25. A particular compound microscope, with the 43x objective in position, has a diameter of thefield of view of approximately 0.31 mm and an area of approximately 0.1 mm2. The diameter of thefield of view is inversely proportional to the magnification of the objective. Counting left to rightone observes 15 rectangular onion epidermal cells across the field of view with the 43x objective inplace. Counting from top to bottom it can be observed that four cells span the diameter of the field.Using the 43x objective, one sees an average of 12 stomates per field of view. What is the bestestimate of the density of stomates per sq mm of epidermal surface?A. 12B. 15C. 36D. 120E. 150
Question
- A particular compound microscope, with the 43x objective in position, has a diameter of thefield of view of approximately 0.31 mm and an area of approximately 0.1 mm2. The diameter of thefield of view is inversely proportional to the magnification of the objective. Counting left to rightone observes 15 rectangular onion epidermal cells across the field of view with the 43x objective inplace. Counting from top to bottom it can be observed that four cells span the diameter of the field.Using the 43x objective, one sees an average of 12 stomates per field of view. What is the bestestimate of the density of stomates per sq mm of epidermal surface?A. 12B. 15C. 36D. 120E. 150
Solution
To solve this problem, we need to find the density of stomates per square millimeter.
First, we know that the area of the field of view is approximately 0.1 mm^2 and there are 12 stomates in this area.
Density is defined as the number of objects per unit area. So, to find the density of stomates, we divide the number of stomates by the area of the field of view.
Density = Number of stomates / Area of field of view = 12 stomates / 0.1 mm^2 = 120 stomates/mm^2
So, the best estimate of the density of stomates per square millimeter of epidermal surface is 120 stomates/mm^2.
Therefore, the answer is D. 120.
Similar Questions
A student observed 24cells using microscope whose eyepiece lens magnification is x10 and lower power objective lens magnification x35. Calculate the number of cells he likely to observe under high power objective lens magnification of x45
The diameter of a field of view is 4.7mm. 4 cells fit across the diameter under 4x objective magnification.What is the actual size of each cell?2019 Atomi QuestionA(Choice A)2.94μmB(Choice B)29.4μmC(Choice C)0.0294μmD(Choice D)0.294μm
x40 objective lens and estimated the field of view to be 480 um. The student also estimated that 12 cells of this specimen fit across the field of view. The student completed a scientific drawing of one of the cells, and the drawing of the cell measured 16 cm in length.What is the size of each cell in um?
A student's microscope has an ocular lens with a 10x magnification. Calculate the total magnification, if the objective lens is set to 30x.
Magnification can be calculated using the equation: magnification = size of image / actual size of the specimen. A bacterial cell is 0.003mm long. A magnified image of the cell is 21mm long. The magnification of the image is ×_____.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.