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InterviewSolution
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 solar spectrum during a complete solar eclipse isA. ContinuousB. Emission lineC. Dark lineD. Dark band |
| Answer» