Tiêu đề 10. WAVE OPTICS.pdf ✅

8. () : Explana on 9. () : Explana on Path length of ray Path length of ray Path difference 10. () : Explana on Given that Fringe width becomes 4 mes 11. () : Explana on Minimum angular separa on ( = Resolving power) 12. () : Explana on For both . For light, we denote is very high. The Doppler effect is more readily observed in sound waves due to their larger wavelengths, while the same is not the case with light waves because they have much shorter wave‐ lengths in everyday life. So correct op on is (2). 13. () : Explana on fringe visibility Where and are amplitudes. For best visibility, So correct op on is (3). 14. () : Explana on Intensity of unpolarized light Intensity of light emerging from Intensity of light emerging from , . Intensity of light emerging from , is given by 15. () : Explana on For maximum intensity on the screen, or Thus, only seven maxima can be obtained on both sides of the screen. 16. () : Explana on When the slits are of unequal widths, the source intensi es are also unequal, i.e. Minimum intensity i.e., the dark fringes are not perfectly dark. 17. () : Explana on A broad source is equivalent to a large num‐ ber of narrow sources arranged side by side. Each of these sets of sources generates its own dis nct interference pa ern resul ng in general illumina on. This situa on under‐ scores the necessity of employing a narrow slit for interference experiments. So correct op on is . 18. () : Explana on 19. () : Explana on At central point on screen, path difference is zero for all wavelengths. So, central bright fringe is white and other fringes depend on wavelength as . Therefore, other fringes will be coloured. Correct op on is (3). I = I0cos 2θ = I0cos 230 ∘ = 3I0 4 1 = μ1 (S1P − t) + t 2 = μ1S2P = μ1S2P − [μ1 (S1P − t) + t] = μ1 (S2P − S1P) − t(1 − μ1) = μ1 (S2P − S1P) − t(μ1 − 1) = μ1d sin θ + t(μ1 − 1) = μ1 + t(μ1 − 1) yd D β = λD d β ′ = λD′ d′ D′ = 2D, d ′ = d 2 β ′ = = λ×2D d/2 4λD d β ′ = 4β dθ = = 1 R.P 1.22λ d R. P = = 0.3 × 10 −6 rad 1.22×5000×10 −10 2 v = nλ v = c, c V = [ ] Imax − Imin Imax + Imin I ∝ a 2 , Imax ∝ (a1 + a2) 2 Imin ∝ (a1 − a2) 2 a1 a2 a1 = a2 = I0 A, IA = I0 2 C IC = IA cos 2 45 ∘ = I0 4 B IB = IC cos 2 45 ∘ = × = I0 4 1 2 I0 8 d sin θ = nλ sin θ = = = nλ d (n)(2000) (7000) n 3.5 sin θ ≥ 1 ⇒ n = 0, 1, 2, 3 I1 ≠ I2 ∴ Imin = (√I1 − √I2) 2 ≠ 0 (1) = ⇒ d = 1.22λ a x d x×a 1.22λ = 1×10 = 5m −3×3×10 −3 1.22×500×10 −9 β = λD d
20. () : Explana on The condi on for first minima is . The diffrac on of sound is only possible when the size of opening should be of the same or‐ der as its wavelength and the wavelength of sound is of the order of hence, for a very small opening no diffrac on is produced in sound since . So op on (1) is correct. 21. () : Explana on Correct op on is (1). 22. () : Explana on Path length of ray Path length of ray Path difference 23. () : Explana on Given that Let 24. () : Explana on According to Newton’s corpuscular theory of light, the light should travel faster in denser than in rarer media. It is contrary to present theory of light which explains the light travels faster in air (rarer) and slow in water (denser). So op on (1) is correct. 25. () : Explana on The maximum intensity is The maximum intensity is 26. () : Explana on 27. () : Explana on 28. () : Explana on When a thin film of oil spreads across the sur‐ face of water and is viewed in broad daylight, it displays brilliant colors. A similar colorful pa ern is observed in soap bubbles. These vi‐ brant colors arise from the interference of sunlight that is reflected from both the upper and lower surfaces of the thin film. So correct op on is (3). 