Exercises — Geometric optics — rectilinear propagation, reflection, refraction

KYA: Ek single uniform medium ke andar ray straight line mein travel karti hai. KYUN: Ek homogeneous medium mein speed $v$ har point pe same hoti hai. Do points ke beech constant speed pe least-time path shortest path hota hai, jo ek straight line hoti hai. Koi bhi bend length, isliye time badha dega. Yahi rectilinear propagation hai.

Trap-aware step: $37^{\circ }$ surface se diya gaya tha, lekin optics normal use karta hai. Convert karo: ${\theta }_i=90^{\circ }-37^{\circ }=53^{\circ }(\text{from the normal}).$ Law of reflection: ${\theta }_r={\theta }_i=53^{\circ }$.

KYUN Snell: boundary cross karne par speed change hoti hai, isliye ray bend karti hai; quantity $n\sin \theta$ surface ke across conserved hoti hai. $n_1\sin {\theta }_1=n_2\sin {\theta }_2\Rightarrow \sin {\theta }_2=\frac{1.00\cdot \sin 40^{\circ }}{1.33}.$ $\sin 40^{\circ }=0.6428$, toh $\sin {\theta }_2=\frac{0.6428}{1.33}=0.4833$. ${\theta }_2=\arcsin (0.4833)=28.9^{\circ }.$ Sanity check: water air se denser hai, isliye ray normal ki taraf bend karti hai: ${\theta }_2<{\theta }_1$. ✔

$n=\frac{c}{v}=\frac{3.00\times 10^8}{2.00\times 10^8}=\mathrm{1.50.}$ Padho ise: is plastic mein light vacuum ke comparison mein $1.5\times$ slow hai.

Frequency source se set hoti hai aur refraction par change nahi hoti: $f=\frac{c}{{\lambda }_0}=\frac{3.00\times 10^8}{500\times 10^{-9}}=6.00\times 10^{14}\text{Hz}.$ Wavelength $n$ ke factor se shrink hoti hai (speed drop hui, frequency fixed rahi, toh $v=f\lambda$ ne $\lambda$ ko force kiya): ${\lambda }_{\text{med}}=\frac{{\lambda }_0}{n}=\frac{500}{1.50}=333\text{nm}.$

KYUN: Jaise ${\theta }_1$ badhta hai, ${\theta }_2$ faster badhta hai (rarer, faster medium mein jaana ray ko normal se door bend karta hai). Critical angle par refracted ray exactly surface ke saath graze karti hai: ${\theta }_2=90^{\circ }$ — yahi figure s01 mein flat yellow ray hai. $n_1\sin {\theta }_c=n_2\sin 90^{\circ }=n_2\Rightarrow \sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.00}{1.50}=\mathrm{0.6667.}$ ${\theta }_c=\arcsin (0.6667)=41.8^{\circ }.$ $41.8^{\circ }$ ke aage, Snell demand karta $\sin {\theta }_2>1$ (impossible) — isliye light saari internally reflect ho jaati hai. Dekho Total internal reflection aur Optical fibres.

Air index $n_a$, glass index $n_g$ lo. Top face par: $n_a\sin {\theta }_1=n_g\sin {\theta }_2.$ Dono faces parallel hain, isliye bottom face par normal, top pe normal ke parallel hai; internal ray bottom face ko same angle ${\theta }_2$ par meet karta hai. Bottom face par: $n_g\sin {\theta }_2=n_a\sin {\theta }_3.$ Chain karo: $n_a\sin {\theta }_1=n_g\sin {\theta }_2=n_a\sin {\theta }_3$, isliye $\sin {\theta }_1=\sin {\theta }_3$, giving ${\theta }_3={\theta }_1$. Exit ray same angle par nikali jis par enter ki thi → parallel (bas sideways shift hui). Glass andar $n_g$ "lete" hai lekin bahar jaate waqt wapas de deta hai.

Step 1 — kya $60^{\circ }$ critical angle se past hai? Water se air mein: $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.00}{1.33}=0.7519\Rightarrow {\theta }_c=48.8^{\circ }.$ Kyunki $60^{\circ }>48.8^{\circ }$, hum critical se aage hain. Step 2 — Snell directly check karo: $\sin {\theta }_2=\frac{1.33\sin 60^{\circ }}{1.00}=1.33\times 0.8660=1.152>1$: impossible. Conclusion: Koi light escape nahi hoti; woh paani mein wapas totally internally reflect ho jaati hai. Isliye ek calm water surface ka underside ek mirror ki tarah kaam karta hai jab tum steep angle par upar dekhte ho.

