Exercises — Polarization — Malus's law, Brewster's angle derivation

Input kis type ka hai? Unpolarized — $\vec{E}$ har transverse direction mein randomly point karta hai. Malus directly kyun nahi? Malus ke liye ek single angle $\theta$ chahiye. Unpolarized light ka koi single angle nahi hota; yeh sabhi angles ka even mix hai. Toh hum ${\cos }^2\theta$ ko ek full turn par average karte hain, jo $\frac{1}{2}$ deta hai. $I=\frac{1}{2}{I}_0.$ Output ab is polarizer ke axis ke along polarized hai (yeh agli problems ke liye important hai).

Brewster's law hai $\tan {\theta }_B=n_2/n_1$, jahan $n_2$ woh medium hai jisme light jaati hai aur $n_1$ woh medium hai jisse woh aati hai. $\tan {\theta }_B=\frac{2.42}{1}=\mathrm{2.42.}$ Yeh rahi equation. (Solve karne par: ${\theta }_B=\arctan 2.42\approx 67.5^{\circ }$ — L2 mein puri tarah kiya gaya hai.)

$I=I_0{\cos }^{2}90^{\circ }=I_0\cdot 0^2=0.$ Zero kyun? $\cos 90^{\circ }=0$: wiggle poori tarah slot ke perpendicular hai, isliye axis ke along koi bhi component nahi hai. Yeh ek crossed polarizer hai.

Step 1: $\tan {\theta }_B=2.42\Rightarrow {\theta }_B=\arctan (2.42)=67.54^{\circ }.$ Step 2 (WHY hum ${\theta }_r$ free mein le sakte hain): Brewster's angle par reflected aur refracted rays perpendicular hote hain, isliye ${\theta }_r=90^{\circ }-{\theta }_B=22.46^{\circ }.$ Snell se check: $\frac{\sin 67.54^{\circ }}{\sin 22.46^{\circ }}=\frac{0.924}{0.382}=2.42$ ✓.

A se: angle $0^{\circ }$ hai, isliye $I_A=I_0{\cos }^2{0}^{\circ }=I_0$ (A aligned hai — kuch nahi gaya). B se: A ko leave karne wali light A ke axis ke along polarized hai, toh woh B se $60^{\circ }$ par milti hai: $I_B=I_A{\cos }^{2}60^{\circ }=I_0\cdot {\left(\frac{1}{2}\right)}^2=\frac{1}{4}{I}_0=0.25I_0.$

Step 1 (A): input unpolarized → halve karo: $I_A=\frac{1}{2}{I}_0$, ab A ke along polarized. Step 2 (B): polarized light B se $30^{\circ }$ par milti hai: $I_B=I_A{\cos }^{2}30^{\circ }=\frac{1}{2}{I}_0\cdot {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{2}{I}_0\cdot \frac{3}{4}=\frac{3}{8}{I}_0=0.375I_0.$

Consecutive polarizers ke beech ke angles matter karte hain — yeh upar wali figure mein key idea hai. A ke baad: unpolarized halved → $I_A=\frac{1}{2}{I}_0$, $0^{\circ }$ par polarized. B ke baad: A se B tak angle $30^{\circ }$ hai: $I_B=I_A{\cos }^{2}30^{\circ }=\frac{1}{2}{I}_0\cdot \frac{3}{4}=\frac{3}{8}{I}_0.$ Light ab B ke along polarized hai ($30^{\circ }$ par). C ke baad: B ($30^{\circ }$) se C ($75^{\circ }$) tak angle $75^{\circ }-30^{\circ }=45^{\circ }$ hai: $I_C=I_B{\cos }^{2}45^{\circ }=\frac{3}{8}{I}_0\cdot \frac{1}{2}=\frac{3}{16}{I}_0=0.1875I_0.$

A → B: angle $\varphi$ hai, isliye $I_B=I_0{\cos }^2\varphi$ (A aligned hai, A par koi loss nahi). B → C: C $90^{\circ }$ par hai, B $\varphi$ par hai, toh unke beech angle $90^{\circ }-\varphi$ hai: $I_C=I_B{\cos }^2(90^{\circ }-\varphi )=I_0{\cos }^2\varphi {\sin }^2\varphi .$ Light kyun wapas aati hai: bina middle filter ke C ke axis ke along field zero hai. Middle filter field ko ek slanted axis par re-project karta hai, jisse C ke along ek nonzero component mil jaata hai. Maximize: $\cos \varphi \sin \varphi =\frac{1}{2}\sin 2\varphi$ use karo, isliye $I_C=\frac{1}{4}{I}_0{\sin }^{2}2\varphi$. Yeh tab peak karta hai jab $\sin 2\varphi =1$, yaani $2\varphi =90^{\circ }\Rightarrow \varphi =45^{\circ }$, jo deta hai $I_C^{\max }=\frac{1}{4}{I}_0=0.25I_0.$

