Evaluating claims about imaging and fiber optics

A fiber waveguide keeps light inside by total internal reflection: at the core–cladding boundary a ray is reflected when the angle of incidence exceeds the critical angle $\theta_c$, where $\sin\theta_c=n_2/n_1$.

The numerical aperture $NA=\sqrt{n_1^2-n_2^2}$ sets the maximum launch angle from air: $\sin\theta_{max}=NA$, so a claim about a “wide acceptance angle” can be checked by calculation.

Loss claims are expressed in decibels: $L\,[\text{dB}]=10\log_{10}(P_{in}/P_{out})$, attenuation per unit length is measured in dB/km, and the power budget is the sum of fiber, connector and splice losses.

Imaging claims are checked differently: magnification is not resolution, and the resolution limit is set by diffraction, $\theta_{min}\approx 1.22\lambda/D$ or $d\approx 0.61\lambda/NA$.

A useful claim is testable: it names a measurable quantity, its units and the conditions of the measurement; otherwise no check is possible.

The general evaluation routine is: pick the quantity, choose the formula from the prerequisites (reflection, refraction, total internal reflection), substitute reasonable values and compare with the claim — a discrepancy by a factor of several means the claim is wrong.

A step-index fiber has core refractive index $n_1=1.50$ and cladding refractive index $n_2=1.48$. Find the critical angle $\theta_c$ and the numerical aperture $NA$. At what largest angle to the axis can light be launched from air?

A manufacturer claims: “A fiber with $n_1=1.50$ and $n_2=1.49$ accepts light from air at angles up to $60^\circ$.” Test the claim in two ways: via the numerical aperture, and via the angle of incidence on the core–cladding boundary for a ray entering at $60^\circ$.

A supplier claims: “A $25$ km link with $0.40$ dB/km attenuation and two connectors of $0.75$ dB each works with margin under a $12$ dB power budget.” Check the calculation, judge how reliable the claim is, and propose an independent measurement that would support or refute it.

Self-check without hazardous light sources.

1. Open a refraction and total internal reflection simulation (for example PhET “Bending Light”), set $n_1=1.50$ and $n_2=1.48$. Slowly increase the angle of incidence and find the value at which light stops leaving the medium; compare it with $\theta_c=\arcsin(n_2/n_1)$.

2. Build a table and enter three “advertising” claims:

• acceptance angle $30^\circ$ for $n_1=1.50$, $n_2=1.48$;
• $40$ km with $0.5$ dB/km, two connectors of $0.5$ dB each and a $25$ dB budget;
• resolution $200$ nm at $NA=0.5$, $\lambda=550$ nm.

3. For each row compute $NA$, the total loss $L$ and the limit $d\approx 0.61\lambda/NA$, then mark the claim as “supported”, “refuted” or “insufficient data”, naming the decisive quantity.

Safety: never look into the end face of a fiber and never point a laser at your eyes; replace all measurements with a simulation or a low-power LED.

To show why light stays in the core and where exactly a claim breaks down: a ray inside the acceptance cone undergoes total internal reflection, while a ray outside the cone escapes through the cladding.

Diagram (fiber cross-section). On the left, the end face and the acceptance cone in air with half-angle $\theta_{max}$, where $\sin\theta_{max}=NA$. On the right, a longitudinal section of the core $n_1$ and cladding $n_2$ with three rays: ray 1 enters at an angle smaller than $\theta_{max}$ and undergoes total internal reflection ($\alpha>\theta_c$); ray 2 travels along the axis; ray 3 enters at an angle larger than $\theta_{max}$, and at the boundary $\alpha<\theta_c$, so it refracts into the cladding and is drawn escaping. Below, a horizontal power-budget bar: a segment of $10.0$ dB (fiber) plus $1.5$ dB (connectors), with a vertical mark at the $12$ dB budget.

A phone advertises 108 MP. Its main sensor is about $8.8\,mm \times 6.6\,mm = 58\,mm^2$, so each pixel is roughly $0.8\,\mu m$ wide. The lens has aperture f/1.7, and light has wavelength about 550 nm. The smallest bright spot the lens can form has radius about $r \approx 1.22 \lambda \cdot 1.7 = 1.22(0.55\,\mu m)(1.7)\approx1.1\,\mu m$. The optical spot covers more than one pixel. Claim: "108 MP gives three times finer real detail than 12 MP on the same sensor." Check: at 12 MP the pixel width is about $\sqrt{58\,mm^2/(12\times10^6)}\approx2.2\,\mu m$; at 108 MP it is about $0.8\,\mu m$. The 108 MP sensor samples the optical image more finely, but it does not make the lens resolve three times smaller detail if the optical spot is $1.1\,\mu m$.

The scenario uses refraction at the curved lens surfaces to focus light, and then checks whether the claimed pixel count can exceed the optical spot size. Evaluating the claim means comparing the pixel pitch $p$ with the focused spot scale, not accepting megapixels as resolution by themselves.

A convex passenger-side mirror has radius of curvature $R=1200\,mm$, so the magnitude of its focal length is $|f|=R/2=0.60\,m$. A car is $d_o=20\,m$ behind. The mirror equation $1/f=1/d_o+1/d_i$ with $f=-0.60\,m$ gives $d_i=-0.62\,m$, and the magnification is $m=-d_i/d_o=0.031$. The image is virtual, upright, and only 3.1% of the object size. The warning is true: the small image makes the car look farther away than it is.

The scenario uses the reflection law at a convex mirror and the mirror equation; evaluating the claim relies on reflection, image distance, and magnification.

A fibre patch cable advertises: "1 Gbit/s and can bend 90° around a door frame without loss." Core $n_1=1.48$, cladding $n_2=1.46$. The critical angle for total internal reflection is $\theta_c=\arcsin(n_2/n_1)=\arcsin(1.46/1.48)\approx80.6^\circ$ from the normal to the core-cladding boundary. A ray must therefore hit the boundary at more than $80.6^\circ$ from the normal, which means it must travel within about $90^\circ-80.6^\circ=9.4^\circ$ of the fibre axis. In a straight cable this is easy. If the cable is bent with radius $R=10\,mm$, a ray can change its angle to the axis by roughly $\Delta\theta \approx s/R$ over a distance $s$. After just $s=1\,mm$, $\Delta\theta \approx 1\,mm/10\,mm=0.1\,rad\approx5.7^\circ$. A ray that started $5^\circ$ off axis can exceed the $9.4^\circ$ limit and refract out, so the "without loss" part is false for a very tight bend.

The scenario uses total internal reflection and the critical angle to test the claim; the numerical check shows why a fibre has a minimum bend radius.