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: low power diameter of field / high power diameter of field = high power magnification / low power magnification 1. calculate: high power field diameter (give answer in both mm and μm given: a. the above formula. b. 100x field diameter = 2.3 mm c. high power = 800x

Explanation:

Step1: Calculate the low - power magnification

The low - power magnification is the product of the ocular and the low - power objective. So, $M_{low}=10\times20 = 200x$.

Step2: Calculate the high - power magnification

The high - power magnification is the product of the ocular and the high - power objective. So, $M_{high}=10\times40 = 400x$.

Step3: Find the size of the cell at low - power

The low - power field of view diameter is $d_{low}=2\ mm = 2000\ \mu m$. Assume the cell takes up about $\frac{1}{10}$ of the diameter of the low - power field of view. So the size of the cell $s\approx\frac{2000}{10}=200\ \mu m$.

Step4: Calculate the high - power field of view diameter

Using the formula $\frac{d_{low}}{d_{high}}=\frac{M_{high}}{M_{low}}$, we substitute $d_{low} = 2\ mm$, $M_{low}=200x$, and $M_{high}=400x$. Then $d_{high}=\frac{M_{low}\times d_{low}}{M_{high}}=\frac{200\times2}{400}=1\ mm = 1000\ \mu m$.

Step5: Calculate the number of cells that can fit in the high - power field of view

The number of cells $n=\frac{d_{high}}{s}$. Substituting $d_{high}=1000\ \mu m$ and $s = 200\ \mu m$, we get $n=\frac{1000}{200}=5$.

For the additional problem:
Using the formula $\frac{d_{low}}{d_{high}}=\frac{M_{high}}{M_{low}}$, where $d_{low}=2.3\ mm$, $M_{low}=100x$, $M_{high}=800x$. Then $d_{high}=\frac{M_{low}\times d_{low}}{M_{high}}=\frac{100\times2.3}{800}=0.2875\ mm=287.5\ \mu m$

Answer:

a) $200\ \mu m$
b) $1\ mm$ (or $1000\ \mu m$)
c) $5$

1. $0.2875\ mm$, $287.5\ \mu m$