Polarization — Malus's law, Brewster's angle derivation: Worked examples

2.5.17 · D3 · Physics › Optics › Polarization — Malus's law, Brewster's angle derivation

Yeh page har tarah ke questions parent note se dhundh ke laati hai. Pehle hum ek scenario matrix banate hain — saare case-classes ka ek grid — phir har cell ke liye ek poora example karte hain taaki koi bhi aisi situation na aaye jo tumne pehle dekhi na ho.

Do tools yahan saara kaam karte hain, aur dono humne parent note mein earn kiye hain:

Malus's law $I = I_{0} cos^{2} θ$ — pehle se polarized light mein se kitna bach jaata hai jab polarizer $θ$ angle par tilt ho.
Brewster's law $tan θ_{B} = n_{2} / n_{1}$ — wo incidence angle jo reflected light ko perfectly polarized bana deta hai.

Agar neeche koi symbol unfamiliar lage, to pehle Unpolarized vs Polarized Light aur Wave Nature of Light dobara padho — hum assume karte hain ki tum jaante ho ki light ek wiggling transverse wave hai aur ek polarizer sirf apni axis ke saath wiggle ko rakhta hai.

Scenario matrix

Har polarization problem in cells mein se ek (ya ek chain) hota hai:

#	Cell class	Isme kya tricky hai	Example jo ise cover karta hai
A	Unpolarized → 1 polarizer	$\frac{1}{2}$ rule use karo, $cos^{2}$ nahin	Ex 1
B	Polarized → polarizer at angle $θ$	Plain Malus, square yaad rakho	Ex 2
C	Degenerate angles: $θ = 0^{\circ}$ aur $θ = 9 0^{\circ}$	Full pass vs. total block (crossed)	Ex 3
D	Chain of $N$ polarizers, stepped angles	Malus stage-by-stage lagao; light wapas aati hai	Ex 4
E	"Kaunsa angle ek target fraction deta hai?" (inverse Malus)	$cos^{2} θ = k$ solve karo, dono roots dekho	Ex 5
F	Brewster's angle, air → dense medium	$tan θ_{B} = n_{2} / n_{1}$ , reflected ⟂ refracted check karo	Ex 6
G	Brewster reversed: dense → rare medium	Ratio flip hota hai; $θ_{B}$ chhota ho jaata hai	Ex 7
H	Real-world word problem (glare / sunglasses)	Physics words ko $θ$ aur axes mein translate karo	Ex 8
I	Exam twist: Brewster aur Malus combine karo	Ek chain mein do tools	Ex 9

Ab hum poora grid clear karte hain.

Cell A — Unpolarized light, ek polarizer

Worked example Ex 1 · The

$\frac{1}{2}$ rule (Cell A) Unpolarized sunlight jiska intensity $I_{0} = 800 W/m^{2}$ hai, ek single polarizer se guzarti hai. Kya nikalta hai, aur kya woh polarized hai?

Forecast: Padhne se pehle guess karo — kya yeh half se kam, exactly half, ya angle-dependent hoga?

Pehchano ki input unpolarized hai. Yeh step kyun? Unpolarized light har wiggle-angle $θ$ ka ek random mix hai, isliye koi ek $θ$ nahin hai jo $cos^{2} θ$ mein plug kiya ja sake. Hume saare angles par average karna hoga.
Averaged result use karo $⟨ cos^{2} θ ⟩ = \frac{1}{2}$ : $I = \frac{1}{2} I_{0} = \frac{1}{2} (800) = 400 W/m^{2} .$ Yeh step kyun? Parent ne derive kiya tha $\frac{1}{2 π} \int_{0}^{2 π} cos^{2} θ d θ = \frac{1}{2}$ — aadhi energy survive karti hai chahe tum is pehle filter ko kitna bhi rotate karo.
Output polarization batao. Kyun? Jo bhi survive karta hai woh transmission axis ke saath aligned hota hai, isliye beam ab fully polarized hai us axis ke saath.

Verify: Units $W/m^{2}$ andar, $W/m^{2}$ bahar ✓. Is single polarizer ko rotate karne se reading nahin badlegi (abhi bhi $400$ rahega) — yeh unpolarized input ki tell-tale signature hai. Agar badalta, to input pehle se polarized tha.

