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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
The spectral composition of solar radiation is much the same as that of a black whose maximum emission corresponds to the wavelength `0.48mu m`. Find the mass lost by the sun every second due to radiation. Evaluate the time interval during which the mass of the sun diminishes y `1` per cent. |
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Answer» The black body temperature of the sum may be taken as `T_(Theta) = (0.29)/(0.48 xx 10^(-4)) = 6042K` Thus the radiosity is `M_(e Theta) = 5.67 xx 10^(-8) (6042)^(4) = 0.7555 xx 10^(8)W//m^(2)` Energy lost by sun is `4pi(6.95)^(2) xx 10^(16) xx 0.7555 xx 10^(8) = 4.5855 xx 10^(26)watt` This corresponds to a mass loss of `(4.5855 xx 10^(26))/(9xx 10^(16))kg//sec = 5.1 xx 10^(9) kg//sec` The sun loses `1%` of its mass in `(1.97 xx 10^(30) xx 10^(-2))/(5.1 xx 10^(9)) sec = 1.22 xx 10^(11)` years. |
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| 52. |
A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and `PO=OQ`. The distance `PO`A. `5R`B. `3R`C. `2R`D. `1.5R` |
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Answer» Correct Answer - A |
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| 53. |
A spherical convex surface separates object and image space of refractive index `1.0` and `4//3`. If radius of curvature of the surface is `10 cm`, find its power. |
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Answer» Correct Answer - `2.5 D` Here, `mu_(1) = 1.0, mu_(2) = 4//3, R = 10 cm. P = ?` `-(mu_(1))/(u) + (mu_(2))/(v) - (mu_(2) - mu_(1))/(R )` when `u = oo, v = f` `:. -(1)/(oo) + (4)/(3 f) = (4//3 - 1)/(10) = (1)/(30) , f = 40 cm`. `P = 100// f = (100)/(40) = 2.5 D` |
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| 54. |
A thin concavo-convex lens ha two surface of radii of curvature `R` and `2R`. The mateial of the lens has a refractive index `mu`. When kept in air, the focal length of the lensA. wil depend on the direction from which light is incident on itB. will be the same, irrespective of the direction from which light is incident on itC. will be equal to `R/(mu-1)`D. will be equal to `(2R)/(mu-1)` |
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Answer» Correct Answer - B::D |
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| 55. |
A small point objects is placed in air at a distance of `60 cm` from a convex spherical refractive surface of `mu = 1.5`. If radius of curvature of spherical surface is `25 cm`, calculate the position of the image and the power of the refracting surface. |
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Answer» Here, `u = -60 cm, mu_(1) = 1, mu_(2) = 1.5, R = + 25 cm, v = ?` As refraction occurs from rarer ti denser medium, therefore `(-mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )` `(-1)/(-60) + (1.5)/(v) = (1.5 - 1)/(25)` `(3)/(2v) = (1)/(50) - (1)/(60) = (1)/(300)` As `v` positive, image formed on the other side of the object, i.e., in refracting denser medium at `450 cm` from the pole. Power of teh refracting surface, `P = (mu_(2) - mu_(1))/(R ) = (1.5 - 1)/(0.25) = (0.5)/(0.25) = 2 D` |
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| 56. |
Light from a point source in air falls on a convex spherical glass surface `(mu = 1.5 and R = 20 cm)`. Calculate position of the image when the light source is at `1m` from the glass surface. |
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Answer» Here, `mu_(1) = 1` (for air), `mu_(2) = 1.5` (for glass), `R = + 20 cm, u =-m = -100 cm, v = ?` As refraction occurs from rarer to denser medium, therefore `-(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )` `-(1)/(-100) + (1.5)/(v) = (1.5 - 1)/(20) = (1)/(40)` `(3)/(2v) = (1)/(40) - (1)/(100) = (5 - 2)/(200) = (3)/(200)` `v = 100 cm` `:.` Image is formed at `100 cm` from the pole of the surface in the direction of incidence of light. |
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| 57. |
Light from a point source in air falls on a spherical glass surface. If `mu = 1.5`, and radius of curvature `= 20 cm`, the distance of light source from the glass surface is `100 cm` , at what position will the image be formed ? (NCERT Solved Example) |
