five
The answer is 1.6mm.
The magnification is inversely proportional to field of view.
Therefore, the calculation should be as follow:
Diameter of the f.o.v of low power lens (10x) / Diameter of the f.o.v of scanning lens (4x) = Scanning lens magnification (4x) / Low power lens magnification (10x)
k/4=4/10
k= (4/10) x 4
k= 1.6mm
Note: f.o.v= field of view
I am almost positive it is 1,000 millimeters. I am not promising you that is the right answer. I am 95% positive.
0.5 mm
The sun is not millimeters. It is a G-type star, on the Main Sequence, in mid-life. For its spectral classification, its size is average, with a diameter of about 1,390,000,000,000 millimeters
155 mm
The millimeter should be sufficient. If you want greater precision, measure with micrometers. A nickel made in the USA is about 2 millimeters thick, which is equivalent to 2,000 micrometers. However, keep in mind that the precise thickness will vary from nickel to nickel, depending on the amount of wear.
Objective reasoning
noun1. something that one's efforts or actions are intended to attain or accomplish; purpose; goal; target: the objective of a military attack; the objective of a fund-raising drive. 2. Grammar. a. Also called objective case. (in English and some other languages) a case specialized for the use of a form as the object of a transitive verb or of a preposition, as him in The boy hit him, or me in He comes to me with his troubles. b. a word in that case. his objective was to score a goal in soccer practice.
the naswe is 400
0.75 mm way to get this answer........... (diameter of field A X total magnification of field A) / total magnification of field B so start by finding the diameter of field A= which is the 1.5 next figure out what the total magnification of field A is= 150 (you get this answer by multiplying the ocular # which is 10x by the objective # which is 15x. (10 x 15= 150) next figure out what the total magnification of field B is =300 (you get this answer by multiplying the ocular # which is 10x by the other higher objective # which is 30x. (10 x 30 = 300) then you can use the formula and plug in all the answers you got to get the answer (1.5mm x 150)/300=.75mm
yes it is.
Using 3.14 as Pi the area of circle is: 961.625
You need to measure the diameter of the cent. Then divide the diameter in half to get the radius. The formula for area of a circle = ?r2, where ? is 3.14159. I think you would be better off using millimeters or centimeters instead of ft.
Using 3.14 as Pi the area of circle is: 2461.76
To locate the exact site in the microscope field that you want to magnify.
Please clarify your objective with an example.
You use the low power objective lens first to get your sample centered in the field of view.
We don't think you can do it with that information. 'f-stop' = (focal length of the objective lens) divided by (its diameter) Magnification of the scope = (focal length of the objective) divided by (focal length of the eyepiece) Looks like in order to calculate the 'f-stop', you need to estimate or measure the focal length of either the objective or the eyepiece. Here's an idea: If you can temporarily separate the objective from the tube, use the objective to focus an image of the sun on the sidewalk. (Not on anything flammable.) Measure the distance from the lens to the sharpest image. With the 'object' at infinity, the image is at the focal length.
you use centemeters
Since the field of view is a circle, the size of the field of view is it's area. You would need to find the diameter of the field of view, using a transparent ruler or a micrometer. Divide the diameter measurement by 2 to get the radius. Then use the formula for the area of a circle, Area = πr2. For example, you measure the diameter of the field of view to be 2.14mm. Divide 2.14mm by 2 to get the radius, and you get 1.07mm. Square 1.07mm, which is 1.14mm2. Multiply x 3.14 (pi), and you get 3.58mm2. So the field of view for this example would be 3.58mm2.The field of view differs with different magnifications. The lower the magnification, the larger the field of view.