Cell B — Polarized light, ek polarizer

Worked example Ex 2 · Plain Malus with the square (Cell B)

Light jo pehle se vertically polarized hai, intensity $I_{0} = 200 W/m^{2}$ , ek aisi polarizer se milti hai jiska axis vertical se $6 0^{\circ}$ tilt hai. Transmitted intensity nikalo.

Forecast: $6 0^{\circ}$ par — full pass ke zyada karib ya blocked ke zyada karib?

$θ = 6 0^{\circ}$ identify karo = incoming polarization aur axis ke beech ka angle. Yeh step kyun? Malus's law ko field aur axis ke beech ka angle chahiye, kisi aur cheez se nahin.
Malus square ke saath apply karo: $I = I_{0} cos^{2} 6 0^{\circ} = 200 (\frac{1}{2})^{2} = 200 \cdot \frac{1}{4} = 50 W/m^{2} .$ Square kyun? Ek detector intensity padhta hai, aur intensity $\propto E^{2}$ hoti hai. Projection se amplitude $E_{0} cos θ$ milta hai; square karne se $cos^{2} θ$ milta hai. Ise bhoolne par galat $100$ milta.

Verify: $cos 6 0^{\circ} = 0.5$ , to $cos^{2} = 0.25$ ; $200 \times 0.25 = 50$ ✓. Sanity check: $6 0^{\circ}$ , $4 5^{\circ}$ se zyada hai, isliye half se zyada lost hai — $50 < 100$ , consistent.

Cell C — Degenerate angles: $0^{\circ}$ aur $9 0^{\circ}$

Cell D — Chain of polarizers (light wapas aati hai)

Worked example Ex 4 · Teen polarizers, stepped angles (Cell D)

Polarized light $I_{0} = 100 W/m^{2}$ polarizer P1 ki axis ke saath enter karti hai. P2, P1 se $3 0^{\circ}$ par hai; P3, P1 se $6 0^{\circ}$ par hai. Final intensity nikalo.

Forecast: Kya answer tiny hoga, ya surprisingly large jab total turn $6 0^{\circ}$ hai?

Figure s01 — Ex 4: light teen polarizers P1, P2, P3 se step karti hai jinki axes $0^{\circ}, 3 0^{\circ}, 6 0^{\circ}$ tilt karti hain; intensity (W/m²) har beam segment ke upar print hai.

Figure padhna: beam left → right travel karti hai (yellow arrows, har arrow ke upar number $W/m^{2}$ mein intensity hai). Har shaded block ek polarizer hai; har block ke andar dark line uski transmission axis hai, P1, P2, P3 ke liye vertical se $0^{\circ}$ , $3 0^{\circ}$ , $6 0^{\circ}$ tilt ki hui. Notice karo ki axis block se block mein barabar $3 0^{\circ}$ steps mein rotate karti hai — yahi stepping poora point hai.

P1 → P2, step angle $= 3 0^{\circ}$ (figure mein P1 ki axis aur P2 ki axis ke beech ka tilt): $I_{2} = I_{0} cos^{2} 3 0^{\circ} = 100 (\frac{3}{2})^{2} = 100 \cdot \frac{3}{4} = 75 W/m^{2} .$ Step-by-step kyun? Har polarizer light ko apni axis ke saath repolarize kar deta hai, isliye agla angle pichhli axis se measure hota hai, P1 se nahin.
P2 → P3, step angle $= 6 0^{\circ} - 3 0^{\circ} = 3 0^{\circ}$ : $I_{3} = I_{2} cos^{2} 3 0^{\circ} = 75 \cdot \frac{3}{4} = 56.25 W/m^{2} .$ Phir $3 0^{\circ}$ kyun? P2 ke baad light P2 ki taraf point karti hai. P3, P2 se $3 0^{\circ}$ aage hai (figure mein dono axis lines padho), isliye relevant tilt $3 0^{\circ}$ hai.

Verify: Agar hum (galti se) P2 hata dete, to P1→P3 direct $100 cos^{2} 6 0^{\circ} = 25 W/m^{2}$ deta. Middle filter insert karne se output $56.25 W/m^{2}$ tak badh gaya, jo $25 W/m^{2}$ se zyada hai — parent se reappearing-light surprise, kyunki rotation ko step karna ek bade turn se kam waste karta hai ✓.

