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Microscope and Microscopy - How Does a Microscope Work?-advanced

Grade 9CBSE

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

🔑Concepts

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A Compound Microscope uses two or more double convex lenses to magnify small objects. The primary principle is the refraction of light as it passes through different media (glass and air).

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The Objective Lens (near the object) forms a real, inverted, and magnified image of the specimen inside the body tube.

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The Ocular Lens (eyepiece) acts as a simple magnifier, enlarging the primary image to produce a final virtual image that is inverted relative to the original object.

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Magnification (MM) is the ability of a microscope to enlarge an object. It is a dimensionless ratio of the image size to the actual object size.

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Resolving Power (RPRP) or Resolution is the ability to distinguish two closely placed points as separate entities. It is inversely proportional to the limit of resolution (dd).

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Numerical Aperture (NANA) measures the light-gathering capacity of the lens. It depends on the refractive index (nn) of the medium between the lens and the specimen and the half-angle (θ\theta) of the cone of light entering the lens.

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Oil Immersion technique involves placing a drop of oil with a refractive index similar to glass (n≈1.51n \approx 1.51) between the slide and the objective lens. This prevents light rays from bending away, increasing the NANA and improving resolution.

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The Electron Microscope uses beams of electrons instead of light. Since the wavelength (λ\lambda) of electrons is significantly smaller than that of visible light, it provides much higher resolution (up to 0.20.2 nm).

📐Formulae

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

NA=nsin⁡θNA = n \sin \theta

d=0.61λNAd = \frac{0.61 \lambda}{NA}

RP=1d=NA0.61λRP = \frac{1}{d} = \frac{NA}{0.61 \lambda}

💡Examples

Problem 1:

A student is using a compound microscope with an eyepiece marked 15×15 \times and an objective lens marked 45×45 \times. Calculate the total magnification of the specimen being observed.

Solution:

Given: Magnification of Eyepiece (MeM_e) = 15×15 \times Magnification of Objective (MoM_o) = 45×45 \times

Using the formula: Mtotal=Mo×MeM_{total} = M_o \times M_e Mtotal=45×15M_{total} = 45 \times 15 Mtotal=675×M_{total} = 675 \times

Explanation:

The total magnification is the product of the individual magnifying powers of the two lens systems.

Problem 2:

Compare the limit of resolution (dd) of a microscope using blue light (λ=450\lambda = 450 nm) versus red light (λ=700\lambda = 700 nm), assuming the Numerical Aperture (NANA) remains constant at 1.251.25.

Solution:

For Blue Light: dblue=0.61×4501.25=274.51.25=219.6 nmd_{blue} = \frac{0.61 \times 450}{1.25} = \frac{274.5}{1.25} = 219.6 \text{ nm}

For Red Light: dred=0.61×7001.25=4271.25=341.6 nmd_{red} = \frac{0.61 \times 700}{1.25} = \frac{427}{1.25} = 341.6 \text{ nm}

Since dblue<dredd_{blue} < d_{red}, blue light provides better resolution.

Explanation:

Resolution is directly proportional to the wavelength. Shorter wavelengths (like blue light) result in a smaller limit of resolution, which means higher detail can be seen.