New sheet - 10/27 16:15

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Okay, sorry! Main poora cheat sheet ab Hindi aur English mix (Hinglish) mein, jaise aapne bola hai, deta hoon. Koi bhi topic skip nahi karenge aur jahan diagrams hain, unko bhi explain karenge. Mathematical formulas KaTeX mein hi rahenge. Ray Optics: Introduction Optics physics ki woh branch hai jo light aur dusri electromagnetic waves ke behaviour ko study karti hai. Aam mirrors aur lenses mein, light ki wavelength device ke dimensions ke saamne bohot chhoti hoti hai. Toh light ko ek ray ki tarah treat kiya ja sakta hai, jiska propagation simple geometric rules se govern hota hai. Iss phenomenon se deal karne wale optics ke part ko geometric optics kehte hain. Propagation of Light Light kisi medium ya vacuum mein seedhi line mein chalti hai. Light ka path tabhi change hota hai jab uske raaste mein koi object ho ya medium change ho. Isse rectilinear (straight-line) propagation of light kehte hain. Light rays kisi source ke har point se seedhi lines mein nikalte hain jab tak woh kisi object ya do media ko alag karne wali surface par na gir jayein. Light rays ke bundle ko beam of light kehte hain. Vacuum aur gases ke alawa, light kuch liquids aur solids se bhi travel kar sakti hai. Jo medium light ko lambi doori tak aazadi se travel karne deta hai, use transparent medium kehte hain. Jaise paani, glycerine, glass, aur saaf plastics. Jo medium light ko travel nahi karne deta, use opaque medium kehte hain. Jaise wood, metals, bricks. Kuch materials, jaise tel, light ko thodi doori tak travel karne dete hain, lekin unki intensity jaldi kam ho jati hai. Inhe translucent materials kehte hain. Reflection of Light Jab light rays do media (jaise hawa aur glass) ke boundary se takrati hain, toh light ka kuch hissa usi medium mein wapas bounce ho jata hai. Is process ko Reflection of light kehte hain. (i) Regular / Specular Reflection Jab reflection ek perfect plane surface se hota hai, toh rays reflect hone ke baad bhi parallel rehti hain. Ise Regular reflection kehte hain. Mirror Incident Rays Reflected Rays (ii) Diffused Reflection Jab surface rough hoti hai, toh light uski plane surfaces ke chhote-chhote hisson se irregular directions mein reflect hoti hai. Ise diffused reflection kehte hain. Is process ki wajah se hum kisi object ko kisi bhi position se dekh paate hain. \ | / \ | / \ | / -------/_\------- Surface / | \ / | \ / | \ Laws of Reflection Laws of Reflection: The incident ray, the reflected ray, and the normal to the surface at the point of incidence all lie in the same plane. The angle of reflection ($ \angle r $) is equal to the angle of incidence ($ \angle i $), i.e., $ \angle i = \angle r $. In vector form: $ \vec{r} = \vec{e} - 2(\vec{e} \cdot \vec{n}) \vec{n} $ Reflecting Surface Normal (N) Incident Ray A Reflected Ray B O $i$ $r$ Golden Key Points Rectilinear propagation of light: Ek homogeneous transparent medium mein light seedhi line mein chalti hai. Jab ek ray normally kisi boundary par incident hoti hai, toh reflection ke baad woh apna path retraces karti hai. $i=0, r=0$ plane mirror $i=0, r=0$ concave mirror $i=0, r=0$ convex mirror Reflection par frequency, wavelength, aur speed change nahi hote. Aankh yellow color ke liye sabse zyada sensitive hoti hai, aur violet aur red color ke liye sabse kam. Isi wajah se: Commercial vehicles ko yellow color se paint kiya jata hai. Road lights mein Sodium lamps (yellow color) use kiye jaate hain. Reflection from Plane Mirror Plane mirror object aur image ko jodne wali line ka perpendicular bisector hota hai. Plane mirror se bani image mein lateral inversion hoti hai, yani image mein left right ho jata hai aur vice-versa. back mirror front Jab ek ghadi ko plane mirror ke saamne rakha jata hai, toh ghadi ka time object time hota hai aur mirror ke saamne khade person ko jo time dikhta hai woh image time hota hai: Object Time = AH Image Time = $12 - AH$ Object Time = AH BM Image Time = $11 - 60' - AH BM$ Object Time = AH BMCS Image Time = $11 - 59' - 60'' - AHBMCS$ Example: Its 01:45 -> Its 10:15 Ek plane mirror virtual world ki window ki tarah behave karta hai. Real Space AB $ \rightarrow $ Plane mirror Virtual Space Ek plane mirror mein complete image dekhne ke liye, mirror ki minimum length person ki height ki aadhi honi chahiye. Figure se, $ \Delta HNM $ aur $ \Delta ENM $ congruent hain. $ \therefore EN = HN $ $ \therefore MD = \frac{1}{2} EN $ Similarly $ \Delta EN'M' $ aur $ \Delta LN'M' $ congruent hain. Mirror ki length $ MM' = MD + M'D = \frac{1}{2} HE + \frac{1}{2} EL $ $ = \frac{1}{2} (HE + EL) = \frac{1}{2} HL $ Minimum mirror length person ki height ki just aadhi hoti hai. Yeh result aankh ki position (ground se height) par depend nahi karta. Yeh result person ki mirror se doori par depend nahi karta. H L E M M' H' L' Single mirror ke liye deviation $ \delta $: $ \delta = 180^\circ - (i+r); \angle i = \angle r; \delta = 180^\circ - 2i $ $i$ $r$ $\delta$ Do plane mirrors ke combination se total deviation, jo ek doosre se $ \theta $ angle par jhuke hue hain, aise determine hota hai: $ \delta = \delta_1 + \delta_2 = 180^\circ - 2\alpha + 180^\circ - 2\beta = 360^\circ - 2(\alpha + \beta) \dots(i) $ $ \Delta QAB $ se, $ 0 + 90 - \alpha + 90 - \beta = 180 \Rightarrow 0 = \alpha + \beta \dots(ii) $ $ \theta $ ki value (i) mein (ii) se rakhne par, $ \delta = 360^\circ - 2\theta $ M1 M2 $\theta$ Incident N1 R1 $\alpha$ $\alpha$ N2 R2 $\beta$ $\beta$ $\delta$ Agar do plane mirrors ek doosre se $ \theta $ angle par jhuke hue hain, toh ek point object ki images ki sankhya (n) aise determine ki jaati hai: (a) Agar $ \frac{360^\circ}{\theta} = m $ even hai, toh images ki sankhya $ n = m - 1 $ hogi. (b) Agar $ \frac{360^\circ}{\theta} = m $ odd hai, toh do cases honge: Jab object bisector par nahi hota, toh images ki sankhya $ n = m $ hoti hai. Jab object bisector par hota hai, toh images ki sankhya $ n = m - 1 $ hoti hai. (c) Agar $ \frac{360^\circ}{\theta} = m $ ek fraction hai, aur object symmetrically rakha gaya hai toh images ki sankhya $ n = $ nearest even integer hoti hai. S.No. $ \theta $ in degree $ m = \frac{360^\circ}{\theta} $ Number of images formed if object is placed asymmetrically symmetrically 1. 0 $ \infty $ $ \infty $ $ \infty $ 2. 30 12 11 11 3. 45 8 7 7 4. 60 6 5 5 5. 72 5 5 4 6. 75 4.8 4 4 7. 90 4 3 3 8. 112.5 3.2 3 2 9. 