Exercises — Resolving power — Rayleigh criterion

KYA HAI: Hum bas Rayleigh angle ka naam le rahe hain. ${\theta }_R=1.22\frac{\lambda }{D}$ Units: radians (yeh ek pure ratio $\lambda /D$ times ek number hai, toh koi length units bachti nahi — woh ratio dimensionless hai, aur ek dimensionless "angle" by definition radians mein hota hai).

1.22 kyun nahi: 1.22 factor Bessel function $J_1$ ka pehla zero hai, jo diffraction ko ek disc par integrate karne se aata hai. Slit ek strip hai, isliye hum seedha single-slit result use karte hain (dekho Single-slit diffraction). ${\theta }_R=\frac{\lambda }{a}=\frac{600\times 10^{-9}}{0.10\times 10^{-3}}=6.0\times 10^{-3}\text{ rad}$

Direct substitution, sab kuch metres mein rakhte hain taaki ratio dimensionless rahe: ${\theta }_R=1.22\times \frac{550\times 10^{-9}}{2.0\times 10^{-3}}=3.4\times 10^{-4}\text{ rad}$

Arc length kyun: ek tiny angle ke liye, small-angle approximation (Small angle approximation) kehta hai ki distance $L$ par arc bas $s=\theta L$ hai — koi trig nahi chahiye kyunki $\tan \theta \approx \theta$ jab $\theta$ ek milliradians ka fraction ho. $s={\theta }_{R}L=3.4\times 10^{-4}\times 0.25=8.5\times 10^{-5}\text{ m}\approx 0.085\text{ mm}$ ~0.09 mm se closer do dots merge ho jaate hain — isliye arm's length par barik print blur hoti hai.

${\theta }_R=1.22\times \frac{500\times 10^{-9}}{1.0}=6.1\times 10^{-7}\text{ rad}$ Yeh aankhon se ~560× finer hai — bade telescopes ka main reason resolution hai, sirf light gathering nahi.

${\theta }_R\propto \lambda /D$.

• $\lambda$ halve karna: ${\theta }_R\to \frac{1}{2}{\theta }_R$ (factor 0.5).
• $D$ double karna: ${\theta }_R\to \frac{1}{2}{\theta }_R$ (factor 0.5).

Dono identical hain. Dono resolution exactly factor 2 se improve karte hain. Koi nahi jeeता — formula "$\lambda /D$ kaise shrink karo" mein symmetric hai.

${\theta }_R^{\prime }=1.22\frac{2\lambda }{2D}=1.22\frac{\lambda }{D}={\theta }_R$ Unchanged — dono effects exactly cancel ho jaate hain. Yeh parent note ka Forecast-then-Verify result hai: hamesha ratio $\lambda /D$ se reason karo, do numbers alag alag se nahi.

Step 1 — resolution angle: ${\theta }_R=1.22\times \frac{550\times 10^{-9}}{5.0\times 10^{-3}}=1.342\times 10^{-4}\text{ rad}$ Step 2 — arc relation ko invert karo. Hum jaante hain ki subtended actual angle $\theta =s/L$ hai. Resolve hone ke liye $\theta \ge {\theta }_R$ chahiye, toh limiting distance woh hai jahan $\theta ={\theta }_R$: $L_{\max }=\frac{s}{{\theta }_R}=\frac{1.5}{1.342\times 10^{-4}}\approx 1.12\times 10^4\text{ m}\approx 11\text{ km}$ ~11 km se aage (ideal, diffraction-limited conditions mein) do lights ek point mein fuse ho jaati hain. Real aankhein aberrations ki wajah se worse karti hain, lekin yeh hard ceiling hai.

Numerical aperture kyun, $D$ nahi: ek near object ke liye lens light ka ek cone collect karta hai; woh cone jitna wider ho, utna finer detail. Cone width $\text{N.A.}=n\sin \beta$ se capture hoti hai (dekho Numerical aperture), aur $d_{\min }=0.61\lambda /\text{N.A.}$

Pehle, $\sin 65^{\circ }=0.9063$.

