Question

1. a student was using microscope with a 10x eyepiece and 10x and 40x objective lenses. he viewed the edge of a metric rule under low power and observed the following field of vision. what is the diameter of the low - power field of vision in micrometers? a) 1 μm b) 2μm c) 1,000 μm d) 2,000 μm 10.a student using a compound light microscope estimated the diameter of a white blood cell to be 12 micrometers. what is the diameter of this white blood cell in millimeters? 1) 12mm b) 0.120 mm c) 0.012 mm d) 1.2 mm 11. a student was observing cells with the microscope and noted that one cell occupied one - fourth of the diameter of the field of view. if the diameter of the field was 1.5 millimeters, what was the approximate length of the cell? a) 1.5 μm b) 1,500 μm c) 375 μm d) 38 μm 12.a student determined that exactly 10 pollen grains could fit along the diameter of the field of view of a microscope. if each pollen grain has a diameter of 200 micrometers, what is the diameter in millimeters of this microscopes field of view? a) 2 mm b) 5 mm c) 2,000 mm d) 500 mm 13.if the diameter of the low power field of your microscope was fond to be 2 millimeters and the ratio between the low power and high power fields was 10, what would the diameter of the high power field be in micrometers? a. 0.2 b.20 c. 40 d. 200

Explanation:

Step1: Convert millimeters to micrometers

We know that \(1\) millimeter \(= 1000\) micrometers.

Step2: Calculate the diameter of the high - power field

Given the diameter of the low - power field \(d_{low}=2\) mm. First, convert it to micrometers: \(d_{low}=2\times1000 = 2000\) \(\mu m\).
The ratio of low - power to high - power fields \(r = 10\). Let the diameter of the high - power field be \(d_{high}\).
We use the formula \(r=\frac{d_{low}}{d_{high}}\), so \(d_{high}=\frac{d_{low}}{r}\).
Substitute \(d_{low} = 2000\) \(\mu m\) and \(r = 10\) into the formula: \(d_{high}=\frac{2000}{10}=200\) \(\mu m\)

Answer:

d. \(200\)