Cell E — Inverse Malus (angle nikalo)

Worked example Ex 5 · Tilt ke liye solve karo (Cell E)

Polarized light ko ek polarizer se exactly apni intensity ke $\frac{1}{4}$ tak dim karna hai. Axis kitne angle par set karni chahiye?

Forecast: Ek answer, ya kai?

Target ko equation ke roop mein likho: $\frac{I}{I _{0}} = cos^{2} θ = \frac{1}{4} .$ Yeh step kyun? "Intensity ka fraction" hi $cos^{2} θ$ hai; unknown ab $θ$ hai, isliye hum Malus ko invert karte hain.
Square root lo, dono signs rakhke: $cos θ = \pm \frac{1}{2} .$ $\pm$ kyun? Squaring ne sign mita diya; $+ \frac{1}{2}$ aur $- \frac{1}{2}$ dono equation satisfy karte hain, aur physical tilts kisi bhi taraf ho sakti hain.
$[0^{\circ}, 18 0^{\circ})$ mein saare angles padho: $θ = 6 0^{\circ} or θ = 12 0^{\circ} .$ Do kyun? $cos 6 0^{\circ} = + \frac{1}{2}$ aur $cos 12 0^{\circ} = - \frac{1}{2}$ ; dono $cos^{2} θ = \frac{1}{4}$ dete hain. Ek polarizer projection ke sign ki parwah nahin karta, sirf uske square ki.

Verify: $cos^{2} 6 0^{\circ} = 0.25$ ✓ aur $cos^{2} 12 0^{\circ} = (- 0.5)^{2} = 0.25$ ✓. Dono quarter tak dim karte hain — matrix cell ka "watch both roots" satisfy ho gaya.

Cell F — Brewster's angle, air → dense medium

Worked example Ex 6 · Diamond ke liye Brewster (Cell F)

Air ( $n_{1} = 1$ ) mein light diamond ( $n_{2} = 2.42$ ) se takraati hai. Brewster's angle nikalo aur confirm karo ki reflected aur refracted rays perpendicular hain.

Forecast: Glass ke $56. 3^{\circ}$ se bada ya chhota?

Figure s02 — Ex 6: Brewster's angle par reflected ray (coral) aur refracted ray (mint) normal se $θ_{B}$ aur $θ_{r}$ par milti hain, aur dono rays bilkul $9 0^{\circ}$ apart hain.

Figure padhna: horizontal line air–diamond surface hai; dashed vertical line normal hai (surface ke perpendicular). Upper left se neeche aata yellow arrow incident ray hai, upper right ko jaata coral arrow reflected ray hai, aur diamond mein neeche jaata mint arrow refracted ray hai. Har ray jo normal se angle banati hai woh marked hai: incident aur reflected ke liye $θ_{B}$ (reflection ke law se dono equal hain), refracted ke liye $θ_{r}$ . Dayi taraf ka label reflected aur refracted arrows ke beech $9 0^{\circ}$ dikhata hai.

Brewster's law apply karo: $tan θ_{B} = \frac{n _{2}}{n _{1}} = \frac{2.42}{1} = 2.42.$ $tan$ kyun, $sin$ nahin? Parent ki derivation ne Snell's law ko $θ_{r} = 9 0^{\circ} - θ_{B}$ ke saath combine kiya; $cos θ_{B}$ se divide karne par ratio tangent ban gaya. ("Brewster Tangles.")
Arctan se invert karo (woh angle jiska tangent $2.42$ hai): $θ_{B} = arctan (2.42) = 67. 5^{\circ} .$
$θ_{r} = 9 0^{\circ} - θ_{B}$ kahan se aata hai? (mini-derivation, figure se padho). Us point ko dekho jahan beam surface se takraati hai. Teen cheezein normal ke saath upar aur neeche right-hand side mein fan out karti hain: reflected ray normal se $θ_{B}$ par hai (upper right), refracted ray normal se $θ_{r}$ par hai (lower right), aur reflected aur refracted rays ke beech $9 0^{\circ}$ Brewster corner hai. Yeh teen angles straight normal line ke around stack hain, jo straight upar se straight neeche tak $18 0^{\circ}$ span karti hai. Isliye inhe ek straight angle mein add hona chahiye: $θ_{B} + 9 0^{\circ} + θ_{r} = 18 0^{\circ} \Rightarrow θ_{r} = 9 0^{\circ} - θ_{B} = 9 0^{\circ} - 67. 5^{\circ} = 22. 5^{\circ} .$ Yeh kyun matter karta hai? Yahi geometric fact hai jo Snell ko Brewster mein convert karti hai; iske bina hum $θ_{r}$ eliminate nahin kar sakte.

