Exercises — Optical instruments — human eye, simple microscope, compound microscope, telescope

WHAT we want: the number that says "how many times bigger the angle looks." WHY this formula: relaxed eye means the image sits at infinity, which happens when the object is exactly at the focus. As derived above, the angular power is then $M_{\infty }=\frac{D}{f}.$ Plug in: $M=\frac{25}{5}=\boxed{5}$. The object looks $5\times$ larger in angle than at the naked-eye near point.

WHY these formulas: a telescope views objects at infinity, so the reference angle is the real angle the object subtends — the actual angle $\alpha$ between rays from its top and bottom, roughly $h/f_o$ once focused — not $h/D$ (there is no bringing a star to the near point). Here $h$ is the height of the intermediate image and $\alpha ,\beta$ are the incoming and outgoing angles at the eye. Normal adjustment ⇒ the two foci coincide. $M=\frac{f_o}{f_e}=\frac{120}{3}=\boxed{40}.$ WHY the tube length is $f_o+f_e$ (not a black box): parallel starlight is focused by the objective at its focal point, a distance $f_o$ behind the objective. For a relaxed eye the eyepiece must send rays out parallel, which happens only when its focal point sits on that very same image — i.e. the eyepiece is a further $f_e$ down the tube. So the lens-to-lens gap is $L=f_o+f_e=120+3=\boxed{123\text{ cm}}.$

Magnification — WHY this formula: placing the virtual image at $D$ makes the object subtend the largest angle the eye can still focus, so this is the maximum magnifier setting. Using the derived result, $M_D=1+\frac{D}{f}=1+\frac{25}{5}=\boxed{6}.$ The extra "$+1$" over the relaxed value $5$ is the reward for straining the eye.

Object distance — WHY the lens equation: we know where the image is ($v=-25$, virtual ⇒ negative per our convention) and want where to put the object ($u$). The lens equation links exactly those two with $f$. Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with $v=-25\text{ cm}$, $f=+5\text{ cm}$: $\frac{1}{-25}-\frac{1}{u}=\frac{1}{5}\Rightarrow \frac{1}{u}=-\frac{1}{25}-\frac{1}{5}=-\frac{6}{25}.$ So $u=-\frac{25}{6}\approx \boxed{-4.17\text{ cm}}$. The minus confirms a real object; $|u|=4.17<f=5$, so the object sits inside the focal length, exactly as a magnifier requires.

Objective — WHY $m_o\approx L/f_o$ (deriving the approximation): the object sits just outside $f_o$, so $|u_o|\approx f_o$. The real image forms near the far focus of the tube, at $v_o\approx L$. Then the linear magnification magnitude ($m_o=v_o/u_o$, a height ratio) is $|m_o|=\frac{|v_o|}{|u_o|}\approx \frac{L}{f_o}=\frac{15}{1.5}=10.$ The image is inverted, so with sign $m_o\approx -10$. Eyepiece — WHY $D/f_e$: the eyepiece is a simple magnifier acting on that intermediate image, relaxed eye ⇒ image at infinity. Its factor is the magnifier's angular power $M_{\infty }=D/f_e$; we write it $m_e$ but remember it is a capital-$M$ (angular) quantity: $m_e=\frac{D}{f_e}=\frac{25}{5}=5.$ They multiply (the eyepiece enlarges the already-enlarged picture), and the signs multiply too: $M=m_o{m}_e\approx (-10)(5)=-50\Rightarrow |M|=\boxed{50}.$ The minus tells you the final image is inverted — which is fine for looking at cells.

WHY: $M=f_o/f_e$ grows when the objective is long and the eyepiece is short. Correct choice: objective $f_o=80\text{ cm}$, eyepiece $f_e=2\text{ cm}$: $M=\frac{80}{2}=\boxed{40}.$ Swapped (objective $2$, eyepiece $80$): $M=\frac{2}{80}=0.025(<1),$ which shrinks the angle — a "wrong-way" telescope, exactly what you see looking through binoculars backwards.