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Answer» Here, `mu_(2) = 1.5, mu_(1) = 1` , `R = 20 cm`, `u = -100 cm, v = ?` As light goes from air (rarer medium) to glass (denser medium), therefore, `-(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )` `-(1)/(-100) + (1.5)/(v) = (1.5 - 1)/(20) = (1)/(40)` `(3)/(2 v) = (1)/(40) - (1)/(100) = (5 - 2)/(200) = (3)/(200) = 100cm` `:.` The image is formed at a distance of `100cm` from the glass surface, in the direction of incident light. |
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| 58. |
What curvature must be given to the bounding surface of `mu = 1.5` for virtual image of an object in the medium of `mu = 1 at 10 cm` to be formed at a distance of `40 cm`. Calculate power of the refracting surface and also two principal focal lengths of the surface. |
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Answer» `R = ?, mu_(2) = 1.5, mu_(1) = 1` `u = - 10 cm, v = -40 cm` As object is placed in rarer medium, `:. -(mu_(1))/(u) + (mu_(2))/(v) = (mu_(@) - mu_(1))/(R )` `-(1)/(-10) + (1.5)/(-40) = (1.5 - 1)/(R ) = (1)/(2R)` `(1)/(2R) = (1)/(10) - (3)/(80) = (5)/(80) = (1)/(16)` `R = (16)/(2) = 8cm` As `R` is positive, the refracting surface is convex. Power of surface, `P = (mu_(2) - mu_(1))/(R ) = (1.5 - 1)/(8//100) = 6.25 D` |
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| 59. |
What curvature must be given to the bounding concave surface of refracting medium `(mu = 3//2)` for a virtual image at `40 cm` of an object in this medium at a distance of `60 cm`. The adjoining medium is air `(mu = 1)`. |
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Answer» Here, `R = ? Mu = 3//2, v = -40 cm, u = -60 cm, mu_(1) = 1` As refracting occurs from denser to rarer medium, therefore, ` -(mu_(2))/(u) + (mu_(1))/(v) = (mu_(1) - mu_(2))/(R )` `-(3)/(2(- 60)) + (1)/(-40) = (1 - 3//2)/(R ) , -(1)/(40) - (1)/(40) = -(1)/(2R), -(1)/(20) = -(1)/(2R) = R = 10 cm` |
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| 60. |
Photograph of the ground are taken form an air-craft ,flying at an altitude of 2000 m by a camera with a lens of focal length `50 cm`. The size of the film in the camera is `18xx18cm`.What area of the ground can be photography by this camera at any one time.A. `720 m xx 720 m`B. `240 m xx 240 m`C. `1080 m xx 1080 m`D. `100 m xx 100 m` |
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Answer» Correct Answer - C As ground is at very large distance from camera, thereofore image of ground can be assumed to be formed at the focus of the camera lens. `:. V = 50 m = 0.5 m, u = - 3km = - 3000 m`. If `x` is length of ground photographed, then `|m|= |(h_(i))/(h_(0))| = |(v_(0))/(u)|` or `(0.18)/(x) = (0.5)/(3000)` `:. x = (3000 xx 0.8)/(0.5) = 1080 m` `:.` Area of ground photographed `= 1080 m xx 1080 m` |
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| 61. |
For a person seeing an object placed in optically rarer medium,(A) apparent depth of the object is more than real depth(B) apparent depth is smaller than the real depth.(C) apparent depth might be smaller or greater depending on the position of the person.(D) nothing can be concluded about the depth of object from given data. |
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Answer» (A) apparent depth of the object is more than real depth |
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| 62. |
Determine refractive index of a substance if critical angle is `45^(@)`. |
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Answer» Here, `C = 45^(@), mu = ?` `mu = (1)/(sin C) = (1)/(sin 45^(@)) = (1)/(1//sqrt(2)) = sqrt(2)` |
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| 63. |
Calculate the speed of light in a medium whose critical angle is `45^(@)`. |
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Answer» Correct Answer - `2.12 xx 10^(8) m//s` Here, `v = ?, C = 45^(@)` `mu = (1)/(sin C) = (1)/(45^(@)) = sqrt(2)` `v = ( c)/(mu) = (3 xx 10^(8)m//s)/(sqrt(2)) = 2.12 xx 10^(8)m//s` |
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| 64. |
