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Light & Interference 1. (B) According to lens maker's formula, when the lens in the air. 1 f = ( 3 2 − 1) ( 1 R1 − 1 R2 ) ⇒ 1 f = 1 2x ⇒ f = 2x Here, 1 x = 1 R1 − 1 R2 , x is positive for convex lens. In case of liquid, where refractive indices are 4 3 and 5 3 , we get : Focal length in first liquid, 1 f1 = ( μs μs1 ) ( 1 R1 − 1 R2 ) 1 f1 = ( 3/2 4/3 − 1) 1 x ⇒ f1 is positive, Nature of lens does not change, 1 f1 = 1 8x = 1 4(2x) = 1 4f = f1 = 4f 1 f2 = ( μs μ/2 − 1) ( 1 R1 − 1 R2 ) ⇒ 1 f2 = ( 3 2 5 3 − 1)( 1 x ) ⇒ f2 is negative. Nature of lens change i.e. convex behave as diverging lens. 2. (B) As beam of light in incident normally on the face AB of the right angled prism ABC, so no refraction occurs at face AB and it passes straight and strikes the face AC at an angle of incidence i = 45∘ . For total reflection to take place at face AC, i > ic or sin i > sin ic Where ic is the critical angle. ∴ sin 45∘ > 1 μ or 1 √2 > 1 μ Or μ > √2 = 1.414 As μred (= 1.39) < μ(= 1.414) while μgreen (= 1.44) and μblue (= 1.47) > μ(= 1.414), so only red colour will be transmitted through face AC while green and blue colour will suffer total internal reflection. So the prism will separate red colour from the green and blue colours as shown in the following figure. 3. (B) I = 4I0cos2 φ/2 ∴ I1 = 4I0cos2 0/2 = 4I0 (∵ Δx = 0) Now, ∵ φ = 2π λ × Δx = 2π λ × λ 4 ⇒ φ = π 2 and I2 = 4I0cos2 π 4 = 2I0 [φ = 2π λ Δx] ∴ I1:I2 = 2: 1 4. (A) The situation is shown in the figure. In figure A and B represent the edges of the slit AB of width a and C represents the midpoint of the slit. For the first minimum at P, asin θ = λ Where λ is the wavelength of light. The path difference between the wavelets from A to C is Δx = a 2 sin θ = 1 2 (asin θ) = λ 2 Screen The corresponding phase difference Δφ is Δφ = 2π λ Δx = 2π λ × λ 2 = π 5. (B) The situation is shown in the figure. Let fo and fe be the focal lengths of the objective and eyepiece respectively. For normal adjustment distance of the objective from the eyepiece (tube length) = fo + fe . Treating the line on the objective as the object and eyepiece as the lens. ∴ u = −(fo + fe ) and f = fe
As 1 v − 1 u = 1 f ∴ 1 v − 1 −(fo + fe ) = 1 fe 1 v = 1 fe − 1 fo + fe = fo + fe − fe fe (fo + fe ) = fo fe (fo + fe ) Or v = fe (fo+fe ) fo Thus, I L = v |u| = fe(fo+fe) fo (fo+fe ) = fe fo Or fo fe = L I ∴ The magnification of the telescope in normal adjustment is m = fo fe = L I (Using (i)) 6. (C) Δhω = (1 − 1 μ1 ) h1; Δhk = (1 − 1 μ2 ) h2 ∴ Δh = Δhω + Δhk = (1 − 1 μ1 ) h1 + (1 − 1 μ2 ) h2 7. (C) Given situation is shown in the figure. For central maxima, sin θ = λ a Also, θ is very -very small so sin θ ≈ tan θ = y D ∴ y D = λ a , y = λD a Width of central maxima = 2y = 2λD a 8. (A) 9. (C) Scattering for blue light is largest and it is polarized by scattering. Also for polarized light I = I0cos2 θ 10. (D) As μ = sin ( A+δ 2 ) sin ( A 2 ) cot A 2 = sin ( A+δ 2 ) sin ( A 2 ) [∵ μ = cot A 2 ] cos ( A 2 ) sin ( A 2 ) = sin ( A + δ 2 ) sin ( A 2 ) sin ( π 2 − A 2 ) = sin ( A 2 + δ 2 ) ; π 2 − A 2 = A 2 + δ 2 δ = π − 2 A = 180∘ − 2 A 11. (C) For double slit experiment, d = 1 mm = 1 × 10−3 m,D = 1 m, λ = 500 × 10−9 m Fringe width β = Dλ d Width of central maxima in a single slit As per question, with of central maxima of single slit pattern = width of 10 maxima of double slit pattern 2λD a = 10 ( λD d ) a = 2d 10 = 2 × 10−3 10 = 0.2 × 10−3 m = 0.2 mm 12. (B) Here, sin C = 1 nwater and nwater = a + b λ 2 If frequency is less, then λ is greater and hence RI n(water) is less and therefore critical angle increases. So, they do not suffer total internal reflection and come out at angle less than 90∘ in the air medium. 13. (C) Intensity at any point on the screen is I = 4I0cos2 φ 2 Where I0 is the intensity of either wave and φ is the phase difference between two waves. Phase difference, φ = 2π λ × path difference When path difference is λ, then φ = 2π λ × λ = 2π ∴ I = 4I0cos2 ( 2π 2 ) = 4I0cos2 (π) = 4I0 = K When path difference is λ 4 , then φ = 2π λ × λ 4 = π 2 ∴ I = 4I0cos2 ( π 4 ) = 2I0 = K 2 14. (A) Note As refractive index for z > 0 and z ≤ 0 is different X − Y plane should be boundary between