KYUN inverted: ped ka har point single pinhole se seedhi ray bhejta hai; top se ray neeche-aur-through travel karti hai, screen par low land karti hai, jabki bottom ray high land karti hai — top aur bottom swap ho jaate hain → inverted (figure s02 mein crossed chalk rays). Similar triangles pinhole vertex share karte hain, isliye corresponding sides proportional hain. Standard pinhole sign convention mein hum inversion mark karne ke liye minus lagate hain: $m=\frac{h}{H}=-\frac{v}{u}\Rightarrow h=-H\cdot \frac{v}{u}=-6.0\times \frac{0.30}{2.0}=-0.90\text{m}.$ Magnitude $0.90\text{m}$ hai; negative sign algebra ka tarika hai yeh kehne ka ki "inverted" hai. Image: $0.90\text{m}$ tall, inverted.

KYUN least length: mirror se speed constant hai, isliye least time = least path length $L(x)=\sqrt{x^2+3^2}+\sqrt{(8-x{)}^2+1^2}.$ Image trick (answer tak sabse tez raasta): $B=(8,1)$ ko mirror ke across $B^{\prime }=(8,-1)$ par reflect karo. Koi bhi path $A\to P\to B$ ki length utni hi hai jitni $A\to P\to B^{\prime }$ ki (doosra leg mirrored hai), aur shortest $A\to B^{\prime }$ straight line hai — figure s03 mein yellow dashed line. Toh $P$ wahan hai jahan line $A{B}^{\prime }$, $x$-axis ko cross karti hai. Line from $A=(0,3)$ to $B^{\prime }=(8,-1)$: slope $=\frac{-1-3}{8-0}=-\frac{1}{2}$, toh $y=3-\frac{1}{2}x$. Set $y=0$: $0=3-\frac{1}{2}x\Rightarrow x=6.0\text{m},P=(6,0).$ Equal angles verify karo. Dono ko $P$ par vertical normal se measure karo.

• Incident side ($A$ to $P$): horizontal run $=6-0=6$, vertical drop $=3$. $\tan {\theta }_i=\frac{6}{3}=2$, toh ${\theta }_i=\arctan 2=63.43^{\circ }$.
• Reflected side ($P$ to $B$): horizontal run $=8-6=2$, vertical rise $=1$. $\tan {\theta }_r=\frac{2}{1}=2$, toh ${\theta }_r=63.43^{\circ }$. ${\theta }_i={\theta }_r$ ✔ — Fermat ka minimum exactly law of reflection reproduce karta hai.

KYUN ab least-length nahi: dono speeds differ karti hain, isliye time $\ne$ distance. Time minimize karo: $T(x)=\frac{\sqrt{x^2+9}}{v_1}+\frac{\sqrt{(6-x{)}^2+9}}{v_2}.$ $\frac{dT}{dx}=0$ set karne se milta hai $\frac{x}{v_1\sqrt{x^2+9}}=\frac{6-x}{v_2\sqrt{(6-x{)}^2+9}},$ jo exactly $\frac{\sin {\theta }_1}{v_1}=\frac{\sin {\theta }_2}{v_2}$ hai, yaani Snell's law, kyunki har fraction (horizontal offset)/(hypotenuse) = vertical normal se measure kiya gaya sine hai. Numerically solve karo. Balance condition ko cross-multiply karo. $v_1=3$ aur $v_2=2$ ke saath ($10^8\text{m/s}$ ki units mein), $v_1=1.5v_2$, toh $v$-factors dete hain $v_{2}x\sqrt{(6-x{)}^2+9}=v_1(6-x)\sqrt{x^2+9}\Rightarrow x\sqrt{(6-x{)}^2+9}=1.5(6-x)\sqrt{x^2+9}.$ $0<x<6$ mein solve karne par $x\approx 4.09\text{m}$ milta hai (physical root). Snell confirm karo. $x=4.09$ ke saath:

• $\sin {\theta }_1=\frac{x}{\sqrt{x^2+9}}=\frac{4.09}{\sqrt{4.09^2+9}}=0.8062$, toh $n_1\sin {\theta }_1=1\times 0.8062=0.8062$.
• $\sin {\theta }_2=\frac{6-x}{\sqrt{(6-x{)}^2+9}}=\frac{1.91}{\sqrt{1.91^2+9}}=0.5375$, toh $n_2\sin {\theta }_2=1.5\times 0.5375=0.8062$. Dono sides $0.8062$ equal hain ✔. Ray slower medium mein enter karte waqt normal ki taraf bend karti hai: ${\theta }_2<{\theta }_1$.

• Angles hamesha normal se hote hain.
• Denser medium ($\uparrow n$) → slower light → normal ki taraf bend hoti hai.
• Critical angle sirf rarer medium ki taraf jaane par exist karta hai: $\sin {\theta }_c=n_2/n_1$.
• Parallel-faced slab → exit ray entry se parallel hoti hai.
• Pinhole: $|m|=v/u$, image inverted hoti hai ($m$ ka sign negative hota hai).
• Fermat: reflection path length minimize karta hai; refraction time minimize karta hai (legs ko $1/v$ se weight karo).