Exercises — Total internal reflection — critical angle derivation

KYA: TIR ke liye upar wali checklist ki dono conditions chahiye. KYUN: Condition 1 hai denser→rarer, yani $n_1>n_2$. Yahan ray shuru hoti hai air mein ($n_1=1.00$) aur glass mein enter karti hai ($n_2=1.50$), toh $n_1<n_2$ — yeh rarer→denser ja rahi hai. Condition 1 already fail ho gayi, toh angle check karne ki zaroorat hi nahi. Result: $\boxed{\text{No.}}$ Denser medium mein jaate waqt, $\sin {\theta }_2=\frac{n_1}{n_2}\sin {\theta }_1<1$ hamesha hoga, toh refracted ray hamesha exist karegi. Is direction mein TIR impossible hai.

KYA: Calculator ke bina do indices ke liye ${\theta }_c$ compare karo. KYUN: $\sin {\theta }_c=1/n$. Bada $n$ fraction $1/n$ ko chhota banata hai, aur chhota sine matlab chhota angle (dono angles $[0°,90°]$ mein hain, jahan sine strictly increase karta hai). Diamond ka $n$ bada hai, toh uska ${\theta }_c$ chhota hoga. Isliye glass ka ${\theta }_c$ sach mein bada hai. Result: $\boxed{\text{True.}}$ (Numerically $41.8°>24.4°$ — neeche verify kiya gaya hai.)

Result: $\boxed{\text{(b) }\sin {\theta }_c=n_2/n_1.}$ KYUN: Snell's law mein ${\theta }_2=90°$ rakho: $n_1\sin {\theta }_c=n_2\sin 90°=n_2$. $n_1$ se divide karo. Kyunki ray denser→rarer jaati hai, $n_2<n_1$, toh fraction $<1$ hai — exactly jaisa ek sine hona chahiye. Option (a) $>1$ value deta, jo koi sine equal nahi kar sakta: yeh impossibility built-in "tumne ulta kar diya" alarm hai.

KYA: direct plug-in. Rarer medium air hai ($n_2=1$), denser water hai ($n_1=1.33$). $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1}{1.33}=0.7519$ KYUN sine invert karein: hume angle ka sine pata hai; angle khud us number ka ${\sin }^{-1}$ hai (principal value $[0°,90°]$ mein). ${\theta }_c={\sin }^{-1}(0.7519)=\boxed{48.8°}$

KYA: unknown denser index hai; air $n_2=1$ hai. $\sin {\theta }_c=\frac{1}{n}\Rightarrow n=\frac{1}{\sin {\theta }_c}=\frac{1}{\sin 40.0°}=\frac{1}{0.6428}$ $n=\boxed{1.556}$ Check: $n>1$ jaisa kisi bhi real medium ke liye hona chahiye, aur ${\theta }_c<90°$ automatically $n>1$ deta hai. ✓

KYA: koi bhi medium air nahi hai — dono indices use karo. $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.33}{1.62}=0.8210$ ${\theta }_c={\sin }^{-1}(0.8210)=\boxed{55.2°}$ KYUN glass–air case se bada hai: dono indices zyada close hain, toh fraction $n_2/n_1$ $1$ ke zyada paas hai, jo ${\theta }_c$ ko $90°$ ke paas push karta hai.

$\sin {\theta }_c=\frac{1}{2.42}=0.4132\Rightarrow {\theta }_c={\sin }^{-1}(0.4132)=\boxed{24.4°}$ Matlab: lagbhag koi bhi internal ray ($>24.4°$) trap ho jaati hai, toh light bahar nikalne se pehle kai baar bounce karti hai — yahi diamond ki chamak ka source hai.

Har incidence angle ko ${\theta }_c=41.8°$ se compare karo:

• $30°$ (mint ray): $30°<{\theta }_c$, toh ek real refracted ray exist karti hai — light air mein refract ho jaati hai, normal se door bend karti hai (figure mein mint arrow upar escape karta hua dikhta hai).
• $41.8°$ (butter ray): exactly ${\theta }_c$, toh ${\theta }_2=90°$ — refracted ray graze karti hai surface ke flat along (butter arrow boundary ke saath leta hua).
• $55°$ (coral ray): $55°>{\theta }_c$, toh Snell ko $\sin {\theta }_2=1.5\sin 55°=1.229>1$ chahiye, jo impossible hai — ray totally internally reflect hoti hai aur glass mein wapas jaati hai (coral arrow neeche bounce karta hua). $\boxed{30°\to \text{refracts},41.8°\to \text{grazes},55°\to \text{TIR}}$

