Optical instruments

Whether an object 'looks big' has nothing to do with its actual size alone -- a huge, distant mountain can appear smaller than a nearby small tree, because APPARENT size depends on the VISUAL ANGLE the object subtends at the eye, and this visual angle shrinks steadily with increasing distance. To make something look bigger, we must either physically bring it closer, or use an optical instrument that increases the visual angle without actually moving the object at all -- this increase is called ANGULAR MAGNIFICATION, or MAGNIFYING POWER. The unaided human eye, however, cannot focus clearly on anything closer than the LEAST DISTANCE OF DISTINCT VISION, $D=25$ cm for a normal eye; trying to bring an object closer than $D$, or trying to resolve something too small even when held at $D$, both fail without optical help -- which is precisely why a microscope (for small nearby objects) or a telescope (for objects too distant to ever be brought closer at all) becomes necessary.

Magnifying power is formally defined as the ratio of the visual angle $\beta$ subtended by the image formed by the optical instrument, to the visual angle $\alpha$ that the SAME object would subtend if it were instead placed at the least distance of distinct vision $D$ (for telescopes specifically, $\alpha$ is instead the angle the object subtends from its own actual position, since a star or planet can never physically be brought any closer).

SIMPLE MICROSCOPE (reading glass): a single convex lens, used with the object placed WITHIN its focal length, so that it forms an erect, virtual, magnified image. Since for small angles $\tan\approx$ the angle itself, $M=\dfrac{\tan\beta}{\tan\alpha}=\dfrac{D}{|u|}$ (numerically). For MAXIMUM magnifying power the image should be brought as close as the eye can still focus on, i.e. $v=-D$; substituting this into $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ gives $M_{max}=1+\dfrac{D}{f}$. For MINIMUM magnifying power the image instead forms at infinity (object exactly at the focus, $u=f$ numerically), giving $M_{min}=\dfrac{D}{f}$. So a lens of focal length $f$ provides angular magnification somewhere between $D/f$ and $1+D/f$ only; for the standard $D=25$ cm and, say, $f=5$ cm, this gives $M_{min}=5$ and $M_{max}=6$ -- the image looks only 5 to 6 times bigger, even though the corresponding LINEAR (lateral) magnification $m=v/u$ can in principle be made arbitrarily large (even effectively infinite, as $v\to\infty$ at $u=f$).

Worked illustration: a 10 cm focal length magnifying glass views letters 0.5 mm thick, held 8 cm from the lens. $\frac{1}{10}=\frac{1}{v}-\frac{1}{-8}\Rightarrow v=-40$ cm; $M=D/|u|=25/8=3.125$, so the image is $3.125\times0.5=1.5625$ mm and the letters appear about 3.125 times bigger. Held instead at only 5 cm (closer than the 10 cm focal length): solving gives $v=-10$ cm, which is CLOSER than the eye's least distance of distinct vision (25 cm), so the resulting (magnified) image cannot actually be brought into clear focus by a normal eye -- the letters cannot be comfortably read from 5 cm.

COMPOUND MICROSCOPE: since a simple microscope's power can only be raised by shrinking $f$, and a very small $f$ makes the lens thick (worsening both spherical and chromatic aberration), still higher magnifying power instead uses TWO convex lenses mounted in a tube -- a small-aperture OBJECTIVE facing the tiny object, and a somewhat larger EYE LENS the observer actually looks through. The object is placed between $f$ and $2f$ of the objective, which forms a real, inverted, magnified intermediate image positioned just inside the focus of the eye lens; the eye lens then acts on this intermediate image exactly as a simple microscope would, producing a final virtual, further-magnified image (inverted relative to the original object). The overall magnifying power is the PRODUCT $M=m_o\times M_e$, where $m_o=v_o/u_o$ is the objective's ordinary LINEAR magnification and $M_e=D/u_e$ is the eye lens's own ANGULAR magnifying power (behaving as a simple microscope on the intermediate image); the microscope's overall length is $L=v_o+u_e$. Raising $m_o$ (by pushing $u_o$ closer to the objective's focus) simultaneously increases $v_o$, and hence $L$ -- so $m_o$ can only be pushed so far before the instrument becomes impractically long. The eye lens's own $M_e$ similarly ranges between $D/f_e$ (final image at infinity) and $1+D/f_e$ (final image at $D$).

Worked illustration: eye lens $f_e=6.25$ cm, objective $f_o=2$ cm, overall microscope length 15 cm, with the final image at its biggest (i.e. at $D=25$ cm, so $v_e=-25$ cm). From $\frac{1}{6.25}=\frac{1}{-25}-\frac{1}{u_e}\Rightarrow u_e=-5$ cm; then $v_o=L-|u_e|=15-5=10$ cm; from $\frac{1}{2}=\frac{1}{10}-\frac{1}{u_o}\Rightarrow u_o=-2.5$ cm; the overall magnifying power is $M=m_o\times M_e=\dfrac{v_o}{u_o}\times\dfrac{D}{u_e}=\dfrac{10}{2.5}\times\dfrac{25}{5}=4\times5=20$. …