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Microscope and Microscopy - What is a Microscope?-advanced

Grade 9CBSE

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

🔑Concepts

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Definition: A microscope is a high-precision optical instrument designed to produce magnified images of small objects. Microscopy is the technical field of using microscopes to view samples and objects that cannot be seen with the unaided eye.

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Compound Microscope: This instrument uses a system of two or more lenses. The objective lens (near the specimen) forms a real image, which is then further magnified by the eyepiece (ocular lens) to form a virtual image for the observer.

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Magnification (MM): It is the process of enlarging the apparent size of an object. The total magnification of a compound microscope is calculated as the product of the magnifying powers of the objective and eyepiece lenses.

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Resolving Power (Resolution): The ability of an optical system to distinguish two close points as separate entities. The limit of resolution (dd) is determined by the wavelength of light (λ\lambda) and the numerical aperture (NANA). A smaller value of dd indicates higher resolution.

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Numerical Aperture (NANA): A measure of the light-gathering capacity of the lens system, defined by the formula NA=nsin⁡θNA = n \sin \theta, where nn is the refractive index of the medium.

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Electron Microscope: An advanced microscope that uses a beam of accelerated electrons instead of light. Because the wavelength of electrons is much shorter than that of visible light, electron microscopes achieve magnifications up to 1,000,000×1,000,000\times and much higher resolution (0.20.2 to 0.50.5 nmnm) compared to light microscopes (200200 nmnm).

📐Formulae

Mtotal=Mobjective×MeyepieceM_{total} = M_{objective} \times M_{eyepiece}

d=0.61λNAd = \frac{0.61 \lambda}{NA}

NA=nsin⁡θNA = n \sin \theta

Actual Size=Image SizeMagnification\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}

💡Examples

Problem 1:

A student is observing a plant cell using a compound microscope. The eyepiece lens has a magnification of 10×10\times and the objective lens has a magnification of 40×40\times. Calculate the total magnification. If the cell appears to be 44 mmmm long under this microscope, what is its actual size in micrometers (μm\mu m)?

Solution:

Step 1: Calculate Total Magnification (MtotalM_{total}). Mtotal=10×40=400×M_{total} = 10 \times 40 = 400\times Step 2: Calculate the actual size of the cell. Actual Size=Apparent SizeMagnification\text{Actual Size} = \frac{\text{Apparent Size}}{\text{Magnification}} Actual Size=4 mm400=0.01 mm\text{Actual Size} = \frac{4 \text{ mm}}{400} = 0.01 \text{ mm} Step 3: Convert mmmm to μm\mu m. Since 1 mm=1000μm1 \text{ mm} = 1000 \mu m: 0.01×1000=10μm0.01 \times 1000 = 10 \mu m

Explanation:

The total magnification is the product of the power of the two lenses. To find the actual size, we divide the observed size by the magnification factor and then convert the units to standard biological scales (micrometers).

Problem 2:

Compare the resolution of a light microscope using light of wavelength λ=550\lambda = 550 nmnm and NA=0.25NA = 0.25 with an electron microscope. Why can the electron microscope see smaller structures?

Solution:

For the light microscope: d=0.61×550 nm0.25d = \frac{0.61 \times 550 \text{ nm}}{0.25} d=335.50.25=1342 nmd = \frac{335.5}{0.25} = 1342 \text{ nm} In contrast, an electron microscope uses electron beams with wavelengths as small as 0.0050.005 nmnm.

Explanation:

Resolution (dd) is directly proportional to the wavelength (λ\lambda). Since electrons have a significantly smaller wavelength than photons of visible light, the limit of resolution (dd) for an electron microscope is much smaller, allowing it to distinguish structures that are nanometers apart.