the high power field of view? for all of the following problems, you will need this formula: \\(\frac{low\\ power\\ diameter\\ of\\ field}{high\\ power\\ diameter\\ of\\ field}=\frac{high\\ power\\ magnification}{low\\ power\\ magnification}\\). 1. calculate: high power field diameter (give answer in both mm and \\(\mu m\\)). given: a. the above formula. b. 100x field diameter = 2.3 mm c. high power = 800x

Explanation:

Step1: Convert units

We know that 1 mm = 1000 μm. The low - power field diameter is 2 mm, so in micrometers it is $2\times1000=2000$ μm.

Step2: Calculate magnification ratio

The magnification of the low - power objective is 40x and the high - power objective is 20x. The ratio of low - power magnification to high - power magnification is $\frac{40}{20}=2$.

Step3: Calculate high - power field diameter

Using the formula $\frac{\text{Diameter of low - power field}}{\text{Diameter of high - power field}}=\frac{\text{High - power magnification}}{\text{Low - power magnification}}$. Let the high - power field diameter be $d$. We have $\frac{2000}{d}=\frac{40}{20}$, so $d = 1000$ μm or 1 mm.

Step4: Estimate cell size

If the cell takes up about $\frac{1}{4}$ of the diameter of the low - power field of view which is 2 mm (2000 μm), the approximate size of the cell is $\frac{2000}{4}=500$ μm.

Step5: Answer part b

The picture in the high - power field of view would show a smaller area. Since the high - power field diameter is 1 mm and the cell size is 500 μm, about 2 cells could fit in the high - power field of view (1000 μm/500 μm = 2).

Step6: Answer part c

We know that $\frac{\text{Diameter of low - power field}}{\text{Diameter of high - power field}}=\frac{\text{High - power magnification}}{\text{Low - power magnification}}$. Given low - power field diameter $D_{l}=2.3$ mm, low - power magnification $M_{l}=100x$, high - power magnification $M_{h}=800x$. Let high - power field diameter be $D_{h}$. Then $\frac{D_{l}}{D_{h}}=\frac{M_{h}}{M_{l}}$, so $D_{h}=\frac{D_{l}\times M_{l}}{M_{h}}=\frac{2.3\times100}{800}=0.2875$ mm or 287.5 μm.

Answer:

a. The approximate size of the cell is 500 μm.
b. 2 cells could fit in the high - power field of view.
c. The high - power field diameter is 0.2875 mm or 287.5 μm.