Air: $\text{N.A.}=1.00\times 0.9063=0.9063$ $d_{\min }^{\text{air}}=\frac{0.61\times 500\times 10^{-9}}{0.9063}=3.37\times 10^{-7}\text{ m}\approx 337\text{ nm}$

Oil: $\text{N.A.}=1.515\times 0.9063=1.373$ $d_{\min }^{\text{oil}}=\frac{0.61\times 500\times 10^{-9}}{1.373}=2.22\times 10^{-7}\text{ m}\approx 222\text{ nm}$

Oil resolution ko $n=1.515$ ke factor se improve karta hai — kyunki immersion oil $n$ badhata hai, effective cone wide karta hai. Yahi pura reason hai ki oil-immersion lenses finer resolve karte hain.

(a) Telescope: ${\theta }_R^{\text{tel}}=1.22\times \frac{550\times 10^{-9}}{0.15}=4.47\times 10^{-6}\text{ rad}$ Stars ka separation $3.0\times 10^{-6}$ rad ${\theta }_R^{\text{tel}}$ se chhhota hai, toh resolved nahi — yeh telescope bhi unhe split nahi kar sakta. (b) Eye: ${\theta }_R^{\text{eye}}=1.22\times \frac{550\times 10^{-9}}{2.0\times 10^{-3}}=3.36\times 10^{-4}\text{ rad}$ Yeh aur bhi zyada bada hai ⇒ aankhein itna fine kuch resolve nahi kar sakti. Dono fail. Sirf bada $D$ (ya chhhota $\lambda$) help karta; magnification akela nahi — woh merged blob ko bada karta hai.

Step 1 — required angle (small-angle law): ${\theta }_{\text{req}}=\frac{s}{L}=\frac{0.30}{3.0\times 10^5}=1.0\times 10^{-6}\text{ rad}$ Step 2 — Rayleigh invert karo $D$ solve karne ke liye. Humein ${\theta }_R\le {\theta }_{\text{req}}$ chahiye, matlab mirror itna bada hona chahiye ki uska resolution angle itna chhhota ho: $D_{\min }=\frac{1.22\lambda }{{\theta }_{\text{req}}}=\frac{1.22\times 500\times 10^{-9}}{1.0\times 10^{-6}}=0.61\text{ m}$ Kam se kam ~0.61 m ka mirror chahiye. Diffraction, lens quality nahi, yeh floor set karta hai.

• Same idea: dono mein, resolution tab improve hoti hai jab wave aperture ka ek wider extent sample kare — zyada slits $N$ (grating) ya bada diameter $D$ (lens). Zyada illuminated region ka matlab hai ek narrower diffraction feature, toh do nearby peaks/wavelengths alag ho jaate hain. Dekho Diffraction grating — resolving power.
• Different costume: grating do wavelengths alag karta hai (spectral), aperture do directions/points alag karta hai (spatial); lekin dono is baat se limited hain ki wavefront kitna capture kiya. Same physics (Huygens principle: wavefront ka har point re-emit karta hai, aur secondary sources ka wider set milke ek sharper combined pattern banata hai).

(a) ${\theta }_R=1.22\times \frac{550\times 10^{-9}}{3.0\times 10^{-3}}=2.237\times 10^{-4}\text{ rad}$ $s={\theta }_{R}L=2.237\times 10^{-4}\times 0.30=6.71\times 10^{-5}\text{ m}\approx 0.067\text{ mm}$ (b) Pixels per inch = ek inch (25.4 mm) ko pitch se divide karo: $\text{ppi}=\frac{25.4\text{ mm}}{0.0671\text{ mm}}\approx 378\text{ ppi}$ Typical viewing distance par 300–400 ppi wale displays aankhon ke diffraction+cone limit ke bilkul paas baithe hain — isse kaafi upar jaana viewer ko invisible hai, exactly wahi "empty magnification" idea screens par apply hota hai.

Aperture se angle ::: ${\theta }_R=1.22\lambda /D$ (circle) ya $\lambda /a$ (slit). Angle ↔ separation ::: $s=\theta L$, toh $L=s/{\theta }_R$ aur $D=1.22\lambda L/s$. Microscope ::: $d_{\min }=0.61\lambda /\text{N.A.}$, N.A. $=n\sin \beta$; oil $n$ badhata hai. Sab kuch improve hota hai ::: bade aperture / cone se, shorter wavelength se — kabhi magnification se nahi.