KYA: ${\theta }_c={\sin }^{-1}(n_2/n_1)$ ko study karo jaise uska input barhta hai. KYUN: ${\sin }^{-1}$ $[0,1]$ par ek increasing function hai (principal angle $[0°,90°]$ mein return karta hai) — bada input bada angle deta hai. Toh jaise $n_2/n_1$ badhta hai, ${\theta }_c$ badhta hai. Limit: jab $n_2\to n_1$, $n_2/n_1\to 1$ aur ${\theta }_c\to {\sin }^{-1}(1)=90°$. Physical matlab: agar dono media almost identical hain, toh light barely bend karti hai, toh TIR set hone se pehle surface ke almost parallel ($\approx 90°$) strike karna padega. $\boxed{{\theta }_c\text{badhta hai, }90°\text{ tak pahunchta hai jab }{n}_2\to n_1.}$

Algebra sahi hai: $n_2>n_1$ ke saath, $\sin {\theta }_c=n_2/n_1>1$ ka koi real solution nahi hai. Interpretation asli point hai: "no critical angle" ka matlab hai ki koi aisa incidence angle nahi hai jis par refraction band ho jaaye — rarer→denser jaate waqt, har ${\theta }_1$ ke liye $90°$ tak refracted ray hamesha exist karti hai. Toh us direction mein TIR simply ho hi nahi sakti. $\boxed{\text{Reasoning valid hai; yeh sahi signal karta hai ki TIR rarer}\to \text{denser mein impossible hai.}}$

KYA: hypotenuse par TIR ke liye incidence angle ($45°$) critical angle se zyada hona chahiye: $45°>{\theta }_c\iff \sin 45°>\sin {\theta }_c=\frac{1}{n}$ KYUN inequality ki direction sahi hai: sine $[0°,90°]$ par increasing hai, toh $45°>{\theta }_c$ same hai jaise $\sin 45°>\sin {\theta }_c$. $n$ ke liye solve karo: $\frac{1}{n}<\sin 45°=0.7071\Rightarrow n>\frac{1}{0.7071}=1.414$ $\boxed{n>1.414}$ Matlab: ordinary glass ($n=1.5$) yeh aasani se satisfy karta hai, isliye binoculars aur periscopes mein right-angle prisms bina kisi silvering ke reflect karte hain. Prisms and Total Internal Reflection dekho.

KYA: boundary core (denser) → cladding (rarer) hai, toh $n_1=1.50$, $n_2=1.45$. $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.45}{1.50}=0.9667$ ${\theta }_c={\sin }^{-1}(0.9667)=\boxed{75.2°}$ Matlab: koi bhi ray jo wall se $75.2°$ se zyada (normal se) takraaye woh trap ho jaati hai aur fibre ke neeche guide hoti hai. Kyunki ${\theta }_c$ itna $90°$ ke paas hai, sirf woh rays jo axis ke lagbhag parallel travel karti hain woh guided rehti hain — yahi Optical Fibres ka design idea hai.

(a) Water→oil $n=1.33\to 1.46$ jaata hai, yani rarer→denser. TIR ke liye denser→rarer chahiye, toh $\boxed{\text{water–oil boundary par TIR nahi.}}$ (b) Oil→air denser→rarer hai jisme $n_1=1.46$, $n_2=1$: $\sin {\theta }_c=\frac{1}{1.46}=0.6849\Rightarrow {\theta }_c={\sin }^{-1}(0.6849)=\boxed{43.2°}$

Step 1 — critical angle. KYA/KYUN: sirf woh rays escape karti hain jo half-angle ${\theta }_c$ wale cone ke andar hain, toh pehle water ke liye air ke against ${\theta }_c$ nikalo: $\sin {\theta }_c=\frac{1}{1.33}=0.7519,{\theta }_c={\sin }^{-1}(0.7519)=48.75°$ Step 2 — kyun tangent, aur woh kahan se aata hai. Figure mein shaded right triangle dekho. Uska vertical leg depth $h$ hai (yeh woh side hai jo angle ${\theta }_c$ ke adjacent hai, kyunki ${\theta }_c$ vertical normal se measure hoti hai). Uska horizontal leg radius $R$ hai (angle ${\theta }_c$ ke opposite wali side). Hume $h$ aur angle se $R$ chahiye, yani hume opposite-from-adjacent chahiye — aur woh trig ratio jo ek angle ko opposite-over-adjacent se connect karta hai woh exactly tangent hai: $\tan {\theta }_c=\frac{\text{opposite}}{\text{adjacent}}=\frac{R}{h}.$ Isliye tangent use kiya naa ki sine ya cosine: sine ke liye hypotenuse (ray length) chahiye hoti hai, jo humhare paas nahi hai aur chahiye bhi nahi, jabki tangent sirf dono legs use karta hai jo humhare paas hain. Step 3 — $R$ ke liye solve karo. Dono sides ko $h$ se multiply karo: $R=h\tan {\theta }_c=2.0\times \tan (48.75°)=2.0\times 1.1391=\boxed{2.28\text{m}}$ Toh surface par lit circle ki radius lagbhag $2.28$ m hai; uske bahar sab kuch neeche se mirror jaisa dikhta hai.