WHY start from $M_{\infty }=D/f$: it is the only relation linking the given magnification to the unknown focal length. Invert it: $f=\frac{D}{M_{\infty }}=\frac{25}{4}=6.25\text{ cm}=0.0625\text{ m}.$ Power is $P=1/f$ with $f$ in metres (the definition of the dioptre): $P=\frac{1}{0.0625}=\boxed{16\text{D}}.$

WHY this formula: relaxed-eye compound power is $M=\frac{L}{f_o}\cdot \frac{D}{f_e}$ (objective size factor × eyepiece angular factor). We fix everything except $f_e$ and solve for it — that is how you design to a target. $300=\frac{20}{1}\cdot \frac{25}{f_e}=\frac{500}{f_e}.$ Invert for $f_e$: $f_e=\frac{500}{300}=\boxed{1.67\text{ cm}}(=5/3\text{ cm}).$ Sanity check: $m_o=20/1=20$, $m_e=25/(5/3)=15$, and $20\times 15=300$. ✓

First fix the symbols on the ray picture. Let $h$ be the height of the intermediate image formed by the objective in its focal plane. Let $\alpha$ be the angle the distant object subtends at the eye (the "without instrument" angle), and $\beta$ the angle the final image subtends (the "with instrument" angle). Both are small angles measured at the axis.

(a) Normal adjustment — WHY $f_o/f_e$: parallel rays from the distant object arrive at angle $\alpha$; the objective focuses them to the image of height $h$ in its focal plane, so the geometry of that right triangle gives $\alpha \approx \frac{h}{f_o}$ (opposite $h$ over adjacent $f_o$). The eyepiece has this same image at its focus, so it sends rays out parallel at angle $\beta \approx \frac{h}{f_e}$. Therefore $M=\frac{\beta }{\alpha }=\frac{h/f_e}{h/f_o}=\frac{f_o}{f_e}=\frac{100}{5}=\boxed{20}.$ The $h$ cancels — which is exactly why the object's real height never appears in the answer.

(b) Near point — WHY the extra factor $\left(1+\frac{f_e}{D}\right)$, derived: now the eyepiece must throw its final image to $D$ (not infinity). By the magnifier near-point rule applied to the eyepiece, its angular factor rises from $\frac{1}{f_e}$-type to include the "$+$" term: the exit angle becomes ${\beta }^{\prime }=\frac{h}{f_e}\left(1+\frac{f_e}{D}\right)$. Dividing by the same $\alpha =h/f_o$, $M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=20\left(1+\frac{5}{25}\right)=20\times 1.2=\boxed{24}.$ Fractional gain: $(24-20)/20=0.2=20\%$. Note this is a smaller relative boost than a microscope's "$+1$", because for a telescope the extra factor is $f_e/D$, and $f_e\ll D$.

(a) Objective image (the yellow-ray crossing in Figure 1). Real object ⇒ $u_o=-2.5\text{ cm}$, $f_o=+2\text{ cm}$. Solve the lens equation for $v_o$ using the rearranged form $\frac{1}{v_o}=\frac{1}{f_o}+\frac{1}{u_o}$ (same equation, add $\frac{1}{u_o}$ to both sides): $\frac{1}{v_o}=\frac{1}{2}+\frac{1}{-2.5}=0.5-0.4=0.1,$ so $v_o=+10\text{ cm}$. The positive sign means it is where the red arrow sits in Figure 1: a real image $10\text{ cm}$ behind the objective. ✓

(b) Objective magnification: $m_o=\frac{v_o}{u_o}=\frac{10}{-2.5}=-4$. The minus is the inversion you see in Figure 1 (red arrow points down). Size factor $|m_o|=4$.

