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Microscope and Microscopy - Slide Preparation and Focusing (temporary mount)-advanced

Grade 9CBSE

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

🔑Concepts

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The compound microscope uses a system of lenses to magnify small objects. The total magnification is the product of the ocular lens (eyepiece) and the objective lens powers: Mtotal=Meyepiece×MobjectiveM_{total} = M_{eyepiece} \times M_{objective}.

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Temporary Mount Preparation: A specimen (e.g., onion peel or human cheek cells) is placed on a glass slide in a drop of water or mounting medium.

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Staining: Biological stains like Safranin (for plant cells) or Methylene Blue (for animal cells) are used to provide contrast to transparent cell organelles. The stain reacts with specific chemical components like DNA or pectin.

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Mounting Medium: Glycerine is used as a mounting medium because it has a high refractive index and prevents the specimen from drying out (desiccation) during observation.

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Coverslip Placement: The coverslip is lowered at a 45∘45^{\circ} angle using a mounting needle to avoid the entrapment of air bubbles, which can obstruct the view and create artifacts.

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Focusing Technique: Observation should always begin with the low-power objective lens (10×10 \times) using the coarse adjustment knob. Once the specimen is located, the high-power objective (40×40 \times or 100×100 \times) is used with the fine adjustment knob for sharp focusing.

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Numerical Aperture (NANA): This represents the light-gathering capacity of the lens. The resolving power of a microscope is inversely proportional to the wavelength of light (λ\lambda) used.

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Refractive Index (nn): The ratio of the speed of light in a vacuum to the speed of light in the medium. Matching the refractive index of the mounting medium with the glass slide minimizes light reflection and improves image clarity.

📐Formulae

Mtotal=Mocular×MobjectiveM_{total} = M_{ocular} \times M_{objective}

Resolving Power∝1d=2⋅n⋅sin⁡θλ\text{Resolving Power} \propto \frac{1}{d} = \frac{2 \cdot n \cdot \sin \theta}{\lambda}

d=0.61λNAd = \frac{0.61 \lambda}{NA}

Field of View (FOV)∝1Magnification\text{Field of View (FOV)} \propto \frac{1}{\text{Magnification}}

💡Examples

Problem 1:

A student observes a specimen using an eyepiece marked 15×15 \times and an objective lens marked 40×40 \times. Calculate the total magnification of the specimen.

Solution:

Mtotal=15×40=600M_{total} = 15 \times 40 = 600

Explanation:

The total magnification is calculated by multiplying the magnifying power of the ocular lens by the magnifying power of the objective lens. Therefore, the image appears 600600 times larger than the actual specimen.

Problem 2:

During the preparation of an onion peel slide, a student uses a mounting needle to lower the coverslip at an angle. If the student fails to do this and drops the coverslip vertically, what is the most likely observation?

Solution:

The observation will show dark, circular outlines or 'blobs' which are air bubbles.

Explanation:

Lowering the coverslip at a 45∘45^{\circ} angle allows the liquid to spread evenly and pushes air out. A vertical drop traps air, creating bubbles that have thick dark boundaries due to the difference in refractive index between air and the mounting medium, obscuring the cellular details.

Problem 3:

If the diameter of the field of view is 2 mm2 \text{ mm} at 100×100 \times magnification, what will be the diameter of the field of view at 400×400 \times magnification?

Solution:

New FOV=Original FOV×(Original MagnificationNew Magnification)\text{New FOV} = \text{Original FOV} \times \left( \frac{\text{Original Magnification}}{\text{New Magnification}} \right) New FOV=2 mm×(100400)=0.5 mm\text{New FOV} = 2 \text{ mm} \times \left( \frac{100}{400} \right) = 0.5 \text{ mm}

Explanation:

The Field of View (FOV) is inversely proportional to the magnification. As magnification increases by a factor of 44 (from 100×100 \times to 400×400 \times), the visible area (diameter) decreases by the same factor.