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Cells and Life Processes - Microscopy Skills

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The total magnification of a light microscope is the product of the eyepiece lens magnification and the objective lens magnification: Total Magnification=Eyepiece×ObjectiveTotal\ Magnification = Eyepiece \times Objective.

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Units of measurement in microscopy must be consistent. The standard units include millimeters (mmmm), micrometers (μm\mu m), and nanometers (nmnm). Note that 1 mm=1000 μm1\ mm = 1000\ \mu m and 1 μm=1000 nm1\ \mu m = 1000\ nm.

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The relationship between the size of a drawing or photograph (Image size), the real size of the object (Actual size), and the Magnification is represented by the formula I=A×MI = A \times M.

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Resolution is the ability to distinguish between two points that are very close together; higher resolution allows for more detail to be seen in the specimen.

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Field of View (FOVFOV) decreases as magnification increases. If the magnification is doubled, the diameter of the FOVFOV is halved.

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Staining is used to enhance visualization; for example, Iodine solution is often used for plant cells (staining starch), while Methylene Blue is used for animal cell nuclei.

📐Formulae

Total Magnification=Magnificationeyepiece×MagnificationobjectiveTotal\ Magnification = Magnification_{eyepiece} \times Magnification_{objective}

Magnification(M)=Image size(I)Actual size(A)Magnification (M) = \frac{Image\ size (I)}{Actual\ size (A)}

Actual size(A)=Image size(I)Magnification(M)Actual\ size (A) = \frac{Image\ size (I)}{Magnification (M)}

1 mm=103 μm=106 nm1\ mm = 10^3\ \mu m = 10^6\ nm

💡Examples

Problem 1:

A student uses a 10×10\times eyepiece and a 40×40\times objective lens to view a plant cell. What is the total magnification?

Solution:

10×40=400×10 \times 40 = 400\times

Explanation:

To find the total magnification, multiply the power of the eyepiece lens by the power of the objective lens used.

Problem 2:

An image of a bacterial cell is 40 mm40\ mm long. The magnification used was 2000×2000\times. Calculate the actual size of the cell in μm\mu m.

Solution:

A=IM=40 mm2000=0.02 mmA = \frac{I}{M} = \frac{40\ mm}{2000} = 0.02\ mm. Converting to μm\mu m: 0.02×1000=20 μm0.02 \times 1000 = 20\ \mu m.

Explanation:

First, divide the image size (40 mm40\ mm) by the magnification (20002000). Then, convert the result from millimeters to micrometers by multiplying by 10001000.

Problem 3:

If a cell has an actual length of 0.05 mm0.05\ mm and its image in a textbook is 10 cm10\ cm long, what is the magnification used?

Solution:

I=10 cm=100 mmI = 10\ cm = 100\ mm. M=IA=100 mm0.05 mm=2000×M = \frac{I}{A} = \frac{100\ mm}{0.05\ mm} = 2000\times.

Explanation:

First, ensure both measurements are in the same units (mmmm). Then divide the image size by the actual size to find the magnification factor.