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Microscope and Microscopy - Light Microscope – working-advanced

Grade 9CBSE

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

🔑Concepts

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The Compound Light Microscope uses visible light and two sets of glass lenses (the ocular and the objective) to magnify small specimens like cells and tissues.

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Total Magnification is the product of the magnifying powers of the ocular lens (eyepiece) and the objective lens. In Grade 9 labs, common ocular lenses are 10×10\times or 15×15\times.

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Resolving Power (Resolution) is the ability to distinguish two closely spaced points as separate entities. It is determined by the wavelength of light (λ\lambda) and the numerical aperture (NANA). The limit of resolution for a light microscope is approximately 0.2μm0.2 \mu m.

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Refractive Index (nn) is a measure of how much light bends when entering a medium. When using an oil immersion lens (100×100\times), immersion oil is used because its refractive index is similar to glass (n≈1.51n \approx 1.51), preventing the loss of light due to refraction.

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Numerical Aperture (NA) describes the light-gathering ability of the objective lens. A higher NANA results in better resolution and a brighter image.

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Path of Light: Light source →\rightarrow Condenser →\rightarrow Specimen →\rightarrow Objective lens →\rightarrow Body tube →\rightarrow Ocular lens →\rightarrow Human eye.

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Parfocal Property: A microscope is said to be parfocal if the specimen remains nearly in focus when the user switches from a lower power objective to a higher power objective.

📐Formulae

Mtotal=Mobjective×MocularM_{total} = M_{objective} \times M_{ocular}

d=λ2⋅NAd = \frac{\lambda}{2 \cdot NA}

NA=n⋅sin⁡(θ)NA = n \cdot \sin(\theta)

Actual Size=Observed SizeMagnificationActual \text{ } Size = \frac{Observed \text{ } Size}{Magnification}

💡Examples

Problem 1:

A student observes a plant cell using a microscope. The eyepiece is marked 10×10\times and the objective lens being used is marked 40×40\times. Calculate the total magnification. If the cell appears to be 4mm4 mm long under the microscope, what is its actual size in micrometers (μm\mu m)?

Solution:

First, calculate total magnification: Mtotal=10×40=400×M_{total} = 10 \times 40 = 400\times

Now, calculate actual size using the formula: Actual Size=4 mm400=0.01 mmActual \text{ } Size = \frac{4 \text{ } mm}{400} = 0.01 \text{ } mm

Convert to micrometers: 0.01 mm×1000=10μm0.01 \text{ } mm \times 1000 = 10 \mu m

Explanation:

The total magnification is found by multiplying the power of the two lenses. The actual size is the observed size divided by that magnification factor. Since 1mm=1000μm1 mm = 1000 \mu m, we multiply by 10001000 to get the final answer in μm\mu m.

Problem 2:

Why is immersion oil used with a 100×100\times objective lens? Explain using the concept of refractive index (nn).

Solution:

The refractive index of air is approximately 1.01.0, while the refractive index of glass is approximately 1.51.5. When light passes from the glass slide into the air, it bends (refracts) away from the objective lens. By placing immersion oil (n≈1.51n \approx 1.51) between the slide and the lens, the light does not refract as much, and more light is captured by the objective.

Explanation:

Matching the refractive index of the medium to the glass prevents light scattering, which increases the Numerical Aperture (NANA) and improves the resolution of the image.