Tiêu đề 10. WAVE OPTICS(H).pdf ✅

NEET REVISION 10. WAVE OPTICS(H) NEET REVISION Date: March 18, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on Given that 2. () : Explana on When rays of monochroma c light of wavelength are incident on a diffrac on gra ng in which slit separa on is , then for angle of diffrac on , the following rela on holds true. where, is called the spectrum order. Given, and 3. () : Explana on 4. () : Explana on Let the intesi es of two waves be and . (Given) 5. () : Explana on Wavelength, slit width Angular spread of central maxima Using condi on of diffrac on minima, , where Diffrac on angle So, angular spread So, correct op on is (1). 6. () : Explana on ( ) 2 = 16 A1+A2 A1−A2 = 4 ⇒ = A1+A2 A1−A2 A1 A2 25 9 λ d θ d sin θ = nλ n n = 1, λ = 6500 ∘ A = 6500 × 10 −10 m sin 30 ∘ = 1 2 ⇒ d = = nλ sin 30 ∘ 6500×10−10 (1/2) ⇒ d = 1.3 × 10 −6 m √ = √ = I1 I2 9 4 3 2 √I2 = √I1 2 3 ⇒ = ( ) 2 Imax Imin √I1+√I2 √I1−√I2 = [ ] 2 √I1+ √I1 2 3 √I1− √I1 2 3 = ( ) 2 = 25 √I1 √I1 5 1 I1 I2 I1 : I2 = 49 : 16 ⇒ = ( ) 2 Imax Imin √I1 + √I2 √I1 + √I2 ⇒ = ( ) 2 Imax Imin √49 + √16 √49 − √16 ⇒ = ( ) 2 ⇒ = ⇒ Imax : Imin = 121 : 9 Imax Imin 7 + 4 7 − 4 Imax Imin 121 9 λ = 2 cm, b = 4 cm. = 2θ b sin θ = λ θ = sin θ = = = λ b 2 × 10 −2 4 × 10 −2 1 2 ⇒ θ = sin −1( ) = 30 1 ∘ 2 = 2θ = 2 × 30 ∘ = 60 ∘ = ⇒ x = x d 1.22λ a 1.22λd a = = 50 m 1.22×500×10 −9×400×10 3 5×10 −3

NEET REVISION 15. () : Explana on (2) Take base 16. () : Explana on As the light is ge ng red shi ed the wavelength increases. - refrac ve velocity of the star w.r.to the earth 17. () : Explana on For dark fringe, For first dark fringe, 18. () : Explana on The resolving angle of eye is Where is the fringe width 19. () : Explana on Op cal path for Op cal path for Phase difference, 20. () : Explana on Given, distance between source and observer, Distance of 5th dark fringe from centre Wavelength, Distance of 5th dark fringe from centre where, fringe width. According to ques on, 21. () : Explana on Given that For first diffrac on minima 22. () : Explana on Let be the point on -axis, where construc ve interference is obtained and AP be . For first construc ve interference, Further, , 23. () : Explana on 24. () : Explana on Using law of Malus resultant intensity can be cal‐ culated as S1P − S2P = λ/6 ∴ SS1P − SS2P = λ/3 SS1P − SS3P = 4λ/3 ⇒ Δφ = × 2π λ 4λ 3 (−1) SS2P − SS3P = λ ⇒ Δφ = × λ = 2π 2π λ SS3P Inet = (2√I) 2 + (√I) + 2.2√I√I cos 120 ∘ Inet = 3I = Δλ λ u c u u = = = 5 × 10 5m/s Δλc λ 10×3×10 8 6000 n th (2n − 1) = λD 2d d 2 ⇒ λ = d 2 (2n − 1)D n = 1 ⇒ λ = d 2 D β tan ≈ = ⇒ θ = θ 2 θ 2 β/2 D β D θ = = λD d(D) λ d × = ⇒ d = = 2 mm 1 60 π 180 λ d λ×60×180 3.14 1 st ray = n1l1 2 nd ray = n2l2 Δφ = Δx 2π λ = (n1l1 − n2l2) 2π λ D = 2m = 4 × 10 −3 m λ = 600 nm = 6 × 10 −7 m = β + β + β + β + = β 2 9β 2 β = = 4 × 10 −3 9β 2 × = 4 × 10 9 −3 2 Dλ d × = 4 × 10 −3 9 2 2 × 6 × 10 −7 d ⇒ d = 1.35 × 10 −3 m = 1.35 mm λ = 6000 × 10 −8cm sin θ1 = λ a sin θ2 = 2λ a sin θ1 = = ⇒ θ1 = 25 sin θ2 ∘ 2 √3 4 P X x BP − AP = λ √9λ 2 + x 2 − x = λ ∴ 9λ 2 + x 2 = x 2 + λ 2 + 2xλ x = 4λ = = Imax Imin (√16+√9) 2 (√16−√9) 2 49 1 Ires = cos 2 θ sin 2 θ I0 2 = cos 2 30 ∘ sin 2 30 ∘ = I0 2 3I0 32