29. () : Explana on For minima, as So, we will have 3 dark fringe on one side above central maxima. If we consider both side it will be , , 6 30. () : Explana on Both asser on and reason are correct and rea‐ son is the correct explana on for asser on. We know that earth's atmosphere has some par cles which can sca er the light. Hence due to these par cles the incoming light re‐ flected by earth is par ally polarised. sin θ = λ a 1.0 m, sin θ > 1 I = 4I0 cos 2( x) π β ⇒ I = 4I0cos 2 ( ) π 3 ⇒ I = 4 × 2 × 10 −2 × 1 4 ⇒ I = 2 × 10 −2 W/m2 . 1 = (S1P − t1) + t1μ1 2 = (S2P − t2) + t2μ2 = [S2P − t2 + μ2t2 ] − [(S1P − t1) + μ1t1 ] = [(S2P − S1P) + t2(μ2 − 1)] − t1(μ1 − 1) = [d sin θ + t2(μ2 − 1)] − t1(μ1 − 1) = [ + t2(μ2 − 1)] − t1(μ1 − 1) yd D = n I1 I2 I2 = I0 ⇒ I1 = nI0 Imin = (√I1 + √I2) 2 = (√I0 + √n√I0) 2 Imax = (√I1 − √I2) 2 = (√I0 − √n√I0) 2 = Imax−Imin Imax+Imin (√I0+√n√I0) 2−(√I0−√n√I0) 2 (√I0+√n√I0) 2+(√I0−√n√I0) 2 = 2√n 1+n 1st I1 = (N 2I0) 4 9π 2 2nd I2 = (N 2I0) 4 25π2 = I2 I1 9 25 I = I0cos 2θ I = ⇒ × 100 = 75% 3I0 4 I I0 = = ∴ = I1 I2 a 2 1 a 2 2 3 1 a1 a2 √3 1 d sin θ = (n − ) λ 1 2 ⇒ sin θ = (n − ) 1 2 λ d sin θ ≤ 1 ⇒ (n − ) ⩽ 1 i. e. , (n − ) ⩽ 1 2 λ d 1 2 d λ i. e n ≤ + d λ 1 2 i. e n ≤ + 3λ λ 1 2 i. e n ≤ 7 2 i. e n = 1, 2, 3 (3 + 3) i. e
31. () : Explana on According to Malus’ law, the intensity of emergent polarized light from second polar‐ izer is Here, is the angle between pass axes of two polarizers and is the intensity of polarized light a er passing through the first polarizer. Given, 32. () : Explana on 33. () : Explana on For common maxima, 34. () : Explana on Distance between slits and screen Separa on between the two slits Thickness of mica sheet Refrac ve index Shi in the fringe pa ern, 35. () : Explana on If is the half angular width of principal maxi‐ mum, then Thus, the angular width of principal maximum 36. () : Explana on The fringe width, represented by and calcu‐ lated using the expression , is unaf‐ fected by the specific posi on of the fringe. It gives consistent fringe width across the inter‐ ference pa ern. So correct op on is (1). 37. () : Explana on Distance between slits, Distance of the screen from the slits, Wavelengths, and Distance of bright fringe from the centre Thus, the separa on between the bright fringes due to 38. () : Explana on Conceptual Ques on 39. () : Explana on The colors of a soap film depend on its thick‐ ness and the wavelength of incident light, causing interference pa erns. Minimum thick‐ ness of soap-bubble for construc ve interfer‐ ence in reflected light is In a ver cal arrangement, the film's thickness varia ons lead to downward shi ing interference bands, and when the up‐ per part a ains a thickness equal to one-quar‐ ter of the shortest wavelength, destruc ve in‐ terference occurs, making it appear dark. So correct op on is (3). 40. () : Explana on In this process wavelength and speed changes but frequency remains the same. Wavelength of refracted light Velocity of refracted light I = I0cos 2θ θ I0 θ = 60 ∘ ∴ I = I0cos 260 ∘ = I0 × ( ) 2 = 1 2 I0 4 β = λD = m d 5000×10 −10×0.9 3×10 −3 = 0.15mm n1λ1 = n2λ2 ⇒ = = = n1 n2 λ2 λ1 520×10 −9 650×10 −9 4 5 For λ1 y = = = 7.8mm n1λ1D d 4×650×10 −9×1.5 0.5×10−3 D = 1.2 m, d = 2.4 mm = 2.4 × 10 −3 m t = 1 × 10 −6m μ = 1.5 ∴ Δy = (μ − 1)t = (1.5 − 1) × 1 × 10 D −6 d 1.2 2.4×10 −3 = 0.25 × 10 −3 m = 0.25 mm θ θ = λ a =2θ = 2λ a β β = λD d d = 2mm = 2 × 10 −3m D = 1.2m λ1 = 6500 ∘ A = 6500 × 10 −10 = 6.5 × 10 −7m λ2 = 5200 ∘ A = 5200 × 10 −10 = 5.2 × 10 −7m n th xn = nλD d 4 th λ1 and λ2, Δx = − = 4λ1D d 4λ2D d 4D(λ1−λ2) d = 4(1.2m)(6.5×10 −7m−5.2×10 −7m) (2×10 −3m) = 3.12 × 10 −4 m = 0.312 mm t = λ 4μ (∵ 2μt = ) λ 2 λ ′ = = = 4.42 × 10 μ λ −7m 589×10−9 1.33 v = = = 2.25 × 10 8m/s c μ 3×10 8 1.33