120 3 3 2 Agar object do plane mirrors ke beech rakha ho toh multiple reflections ki wajah se images bante hain. Har reflection par light energy ka kuch हिस्सा absorb ho jata hai. Isliye, door ki images fainter ho jaati hain. Agar mirror fixed ho aur incident ray ko kuch angle se rotate kiya jaye, toh reflected ray bhi same angle se rotate hoti hai lekin opposite sense mein. (Fig. 1 dekhein) Mirror N I R I' $\theta$ R' $\theta$ (Fig. 1) Mirror Fixed, Incident Ray Rotated Incident ray fixed ho aur mirror ko kuch angle se rotate kiya jaye, toh reflected ray double angle se rotate hoti hai aur same sense mein. (Fig. 2 dekhein) M N I R M' N' R' $2\theta$ $\theta$ (Fig. 2) Incident Ray Fixed, Mirror Rotated Object aur image ki speed same hoti hai. $ V_{op} $ = mirror ke parallel object ki velocity ka component. $ V_{on} $ = mirror ke normal object ki velocity ka component. $ V_{ip} $ = mirror ke parallel image ki velocity ka component. $ V_{in} $ = mirror ke normal image ki velocity ka component. $ V_{ip} = V_{op} $ aur $ (V_{im})_n = -(V_{om})_n $ Agar mirror move kar raha ho: $ V_{ip} = V_{op} $ aur $ (V_{im})_n = -(V_{om})_n + 2(V_{mn})_n $ $ (V_{mn})_n $ = mirror ke normal mirror ki velocity ka component. Example: Image ki velocity batao. Solution: $ V_{ox} = (-10 \cos 37^\circ) \hat{i} = -8\hat{i} $ $ V_{oy} = (10 \sin 37^\circ) \hat{j} = 6\hat{j} $ $ V_{ix} = 2V_{mx} - V_{ox} = 2(-2\hat{i}) - (-8\hat{i}) = 4\hat{i} $ $ V_{iy} = V_{oy} = 6\hat{j} $ Toh image ki velocity $ \vec{V}_i = 4\hat{i} + 6\hat{j} $ m/s hogi. Object 10m/s $37^\circ$ 2m/s Real and Virtual Spaces Ek mirror, chahe plane ho ya spherical, space ko do parts mein divide karta hai: Real space: Woh side jahan reflected rays exist karti hain. Virtual space: Woh side jahan reflected rays exist nahi karti hain. Real Space Virtual Space Virtual Space Object Object sirf incident rays se decide hota hai. Point object woh point hota hai jahan se incident rays actually diverge ( Real object ) ya jahan par incident rays converge hoti hui appear hoti hain ( Virtual object ). Point Object (Real) Point Object (Virtual) Image Image reflected ya refracted rays se decide hoti hai. Point image woh point hota hai jahan reflected/refracted rays actually converge ( real image ) ya jahan se reflected/refracted rays diverge hoti hui appear hoti hain ( virtual image ). (Real image) (Virtual image) Do yourself - 1: Ek light ray 'i' angle par incident hone ke baad reflect hoti hai, toh angle of deviation $ \delta = \pi - 2i $ ya $ \pi + 2i $ prove karo. Ek object right ki taraf 5 m/s se move kar raha hai jabki mirror left ki taraf 1 m/s se move kar raha hai jaisa ki figure mein dikhaya gaya hai. Image ki velocity batao. Figure mein dikhayi gayi situation mein image ki velocity batao. Ek person ki aankh 1.5 m ki height par hai. Woh zameen se 0.8 m upar rakhe 0.3 m lambe plane mirror ke saamne khada hai. Uski apni image ki jo length use dikhti hai, woh kitni hai: (A) 1.5m (B) 1.0m (C) 0.8m (D) 0.6m Object 5 m/s $ \rightarrow $ $ \leftarrow $ Mirror 1 m/s X 5 m/s $30^\circ$ 10 m/s $60^\circ$ y Spherical (Curved) Mirror Curved mirror ek hollow sphere ka part hota hai. Agar reflection inner surface se hota hai toh mirror ko concave kehte hain, aur agar outer surface reflector ki tarah act karta hai toh use convex kehte hain. principal axis spherical surface C P F M concave mirror C P F M convex mirror C P F M Thin Spherical Mirrors ke liye Definitions Pole (P): Mirror ki reflecting surface par koi bhi point hota hai. Convenience ke liye, hum ise mirror ka midpoint maante hain. Principal-section: Mirror ka koi bhi section, jaise MM', jo pole se guzarta hai, use principal-section kehte hain. Centre of curvature (C): Us sphere ka centre jiska mirror ek part hai. Radius of curvature (R): Us sphere ki radius jiska mirror ek part hai. Principal-axis: Line CP, jo pole aur centre of curvature ko jodi hai. Principal-focus (F): Ek image point F jo principal axis par hota hai, jab object infinity par ho. parallel to axis M C P F focus centre of curvature parallel to axis M C P F focus centre of curvature Focal-length (f): Pole P aur focus F ke beech ki distance principal axis ke along. Aperture: Mirror ke reference mein, light reflecting area ka effective diameter. Focal Plane: Woh plane jo focus se guzarta hai aur principal axis ke perpendicular hota hai. Focal plane P F C A B Focal plane P F C A B Paraxial Rays: Woh rays jo normal ke saath chhota angle banati hain incidence point par, aur isliye principal axis ke kareeb hoti hain. n $\theta$ P C n $\theta$ P C Marginal rays: Woh rays jo incidence point par normal ke saath bada angle banati hain ($ \theta $ bada hota hai). n $\theta$ P C Sign-Convention negative positive light P F C positive negative light P F C negative positive light P F C Principal axis ke along distances pole se measure ki jaati hain (pole ko origin maana jata hai). Light ki direction mein distances ko positive maana jata hai, aur opposite direction mein negative. Principal axis ke upar ki distances ko positive maana jata hai, aur neeche ki negative. Jab bhi possible ho, light ray ko left se right travel karte hue maana jata hai. Rules for Image Formation (Paraxial Rays Only) (Ye rules reflection ke laws $ \angle i = \angle r $ par based hain) Ek ray jo principal axis ke parallel hoti hai, reflection ke baad mirror ke focus se guzarti hai ya guzarti hui dikhayi deti hai (focus ki definition se). M P F C M P F C Ek ray jo focus se guzarti hai ya focus ki taraf directed hoti hai, reflection ke baad principal axis ke parallel ho jaati hai. M P F C parallel M P F C parallel Ek ray jo centre of curvature se guzarti hai ya centre of curvature ki taraf directed hoti hai, reflection ke baad apna path retraces karti hai ($ \angle i = 0 $ aur $ \angle r = 0 $). M P F C directed towards center of curvature M P F C directed towards center of curvature Mirror ke pole par incident aur reflected rays principal axis ke around symmetrical hoti hain ($ \angle i = \angle r $). M P $i$ $r$ Relations for Spherical Mirrors $f$ aur $R$ ke beech relation spherical mirror ke liye: Marginal rays ke liye: $ \Delta ABC $ mein, $ AB = BC $ $ AC = CD + DA = 2BC \cos \theta \Rightarrow R = 2BC \cos \theta $ $ \Rightarrow BC = \frac{R}{2 \cos \theta} $ aur $ BP = PC - BC = R - \frac{R}{2 \cos \theta} $ Note: B focus nahi hai; yeh bas woh point hai jahan reflection ke baad marginal ray meet karti hai. M P B C A $2\theta$ $\theta$ D Paraxial rays (principal axis ke parallel) ke liye: ($ \theta $ chhota hai toh $ \sin \theta \approx \theta $, $ \cos \theta \approx 1 $, $ \tan \theta \approx \theta $). Isliye $ BC = \frac{R}{2} $ aur $ BP = \frac{R}{2} $. Toh, point B, PC ka midpoint hai (yani radius of curvature). Aur ise FOCUS define kiya jata hai, isliye $ BP = f = \frac{R}{2} $. (Definition: Paraxial rays jo principal axis ke parallel hote hain, reflection ke baad mirror ke focus par milte hain.) BP focal length (f) hai: $ f = \frac{R}{2} $. M P F C A $\theta$ Paraxial rays (jo principal axis ke parallel nahi hain) ke liye: Aisi rays reflection ke baad focal plane (F') mein ek point par milti hain, jahan $ \frac{FF'}{FP} = \tan \theta \approx \theta \Rightarrow \frac{FF'}{f} = \theta \Rightarrow FF' = f\theta $. M P F C A $\theta$ Curved Mirror ke liye u,v aur f ka Relation Ek object mirror ke pole se $u$ distance par rakha hai aur uski image