KYA: hum $\tan {\theta }_c$ ko $n$ mein expression mein convert karte hain taki koi calculator angle na chahiye. KYUN / uske peeche ka triangle: $\sin {\theta }_c=1/n$ se ek right triangle banao jiska hypotenuse $n$ hai aur ${\theta }_c$ ke opposite wali side $1$ hai (woh pair $\sin {\theta }_c=1/n$ realise karta hai). Pythagoras phir adjacent side fix karta hai: $\text{adjacent}=\sqrt{{\text{hyp}}^2-{\text{opp}}^2}=\sqrt{n^2-1}.$ Isliye $\tan {\theta }_c=\frac{\text{opposite}}{\text{adjacent}}=\frac{1}{\sqrt{n^2-1}}.$ $R=h\tan {\theta }_c$ mein substitute karo: $R=\frac{h}{\sqrt{n^2-1}}=\frac{2.0}{\sqrt{1.33^2-1}}=\frac{2.0}{\sqrt{0.7689}}=\frac{2.0}{0.8769}=\boxed{2.28\text{m}}✓$ Dono methods agree karte hain, clean formula confirm karta hai.

Step 1 — top face par refraction (air→glass, toh ray normal ki taraf bend karti hai). KYA/KYUN: top interface par Snell's law apply karo, air $n=1$ bahar, glass $n=1.5$ andar: $1\cdot \sin \alpha =1.5\sin r\Rightarrow \sin r=\frac{\sin 40°}{1.5}=\frac{0.6428}{1.5}=0.4285,r={\sin }^{-1}(0.4285)=25.37°$ Yahan $r$ glass ke andar refraction angle hai, top face ke normal (vertical) se measure kiya gaya. Step 2 — top normal se side normal tak geometry. KYA: top face horizontal hai, toh uska normal vertical hai; side face vertical hai, toh uska normal horizontal hai. Yeh do normals perpendicular hain. KYUN side par incidence $90°-r$ hai: refracted ray vertical se $r$ door lean karti hai. Kyunki side-face normal horizontally point karta hai (vertical se $90°$ rotated), wahi ray usse $90°-r$ angle banati hai — ek single straight line do perpendicular reference lines ke saath jo do angles banati hai woh hamesha $90°$ mein add hote hain. Toh ${\theta }_{\text{side}}=90°-r=90°-25.37°=64.63°.$ Kaisa dikhta hai: block ke andar ray ko ek slightly-tilted spear ki tarah socho; woh almost vertical hai ($25°$ top normal se), isliye horizontal side normal ke almost perpendicular hai, yani bada ${\theta }_{\text{side}}$. Step 3 — ${\theta }_c$ se compare karo. Glass ke liye, ${\theta }_c={\sin }^{-1}(1/1.5)=41.8°$. Kyunki ${\theta }_{\text{side}}=64.63°>41.8°={\theta }_c$, ray side face par TIR undergo karti hai. $\boxed{{\theta }_{\text{side}}=64.6°>41.8°\Rightarrow \text{TIR hoti hai.}}$

Recognition — Kaun si direction TIR allow karti hai? ::: Sirf denser se rarer ($n_1>n_2$). Application — $n=1.33$ vs air ke liye ${\theta }_c$? ::: ${\sin }^{-1}(1/1.33)\approx 48.8°$. Analysis — Exactly ${\theta }_c$ par kya hota hai? ::: Refracted ray $90°$ par graze karti hai; TIR ke liye strictly ${\theta }_1>{\theta }_c$ chahiye. Synthesis — $45°$ prisms kyun reflect karte hain? ::: $45°>{\theta }_c$ glass ke liye (${\theta }_c\approx 41.8°$), toh silvering ke bina TIR. Mastery — Escape-window radius formula? ::: $R=h/\sqrt{n^2-1}$.