A beam of light, consisting of red, green and blue colours, is incident on a right-angled prism, as shown. The refractive indices of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will (a) separate part of the red colour from the green and blue colours (b) separate part of the blue colour from the red and green colours (c) separate all the three colours from one another (d) not separate even partially any colour from the other two colours |
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Answer» Correct Answer is: (a) The angle of incidence of all the rays is 45° at the hypotenuse. For a critical angle of 45°, the refractive index must be (sin 45°) -1 = √2 = 1.414. For red light, μ = 1.39 < 1.414. Hence, its critical angle is > 45°. Therefore, red light will pass through the surface into air. For green and blue lights, μ > 1.414. Hence, their critical angles are < 45°. They will be reflected internally and emerge from the surface at the bottom. |
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| 65. |
In the Fizeau experiment on measurement of the velocity of light the distance between the gear wheel and the mirror is `l = 7.0 km`, the number of teeth is `z = 720`. Two successive disappearances of light are observed at the folllowing rotation velocities of the wheel: `n_(1) = 283rps` and `n_(2) = 313` rps. Find the velocity of light. |
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Answer» In the Fizean experiment, light disppears when the wheel rotates to bring a tooth in the position fromerly occupied by a gap in the time taken by light to go from the wheel to the mittor and back. Thusdistance travelled `= 2l` Suppose the `n^(th)` tooth after the gap has come in place of the latter. Then time taken `= (2(m-1)+1)/(2zn_(1))sec.` in the first case `= (2m+1)/(2zn_(1))` sec in the second case `= (1)/(z(n_(2)-n_(1)))` Then `c=2l z(n_(1)-n_(1)) = 3.024 xx 10^(8)m//sec` |
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| 66. |
In Fig. 6(b).80, light rays of blue, green and red wavelength are incident on an isoscels right angled prism. Explain with reason which ray of light will be transmitted through the face `AC`. The refractive index of the prism for red, green and blue light are `1.39, 1.424` and `1.476` respectively. |
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Answer» Correct Answer - Red rays will be transmitted As `ABC` is an isoceles right angled prism, angle of incidence of each ray is `45^(@)`. If critical angle `C` is less than `45^(@)`, the ray will be totally internally reflected at `AC`. When `/_C gt 45^(@)`, the ray will be transmitted through the face `AC`. For red ray, `mu = 1.39` `sin C = (1)/(mu) = (1)/(1.39) = 0.719, C = 46.0^(@)` `:.` Red ray will be transmitted. For green ray, `mu = 1.476`, `sin C = (1)/(mu) = (1)/(1.424) = 0.702, C = 44.6^(@)` `:.` green ray will be reflected at face `AC`. For blue ray, `mu = 1.476`, `sin C = (1)/(mu) = (1)/(1.476) = 0.678, C = 42.6^(@)` `:.` Blue ray will also be reflected at face `AC`. |
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| 67. |
The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let δ1 and δ2 be angles of minimum deviation for red and blue light respectively in a prism of this glass, then(a) δ1, can be less than or greater than δ2 depending upon the values of δ1 and δ2 (b) δ1 > δ2 (c) δ1 < δ2 (d) δ1 = δ2 |
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Answer» (c) δ1 < δ2 δ1 = (μR – 1)A, δ2 = (μB – 1)A As, μR < μB ∴ δ1 < δ2 |
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| 68. |
If two slits in `YDSE` have width ratio `4 : 1`, deduce the ratio of maxima and minima in the interference pattern. |
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Answer» Here, `(w_(1))/(w_(2)) = (4)/(1) = (a^(2))/(b^(2))` `:. (a)/(b) = (2)/(1)` `(I_(max))/(I_(min)) = (a + b)^(2)/(a - b)^(2) = (2b + b)^(2)/(2b - b)^(2) = (9)/(1)` |
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| 69. |
Two interfering sources have an intensity ratio `16 : 1`. Deduce ratio and ratio of intensity between the maxima and minima in interference pattern. |
| Answer» Correct Answer - `4 : 1 ; 25 : 9` | |
| 70. |
A light of wavelength `6000 A` in air, enters a medium with refractive index 1.5 Inside the medium its frequency is….Hz and its wavelength is ….`A`A. `v=5xx10^(14)Hz`B. `v=7.5xx10^(14)Hz`C. `lamda=4000Å`D. `lamda=9000Å` |
