# JEE Main & Advanced Physics Ray Optics Microscope

Microscope

Category : JEE Main & Advanced

It is an optical instrument used to see very small objects. It's magnifying power is given by

$m=\frac{\text{Visual angle with instrument(}\beta \text{)}}{\text{Visual angle when object is placed at least distance of distinct vision (}\alpha \text{)}}$

(1) Simple microscope

(i) It is a single convex lens of lesser focal length.

(ii) Also called magnifying glass or reading lens.

(iii) Magnification's, when final image is formed at D and $\infty$

(i.e. ${{m}_{D}}$ and ${{m}_{\infty }}$)      ${{m}_{D}}={{\left( 1+\frac{D}{f} \right)}_{\max }}$ and ${{m}_{\infty }}={{\left( \frac{D}{f} \right)}_{\min }}$

(iv) If lens is kept at a distance a from the eye then ${{m}_{D}}=1+\frac{D-a}{f}$ and ${{m}_{\infty }}=\frac{D-a}{f}$ (2) Compound microscope (i) Consist of two converging lenses called objective and eye lens.

(ii) ${{f}_{\text{eye lens}}}>{{f}_{\text{objective}}}$and ${{(\text{diameter)}}_{\text{eye lens}}}>{{(\text{diameter})}_{\text{objective}}}$

(iii) Intermediate image is real and enlarged.

(iv) Final image is magnified, virtual and inverted.

(v) ${{u}_{o}}=$Distance of object from objective (o), ${{v}_{o}}=$Distance of image $({A}'{B}')$ formed by objective from objective, ${{u}_{e}}=$ Distance of ${A}'{B}'$ from eye lens, ${{v}_{e}}=$ Distance of final image from eye lens, ${{f}_{o}}=$ Focal length of objective, ${{f}_{e}}=$ Focal length of eye lens.

(vi) Final image is formed at D : Magnification ${{m}_{D}}=-\frac{{{v}_{o}}}{{{u}_{o}}}\left( 1+\frac{D}{{{f}_{e}}} \right)$ and length of the microscope tube (distance between two lenses) is ${{L}_{D}}={{v}_{o}}+{{u}_{e}}$.

Generally object is placed very near to the principal focus of the objective hence ${{u}_{o}}\tilde{=}\,{{f}_{o}}.$ The eye piece is also of small focal length and the image formed by the objective is also very near to the eye piece.

So ${{v}_{o}}\tilde{=}{{L}_{D}},$ the length of the tube.

Hence, we can write ${{m}_{D}}=\frac{-L}{{{f}_{o}}}\,\left( 1+\frac{D}{{{f}_{e}}} \right)$

(vii) Final image is formed at $\infty$ : Magnification

${{m}_{\infty }}=-\frac{{{v}_{0}}}{{{u}_{0}}}.\frac{D}{{{f}_{e}}}$and length of tube ${{L}_{\infty }}={{v}_{0}}+{{f}_{e}}$

In terms of length ${{m}_{\infty }}=\frac{({{L}_{\infty }}-{{f}_{o}}-{{f}_{e}})D}{{{f}_{o}}{{f}_{e}}}$

(viii) For large magnification of the compound microscope, both ${{f}_{o}}$ and ${{f}_{e}}$ should be small.

(ix) If the length of the tube of microscope increases, then its magnifying power increases.

(x) The magnifying power of the compound microscope may be expressed as $M={{m}_{o}}\times {{m}_{e}}$; where  ${{m}_{o}}$ is the magnification of the objective and ${{m}_{e}}$ is magnifying power of eye piece.

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