If the diameter of the low-power field is 1.5 mm, an object that occupies approximately a third of that field has an estimated diameter of

d' = 0.86 mm

Explanation:

First we need to find the area of low power field:

A = πd²/4

where,

A = Area = ?

d = diameter = 1.5 mm

Therefore, using the value of diameter, we get:

A = π(1.5 mm)²/4

A = 1.767 mm²

Now, the new object occupies 1/3rd of this area:

A' = (1/3)A

πd'²/4 = (1/3)A

d'² = (4/3π) A

d'² = (4/3π)(1.767 mm²)

d' = √(0.75 mm²)

d' = 0.86 mm