Verify (Snell mein plug karo, rounding dhyan se): $n_{1} sin θ_{B} = 1 \cdot sin 67. 5^{\circ} = 0.9239$ ; $n_{2} sin θ_{r} = 2.42 \cdot sin 22. 5^{\circ} = 2.42 \times 0.3827 = 0.9261$ . Yeh sirf teesre decimal mein differ karte hain ( $0.9239$ vs $0.9261$ , ek $0.2%$ gap) — yeh pure rounding hai $θ_{B}$ ko $67. 5^{\circ}$ likhne se, $arctan 2.42 = 67.52 3^{\circ}$ ki jagah. Un-rounded angle use karne par exactly match karte hain ✓. Denser medium ⇒ glass ke $56. 3^{\circ}$ se bada $θ_{B}$ , jaise forecast kiya tha.

Cell G — Brewster reversed: dense → rare medium

Worked example Ex 7 · Glass se air mein jaati light (Cell G)

Light glass ke andar travel kar rahi hai ( $n_{1} = 1.5$ ) aur glass–air surface ( $n_{2} = 1.0$ ) se takraati hai. Is internal reflection ke liye Brewster's angle nikalo.

Forecast: Air→glass value $56. 3^{\circ}$ se bada ya chhota?

Ratio sahi order mein rakho: $tan θ_{B} = \frac{n _{2}}{n _{1}} = \frac{1.0}{1.5} = 0.667.$ Flip kyun? $n_{2}$ woh medium hai jisme light jaati hai (air), $n_{1}$ wahan hai jahan se woh start karti hai (glass). Ise ulta karna parent ki mistakes mein flagged classic error hai.
Invert karo: $θ_{B} = arctan (0.667) = 33. 7^{\circ} .$

Verify: Notice karo $33. 7^{\circ} + 56. 3^{\circ} = 9 0^{\circ}$ : kisi bhi interface pair ke liye dono Brewster angles complementary hain. Kyun? $tan θ_{B}^{in} = 1/ tan θ_{B}^{out}$ , aur jinke tangents reciprocal hain wo angles $9 0^{\circ}$ mein add hote hain ✓. Reversed direction chhota angle deta hai, jaise forecast kiya tha.

Cell H — Real-world word problem (glare)

Worked example Ex 8 · Fisherman ke sunglasses (Cell H)

Sunlight ek still lake se reflect ho ke tumhari aankhon tak glare ke roop mein pahunchti hai. (a) Vertical se kitne viewing angle par glare fully polarized hota hai? (b) Woh glare kis direction mein polarized hai, aur sunglasses ki axis use kill karne ke liye kaise orient karni chahiye? Water $n = 1.33$ . (Plane of incidence is page ke shuru mein define kiya gaya tha.)

Forecast: Glasses ke liye vertical ya horizontal transmission axis?

Glare Brewster's angle par fully polarized hota hai: $tan θ_{B} = \frac{n _{water}}{n _{air}} = \frac{1.33}{1} = 1.33, θ_{B} = arctan (1.33) = 53. 1^{\circ} .$ Kyun? Kisi surface se reflection sirf Brewster's angle par fully polarized hoti hai — physics word "fully polarized glare" ka code " $θ_{B}$ " hai. (a) ka Answer: $53. 1^{\circ}$ vertical se (yaani water ke normal se).
Glare ki polarization ki direction: reflected light plane of incidence ke perpendicular polarized hai ("s" direction jo upar define ki gayi), jo water surface ke parallel = horizontal rehti hai. Kyun? Parent ka electron-oscillation argument in-plane ("p") part ko $θ_{B}$ par kill kar deta hai, isliye sirf surface-parallel ("s") wiggle reflection mein survive karti hai.
Sunglasses orient karo use block karne ke liye: transmission axis vertical set karo, horizontal glare ke perpendicular. Kyun? Malus with $θ = 9 0^{\circ}$ deta hai $I = I_{0} cos^{2} 9 0^{\circ} = 0$ — horizontal glare completely absorb ho jaata hai. (b) ka Answer: glare horizontal hai, isliye sunglasses ki axis use kill karne ke liye vertical honi chahiye.