two media. Angle of incidence, cos i = | Az Ax 2+Ay 2+Az 2 | = 1 2 ∴ i = 60∘ From Snell's law sin i sin r = √3 2 ⇒ r = 45∘ 15. (D) Magnifying power of a microscope, m = ( L fo ) ( D fe ) Where fo and fe are the focal lengths of the objective and eyepiece respectively and L is the distance between their focal points and D is the least distance of distinct vision. If fo increases, then m will decrease. Magnifying power of a telescope, m = f0 fe Where fo and fe are the focal lengths of the objective and eyepiece respectively. If fo increase, then m will increase. 16. (B) On reflection from silvered surface, the incident ray will retrace its path, if it falls normally on the surface. By geometry, r = A Applying snell's law at surface PQ, 1sin i = μsin rμ = sin i sin r = sin 2A sin A = 2cos A 17. (B) Both statements I and II are correct but statement II does not explain statement I. 18. (A) The combination of two lenses 1 and 2 is as shown in figure. ∴ 1 f = 1 f1 + 1 f2 According to lens maker's formula 1 f1 = (μ1 − 1) ( 1 ∞ − 1 −R ) = (μ1 − 1) R ; 1 f2 = (μ2 − 1) ( 1 −R − 1 ∞ ) = (μ2 − 1) (− 1 R ) = − (μ2 − 1) R ∴ 1 f = (μ1 − 1) R − (μ2 − 1) R 1 f = (μ1 − μ2 ) R f = R (μ1 − μ2 ) 19. (A) Converging power of cornea, Pc = +40D Least converging power of eye lens, Pc = +20D Power of the eye-lens, P = Pc + Pe = 40D + 20D = 60D Power of the eye lens P = 1 Focal length of the lens (f) f = 1 P = 1 60D = 1 60 m = 100 60 cm = 5 3 cm Distance between the retina and cornea-eye lens = focal length of the eye lens = 5 3 cm = 1.67 cm 20. (A) 1 u + 1 v = 1 f ⇒ − 1 u2 du dt − 1 v 2 dv dt = 0 ⇒ dv dt = − v 2 u2 ( du dt) But v u = f u−f ∴ dv dt = − ( f u−f ) 2 ( du dt) = ( 0.2 −2.8−0.2 ) 2 × 15 = 1 15 ms−1 21. (C) 3λ1 = 4λ2 ⇒ λ2 = 3 4 λ1 = 3 4 × 590 = 1770 4 = 442.5 nm 22. (C) For the second minimum, Path difference = 2λ Therefore, corresponding value of phase difference is Δφ = 2π λ × path difference ∴ Δφ = 2π λ × 2λ = 4π 23. (A) As the emergent ray emerges normally from the opposite face,
∴ e = 0, r2 = 0 as r1 + r2 = A ∴ r1 = A Applying snell's law for incident ray 1sin i = μsin r1 = μsin A or μ = sin i sin A For small angle, sin i ≈ i, sin A ≈ A ∴ μ = i A or i = μA 24. (D) By law of Malus i.e., I = I0cos2 θ Now, IA ′ = IAcos2 30∘ IB ′ = IBcos2 60∘ As, IA ′ = IB ′ IAcos2 30∘ = IBcos2 60∘ ⇒ IA 3 4 = IB 1 4 ⇒ IA IB = 1 3 Transmission axis 25. (C) 1 v − 1 u = 1 f = constant, so ( C ) is the correct graph. 26. (C) 27. (A) It is possible when object kept at centre of curvature. u = v u = 2f, v = 2f 28. (B) As μ = sin ( A+δm 2 ) sin ( A 2 ) μ = sin ( A+A 2 ) sin ( A 2 ) = sin A sin ( A 2 ) (∵ δm = A (Given)) = 2sin ( A 2 ) cos ( A 2 ) sin ( A 2 ) = 2cos ( A 2 ) As δ = i + e − A At minimum deviation, δ = δm, i = e ∴ δm = 2i − A; 2i = δm + A i = δm + A 2 = A + A 2 = A(∵ δm = A given )) imin = 0 ∘ ⇒ Amin = 0 ∘ Then, μmax = π 2 ⇒ Amax = π 2 ∵ imax = π 2 ⇒ Amax = π 2 Then, μmin = 2cos 45∘ = 2 × 1 √2 = √2 So refractive index lies between 2 and √2 29. (C) Let μ = 3 2 According to lens maker's formula 1 f = (μ − 1) ( 1 R1 − 1 R2 ) For biconvex lens, R1 = +20 cm,R2 = −20 cm ∴ 1 f = ( 3 2 − 1) ( 1 20 + 1 20) = 1 20 or f = 20 cm According to thin lens formula, 1 f = 1 v = 1 u Here, u = −30 cm ∴ 1 20 = 1 v − 1 −30 ⇒ 1 v = 1 20 − 1 30 v = 60 cm The image is formed at a distance of 60 cm on the right hand side of the lens. It is a real image. Magnification, m = v u = h1 h0 60 cm −30 cm = h1 2 cm ⇒ h1 = −4 cm −ve sign shows that image is inverted. The image is real, inverted and height of 4 cm as shown in figure.
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