(c) Eyepiece (the green rays in Figure 1). The red image is the object for the eyepiece, which throws it to $D$. Using the magnifier near-point rule, $m_e=1+\frac{D}{f_e}=1+\frac{25}{6.25}=1+4=5.$

(d) Total — signs multiply: $M=m_o{m}_e=(-4)(5)=-20\Rightarrow |M|=\boxed{20}.$ The minus says the final image is inverted, matching the downward red arrow feeding the green stage.

Start from $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, rearranged to $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$, with $u<0$, $f>0$. Take $f=5\text{ cm}$ to match Figure 2.

WHERE $M=D/|u|$ comes from and WHY it has a domain. The angle the object subtends through the lens is $\theta \approx h/|u|$ (object height $h$ over its distance $|u|$); the naked-eye reference is ${\theta }_0=h/D$. So the angular power is $M=\frac{\theta }{{\theta }_0}=\frac{h/|u|}{h/D}=\frac{D}{|u|}.$ This is only meaningful while the virtual image lands where the eye can focus it, i.e. $|v|$ between $D$ and $\infty$. Outside that window the image is either un-focusable (closer than $D$) or real (object outside $f$), and the formula no longer describes a usable magnifier. That window is exactly the blue curve of Figure 2 between the green dot and the red line.

(i) As $u\to -f^{-}$ (object approaches focus from inside, $|u|<f$): $\frac{1}{u}\to -\frac{1}{f}$, so $\frac{1}{v}\to 0$ and $|v|\to \infty$ — this is the blue curve in Figure 2 diving downward without limit as it nears the red dashed line at $|u|=5$. The virtual image races out to infinity: the relaxed-eye limit. Concretely, at $|u|=4.9$ the formula gives $v=1/(0.2-1/4.9)\approx -245\text{ cm}$; at $|u|=4.99$, $v\approx -2490\text{ cm}$ — sprinting to $-\infty$.

(ii) Magnification range (reading the two marked lines in Figure 2). At $|u|=f$ (image at $\infty$, the red line) we get the smallest usable value $M_{\infty }=D/f=25/5=5$. Pulling the object slightly inside $f$ brings the image in from $\infty$ toward $D$; at $v=-D$ (the green dot sitting on the yellow $v=-25$ line) we hit the largest usable value $M_D=1+D/f=6$, reached at $|u|=25/6\approx 4.17\text{ cm}$. So the usable magnification spans the closed range $[D/f,1+D/f]=[5,6]$ — a surprisingly narrow band, which the flat blue segment near the green dot makes visual.

(iii) Exactly at $u=-f$: image at infinity, eye relaxed, $M=D/f=5$. Just beyond, $|u|>f$ (object outside focus): now $\frac{1}{f}+\frac{1}{u}$ turns positive, so $v>0$ — the image jumps to the far side and becomes real and inverted (this is why Figure 2's blue curve is only drawn for $|u|<f$; to the right of the red line the device is no longer a magnifier). At $|u|=0$ the other extreme, $M=D/|u|\to \infty$ formally, but the image then sits closer than $D$ and cannot be focused — see the L5 trap. All cases covered: inside focus → usable virtual magnifier ($M\in [5,6]$ near the near-point end, larger but un-focusable as $|u|\to 0$); at focus → image at $\infty$, $M=5$; outside focus → real inverted image, not a magnifier.

Numeric check for $f=5$: at $u=-25/6\approx -4.17\text{ cm}$, $v=-25$ (image at $D$), $M=6$; at $u=-5$, $v\to \infty$, $M=5$. ✓

Simple microscope, relaxed $M$ for $f=5$ ::: $D/f=5$ Simple microscope, near-point $M$ for $f=5$ ::: $1+D/f=6$ Telescope $M$ for $f_o=120,f_e=3$ ::: $40$ Compound scope $|M|$ for $L=15,f_o=1.5,f_e=5$ ::: $50$ (inverted) Dioptres of a $6.25\text{ cm}$ lens ::: $16\text{D}$