Question

1. calculate: high power field diameter (give answer in both mm and μm)

given: a. the above formula.
b. 200 x field diameter = 1.1 mm
c. high power = 400x

1. calculate: high power magnification (give answer in both mm and μm)

given: a. the above formula.
b. 10x field diameter = 40 mm
c. high power field diameter = 10 mm

1. calculate: low power diameter (give answer in both mm and μm)

given: a. the above formula.
b. low power field diameter = 4.5 mm
c. 800x field diameter = 0.35 mm

Explanation:

Step1: Calculate initial field diameter for problem 2

From $200\times\text{Field Diameter}=1.1\text{ mm}$, we can find $\text{Field Diameter}=\frac{1.1}{200}=0.0055\text{ mm}$.

Step2: Calculate high - power field diameter for problem 2

We know that magnification and field diameter are inversely proportional. Let $M_1 = 200$, $M_2=400$, $D_1$ be the initial field diameter and $D_2$ be the high - power field diameter. Then $M_1D_1 = M_2D_2$. So $D_2=\frac{M_1D_1}{M_2}=\frac{200\times0.0055}{400}=0.00275\text{ mm}$. To convert to $\mu\text{m}$, since $1\text{ mm} = 1000\mu\text{m}$, $D_2 = 0.00275\times1000 = 2.75\mu\text{m}$.

Step3: Calculate magnification for problem 3

From $10\times\text{Field Diameter}=40\text{ mm}$, we get $\text{Field Diameter}=4\text{ mm}$. Let $M_1 = 10$, $D_1 = 4\text{ mm}$, $D_2 = 10\text{ mm}$. Using $M_1D_1 = M_2D_2$, we have $M_2=\frac{M_1D_1}{D_2}=\frac{10\times4}{10}=4\text{ mm}$. In $\mu\text{m}$, since $1\text{ mm}=1000\mu\text{m}$, $4\text{ mm}=4000\mu\text{m}$.

Step4: For problem 4, it seems there is an error in the problem setup as we are already given the low - power field diameter as $4.5\text{ mm}$. In $\mu\text{m}$, it is $4.5\times1000 = 4500\mu\text{m}$.

Answer:

1. In mm: $0.00275\text{ mm}$, in $\mu\text{m}$: $2.75\mu\text{m}$
2. In mm: $4\text{ mm}$, in $\mu\text{m}$: $4000\mu\text{m}$
3. In mm: $4.5\text{ mm}$, in $\mu\text{m}$: $4500\mu\text{m}$