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Microscope and Microscopy - Light (Compound) Microscope-advanced

Grade 9CBSE

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

🔑Concepts

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A compound microscope uses two sets of lenses: the Objective lens (placed near the specimen) and the Ocular lens (eyepiece).

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Magnification refers to the increase in the apparent size of an object. The total magnification is calculated by multiplying the power of the ocular lens by the power of the objective lens.

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Resolving Power (Resolution): The ability to distinguish two closely placed points as separate. The resolution of a standard light microscope is limited by the wavelength of visible light to approximately 0.2 μm0.2\ \mu m (200 nm200\ nm).

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Numerical Aperture (NA): A measure of the light-gathering capacity of the lens system. Higher NANA values provide better resolution and brighter images.

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Refractive Index (nn): The ratio of the speed of light in a vacuum to its speed in a medium. Immersion oil (with n≈1.5n \approx 1.5) is used with 100×100\times objectives to prevent light refraction and improve NANA.

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Field of View (FOV): The diameter of the circular area visible through the microscope. As magnification increases, the FOV decreases inversely.

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Working Distance: The distance between the front lens of the objective and the cover slip of the specimen when the image is in focus. It decreases as magnification increases.

📐Formulae

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

d=0.61λNAd = \frac{0.61 \lambda}{NA}

NA=nsin⁡θNA = n \sin \theta

Actual Size=Image SizeMagnificationActual\ Size = \frac{Image\ Size}{Magnification}

FOV2=FOV1×M1M2FOV_2 = \frac{FOV_1 \times M_1}{M_2}

💡Examples

Problem 1:

A student is using a compound microscope equipped with a 15×15\times eyepiece and a 40×40\times objective lens. If they observe a bacterial cell that appears to be 1.2 mm1.2\ mm long under the microscope, what is the actual length of the cell in micrometers (μm\mu m)?

Solution:

  1. Calculate Total Magnification: Mtotal=15×40=600M_{total} = 15 \times 40 = 600

  2. Use the magnification formula to find actual size: Actual Size=Image SizeMagnificationActual\ Size = \frac{Image\ Size}{Magnification} Actual Size=1.2 mm600=0.002 mmActual\ Size = \frac{1.2\ mm}{600} = 0.002\ mm

  3. Convert mmmm to μm\mu m (1 mm=1000 μm1\ mm = 1000\ \mu m): 0.002×1000=2 μm0.002 \times 1000 = 2\ \mu m

Explanation:

The total magnification is the product of the ocular and objective powers. Dividing the apparent (image) size by this total magnification gives the actual size in millimeters, which is then converted to micrometers.

Problem 2:

If the Field of View (FOV) at 100×100\times magnification is 2.0 mm2.0\ mm, calculate the Field of View when the student switches to 400×400\times magnification.

Solution:

Using the inverse relationship between magnification and FOV: FOV1×M1=FOV2×M2FOV_1 \times M_1 = FOV_2 \times M_2 2.0 mm×100=FOV2×4002.0\ mm \times 100 = FOV_2 \times 400 FOV2=200400FOV_2 = \frac{200}{400} FOV2=0.5 mmFOV_2 = 0.5\ mm

Explanation:

Because magnification and the diameter of the field of view are inversely proportional, increasing the magnification by a factor of 44 (from 100100 to 400400) reduces the field of view by a factor of 44 (from 2.02.0 to 0.50.5).