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Microscope and Microscopy - By what factor is the image larger than the actual object?-advanced

Grade 9CBSE

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

🔑Concepts

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Magnification (MM) is defined as the ratio of the size of the image (II) to the actual size of the object (OO). It represents how many times larger the image appears compared to the specimen.

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The total magnification in a compound microscope is the product of the magnifying power of the ocular lens (eyepiece) and the objective lens.

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Resolution (Resolving Power) is the ability of a microscope to distinguish two close points as separate entities. High magnification without high resolution leads to 'empty magnification', where the image is larger but blurry.

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Microscopic measurements often require unit conversions: 1 mm=103μm1 \text{ mm} = 10^3 \mu\text{m} (micrometers) and 1μm=103 nm1 \mu\text{m} = 10^3 \text{ nm} (nanometers).

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The limit of resolution (dd) for a light microscope is approximately 0.2μm0.2 \mu\text{m}. Any object smaller than this cannot be seen clearly regardless of the magnification factor.

📐Formulae

M=Size of Image (I)Actual Size of Object (O)M = \frac{\text{Size of Image (I)}}{\text{Actual Size of Object (O)}}

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

Actual Size (O)=Size of Image (I)Magnification (M)\text{Actual Size (O)} = \frac{\text{Size of Image (I)}}{\text{Magnification (M)}}

1 mm=1000μm1 \text{ mm} = 1000 \mu\text{m}

1μm=1000 nm1 \mu\text{m} = 1000 \text{ nm}

💡Examples

Problem 1:

A student uses a microscope with a 15×15\times eyepiece and a 40×40\times objective lens. What is the total magnification of the specimen?

Solution:

Mtotal=Mocular×MobjectiveM_{\text{total}} = M_{\text{ocular}} \times M_{\text{objective}} Mtotal=15×40=600×M_{\text{total}} = 15 \times 40 = 600\times

Explanation:

The total magnifying power is calculated by multiplying the power of the two lenses used in the optical path.

Problem 2:

A bacterial cell is observed under a microscope with a magnification of 1000×1000\times. The image of the cell measures 5 mm5 \text{ mm} in length. Calculate the actual size of the cell in micrometers (μm\mu\text{m}).

Solution:

First, convert the image size to micrometers: I=5 mm=5×1000μm=5000μmI = 5 \text{ mm} = 5 \times 1000 \mu\text{m} = 5000 \mu\text{m} Now, use the formula for actual size: O=IMO = \frac{I}{M} O=5000μm1000=5μmO = \frac{5000 \mu\text{m}}{1000} = 5 \mu\text{m}

Explanation:

To find the actual size, divide the image size (converted to the required units) by the magnification factor.

Problem 3:

An object has an actual length of 0.02 mm0.02 \text{ mm}. If it appears 4 cm4 \text{ cm} long under a microscope, what is the magnification used?

Solution:

Convert both measurements to the same unit (e.g., mm): Actual size (O)=0.02 mm\text{Actual size (O)} = 0.02 \text{ mm} Image size (I)=4 cm=40 mm\text{Image size (I)} = 4 \text{ cm} = 40 \text{ mm} Now apply the magnification formula: M=IOM = \frac{I}{O} M=40 mm0.02 mmM = \frac{40 \text{ mm}}{0.02 \text{ mm}} M=40002=2000×M = \frac{4000}{2} = 2000\times

Explanation:

Magnification is a dimensionless ratio, so the units of image size and actual size must be identical before division.