For a higher resolving power of a compound microscope, the wavelength of light used should be ______ .

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Q. For a higher resolving power of a compound microscope, the wavelength of light used should be ___________ .

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Concept understanding — Resolving Power of Microscope

The Core Problem: When Two Points Become One Blur

Imagine you are looking at two tiny dots drawn very close together on a piece of paper. From far away, they look like a single dot. As you bring the paper closer, at some point your eye suddenly sees two separate dots. That moment — the threshold where your eye (or a microscope) can just barely tell that there are two objects instead of one — is the heart of resolving power.

A microscope's job is to show you fine detail. But no matter how good the lenses are, there is a fundamental limit: light itself behaves like a wave. When light passes through the circular opening of a lens, it does not travel in perfect straight lines. It spreads out and forms a pattern called an Airy disk — a bright central spot surrounded by faint rings. Every point in your specimen becomes a tiny blurry disk in the image, not a perfect point.

If two points in the specimen are very close, their Airy disks overlap. When they overlap too much, your eye cannot tell them apart — they merge into one blob. The resolving power of a microscope is its ability to show two closely spaced points as distinct.

Note

Resolving power is not about magnification. You can magnify a blurry image as much as you like — it only becomes a bigger blur. Resolution is about separating detail, not enlarging it.

The Precise Criterion: Lord Rayleigh's Condition

Lord Rayleigh proposed a practical rule: two points are just resolved when the centre of one Airy disk falls exactly on the first dark ring of the other. At that point, the combined intensity has a small dip between the two peaks — your eye can just detect that there are two sources.

For a microscope, the smallest distance $d$ between two points that can just be resolved is given by:

d = \frac{1.22 \lambda}{2 n \sin \beta}

where:

$\lambda$ is the wavelength of light used
$n$ is the refractive index of the medium between the specimen and the objective lens
$\beta$ is the half-angle of the cone of light entering the objective

The quantity $n \sin \beta$ is called the numerical aperture (NA) of the objective lens. So the formula is often written as:

d = \frac{1.22 \lambda}{2 \cdot \text{NA}}

\text{Resolving power} = \frac{1}{d} = \frac{2 \cdot \text{NA}}{1.22 \lambda}

What This Tells Us: Two Levers for Better Resolution

1. Shorter wavelength $\lambda$ — Blue light resolves better than red light. Ultraviolet light resolves even better, which is why electron microscopes (using much shorter "wavelengths" of electrons) can see atoms.

2. Larger numerical aperture $n \sin \beta$ — You can increase $n$ by using oil between the slide and the objective (oil immersion). Air has $n \approx 1$ , but special oils have $n \approx 1.5$ . You can increase $\sin \beta$ by using a lens that collects light from a wider cone — a lens with a shorter focal length and larger diameter. …

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The resolving power of a microscope is inversely proportional to the wavelength of light used. To get higher resolving power, we need a shorter wavelength, so the blank should be filled with small or short.

The Concept: What Resolving Power Really Means

When you look through a microscope, you want to see fine details — two tiny dots close together should appear as two separate dots, not one blurry blob. The resolving power is the microscope's ability to distinguish between two closely spaced objects as distinct. It is not the same as magnification; you can magnify a blurry image all you want, but you won't see more detail.

The key formula that governs this is the Abbe diffraction limit for a microscope:

\text{Resolving power} \propto \frac{1}{\text{minimum resolvable distance } d} \propto \frac{1}{\lambda}

More precisely, the minimum distance $d$ that can be resolved is given by:

d = \frac{0.61 \lambda}{n \sin \theta}

where $\lambda$ is the wavelength of light used, $n$ is the refractive index of the medium between the specimen and the objective lens, and $\theta$ is the half-angle of the cone of light entering the objective.

Notice that $d$ is the smallest separation you can see. A smaller $d$ means higher resolving power (you can see finer details). Since $d$ is directly proportional to $\lambda$ , a smaller $\lambda$ gives a smaller $d$ , and thus a higher resolving power.

Step-by-Step Reasoning

Identify the goal: We want higher resolving power. That means we want to see finer details, so the minimum distance $d$ between two distinguishable points must become smaller.
Look at the formula: From $d = \frac{0.61 \lambda}{n \sin \theta}$ , for a fixed microscope (where $n$ and $\theta$ are constant), $d$ is directly proportional to $\lambda$ .
Apply the relationship: If $\lambda$ decreases, $d$ decreases. A smaller $d$ means better resolution — we can distinguish objects that are closer together. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set 55/2/11 mark
$Q. For a higher resolving power of a compound microscope, the wavelength of light used should be ___________ .$
›Reveal solutionSolution
The resolving power of a microscope is inversely proportional to the wavelength of light used. To get higher resolving power, we need a shorter wavelength, so the blank should be filled with small or short.

The Concept: What Resolving Power Really Means

When you look through a microscope, you want to see fine details — two tiny dots close together should appear as two separate dots, not one blurry blob. The resolving power is the microscope's ability to distinguish between two closely spaced objects as distinct. It is not the same as magnification; you can magnify a blurry image all you want, but you won't see more detail.

The key formula that governs this is the Abbe diffraction limit for a microscope:
$\text{Resolving power} \propto \frac{1}{\text{minimum resolvable distance } d} \propto \frac{1}{\lambda}$
More precisely, the minimum distance $d$ that can be resolved is given by:
$d = \frac{0.61 \lambda}{n \sin \theta}$
where $\lambda$ is the wavelength of light used, $n$ is the refractive index of the medium between the specimen and the objective lens, and $\theta$ is the half-angle of the cone of light entering the objective.

Notice that $d$ is the smallest separation you can see. A smaller $d$ means higher resolving power (you can see finer details). Since $d$ is directly proportional to $\lambda$ , a smaller $\lambda$ gives a smaller $d$ , and thus a higher resolving power.

Step-by-Step Reasoning

Identify the goal: We want higher resolving power. That means we want to see finer details, so the minimum distance $d$ between two distinguishable points must become smaller.

Look at the formula: From $d = \frac{0.61 \lambda}{n \sin \theta}$ , for a fixed microscope (where $n$ and $\theta$ are constant), $d$ is directly proportional to $\lambda$ .

Apply the relationship: If $\lambda$ decreases, $d$ decreases. A smaller $d$ means better resolution — we can distinguish objects that are closer together. …
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CBSE 2020Set ANNUAL1 mark
$Q. Mention a method to increase the resolving power of a microscope.$
›Reveal solutionSolution

Resolving power $=\dfrac{2\mu\sin\beta}{\lambda}$ ; increase it by using a shorter wavelength $\lambda$ or a larger numerical aperture $\mu\sin\beta$ (e.g. oil immersion).

Concept. The resolving power of a microscope is

$R.P. = \frac{2\mu\sin\beta}{\lambda},$

where $\lambda$ is the wavelength of light used, $\mu$ the refractive index of the medium between the object and objective, and $\beta$ the half-angle of the cone of light entering the objective. ( $\mu\sin\beta$ is the numerical aperture.)

…

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Q.For a higher resolving power of a compound microscope, the wavelength of light used should be ___________ .

The Concept: What Resolving Power Really Means

Step-by-Step Reasoning

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