pole se $v$ distance par banti hai. Agar angle bohot chhota hai: $ \alpha = \frac{MP}{u} $, $ \beta = \frac{MP}{R} $, $ \gamma = \frac{MP}{V} $ $ \Delta CMO $ se, $ \beta = \alpha + \theta \Rightarrow \theta = \beta - \alpha $ $ \Delta CMI $ se, $ \gamma = \beta + \theta \Rightarrow \theta = \gamma - \beta $ Toh hum likh sakte hain: $ \beta - \alpha = \gamma - \beta $ $ \Rightarrow 2\beta = \gamma + \alpha $ $ \Rightarrow \frac{2}{R} = \frac{1}{V} + \frac{1}{u} $ $ \Rightarrow \frac{1}{f} = \frac{1}{u} + \frac{1}{V} $ Spherical mirrors ke liye object/image ke sign convention: u v Real object -ve Virtual object +ve Real image -ve Virtual image +ve Magnification Transverse ya Lateral Magnification Linear magnification $ m = \frac{\text{height of image}}{\text{height of object}} = \frac{h_i}{h_o} $ $ \Delta ABP $ aur $ \Delta A'B'P $ similar hain, isliye: $ \frac{h_i}{h_o} = \frac{-v}{u} $ Magnification $ m = \frac{h_i}{h_o} = -\frac{v}{u} = \frac{f}{f-u} = \frac{f-v}{f} $ M P F C A B A' B' object $h_o$ image $h_i$ $u$ $v$ Agar ek one-dimensional object principal axis ke perpendicular rakha hai, toh linear magnification ko transverse ya lateral magnification kehte hain. $ m = \frac{h_i}{h_o} = -\frac{v}{u} $ Magnification Image Magnification Image $ |m| > 1 $ enlarged $ |m| diminished $ m inverted $ m > 0 $ erect Longitudinal Magnification Agar ek one-dimensional object principal axis ke along rakha hai, toh linear magnification ko longitudinal magnification kehte hain. $ m_L = \frac{\text{length of image}}{\text{length of object}} = \frac{v_2 - v_1}{u_2 - u_1} $ Chhote objects ke liye: $ m_L = \frac{dv}{du} $ $ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $ ka differentiation karne par $ -\frac{1}{v^2} dv - \frac{1}{u^2} du = 0 $ milta hai. $ \Rightarrow \frac{dv}{du} = -\frac{v^2}{u^2} = -m^2 $ Toh $ m_L = -m^2 $. object image $u_1$ $u_2$ $v_1$ $v_2$ Superficial Magnification Agar ek two-dimensional object apni plane principal axis ke perpendicular rakha hai, toh uske magnification ko superficial magnification kehte hain. Linear magnification $ m = \frac{h_i}{h_o} = \frac{w_i}{w_o} $ $ h_i = m h_o $, $ w_i = m w_o $ Object ka area: $ A_{obj} = h_o \times w_o $ Image ka area: $ A_{image} = h_i \times w_i = (m h_o) \times (m w_o) = m^2 A_{obj} $ Superficial magnification $ m_S = \frac{\text{area of image}}{\text{area of object}} = \frac{(ma) \times (mb)}{(a \times b)} = m^2 $ $h_o$ $w_o$ $h_i = mh_o$ $w_i = mw_o$ Image Formation by Spherical Mirrors Concave mirror Object: Infinity par rakha hai. Image: Real, inverted, diminished, F par banti hai. Object: Infinity aur C ke beech rakha hai. Image: Real, inverted, diminished, C aur F ke beech banti hai ($ |m| u=$ \infty $ P F C M $h_o$ M P F C $h_i$ Object: C par rakha hai. Image: Real, inverted, equal, C par banti hai ($ m = -1 $). Object: F aur C ke beech rakha hai. Image: Real, inverted, enlarged, C ke bahar banti hai ($ |m| > 1 $ aur $ m $h_o$ M P F C $h_i$ $h_o$ M P F C $h_i$ Object: F par rakha hai. Image: Real, inverted, bohot badi (assumed), infinity par banti hai ($ m \ll -1 $). Object: F aur P ke beech rakha hai. Image: Virtual, erect, enlarged, mirror ke peeche banti hai ($ m > +1 $). $h_o$ M P F C $h_o$ M P F C $h_i$ Concave mirror ke liye summary: Object Image Magnification $ -\infty $ F $ m \ll 1 $ & $ m $ -\infty - C $ C - F $ |m| C C $ m = -1 $ C - F $ -\infty - C $ $ |m| > 1 $ & $ m Just before F towards C $ -\infty $ $ m \ll -1 $ Just after F towards P $ +\infty $ $ m \gg 1 $ Convex mirror Image hamesha virtual aur erect banti hai, object ki position kuch bhi ho, aur $ m $ hamesha positive hota hai. Object placed at infinity P F C M at F virtual, erect and very small ($ m \ll +1 $) Object placed in front of mirror P F C M Virtual, erect, diminished ($ m Golden Key Points Real & Virtual Image mein Differences spherical mirror ke liye: Real Image Virtual Image (i) Object ke respect mein inverted hoti hai. (i) Object ke respect mein erect hoti hai. (ii) Screen par obtain ki ja sakti hai. (ii) Screen par obtain nahi ki ja sakti. (iii) Iska magnification negative hota hai. (iii) Iska magnification positive hota hai. (iv) Mirror ke saamne banti hai. (iv) Mirror ke peeche banti hai. Real extended object ke liye, agar single mirror se bani image erect hai toh woh hamesha virtual hogi (yani $m$ positive hoga) aur is situation mein agar image ka size: Object se chhota Object ke barabar Object se bada mirror convex hoga ($m mirror plane hoga ($m = +1$) mirror concave hoga ($m > +1$) O M P I O M P I O M P I Example: Ek concave mirror ki focal length 30cm hai. Mirror ke saamne object ki position batao, taki image ka size teen guna ho. Solution: Object mirror ke saamne hai toh woh real hai. Real object ke liye concave mirror se enlarged image do tarah se ban sakti hai: inverted (real) ya erect (virtual). (a) Agar image inverted ho (yani real): $ m = \frac{f}{f-u} \Rightarrow -3 = \frac{-30}{-30-u} \Rightarrow u = -40 $ cm Object mirror ke saamne 40 cm par hona chahiye (C aur F ke beech). O M P F C I 40cm 120cm O M P F C I 20cm 60cm (b) Agar image erect ho (yani virtual): $ m = \frac{f}{f-u} \Rightarrow 3 = \frac{-30}{-30-u} \Rightarrow u = -20 $ cm Object mirror ke saamne 20 cm par hona chahiye (F aur P ke beech). Example: Ek concave mirror ki focal length $f$ hai, uski principal axis ke along $ \frac{f}{3} $ length ki ek patli rod rakhi hai, is tarah se ki uski real aur elongated image rod ko just touch karti hai. Magnification batao. Solution: Image real aur enlarged hai, toh object C aur F ke beech hona chahiye. Image ka ek end A' rod ke A end se coincide karta hai. $ v_A = u_A = -2f $ (yani A C par hai) Rod ki length $ \frac{f}{3} $ hai. $ u_B = 2f - \frac{f}{3} = \frac{5f}{3} $ Agar end B ki image P se $ v_B $ distance par hai toh: $ \frac{1}{v_B} + \frac{1}{u_B} = \frac{1}{f} \Rightarrow \frac{1}{v_B} + \frac{1}{5f/3} = \frac{1}{f} $ $ \Rightarrow v_B = \frac{5f}{2} $ Toh image ki length $ |v_B| - |v_A| = \frac{5f}{2} - 2f = \frac{f}{2} $ Magnification $ m = \frac{v_B - v_A}{u_B - u_A} = \frac{f/2}{f/3} = \frac{3}{2} $ Negative sign ka matlab hai ki image object ke respect mein inverted hai, aur isliye real hai. A B A' B' P F C rod Example: Ek concave mirror aur ek convex mirror, jinki focal length 10 cm aur 15 cm hai respectively, ek doosre se 40 cm door rakhe hain. Ek point object concave mirror se 15 cm door rakha hai. Pehle concave mirror se aur phir convex mirror se reflection ke baad image ki position batao. Solution: M1 mirror (concave) ke liye: $ u = -15 $ cm, $ f = -10 $ cm $ \frac{1}{v} + \frac{1}{-15} = \frac{1}{-10} \Rightarrow v = -30 $ cm Image I1 mirror M1 se 30 cm door banti hai. Image I1 mirror M2 (convex) ke liye object ki tarah act karegi. M1 aur M2 ke beech distance 40 cm hai, toh M2 se I1 ki distance: $ u_2 = -(40-30) = -10 $ cm. M2 mirror (convex) ke liye: $ f = +15 $ cm $ \frac{1}{v_2} + \frac{1}{-10} = \frac{1}{+15} \Rightarrow v_2 = +6 $ cm Final image I2 convex mirror ke peeche 6 cm par banti hai, aur woh virtual hai. M1 O M2 15cm 40cm I1 6cm I2 Example: Surya ek concave mirror ki focal length $f$ ke pole par $ \theta $ radians ka angle subtend karta hai. Surya ki image ka diameter batao. Solution: Kyuki surya bohot door hai, $u$ bohot bada hoga, toh $ \frac{1}{u} \approx 0 $. $ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} \Rightarrow v = f $. Surya ki image focus par banegi aur woh real, inverted aur diminished hogi. $ A'B' $ = image ki height aur $ \theta = \frac{\text{Arc}}{\text{Radius}} = \frac{d}{FP} $ $ \Rightarrow \theta = \frac{d}{f} \Rightarrow d = f\theta $. P F A' B' $\theta$ Velocity of Image of Moving Object (Spherical Mirror) (a) Velocity component along axis (Longitudinal velocity) Jab ek object concave mirror ke focus ki taraf infinity se aa raha hota hai: $ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $ Iska time ke respect mein differentiation karne par: $ -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0 $ $ \Rightarrow V_{ix} = -\frac{v^2}{u^2} V_{ox} = -m^2 V_{ox} $ Jahan $ V_{ix} $ = principal axis ke along image ki velocity, aur $ V_{ox} $ = principal axis ke along object ki velocity. Object $V_{ox}$ Image $V_{ix}$ (b) Velocity component perpendicular to axis (Transverse velocity) $ m = \frac{h_i}{h_o} = -\frac{v}{u} = \frac{f}{f-u} $ $ \Rightarrow h_i = h_o \frac{f}{f-u} $ Time ke respect mein differentiation karne par: $ \frac{dh_i}{dt} = \frac{f}{(f-u)} \frac{dh_o}{dt} + \frac{f h_o}{(f-u)^2} \frac{du}{dt} $ $ V_{iy} = \frac{f}{f-u} V_{oy} + \frac{f h_o}{(f-u)^2} V_{ox} $ Note: Yahan principal axis ko x-axis ke along maana gaya hai. Power of a Mirror Mirror ki power ko $ P = -\frac{1}{f(m)} = -\frac{100}{f(cm)} $ define kiya jata hai. Newton's Formula Agar object distance ($x_1$) aur image distance ($x_2$) pole ki bajaye focus se measure ki jaati hain, toh $u = -(f+x_1)$ aur $v = -(f+x_2)$. $ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $ mein value rakhne par: $ \frac{1}{-(f+x_2)} + \frac{1}{-(f+x_1)} = \frac{1}{f} $ Solve karne par $ x_1 x_2 = f^2 $ milta hai. Ise Newton's formula kehte hain. O $x_1$ I $x_2$ P F C Golden Key Points Convex mirrors: Erect, virtual aur diminished image banate hain. Convex mirror mein field of view plane mirror ke comparison mein zyada hota hai. Isliye, yeh vehicles mein rear-view mirror ke taur par use hote hain. Concave mirrors: Enlarged, erect aur virtual image banate hain, isliye yeh dentists dwara daant check karne ke liye use hote hain. Apni converging property ki wajah se concave mirrors automobile head lights aur search lights mein reflectors ke taur par bhi use hote hain. Spherical mirror ki focal length $f=R/2$ sirf mirror ki radius par depend karti hai, aur light ki wavelength ya medium ke refractive index par depend nahi karti. Isliye, air ya water mein, ya red ya blue light ke liye focal length same rehti hai. P F C field of view Do yourself - 2: Figure mein ek spherical concave mirror dikhaya gaya hai jiska pole (0, 0) par hai aur principal axis x-axis ke along hai. Ek point object (-40 cm, 1cm) par hai, image ki position batao. Converging rays ek convex spherical mirror par incident hain, jisse unki extensions mirror ke peeche optical axis par 30 cm par intersect karti hain. Reflected rays ek diverging beam banati hain jinki extensions mirror se 1.2 m door optical axis par intersect karti hain. Mirror ki focal length batao. Ek extended object principal axis ke perpendicular 20 cm radius ke concave mirror se 15 cm door rakha hai. Lateral magnification batao. Ek concave mirror se real object ki image object se do guna badi banti hai. Mirror ki focal length 20 cm hai. Mirror se object ki distance kya hai: (A) 10 cm (B) 30 cm (C) 25 cm (D) 15 cm (0, 0) R.O.C. = 10 cm object (-40, 1) (v) Ek square ABCD, jiski side 1 mm hai, concave mirror se 15 cm door rakha hai jaisa ki figure mein dikhaya gaya hai. Mirror ki focal length 10 cm hai. Uski image ke perimeter ki length (lagbhag) kitni hogi: (A) 8 mm (B) 2 mm (C) 12 mm (D) 6 mm B C A D 15cm 12mm (vi) Figure mein dikhayi gayi situation mein do successive reflections ke baad total magnification batao, pehle M1 par aur phir M2 par. (A) +1 (B) -2 (C) +2 (D) -1 M1 f = 10cm M2 f = -20cm 10cm 30cm Refraction Refraction woh phenomenon hai jismein light ki propagation ki direction change ho jaati hai jab woh ek medium se doosre medium mein jaati hai. Refraction mein light ki frequency change nahi hoti. Laws of Refraction Incident ray, refracted ray, aur normal sab ek hi plane mein hote hain. Vector form mein $ (\vec{e} \times \vec{n}) \cdot \vec{f} = 0 $. Refractive index aur angle of incidence ke sine ka product kisi medium mein constant hota hai. $ \mu_1 \sin i = \mu_2 \sin r $ (Snell's law). Vector form mein $ \mu_1 \vec{e} \times \vec{n} = \mu_2 \vec{r} \times \vec{n} $. $\mu_1$ $\mu_2$ normal $i$ $r$ Absolute refractive index Isko free space mein light ki speed 'c' aur given medium mein light ki speed 'v' ke ratio se define kiya jata hai. $ \mu $ ya $ n = \frac{c}{v} $. Medium jitna denser hoga, light ki speed utni kam hogi, aur isliye refractive index utna zyada hoga. Jaise $ V_{glass} \mu_{water} $. Relative refractive index Jab light ek medium se doosre medium mein jaati hai, toh medium 2 ka relative refractive index medium 1 ke respect mein $ {}_1 \mu_2 $ se likha jata hai, aur ise aise define kiya jata hai: $ {}_1 \mu_2 = \frac{\mu_2}{\mu_1} = \frac{c/V_2}{c/V_1} = \frac{V_1}{V_2} $. $\mu_1$ $\mu_2$ $V_1$ $V_2$ Bending of Light Ray Snell's law ke according, $ \mu_1 \sin i = \mu_2 \sin r $. (i) Agar light rarer se denser medium mein jaati hai ($ \mu_1 = \mu_r $ aur $ \mu_2 = \mu_d $): $ \frac{\sin i}{\sin r} = \frac{\mu_d}{\mu_r} > 1 \Rightarrow \sin i > \sin r $ Rarer se denser medium mein jaane par, ray normal ki taraf bend hoti hai. air water rarer medium denser medium normal incident ray refracted ray (ii) Agar light denser se rarer medium mein jaati hai ($ \mu_1 = \mu_d $ aur $ \mu_2 = \mu_r $): $ \frac{\sin i}{\sin r} = \frac{\mu_r}{\mu_d} Denser se rarer medium mein jaane par, ray normal se door bend hoti hai. water air denser medium rarer medium normal incident ray refracted ray Apparent Depth and Normal Shift Agar ek point object denser medium mein hai aur use rarer medium se dekha ja raha hai, aur boundary plane hai, toh Snell's law se $ \mu_d \sin i = \mu_r \sin r $. Agar rays OA aur OB aankh tak pahunchne ke liye kaafi paas hain, toh: $ \sin i = \tan i = \frac{P}{d_{ac}} $ aur $ \sin r = \tan r = \frac{P}{d_{ap}} $ Jahan $ d_{ac} $ = actual depth, $ d_{ap} $ = apparent depth. Toh equation ban jaati hai: $ \mu_d \frac{P}{d_{ac}} = \mu_r \frac{P}{d_{ap}} \Rightarrow \frac{d_{ap}}{d_{ac}} = \frac{\mu_r}{\mu_d} $ Agar $ \mu_r = 1 $ aur $ \mu_d = \mu $, toh $ d_{ap} = \frac{d_{ac}}{\mu} $. Isliye $ d_{ap} Object aur image ke beech ki distance, jise normal shift (x) kehte hain: $ x = d_{ac} - d_{ap} = d_{ac} - \frac{d_{ac}}{\mu} = d_{ac} (1 - \frac{1}{\mu}) $ Agar $ d_{ac} = d $, toh $ x = d(1 - \frac{1}{\mu}) $. N object P $i$ $r$ $d_{ac}$ $d_{ap}$ Agar object rarer medium mein hai aur use denser medium se dekha ja raha hai: $ \frac{d_{ap}}{d_{ac}} = \frac{\mu_d}{\mu_r} $ Agar $ \mu_r = 1 $ aur $ \mu_d = \mu $, toh $ d_{ap} = \mu d_{ac} $. Isliye $ d_{ap} > d_{ac} $. Ek high-flying object reality se zyada upar dikhayi deta hai. $ x = d_{ap} - d_{ac} = (\mu - 1) d_{ac} $. N object P $i$ $r$ $d_{ac}$ $d_{ap}$ Lateral Shift Incident aur emergent ray ke beech ki perpendicular distance ko lateral shift kehte hain. Lateral shift $d = BC$ aur slab ki thickness $t$ hai. $ \Delta BOC $ mein: $ \sin(i-r) = \frac{BC}{OB} = \frac{d}{OB} \dots(i) $ $ \Delta OBD $ mein: $ \cos r = \frac{OD}{OB} = \frac{t}{OB} \dots(ii) $ (i) aur (ii) se $ d = \frac{t \sin(i-r)}{\cos r} $. $i$ $r$ $i$ $t$ $d$ Transparent Glass Slab (Normal Shift) Jab ek object ko glass slab ke saamne rakha jaata hai, toh woh incident light ki direction mein shift ho jaata hai aur $x$ distance par image banti hai. $ x = t(1 - \frac{1}{\mu}) $ incident ray $i$ emergent ray $t$ $x$ Some Illustrations of Refraction Bending of an object: Jab ek object denser medium mein hota hai aur use rarer medium se dekha jaata hai, toh woh bend hua dikhayi deta hai. Twinkling of stars: Atmosphere mein refractive index ke fluctuations ki wajah se refraction irregular ho jaata hai, aur light kabhi aankh tak pahunchti hai aur kabhi nahi. Isliye taare timtimate hue dikhayi dete hain. object image star eye Golden Key Points $ \mu $ ek scalar quantity hai aur iski koi unit ya dimension nahi hoti. Agar $ \epsilon_0 $ aur $ \mu_0 $ free space ki electric permittivity aur magnetic permeability hain, aur $ \epsilon $ aur $ \mu $ kisi medium ki, toh electromagnetic theory ke according: $ c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} $ aur $ v = \frac{1}{\sqrt{\epsilon \mu}} $ $ \mu $ ya $ n = \frac{c}{v} = \frac{\sqrt{\epsilon \mu}}{\sqrt{\epsilon_0 \mu_0}} = \sqrt{\epsilon_r \mu_r} $ Vacuum ya free space mein, light ki speed sabhi wavelengths ke liye maximum hoti hai aur $c$ ke barabar hoti hai. Isliye free space ka refractive index minimum hota hai aur $ \mu = \frac{c}{c} = 1 $ hota hai. Example: Ek light ray refractive index 1.62 ke transparent glass slab par incident hai. Agar reflected aur refracted rays mutually perpendicular hain, toh angle of incidence kya hai? ($ \tan^{-1} (1.62) = 58.3^\circ $) Solution: Given problem ke according: $ r + 90^\circ + r' = 180^\circ \Rightarrow r' = 90^\circ - r $. Kyuki $ \angle i = \angle r $, toh $ r' = 90^\circ - i $. Snell's law ke according: $ 1 \sin i = \mu \sin r' $ $ \sin i = \mu \sin(90^\circ - i) = \mu \cos i $ $ \tan i = \mu \Rightarrow i = \tan^{-1} \mu = \tan^{-1} (1.62) = 58.3^\circ $. $\mu=1$ $\mu$ $i$ $r$ $r'$ Example: Ek 20 cm thick glass slab, jiska refractive index 1.5 hai, ek plane mirror ke saamne rakha hai. Ek object air mein mirror se 40 cm door rakha hai. Object ke paas khade observer ko image ki position batao. Glass slab aur mirror ke beech ke separation ka image par kya asar hoga? Solution: Glass slab ki wajah se object mein shift $ x = d(1 - \frac{1}{\mu}) = 20(1 - \frac{1}{1.5}) = 20(1 - \frac{2}{3}) = \frac{20}{3} $ cm. Mirror se object ki distance (mirror se dekhi gayi) = $ 40 - \frac{20}{3} = \frac{100}{3} $ cm. Image mirror M se $ \frac{100}{3} $ cm door banegi. Glass slab ki wajah se image mein shift = $ \frac{20}{3} $ cm. Toh image ki distance mirror se $ \frac{100}{3} - \frac{20}{3} = \frac{80}{3} $ cm hogi. Actual plane mirror se image ki distance separation 'b' par depend nahi karti. Agar distance zyada hogi toh image ki brightness kam hogi. Slab $\mu = \frac{3}{2}$ M O 40cm 20cm b x M' I Example: Ek prism ka angle 30° hai aur $ \mu=\sqrt{2} $ hai. Agar ek face silvered hai, toh incident ray apna initial path retraces karti hai. Angle of incidence kya hai? Solution: Kyuki incident ray apna path retraces karti hai, iska matlab hai ki ray prism ke silvered face par normally incident hui hai jaisa ki figure mein dikhaya gaya hai. $ \Delta AED $ mein: $ 30^\circ + 90^\circ + \angle D = 180^\circ \Rightarrow \angle D = 60^\circ $. Construction se: $ \angle D + \angle r = 90^\circ \Rightarrow \angle r = 90^\circ - 60^\circ = 30^\circ $. Surface AC par Snell's law lagane par: $ 1 \sin i = \sqrt{2} \sin 30^\circ = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}} $. $ \Rightarrow \sin i = \frac{1}{\sqrt{2}} \Rightarrow i = 45^\circ $. A B C $30^\circ$ $90^\circ$ D E $i$ $r$ $\mu=\sqrt{2}$ Example: Ek object 21 cm door rakha hai ek concave mirror se jiska radius of curvature 20 cm hai. Ek glass slab jiski thickness 3 cm aur refractive index 1.5 hai, object aur mirror ke beech rakhi hai, slab ki nearer surface mirror se 10 cm door hai. Final image ki position batao. Solution: Slab ki wajah se shift $ x = d(1 - \frac{1}{\mu}) = 3(1 - \frac{1}{1.5}) = 3(1 - \frac{2}{3}) = 1 $ cm. Mirror se image banne ke liye object ki distance $ u = -(21-1) = -20 $ cm. Mirror ke liye: $ f = -10 $ cm ($ R = -20 $ cm hai). $ \frac{1}{v} + \frac{1}{-20} = \frac{1}{-10} \Rightarrow v = -20 $ cm. Shift light ki direction mein hoga, toh $ v = -(20+1) = -21 $ cm. 3cm 10cm M O 21cm Example: Ek particle axis ke along height $ \frac{f}{2} $ se drop kiya jata hai ek concave mirror par jiski focal length $f$ hai jaisa ki figure mein dikhaya gaya hai. Image ki maximum speed batao. Solution: $ V_{IM} = -m^2 V_{OM} = -m^2 (gt) $ jahan $ m = \frac{f}{f-u} $. $ u = -gt^2/2 - f $ (initial position) $ m = \frac{-f}{-f - (-gt^2/2)} = \frac{-f}{-f+gt^2/2} = \frac{2f}{2f-gt^2} $. $ V_{IM} = -\left(\frac{2f}{2f-gt^2}\right)^2 (gt) $. Maximum speed ke liye $ \frac{dV_{IM}}{dt} = 0 $. Solve karne par $ V_{IM_{max}} = \sqrt{\frac{3}{4} fg} $. h=$f/2$ Example: Ek concave mirror jiski focal length 10 cm hai aur ek convex mirror jiski focal length 15 cm hai, ek doosre se 40 cm door rakhe hain. Ek point object concave mirror se 15 cm door rakha hai. Pehle concave mirror se aur phir convex mirror se reflection ke baad image ki position batao. Solution: M1 (concave mirror) ke liye: $u = -15$ cm, $f = -10$ cm. $ \frac{1}{v_1} + \frac{1}{-15} = \frac{1}{-10} \Rightarrow v_1 = -30 $ cm. M2 (convex mirror) ke liye: $f = +15$ cm. Image I1, M2 ke liye object ka kaam karegi. $ u_2 = -(40-30) = -10 $ cm. $ \frac{1}{v_2} + \frac{1}{-10} = \frac{1}{+15} \Rightarrow v_2 = +6 $ cm. Final image I2 convex mirror ke peeche 6 cm par banti hai aur virtual hai. M1 f = -10 M2 f = +15 O 15cm 40cm Example: Surya ek concave mirror ki focal length $f$ ke pole par $ \theta $ radians ka angle subtend karta hai. Surya ki image ka diameter batao. Solution: Kyuki surya bohot door hai, $u$ bohot bada hoga, toh $ \frac{1}{u} \approx 0 $. $ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} \Rightarrow v = f $. Surya ki image focus par banegi aur woh real, inverted aur diminished hogi. $ A'B' $ = image ki height aur $ \theta = \frac{\text{Arc}}{\text{Radius}} = \frac{d}{FP} $ $ \Rightarrow \theta = \frac{d}{f} \Rightarrow d = f\theta $. P F A' B' $\theta$ Velocity of Image of Moving Object (Spherical Mirror) (a) Velocity component along axis (Longitudinal velocity) Jab ek object concave mirror ke focus ki taraf infinity se aa raha hota hai: $ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $ Iska time ke respect mein differentiation karne par: $ -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0 $ $ \Rightarrow V_{ix} = -\frac{v^2}{u^2} V_{ox} = -m^2 V_{ox} $ Jahan $ V_{ix} $ = principal axis ke along image ki velocity, aur $ V_{ox} $ = principal axis ke along object ki velocity. Object $V_{ox}$ Image $V_{ix}$ (b) Velocity component perpendicular to axis (Transverse velocity) $ m = \frac{h_i}{h_o} = -\frac{v}{u} = \frac{f}{f-u} $ $ \Rightarrow h_i = h_o \frac{f}{f-u} $ Time ke respect mein differentiation karne par: $ \frac{dh_i}{dt} = \frac{f}{(f-u)} \frac{dh_o}{dt} + \frac{f h_o}{(f-u)^2} \frac{du}{dt} $ $ V_{iy} = \frac{f}{f-u} V_{oy} + \frac{f h_o}{(f-u)^2} V_{ox} $ Note: Yahan principal axis ko x-axis ke along maana gaya hai. Power of a Mirror Mirror ki power ko $ P = -\frac{1}{f(m)} = -\frac{100}{f(cm)} $ define kiya jata hai. Newton's Formula Agar object distance ($x_1$) aur image distance ($x_2$) pole ki bajaye focus se measure ki jaati hain, toh $u = -(f+x_1)$ aur $v = -(f+x_2)$. $ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $ mein value rakhne par: $ \frac{1}{-(f+x_2)} + \frac{1}{-(f+x_1)} = \frac{1}{f} $ Solve karne par $ x_1 x_2 = f^2 $ milta hai. Ise Newton's formula kehte hain. O $x_1$ I $x_2$ P F C Golden Key Points Convex mirrors: Erect, virtual aur diminished image banate hain. Convex mirror mein field of view plane mirror ke comparison mein zyada hota hai. Isliye, yeh vehicles mein rear-view mirror ke taur par use hote hain. Concave mirrors: Enlarged, erect aur virtual image banate hain, isliye yeh dentists dwara daant check karne ke liye use hote hain. Apni converging property ki wajah se concave mirrors automobile head lights aur search lights mein reflectors ke taur par bhi use hote hain. Spherical mirror ki focal length $f=R/2$ sirf mirror ki radius par depend karti hai, aur light ki wavelength ya medium ke refractive index par depend nahi karti. Isliye, air ya water mein, ya red ya blue light ke liye focal length same rehti hai. P F C field of view Do yourself - 2: Figure mein ek spherical concave mirror dikhaya gaya hai jiska pole (0, 0) par hai aur principal axis x-axis ke along hai. Ek point object (-40 cm, 1cm) par hai, image ki position batao. Converging rays ek convex spherical mirror par incident hain, jisse unki extensions mirror ke peeche optical axis par 30 cm par intersect karti hain. Reflected rays ek diverging beam banati hain jinki extensions mirror se 1.2 m door optical axis par intersect karti hain. Mirror ki focal length batao. Ek extended object principal axis ke perpendicular 20 cm radius ke concave mirror se 15 cm door rakha hai. Lateral magnification batao. Ek concave mirror se real object ki image object se do guna badi banti hai. Mirror ki focal length 20 cm hai. Mirror se object ki distance kya hai: (A) 10 cm (B) 30 cm (C) 25 cm (D) 15 cm (0, 0) R.O.C. = 10 cm object (-40, 1) (v) Ek square ABCD, jiski side 1 mm hai, concave mirror se 15 cm door rakha hai jaisa ki figure mein dikhaya gaya hai. Mirror ki focal length 10 cm hai. Uski image ke perimeter ki length (lagbhag) kitni hogi: (A) 8 mm (B) 2 mm (C) 12 mm (D) 6 mm B C A D 15cm 12mm (vi) Figure mein dikhayi gayi situation mein do successive reflections ke baad total magnification batao, pehle M1 par aur phir M2 par. (A) +1 (B) -2 (C) +2 (D) -1 M1 f = 10cm M2 f = -20cm 10cm 30cm Refraction Refraction woh phenomenon hai jismein light ki propagation ki direction change ho jaati hai jab woh ek medium se doosre medium mein jaati hai. Refraction mein light ki frequency change nahi hoti. Laws of Refraction Incident ray, refracted ray, aur normal sab ek hi plane mein hote hain. Vector form mein $ (\vec{e} \times \vec{n}) \cdot \vec{f} = 0 $. Refractive index aur angle of incidence ke sine ka product kisi medium mein constant hota hai. $ \mu_1 \sin i = \mu_2 \sin r $ (Snell's law). Vector form mein $ \mu_1 \vec{e} \times \vec{n} = \mu_2 \vec{r} \times \vec{n} $. $\mu_1$ $\mu_2$ normal $i$ $r$ Absolute refractive index Isko free space mein light ki speed 'c' aur given medium mein light ki speed 'v' ke ratio se define kiya jata hai. $ \mu $ ya $ n = \frac{c}{v} $. Medium jitna denser hoga, light ki speed utni kam hogi, aur isliye refractive index utna zyada hoga. Jaise $ V_{glass} \mu_{water} $. Relative refractive index Jab light ek medium se doosre medium mein jaati hai, toh medium 2 ka relative refractive index medium 1 ke respect mein $ {}_1 \mu_2 $ se likha jata hai, aur ise aise define kiya jata hai: $ {}_1 \mu_2 = \frac{\mu_2}{\mu_1} = \frac{c/V_2}{c/V_1} = \frac{V_1}{V_2} $. $\mu_1$ $\mu_2$ $V_1$ $V_2$ Bending of Light Ray Snell's law ke according, $ \mu_1 \sin i = \mu_2 \sin r $. (i) Agar light rarer se denser medium mein jaati hai ($ \mu_1 = \mu_r $ aur $ \mu_2 = \mu_d $): $ \frac{\sin i}{\sin r} = \frac{\mu_d}{\mu_r} > 1 \Rightarrow \sin i > \sin r $ Rarer se denser medium mein jaane par, ray normal ki taraf bend hoti hai. air water rarer medium denser medium normal incident ray refracted ray (ii) Agar light denser se rarer medium mein jaati hai ($ \mu_1 = \mu_d $ aur $ \mu_2 = \mu_r $): $ \frac{\sin i}{\sin r} = \frac{\mu_r}{\mu_d} Denser se rarer medium mein jaane par, ray normal se door bend hoti hai. water air denser medium rarer medium normal incident ray refracted ray Apparent Depth and Normal Shift Agar ek point object denser medium mein hai aur use rarer medium se dekha ja raha hai, aur boundary plane hai, toh Snell's law se $ \mu_d \sin i = \mu_r \sin r $. Agar rays OA aur OB aankh tak pahunchne ke liye kaafi paas hain, toh: $ \sin i = \tan i = \frac{P}{d_{ac}} $ aur $ \sin r = \tan r = \frac{P}{d_{ap}} $ Jahan $ d_{ac} $ = actual depth, $ d_{ap} $ = apparent depth. Toh equation ban jaati hai: $ \mu_d \frac{P}{d_{ac}} = \mu_r \frac{P}{d_{ap}} \Rightarrow \frac{d_{ap}}{d_{ac}} = \frac{\mu_r}{\mu_d} $ Agar $ \mu_r = 1 $ aur $ \mu_d = \mu $, toh $ d_{ap} = \frac{d_{ac}}{\mu} $. Isliye $ d_{ap} Object aur image ke beech ki distance, jise normal shift (x) kehte hain: $ x = d_{ac} - d_{ap} = d_{ac} - \frac{d_{ac}}{\mu} = d_{ac} (1 - \frac{1}{\mu}) $ Agar $ d_{ac} = d $, toh $ x = d(1 - \frac{1}{\mu}) $. N object P $i$ $r$ $d_{ac}$ $d_{ap}$ Agar object rarer medium mein hai aur use denser medium se dekha ja raha hai: $ \frac{d_{ap}}{d_{ac}} = \frac{\mu_d}{\mu_r} $ Agar $ \mu_r = 1 $ aur $ \mu_d = \mu $, toh $ d_{ap} = \mu d_{ac} $. Isliye $ d_{ap} > d_{ac} $. Ek high-flying object reality se zyada upar dikhayi deta hai. $ x = d_{ap} - d_{ac} = (\mu - 1) d_{ac} $. N object P $i$ $r$ $d_{ac}$ $d_{ap}$ Lateral Shift Incident aur emergent ray ke beech ki perpendicular distance ko lateral shift kehte hain. Lateral shift $d = BC$ aur slab ki thickness $t$ hai. $ \Delta BOC $ mein: $ \sin(i-r) = \frac{BC}{OB} = \frac{d}{OB} \dots(i) $ $ \Delta OBD $ mein: $ \cos r = \frac{OD}{OB} = \frac{t}{OB} \dots(ii) $ (i) aur (ii) se $ d = \frac{t \sin(i-r)}{\cos r} $. $i$ $r$ $i$ $t$ $d$ Transparent Glass Slab (Normal Shift) Jab ek object ko glass slab ke saamne rakha jaata hai, toh woh incident light ki direction mein shift ho jaata hai aur $x$ distance par image banti hai. $ x = t(1 - \frac{1}{\mu}) $ incident ray $i$ emergent ray $t$ $x$ Some Illustrations of Refraction Bending of an object: Jab ek object denser medium mein hota hai aur use rarer medium se dekha jaata hai, toh woh bend hua dikhayi deta hai. Twinkling of stars: Atmosphere mein refractive index ke fluctuations ki wajah se refraction irregular ho jaata hai, aur light kabhi aankh tak pahunchti hai aur kabhi nahi. Isliye taare timtimate hue dikhayi dete hain. object image star eye Golden Key Points $ \mu $ ek scalar quantity hai aur iski koi unit ya dimension nahi hoti. Agar $ \epsilon_0 $ aur $ \mu_0 $ free space ki electric permittivity aur magnetic permeability hain, aur $ \epsilon $ aur $ \mu $ kisi medium ki, toh electromagnetic theory ke according: $ c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} $ aur $ v = \frac{1}{\sqrt{\epsilon \mu}} $ $ \mu $ ya $ n = \frac{c}{v} = \frac{\sqrt{\epsilon \mu}}{\sqrt{\epsilon_0 \mu_0}} = \sqrt{\epsilon_r \mu_r} $ Vacuum ya free space mein, light ki speed sabhi wavelengths ke liye maximum hoti hai aur $c$ ke barabar hoti hai. Isliye free space ka refractive index minimum hota hai aur $ \mu = \frac{c}{c} = 1 $ hota hai. Example: Ek light ray refractive index 1.62 ke transparent glass slab par incident hai. Agar reflected aur refracted rays mutually perpendicular hain, toh angle of incidence kya hai? ($ \tan^{-1} (1.62) = 58.3^\circ $) Solution: Given problem ke according: $ r + 90^\circ + r' = 180^\circ \Rightarrow r' = 90^\circ - r $. Kyuki $ \angle i = \angle r $, toh $ r' = 90^\circ - i $. Snell's law ke according: $ 1 \sin i = \mu \sin r' $ $ \sin i = \mu \sin(90^\circ - i) = \mu \cos i $ $ \tan i = \mu \Rightarrow i = \tan^{-1} \mu = \tan^{-1} (1.62) = 58.3^\circ $. $\mu=1$ $\mu$ $i$ $r$ $r'$ Example: Ek 20 cm thick glass slab, jiska refractive index 1.5 hai, ek plane mirror ke saamne rakha hai. Ek object air mein mirror se 40 cm door rakha hai. Object ke paas khade observer ko image ki position batao. Glass slab aur mirror ke beech ke separation ka image par kya asar hoga? Solution: Glass slab ki wajah se object mein shift $ x = d(1 - \frac{1}{\mu}) = 20(1 - \frac{1}{1.5}) = 20(1 - \frac{2}{3}) = \frac{20}{3} $ cm. Mirror se object ki distance (mirror se dekhi gayi) = $ 40 - \frac{20}{3} = \frac{100}{3} $ cm. Image mirror M se $ \frac{100}{3} $ cm door banegi. Glass slab ki wajah se image mein shift = $ \frac{20}{3} $ cm. Toh image ki distance mirror se $ \frac{100}{3} - \frac{20}{3} = \frac{80}{3} $ cm hogi. Actual plane mirror se image ki distance separation 'b' par depend nahi karti. Agar distance zyada hogi toh image ki brightness kam hogi. Slab $\mu = \frac{3}{2}$ M O 40cm 20cm b x M' I Example: Ek prism ka angle 30° hai aur $ \mu=\sqrt{2} $ hai. Agar ek face silvered hai, toh incident ray apna initial path retraces karti hai. Angle of incidence kya hai? Solution: Kyuki incident ray apna path retraces karti hai, iska matlab hai ki ray prism ke silvered face par normally incident hui hai jaisa ki figure mein dikhaya gaya hai. $ \Delta AED $ mein: $ 30^\circ + 90^\circ + \angle D = 180^\circ \Rightarrow \angle D = 60^\circ $. Construction se: $ \angle D + \angle r = 90^\circ \Rightarrow \angle r = 90^\circ - 60^\circ = 30^\circ $. Surface AC par Snell's law lagane par: $ 1 \sin i = \sqrt{2} \sin 30^\circ = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}} $. $ \Rightarrow \sin i = \frac{1}{\sqrt{2}} \Rightarrow i = 45^\circ $. A B C $30^\circ$ $90^\circ$ D E $i$ $r$ $\mu=\sqrt{2}$ Example: Ek object 21 cm door rakha hai ek concave mirror se jiska radius of curvature 20 cm hai. Ek glass slab jiski thickness 3 cm aur refractive index 1.5 hai, object aur mirror ke beech rakhi hai, slab ki nearer surface mirror se 10 cm door hai. Final image ki position batao. Solution: Slab ki wajah se shift $ x = d(1 - \frac{1}{\mu}) = 3(1 - \frac{1}{1.5}) = 3(1 - \frac{2}{3}) = 1 $ cm. Mirror se image banne ke liye object ki distance $ u = -(21-1) = -20 $ cm. Mirror ke liye: $ f = -10 $ cm ($ R = -20 $ cm hai). $ \frac{1}{v} + \frac{1}{-20} = \frac{1}{-10} \Rightarrow v = -20 $ cm. Shift light ki direction mein hoga, toh $ v = -(20+1) = -21 $ cm. 3cm 10cm M O 21cm Example: Ek particle axis ke along height $ \frac{f}{2} $ se drop kiya jata hai ek concave mirror par jiski focal length $f$ hai jaisa ki figure mein dikhaya gaya hai. Image ki maximum speed batao. Solution: $ V_{IM} = -m^2 V_{OM} = -m^2 (gt) $ jahan $ m = \frac{f}{f-u} $. $ u = -gt^2/2 - f $ (initial position) $ m = \frac{-f}{-f - (-gt^2/2)} = \frac{2f}{2f-gt^2} $. $ V_{IM} = -\left(\frac{2f}{2f-gt^2}\right)^2 (gt) $. Maximum speed ke liye $ \frac{dV_{IM}}{dt} = 0 $. Solve karne par $ V_{IM_{max}} = \sqrt{\frac{3}{4} fg} $. h=$f/2$ Total Internal Reflection (TIR) Jab light ray denser se rarer medium mein jaati hai, toh angle of incidence badhne par angle of refraction bhi badhta hai. Ek particular angle of incidence par, refracted ray normal ke saath 90° angle banati hai. Is angle ko critical angle ($ \theta_c $) kehte hain. Agar angle of incidence isse bhi badh jaye, toh ray usi medium mein wapas aa jaati hai. Is phenomenon ko total internal reflection kehte hain. denser medium rarer medium normal $i_1 $r_1$ $i_2 = \theta_c$ $90^\circ$ $i_3 > \theta_c$ total internal reflection Conditions for TIR Angle of incidence critical angle se zyada hona chahiye ($ i > \theta_c $). Light denser se rarer medium mein travel karni chahiye (jaise Glass se Air, Water se Air, Glass se Water). Boundary XX' par Snell's Law: $ \mu_d \sin \theta_c = \mu_r \sin 90^\circ $ $ \Rightarrow \sin \theta_c = \frac{\mu_r}{\mu_d} $. denser medium rarer medium $\theta_c$ XX' Angle of Deviation ($ \delta $) aur Angle of Incidence ($ i $) ke beech Graph (denser se rarer medium mein rays ke liye) Agar $ i denser medium rarer medium normal $i$ $r$ $\delta$ Agar $ i > \theta_c $, toh $ \delta = \pi - 2i $. denser medium rarer medium normal $i$ $\delta$ Some Illustrations of Total Internal Reflection Sparkling of diamond: Diamond ka chamakna total internal reflection ki wajah se hota hai. Diamond