Question bank — Polarization — Malus's law, Brewster's angle derivation

False — $\frac{1}{2}$ wala rule sirf unpolarized light ke liye hold karta hai. Pehle se polarized light $I=I_0{\cos }^2\theta$ follow karti hai (jahan $\theta$, $\vec{E}$ aur axis ke beech ka angle hai), jo $0$ (crossed) se $I_0$ (aligned) tak range karta hai.

False — Malus ko ek defined polarization direction chahiye hoti hai jisse $\theta$ measure kiya ja sake. Unpolarized light ki koi ek direction nahi hoti, isliye tum ${\cos }^2\theta$ ko sab angles par average karte ho aur $\frac{1}{2}$ milta hai instead.

True — ${\cos }^{2}90^{\circ }=0$, isliye koi surviving component nahi. Lekin unke beech ek teesra polarizer insert karne se kuch light wapas aa sakti hai kyunki woh polarization ko steps mein rotate karta hai.

False — sirf incidence plane ke andar wala polarization component gayab hota hai. Perpendicular component abhi bhi reflect karta hai, isliye reflected ray dim hoti hai lekin 100% polarized hoti hai, gone nahi.

False — law hai $\tan {\theta }_B=n_2/n_1$, jo entering-index aur starting-index ka ratio hai. Sirf jab pehla medium air ho ($n_1=1$) tab yeh $\tan {\theta }_B=n_2$ tak simplify hota hai.

True — $n_1=1$ fixed rakhne par, ${\theta }_B=\arctan (n_2)$ aur $\arctan$ increasing hai, isliye zyada $n_2$ se zyada ${\theta }_B$ milta hai (lekin woh kabhi $90^{\circ }$ tak nahi pahunch sakta). Agar $n_1$ bhi change ho, toh sirf ratio $n_2/n_1$ matter karta.

False — $\theta =0$ par, ${\cos }^{2}0=1$, isliye zyada se zyada $I=I_0$. Ek passive filter kabhi amplify nahi kar sakta; woh sirf pass ya absorb kar sakta hai.

False — woh plane of incidence ke perpendicular polarized hoti hai (surface ke parallel). Isliye water/roads se horizontal glare dominate karta hai.

Error yeh hai ki square bhool gaye. Detectors intensity read karte hain, aur $I\propto E^2$, isliye projection factor square hota hai: $I=I_0{\cos }^2\theta$.

Wrong function — derivation tangent deta hai, $\tan {\theta }_B=n_2$ (jab $n_1=1$). Yeh Snell plus ${\theta }_r=90^{\circ }-{\theta }_B$ se aata hai, jahan $\sin (90^{\circ }-{\theta }_B)=\cos {\theta }_B$ ratio ko $\tan$ mein convert kar deta hai.

$\frac{1}{2}$ sirf pehle polarizer par apply hota hai, kyunki uske baad light polarized ho jaati hai. Har baad wala polarizer pichle axis ke relative ${\cos }^2\theta$ use karta hai.

Inverted ratio. Derivation deta hai $\tan {\theta }_B=n_2/n_1=n_{\text{into}}/n_{\text{from}}$. Air→glass jaate hue, woh $n_{\text{glass}}$ hai, uska reciprocal nahi.

Woh perpendicular hoti hain, ${\theta }_B+{\theta }_r=90^{\circ }$. Wahi right angle poori physical reason hai jisse in-plane reflection cancel hoti hai.

Koi change nahi — ${\cos }^2\theta$ ki period $180^{\circ }$ hai kyunki $\cos (\theta +180^{\circ })=-\cos \theta$ aur squaring sign ko khatam kar deta hai. Axis ek line hai, arrow nahi.

Koi energy add nahi hoti; middle filter polarization ko realign karta hai taaki last filter surviving field ke saath fully crossed na rahe. Har stage sirf absorb karta hai, kabhi amplify nahi.

$\theta$ transmission axis tak measure kiya jaata hai, aur ek vector ka component along ek axis, magnitude times us axis se angle ka cosine hota hai. $\sin$ use karne se woh component measure hota jo absorb ho raha hai.

Ek EM wave ki energy density field amplitude ke square ke saath scale hoti hai, aur intensity energy flow per area hai — isliye $E$ double karne se $I$ chaar guna ho jaata hai.

${\cos }^2\theta$, $\frac{1}{2}$ value ke baare mein symmetrically $0$ aur $1$ ke beech oscillate karta hai (kyunki ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$ aur $\cos 2\theta$ average hokar zero ho jaata hai). Isliye aadhi intensity survive karti hai.

Dono polarization components ek interface par alag-alag strength se reflect karti hain, isliye mix unbalanced hoti hai — partially polarized. Brewster's angle par ek component zero ho jaata hai, jisse woh total ho jaati hai.

Refracted ray glass ke electrons ko apne transverse field ke saath oscillate karne ke liye drive karti hai. Ek charge apne khud ke oscillation axis ke saath kuch bhi radiate nahi karta, aur woh forbidden direction reflected ray ke saath exactly tab align hoti hai jab dono rays perpendicular hoti hain (teesri figure dekho).

Horizontal surfaces (water, roads) se glare horizontally polarized hota hai. Ek vertical axis us glare ke saath crossed hai, isliye ${\cos }^{2}90^{\circ }=0$ use block karta hai jabki useful vertical light through jaane deta hai.

Geometric condition ${\theta }_r=90^{\circ }-{\theta }_B$ akele do angles relate karti hai; Snell un angles ko indices $n_1,n_2$ se link karta hai. Ek ko doosre mein substitute karne se geometry $\tan {\theta }_B=n_2/n_1$ mein convert ho jaati hai.

Zero — ${\cos }^{2}90^{\circ }=0$. Field poori tarah axis ke perpendicular hai, isliye polarizer sab kuch absorb kar leta hai (crossed polarizers).

Formula formally abhi bhi hold karta hai: $\tan {\theta }_B=n_2/n_1=1$, isliye ${\theta }_B=45^{\circ }$. Lekin equal indices matlab koi genuine optical interface nahi hai, isliye koi reflection nahi hoti polarize karne ke liye — number mathematically valid hai lekin act karne ke liye kuch nahi.

Haan — $\tan {\theta }_B=n_2/n_1<1$, ek chhota Brewster angle deta hai. Yeh tab tak exist karta hai jab tak total internal reflection pehle na ho jaye; critical angle alag se check karo.

Nahi — ${\theta }_B=\arctan (n_2/n_1)$ aur $\arctan$ kisi bhi finite index ratio ke liye $90^{\circ }$ ke paas aata hai lekin kabhi pahunchta nahi. Grazing-incidence polarization by reflection impossible hai.

Malus directly apply nahi ho sakta — koi reference direction nahi hai. Tum sab $\theta$ par average karte ho, jisse $\frac{1}{2}$ rule wapas milta hai instead of a ${\cos }^2\theta$ value.

Har step intensity ko ${\cos }^2(\frac{90^{\circ }}{N})\approx 1$ se multiply karta hai, isliye product $[{\cos }^2(\frac{90^{\circ }}{N}){]}^N\to 1$: light gradually poore $90^{\circ }$ rotate ho jaati hai near-zero loss ke saath — crossed-pair revival trick ke peeche yahi limiting behaviour hai (doosri figure dekho).

Full transmission, $I=I_0$, kyunki ${\cos }^{2}0=1$. Koi half-factor apply nahi hota kyunki light polarized hai, unpolarized nahi.