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Answer» Correct Answer - A::C |
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| 71. |
A light of wavelength 6000 Å in air enters a medium of refractive index 1.5. Inside the medium, its frequency is ν and its wavelength is λ. (a) ν = 5 x 1014 Hz (b) ν = 7.5 x 1014 Hz (c) λ = 4000 Å (d) λ = 9000 Å |
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Answer» Correct Answer is: (a) ν = 5 x 1014 Hz , (c) λ = 4000 |
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| 72. |
Why is there no dispersion of light refracted through a rectangular glass slab ? |
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Answer» After refraction at two parallel faces of a glass slab, a ray of light emerges in a direction parallel to the direction of incidence of white light on the slab. As rays of all colours emerge in the same direction (of incidence of white light), hence there is no dispersion, but only lateral displacement. |
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| 73. |
State the applications of convex lenses. |
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Answer» 1. Convex lenses are used as camera lenses. 2. They are used as magnifying lenses. 3. They are used in making microscope, telescope and slide projectors. 4. They are used to correct the defect of vision called hypermetropia. |
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| 74. |
An object is placed at a distance of 30 cm from a concave lens of focal length 15 cm. An erect and virtual image is formed at a distance of 10 cm from the lens. Calculate the magnification. |
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Answer» Type of lens is Cancave lens. Formula: Magnification m = \(\frac{u}{v}\) Object distance u = -30 cm Image distance v = -10 cm m = \(\frac{-10}{-30}\) = \(\frac{1}{3}\) = +0.33 |
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| 75. |
What do you know about lens formula? |
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Answer» The lens formula gives the relationship among distance of the object (u), distance of the image (v) and the focal length (f) of the lens. It is expressed as \(\frac{1}{F} = \frac{1}{v} - \frac{1}{u}\) |
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| 76. |
What are the applications of concave lens? |
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Answer» 1. Concave lenses are used as eye lens of‘Galilean Telescope’. 2. They are used in wide angle spy hole in doors. 3. They are used to correct the defect of vision called ‘myopia’. |
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| 77. |
P is a point on the axis of a concave mirror. The image of P, formed by the mirror, coincides with P. A rectangular glass slab of thickness t and refractive index μ is now introduced between P and the mirror. For the image of P to coincide with P again, the mirror must be moved(a) towards P by (μ - 1)t (b) away from P by (μ - 1)t (c) towards P by t(1 - 1/μ) (d) away from P by t(1 - 1/μ) |
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Answer» Correct Answer is: (d) away from P by t(1 - 1/μ) When the slab is introduced between P and the mirror, the apparent position of P shifts towards the mirror by t(1 - 1/μ). Hence, the mirror must be moved in the same direction through the same distance. |
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| 78. |
P is a point on the axis of a concave mirror. The image of P, formed by the mirror, coincides with P. A rectangular glass slab of thickness t and refractive index `mu` is now introduced between P and the mirror. For the image of P to coincide with P again, the mirror must be movedA. towards `P` by `(mu-1)t`B. away from `P` by `(mu-1)t`C. towares `P` by `t(1-1//mu)`D. away from `P` by `(1-1//mu)` |
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Answer» Correct Answer - D |
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| 79. |
A lens which is thicker in the middle than at the edges is known as: (a) concave lens (b) convex lens (c) bifocal lens (d) cylindrical lens |
