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Long Answer Questions · Q25

Q.Obtain the equation for resolving power of a microscope.

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Step 1. A microscope's resolving power depends on the smallest separation dmin⁡d_{\min} between two object points it can still image as distinct, not merely on how much it magnifies them.

Step 2. By Rayleigh's criterion applied at the image plane, the radius of the (just-resolvable) Airy-disc pattern there is r0=1.22fλ/ar_0=1.22f\lambda/a, where ff is (approximately) the image-side distance and aa the objective's own aperture width -- essentially the same circular-aperture diffraction result used for a telescope, applied here at the microscope's image side.

Step 3. This image-plane separation is related back to the object-plane separation dmin⁡d_{\min} by the objective's own (linear) magnification mm: dmin⁡=r0/md_{\min}=r_0/m.

Step 4. Using the half-angle β\beta the objective's own aperture subtends at the object (via tan⁡β≈sin⁡β=a/f\tan\beta\approx\sin\beta=a/f, relating the aperture size aa and the object-side focal distance ff), and working through the algebra combining Steps 2-3 with this angle, the result simplifies to dmin⁡=1.22λ2sin⁡β\boxed{d_{\min}=\dfrac{1.22\lambda}{2\sin\beta}}. …

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