(a) Air → water: $\tan {\theta }_B=1.33\Rightarrow {\theta }_B=\arctan (1.33)=53.06^{\circ }.$ (b) Glare horizontally polarized hai; glasses vertical pass karte hain. Unke beech angle $90^{\circ }$ hai: $I=I_{\text{glare}}{\cos }^{2}90^{\circ }=0.$ Woh Brewster-reflected glare ka saara hissa block karti hain. (c) ${\theta }_B$ par reflected ray incidence ke plane ke perpendicular 100% polarized hoti hai — horizontal lake ke liye woh plane vertical hai, isliye polarization horizontal hai. Ek vertical filter usse exactly cross karta hai, toh glare khatam ho jaata hai jabki ordinary (partly vertical) scenery light bachti hai.

A ke baad: $\frac{1}{2}{I}_0$. B ke baad: $I=\frac{1}{2}{I}_0{\cos }^2\theta$. Threshold ke equal set karo: $\frac{1}{2}{I}_0{\cos }^2\theta =0.30I_0\Rightarrow {\cos }^2\theta =0.60\Rightarrow \cos \theta =\mathrm{0.7746.}$ $\theta =\arccos (0.7746)=39.23^{\circ }.$ Isse bada koi bhi $\theta$ output ko threshold se neeche dim kar dega, isliye ${\theta }_{\max }=39.23^{\circ }$.

(a) Glass se air tak: $n_2=1$ (air mein), $n_1=1.5$ (glass se): $\tan {\theta }_B^{\prime }=\frac{n_{\text{air}}}{n_{\text{glass}}}=\frac{1}{1.5}=0.6667\Rightarrow {\theta }_B^{\prime }=\arctan (0.6667)=33.69^{\circ }.$ Yeh pehle wale refraction angle ${\theta }_r$ ke barabar hai — upar wali picture dikhati hai kyun internal Brewster ray external refracted ray ko retrace karti hai. (b) ${\theta }_B+{\theta }_B^{\prime }=56.31^{\circ }+33.69^{\circ }=90.00^{\circ }.$ Exactly $90^{\circ }$ kyun: $\tan {\theta }_B=n_2/n_1$ aur $\tan {\theta }_B^{\prime }=n_1/n_2$ reciprocals hain. Kyunki $\tan (90^{\circ }-x)=1/\tan x$, dono angles complementary hain. (c) External refracted ray aur internal Brewster ray ray reversibility se ek hi line hain: Brewster incidence par glass mein refracted ray, ulti chalayi jaaye toh, glass ko apne Brewster angle par leave karti hai.

Reflected beam already polarized hai (horizontal), isliye hum use halve nahi karte — real angles ke saath Malus use karo. Reflected beam: $I_1=0.08I_0$, $0^{\circ }$ par polarized (horizontal). P1 se (at $40^{\circ }$): beam se P1 tak angle $40^{\circ }$ hai: $I_2=I_1{\cos }^{2}40^{\circ }=0.08I_0\cdot (0.766{)}^2=0.08I_0\cdot 0.5868=0.04695I_0.$ Light ab P1 ke along polarized hai ($40^{\circ }$ par). P2 se (at $70^{\circ }$): P1 ($40^{\circ }$) se P2 ($70^{\circ }$) tak angle $30^{\circ }$ hai: $I_3=I_2{\cos }^{2}30^{\circ }=0.04695I_0\cdot 0.75=0.03521I_0.$ Final: $I_3\approx 0.0352I_0.$

Ek polarizer se unpolarized light? ::: $\frac{1}{2}{I}_0$, output axis ke along polarized. Malus angle consecutive filters ke beech, pehle filter se nahi? ::: Sahi — har filter light ko re-polarize karta hai. Brewster angle air→glass ($n=1.5$)? ::: $\arctan 1.5=56.3^{\circ }$. Vertical sunglasses lake glare kyun kill karti hain? ::: Brewster glare horizontal hai, vertical ke saath $90^{\circ }$ par crossed. ${\theta }_B$ aur reverse-direction ${\theta }_B^{\prime }$ ka sum? ::: $90^{\circ }$ (reciprocal tangents). Crossed polarizers ke beech ek middle one se max output? ::: Middle angle $45^{\circ }$ par $\frac{1}{4}{I}_0$.