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Microscope and Microscopy - Microscopy Skills-advanced

Grade 9CBSE

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

🔑Concepts

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Total Magnification is the product of the magnification power of the ocular lens (eyepiece) and the objective lens. For example, if the eyepiece is 10×10\times and the objective is 40×40\times, the total magnification is 400×400\times.

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Resolution (Resolving Power) is the ability of a microscope to distinguish two close points as separate entities. The resolution limit of a standard light microscope is approximately 0.2 μm0.2\ \mu m (200 nm200\ nm).

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Field of View (FOV) is the diameter of the circular area visible through the microscope. As magnification increases, the FOV decreases proportionally: FOV∝1MagnificationFOV \propto \frac{1}{Magnification}.

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Unit conversions are critical in microscopy: 1 mm=103 μm=106 nm1\ mm = 10^3\ \mu m = 10^6\ nm. Biological cells are typically measured in micrometers (μm\mu m).

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Numerical Aperture (NA) is a measure of the light-gathering capacity of the lens. Higher NA values lead to better resolution and brighter images.

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The depth of field refers to the thickness of the specimen that is in sharp focus at one time. Higher magnification results in a shallower depth of field.

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Staining techniques (like using Safranin for plant cells or Methylene Blue for animal cells) increase contrast, making transparent cellular structures like the nucleus visible.

📐Formulae

Total Magnification=Meyepiece×MobjectiveTotal\ Magnification = M_{eyepiece} \times M_{objective}

Actual Size of Specimen=Measured Image SizeTotal MagnificationActual\ Size\ of\ Specimen = \frac{Measured\ Image\ Size}{Total\ Magnification}

FOVhigh=FOVlow×MagnificationlowMagnificationhighFOV_{high} = \frac{FOV_{low} \times Magnification_{low}}{Magnification_{high}}

Cell Size Estimation=Diameter of Field of ViewNumber of cells crossing the diameterCell\ Size\ Estimation = \frac{Diameter\ of\ Field\ of\ View}{Number\ of\ cells\ crossing\ the\ diameter}

💡Examples

Problem 1:

An observer uses a microscope with a 15×15\times eyepiece and a 40×40\times objective lens. If a cell measures 2 cm2\ cm in a drawing made under this microscope, what is the actual size of the cell in micrometers (μm\mu m)?

Solution:

First, calculate the total magnification: Total Magnification=15×40=600×Total\ Magnification = 15 \times 40 = 600\times Next, convert the drawing size to micrometers: 2 cm=20 mm=20,000 μm2\ cm = 20\ mm = 20,000\ \mu m Now, use the actual size formula: Actual Size=20,000 μm600Actual\ Size = \frac{20,000\ \mu m}{600} Actual Size=33.33 μmActual\ Size = 33.33\ \mu m

Explanation:

To find the actual size, divide the observed/drawn size by the total magnification, ensuring all units are converted to the required scale (1 cm=10,000 μm1\ cm = 10,000\ \mu m).

Problem 2:

Under a 100×100\times magnification, the field of view (FOV) is measured to be 1.8 mm1.8\ mm. If the magnification is switched to 400×400\times, calculate the new diameter of the FOV in micrometers.

Solution:

Use the inverse relationship formula: FOVhigh=1.8 mm×100400FOV_{high} = \frac{1.8\ mm \times 100}{400} FOVhigh=180400 mmFOV_{high} = \frac{180}{400}\ mm FOVhigh=0.45 mmFOV_{high} = 0.45\ mm Convert to micrometers: 0.45 mm×1000=450 μm0.45\ mm \times 1000 = 450\ \mu m

Explanation:

The field of view is inversely proportional to the magnification. When magnification increases by 44 times (100100 to 400400), the FOV decreases by 44 times.

Problem 3:

If 88 cells are seen lining up across a field of view that has a diameter of 1.6 mm1.6\ mm, what is the average length of one cell in μm\mu m?

Solution:

Convert the FOV to micrometers: 1.6 mm=1.6×1000=1600 μm1.6\ mm = 1.6 \times 1000 = 1600\ \mu m Calculate the size per cell: Cell size=1600 μm8Cell\ size = \frac{1600\ \mu m}{8} Cell size=200 μmCell\ size = 200\ \mu m

Explanation:

The actual size of a single cell can be estimated by dividing the total diameter of the FOV by the number of cells that fit across that diameter.