ka refractive index 2.5 hai, toh $ \theta_c = 24^\circ $. Diamond ki cutting aisi hoti hai ki $ i > \theta_c $ ho. Isliye TIR baar-baar hota hai. Optical Fibre: Ismein light multiple total internal reflections ke through propagate karti hai fibre ke axis ke along. Fibre ka core ka refractive index surroundings se zyada hota hai. light pipe $\mu_1 > \mu_2$ Mirage and looming: Mirage deserts mein total internal reflection ki wajah se hota hai, jahan garam zameen ke paas hawa ka refractive index upar ki hawa se kam hota hai. Door ke objects se light $ i > \theta_c $ angle par surface par pahunchti hai, jisse TIR hota hai aur hum object ki image dekhte hain. hot surface hot air cold air object image denser rare sky object image Deserts mein 'mirage' ki tarah, polar regions mein 'looming' bhi TIR ki wajah se hota hai. Yahan height ke saath $ \mu $ kam hota hai, aur isliye object ki image air mein banti hai ($ i > \theta_c $ toh). Golden Key Points Paani mein ek diver $d$ depth par, bahar ki duniya ko $ r = d \tan \theta_c $ radius ke horizontal circle se dekhta hai. Total internal reflection mein, incident light ka 100% hissa usi medium mein reflect ho jaata hai, isliye intensity ka koi loss nahi hota. Jabki mirror se reflection ya lenses se refraction mein kuch intensity loss hota hai. Isliye TIR se bani images mirrors ya lenses se bani images se zyada bright hoti hain. Do yourself - 3: Ek light ray 10 cm thickness aur refractive index $n$ ke parallel glass slab se guzarti hai. Angle of incidence air mein 60° aur glass mein 45° hai. Lateral shift batao. Ek concave mirror ko water mein rakha hai, uski shining surface upar ki taraf hai aur principal axis vertical hai. Rays principal axis ke parallel incident hain. Final image ki position batao. Light ray surface se 60° angle par incident hai ice slab par, thickness 1.00 m aur angle of refraction 15° hai. Ray ko slab cross karne mein kitna time lagega? Vacuum mein light ki speed $3 \times 10^8$ m/s hai. Vacuum mein light ki wavelength 6000 A° hai aur medium mein 4000 A° hai. Medium ka refractive index kya hai: (A) 2.4 (B) 1.5 (C) 1.2 (D) 0.67 Ek light ray vacuum se refractive index $n$ ke medium mein jaati hai. Agar angle of incidence angle of refraction ka do guna hai, toh angle of incidence kya hai: (A) $ \cos^{-1} (n/2) $ (B) $ \sin^{-1} (n/2) $ (C) $ 2 \cos^{-1} (n/2) $ (D) $ 2 \sin^{-1} (n/2) $ Ek light ray thickness $t$ aur refractive index $n$ ke parallel slab par incident hai. Agar angle of incidence $ \theta $ chhota hai, toh incident aur emergent ray mein displacement kya hoga: (A) $ \frac{t\theta(n-1)}{n} $ (B) $ \frac{t\theta}{n} $ (C) $ \frac{t\theta n}{n-1} $ (D) none Glass medium ($ \mu = 1.5 $) mein maximum angle kya ho sakta hai agar light ray glass se vacuum mein refract ho rahi ho. Vacuum mein light incident hai medium ($ \mu = 2 $) par, angle of incidence critical angle ka do guna hai. Angle of refraction batao. Air water 4/3 R = 40 cm 30 cm Refraction at Transparent Curved Surface $ \mu_1 $ = us medium ka refractive index jismein actual incident ray hoti hai. $ \mu_2 $ = us medium ka refractive index jismein actual refracted ray hoti hai. O = Object P = pole C = centre of curvature R = PC = radius of curvature O P F C M Image $u$ $v$ $R$ $\alpha$ $\theta_1$ $\theta_2$ $\gamma$ Curved surface se refraction ka formula: $ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $ Agar angle bohot chhota hai: $ \mu_1 \theta_1 = \mu_2 \theta_2 $. Par $ \theta_1 = \alpha + \beta $. $ \beta = \theta_2 + \gamma $. (i), (ii) aur (iii) se $ \mu_1 (\alpha + \beta) = \mu_2 (\beta - \gamma) $. $ \Rightarrow \mu_1 \alpha + \mu_1 \beta = \mu_2 \beta - \mu_2 \gamma \Rightarrow \mu_1 \alpha + \mu_2 \gamma = (\mu_2 - \mu_1) \beta $. $ \frac{\mu_1 PM}{u} + \frac{\mu_2 PM}{v} = \frac{(\mu_2 - \mu_1) PM}{R} $. $ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $. Sign Convention for Radius of Curvature Ye saare single refraction surfaces – convex, concave ya plane ke liye valid hain. Plane refracting surface ke liye $R \rightarrow \infty $. $ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $ Plane surface ke liye $ \frac{\mu_2}{v} - \frac{\mu_1}{u} = 0 \Rightarrow \frac{\mu_2}{v} = \frac{\mu_1}{u} $ Ya $ \frac{d_{ac}}{d_{ap}} = \frac{\mu_1}{\mu_2} $. R = positive R = negative R = $\infty$ Focal Length of a Single Spherical Surface Ek single spherical surface ke do principal focus points hote hain: (i) First focus: First principal focus woh point hai principal axis par jahan object rakhne par image infinity par banti hai. Yaani $ u = f_1, v = \infty $. Toh $ \frac{\mu_2}{\infty} - \frac{\mu_1}{f_1} = \frac{\mu_2 - \mu_1}{R} $. $ \Rightarrow f_1 = \frac{-\mu_1 R}{\mu_2 - \mu_1} $. F1 P F C $\mu_1$ $\mu_2$ (ii) Second focus: Second principal focus woh point hai jahan parallel rays focus hoti hain. Yaani $ u = -\infty, v = f_2 $. Toh $ \frac{\mu_2}{f_2} - \frac{\mu_1}{-\infty} = \frac{\mu_2 - \mu_1}{R} $. $ \Rightarrow f_2 = \frac{\mu_2 R}{\mu_2 - \mu_1} $. F2 P C $\mu_1$ $\mu_2$ (iii) Ratio of Focal length: $ \frac{f_1}{f_2} = -\frac{\mu_1}{\mu_2} $. Do yourself - 4: Figure mein dikhayi gayi situation ke liye image ki position, size aur nature batao. Ek point object 20 cm radius aur $n=2$ refractive index ke transparent solid sphere ke andar hai. Jab object ko nearest surface se air se dekha jaata hai, toh woh surface se 5 cm door dikhta hai. Farthest curved surface se dekhne par object ki apparent distance batao. Ek object 10 cm door rakha hai ek glass piece ($n=1.5$) se jiski length 20 cm hai aur jo 10 cm radius ke spherical surfaces se bounded hai. Do refractions ke baad final image ki position batao. Curved surface se refraction ke baad converging beam ke liye image kahan banti hai (figure mein dikhaya gaya hai): (A) $ x = 40 $ cm (B) $ x = \frac{40}{3} $ cm (C) $ x = -\frac{40}{3} $ cm (D) $ x = \frac{180}{7} $ cm Do refracting media ek spherical interface se separate hain jaisa ki figure mein dikhaya gaya hai. PP' principal axis hai, $ \mu_1 $ aur $ \mu_2 $ incident aur refracted medium ke refractive indices hain. Toh: (A) agar $ \mu_2 > \mu_1 $, toh real object ki real image nahi ban sakti. (B) agar $ \mu_2 > \mu_1 $, toh virtual object ki real image nahi ban sakti. (C) agar $ \mu_1 > \mu_2 $, toh virtual object ki virtual image nahi ban sakti. (D) agar $ \mu_1 > \mu_2 $, toh real object ki real image nahi ban sakti. observer $\mu=1$ $\mu=2$ C 1mm 30cm ROC=20cm ROC=10cm ROC=10cm air n=1.5 air object 10cm n=1 n=3/2 P X 30 R=20cm $\mu_1$ $\mu_2$ P P' Example: Ek air bubble ( $ \mu=1.5 $ ) glass mein hai, jo spherical surface se 3 cm door hai, jiska diameter 10 cm hai. Bubble surface se kitni door dikhega agar surface (a) convex hai, (b) concave hai. Solution: Curved surface se refraction ka formula: $ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $. (a) Convex surface ke liye: $ \mu_1 = 1.5, \mu_2 = 1, R =