Verify: $arctan (1.33) = 53. 1^{\circ}$ ✓; blocking condition $cos^{2} 9 0^{\circ} = 0$ use karti hai ✓. Isi liye polarized sunglasses ki vertical axes hoti hain — parent ke Feynman note se exactly match karta hai.

Cell I — Exam twist: Brewster feeds Malus

Worked example Ex 9 · Ek chain mein do tools (Cell I)

Unpolarized light jiska intensity $I_{0} = 400 W/m^{2}$ hai, glass ( $n = 1.5$ ) se Brewster's angle par reflect hoti hai, fully horizontally polarized ho jaati hai. Phir yeh ek aisi polarizer se guzarti hai jiska axis horizontal se $4 0^{\circ}$ par hai. Final intensity nikalo. (Real reflection mein sirf kuch light reflect hoti hai; yahan assume karo ki jo reflected beam nikalta hai uski intensity $I_{0}$ hai aur fully polarized hai — yeh ek common exam idealisation hai.)

Forecast: Aakhri step mein kaun sa law lagega — $\frac{1}{2}$ rule ya $cos^{2}$ ?

Brewster step polarization state fix karta hai, yahan number nahin. Yeh step kyun? Hume bataya gaya hai ki reflected beam fully horizontally polarized hai intensity $I_{0} = 400$ ke saath. $θ_{B} = arctan (1.5) = 56. 3^{\circ}$ par reflection 100% polarized hai — yahi agle step ko $cos^{2}$ problem banata hai, $\frac{1}{2}$ problem nahin.
Ab light polarized hai, to Malus use karo ( $\frac{1}{2}$ nahin): horizontal polarization aur polarizer axis ke beech angle $θ = 4 0^{\circ}$ hai: $I = I_{0} cos^{2} 4 0^{\circ} = 400 cos^{2} 4 0^{\circ} .$ Halve kyun nahin? $\frac{1}{2}$ rule sirf unpolarized input par apply hota hai; Brewster reflection ke baad beam pehle se polarized hai, isliye $cos^{2} θ$ correct hai.
Compute karo: $cos 4 0^{\circ} = 0.766$ , to $cos^{2} 4 0^{\circ} = 0.587$ : $I = 400 \times 0.587 = 234.7 W/m^{2} .$

Verify: $θ_{B} = arctan 1.5 = 56. 3^{\circ}$ ✓; $400 \times cos^{2} 4 0^{\circ} \approx 234.7 W/m^{2}$ ✓; result $[0, 400]$ mein hai ✓. Agar hum galti se pehle $\frac{1}{2}$ rule apply karte, to $\frac{1}{2} (400) cos^{2} 4 0^{\circ} = 117.4$ milta — correct value ka aadha, exactly woh trap jo matrix mein hai.

Recall

Pehle filter ke liye kaun sa rule — $\frac{1}{2}$ ya $cos^{2}$ ? Agar input unpolarized hai → $\frac{1}{2}$ . Agar pehle se polarized hai → $cos^{2} θ$ . ::: Unpolarized → $\frac{1}{2} I_{0}$ ; polarized → $I_{0} cos^{2} θ$ . Inverse-Malus ek fraction ke liye do angles kyun deta hai? ::: Kyunki $cos^{2} θ = k$ ⇒ $cos θ = \pm k$ , jo $6 0^{\circ}$ aur $12 0^{\circ}$ jaisi supplementary pair deta hai. Glass↔air Brewster pair complementary kyun hai? ::: Tangents reciprocal hain ( $n_{2} / n_{1}$ vs $n_{1} / n_{2}$ ), aur reciprocal-tangent angles $9 0^{\circ}$ mein add hote hain.

Connections

Snell's Law and Refraction — har Brewster verification wapas Snell mein plug hota hai.
Electromagnetic Waves — $I \propto E^{2}$ jo Malus mein square force karta hai.
Reflection and Refraction at Interfaces — reflected ⟂ refracted geometry.
Unpolarized vs Polarized Light — har stage par $\frac{1}{2}$ vs $cos^{2}$ decide karta hai.