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Microscope and Microscopy - Types of Light microscope-advanced

Grade 9CBSE

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

🔑Concepts

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A Compound Microscope uses a two-lens system to achieve high magnification: the objective lens (closer to the object) and the ocular lens or eyepiece (closer to the eye).

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Magnification (MM) is defined as the ratio of the size of the image to the actual size of the object. For a compound microscope, the total magnification is the product of the magnification of the objective (MoM_o) and the eyepiece (MeM_e).

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Resolving Power (RR) is the ability of an optical instrument to show two close objects as separate. It is the reciprocal of the limit of resolution (dd).

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The Limit of Resolution (dd) is the minimum distance between two points at which they can still be seen as distinct. A smaller dd value means higher resolution. It is given by d=0.61λNAd = \frac{0.61 \lambda}{NA}, where λ\lambda is the wavelength of light used.

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Numerical Aperture (NANA) measures the light-gathering capacity of the lens and is given by NA=nsin⁡θNA = n \sin \theta, where nn is the refractive index of the medium between the lens and the specimen.

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Phase-Contrast Microscopy is an advanced technique that allows the visualization of living, unstained cells by converting phase shifts in light passing through a transparent specimen into amplitude (brightness) changes.

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Dark-field Microscopy uses a special condenser that blocks direct light, allowing only light scattered by the specimen to enter the objective. This results in a bright image against a dark background, ideal for observing very thin bacteria.

📐Formulae

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

d=0.61λNAd = \frac{0.61 \lambda}{NA}

NA=nsin⁡θNA = n \sin \theta

R=1d=NA0.61λR = \frac{1}{d} = \frac{NA}{0.61 \lambda}

💡Examples

Problem 1:

Calculate the total magnification of a compound microscope if the eyepiece has a magnifying power of 10x10x and the high-power objective lens has a magnifying power of 45x45x.

Solution:

Given: Meyepiece=10xM_{eyepiece} = 10x Mobjective=45xM_{objective} = 45x Using the formula: Mtotal=Mobjective×MeyepieceM_{total} = M_{objective} \times M_{eyepiece} Mtotal=45×10=450xM_{total} = 45 \times 10 = 450x

Explanation:

The total magnification is the product of the individual magnifications of the lenses in the optical path.

Problem 2:

Determine the limit of resolution (dd) for a microscope using blue light of wavelength λ=450 nm\lambda = 450\text{ nm} and a lens with a numerical aperture (NANA) of 1.251.25.

Solution:

Given: λ=450 nm\lambda = 450\text{ nm} NA=1.25NA = 1.25 Using the formula: d=0.61λNAd = \frac{0.61 \lambda}{NA} d=0.61×4501.25d = \frac{0.61 \times 450}{1.25} d=274.51.25=219.6 nmd = \frac{274.5}{1.25} = 219.6\text{ nm}

Explanation:

The limit of resolution indicates the smallest detail the microscope can resolve. Using shorter wavelengths (like blue light) and higher NANA improves (decreases) the limit of resolution.