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Answer» (b) convex lens |
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| 80. |
For a convex lens the point at which the parallel rays converge is called of the lens. (a) pole (b) centre of curvature (c) principal focus (d) none |
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Answer» (c) principal focus |
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| 81. |
Differentiate convex lens from concave lens. |
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Answer»
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| 82. |
To get a real image using convex lens, the object must be placed at: (a) infinity (b) principal focus (c) beyond principal focus and infinity (d) both (b) and (c) |
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Answer» (d) both (b) and (c) |
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| 83. |
The parallel rays from the outer edge are deviated towards the middle in a: (a) convex mirror (b) concave lens (c) concave mirror (d) convex lens |
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Answer» (d) convex lens |
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| 84. |
Where should an object be placed so that a real and inverted image of same size is obtained by a convex lens: (a)f (b) 2f (c) infinity(d) between f and 2f |
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Answer» Correct answer is (b) 2f |
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| 85. |
Where should an object be placed so that a real and inverted image of the same size is obtained by a convex lens ______.(a) f (b) 2f (c) infinity (d) between f and 2f. |
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Answer» Correct answer is (b) 2f |
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| 86. |
Calculate the deviation produced by a prism of angle `6^@`, given refractive index of the material of the prism is `1.644`. |
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Answer» `delta = ? A = 60^(@), mu = 1.644` As `delta = (mu - 1)A :. delta = (1.644 - 1)6 = 0.644 xx 6 = 3.864^(@)` |
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| 87. |
In Fig., line `AB` represents a lens through which course of rays is as shwon. Is this lens convex or concave ? |
| Answer» The line `AB` represents a concave lens because lens because rays converging towards `O`, diverging to meet at `I`. | |
| 88. |
An illuminated object and a screen are placed `90 cm` apart. What is the focal length and nature of the lens required to produce a clear image on the screen twice the size of the object ? |
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Answer» As image is real, the lens must be convex and it should be placed between the object and screen. Let `x` be distance between the object and convex lens. `:. u = - x, v = 90 - x , m = - 2` As `m = (v)/(u)`, `:. -2 = (90 - x)/(- x), x = 30 cm` `:. u = - x = - 30cm`, `v = 90 - x = 90 - 30 = 60 cm` `(1)/(f) = (1)/(v) - (1)/(u) = (1)/(60) - (1)/(-30) = (3)/(60) = (1)/(20)` `f = 20 cm` |
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| 89. |
Explain with reason how the power of a diverging lens changes when (i) it is kept in a medium of refractive by index greater than that of the lens. (ii) incident red light is replaced by violet light. |
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Answer» (i) Using standard notations, power of lens in air `P_(a) = (1)/(f_(a))=((mu_(g))/(mu_(a))-1)((1)/(R_(1)) - (1)/(R_(2)))` when the lens is immersed in a medium `P_(m) = (1)/(f_(m)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` As lens is diverging, `P_(a)` is negative. As `mu_(m) gtmu_(g)`, therefore `P_(m)` becomes positive. Hence power of the same lens immersed in medium becomes positive, i.e., diverging lens behave as converging lens of smaller power. (ii) When incident red light is replaced by violet light, wavelength decreases. Refractive index increases `(mu_(v) gt mu_(r ))`. Power of the lens increases. |
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| 90. |
A double convex lens of `+ 5 D` is made of glass of refractive index `1.5` with both faces of equal radii of curvature. Find the value of curvature. |
