Resolving power of microscope and telescope — Rayleigh criterion

Question

State Rayleigh’s criterion for resolving power. Derive the expression for the resolving power of (a) a telescope and (b) a microscope. How can the resolving power of each be improved?

(JEE Advanced 2022, similar pattern)

Solution — Step by Step

Two point sources are just resolved when the central maximum of the diffraction pattern of one falls on the first minimum of the other. This is the limit of resolution — any closer and the two images merge into one.

The angular separation at this limit is called the minimum resolvable angle $\theta_{min}$ .

For a telescope with objective aperture diameter $D$ , the minimum resolvable angle (from single-slit diffraction for a circular aperture) is:

\theta_{min} = \frac{1.22\lambda}{D}

The resolving power is the reciprocal of this:

R.P. = \frac{1}{\theta_{min}} = \frac{D}{1.22\lambda}

To improve: use a larger aperture $D$ or observe at shorter wavelength $\lambda$ . This is why radio telescopes are enormous and space telescopes observe in UV/X-ray.

For a microscope, the minimum distance between two resolvable point objects is:

d_{min} = \frac{1.22\lambda}{2n\sin\theta} = \frac{1.22\lambda}{2 \cdot NA}

where $n$ is the refractive index of the medium between the object and objective, $\theta$ is the half-angle of the cone of light, and $NA = n\sin\theta$ is the numerical aperture.

R.P. = \frac{1}{d_{min}} = \frac{2n\sin\theta}{1.22\lambda}

To improve: use shorter wavelength (electron microscopes use this), increase numerical aperture (oil immersion increases $n$ ), or use a wider cone angle.

Why This Works

Every optical instrument has a fundamental resolution limit set by diffraction — not by lens quality or magnification. Even a perfect lens cannot resolve details smaller than about half a wavelength. This is why we moved to electron microscopes for nanoscale imaging — electrons have much shorter de Broglie wavelengths than visible light.

The factor 1.22 comes from the mathematics of diffraction through a circular aperture (the first zero of the Bessel function $J_1$ ). For a rectangular slit, the factor would simply be 1.

Alternative Method — Memorise Using Physical Reasoning

For a telescope: it looks at distant objects (angular separation matters), so R.P. is about angles → $D/\lambda$ .

For a microscope: it looks at nearby objects (linear separation matters), so R.P. is about distances → $NA/\lambda$ .

JEE often asks: “Which has better resolving power — a telescope with 10 cm aperture or one with 20 cm?” The answer is the 20 cm one (R.P. $\propto D$ ). For microscopes, the trick question is: “Does higher magnification mean better resolving power?” No — magnification and resolving power are independent. You can magnify a blurry image without adding detail.

Common Mistake

Students confuse magnifying power with resolving power. Magnifying power makes the image bigger; resolving power determines whether two close objects can be distinguished as separate. A microscope can have 1000x magnification but poor resolving power if the wavelength is large. Always treat these as independent quantities.

Question

Solution — Step by Step

Why This Works

Alternative Method — Memorise Using Physical Reasoning

Common Mistake

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