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Answer» `P = + 5D, mu = 1.5`, `R_(1) = R, R_(2) = -R` From `(1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))` `5 = (1.5 - 1)((1)/(R ) + (1)/(R )) = (1)/(R )` `R = (1)/(5)m = 20 cm` |
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| 91. |
A convex lens of focal length `25 cm` is placed co-axially in contact with a concave lens of focal length `20 cm`. Determine the power of the combination. Will the system be converging or diverging in nature ? |
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Answer» Here, `f_(1) = 25 cm`, `f_(2) = - 20cm, P = ?` `P_(1) = (100)/(f_(1)) = (100)/(25) = 4 D` `P_(2) = (100)/(f_(2)) = (100)/(-20) = -5 D` `P = P_(1) + P_(2) = 4 - 5 = - 1 D` , diverging. |
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| 92. |
The image of a needle placed `45 cm` from a lens is formed on a screen placed `90 cm` on the other side of lens. Find displacement of image if object is moved `5 cm` away from lens. |
| Answer» Correct Answer - `15 cm` | |
| 93. |
Find the focal length and power of a convex lens, which when placed in contact with a concave lens of focal length `25 cm` forms a real image `5` times the size of the object placed `20 cm` from the combination. |
| Answer» Correct Answer - `10 cm, 10 D` | |
| 94. |
A concave lens is placed in contact with a convex lens of focal length `25 cm`. The combination produces a real image at a distance of `80 cm`, when an object is at a distance of `40 cm`. What is the focal length of concave lens ? |
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Answer» Here, `f_(1) = 25cm, f_(2) = ?` For the combination of focal length `F`, `u = - 40 cm, v = +80 cm`. As `(1)/(F) = (1)/(v) - (1)/(u)` `:. (1)/(F) = (1)/(80) - (1)/(-40) = (1 + 2)/(80) = (3)/(80)` As `(1)/(f_(1)) + (1)/(f_(2)) = (1)/(F)` `:. (1)/(f_(2)) = (1)/(F) - (1)/(f_(1)) = (1)/(80) - (1)/(25) = (15 - 16)/(400)` `(1)/(f_(2)) = -(1)/(400), :. f_(2) = - 400 cm` |
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| 95. |
Do interference effects occur for sound waves ? Recall that sound is a longitudinal mechanical wave while light is transverse and non- mechanical ? |
| Answer» Yes, because interference is a wave phenomenon, which takes place for waves which may be longitudinal or tr ansverse , mechanical or non- mechanical. Waves should be of same type and coherent. | |
| 96. |
Which of the following statements indicates that light waves are transverse?A. interferenceB. diffractionC. dispersionD. polarization |
| Answer» Correct Answer - D | |
| 97. |
A ray of light falls on a transparent slab of `mu = 1.0`. If reflected and refracted rays are mutually perpendicular, what is the angle of incidence ?A. `45^(@)`B. `60^(@)`C. `30^(@)`D. `90^(@)` |
| Answer» Correct Answer - A | |
| 98. |
A ray of light falls on a transparent slab of `mu = 1.732`. If reflected and refracted rays are mutually perpendicular, what is the angle of incidence ? |
| Answer» `60^(@)`, using `tan i_(p) = mu = 1.732 = sqrt(3)` | |
| 99. |
A ray of light falls on a transparent glass slab of refractive index `1.62`. If the reflected ray and the refracted rays are mutually perpendicular, what is the angle of refraction ? |
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Answer» Correct Answer - `31.7^(@)` `tan i_(p) = mu = 1.62` `:. i_(p) = tan^(-1) (1.62) = 58.3^(@)` `r = 90^(@) - i_(p) = 90^(@) - 58.3^(@) = 31.7^(@)` |
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| 100. |
An astronomical telesope has objective and eyepiece of focal lengths `40cm` and `4 cm` respectively. To view an object `200cm` away from the objective, the lenses must be separated by a distance `:`A. `46.0 cm`B. `50 cm`C. `54.0 cm`D. `37.3 cm` |
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Answer» Correct Answer - C Here, `f_(0) = 40 cm, f_(e) = 4 cm, u_(0) = - 200 cm` As `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))` `:. (1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0))` `= (1)/(40) - (1)/(200) = (4)/(200) = (1)/(50)` In normal adjustment, distance between objetive and eye piece `= v_(0) + f_(e) = 50 + 4 = 54 cm` |
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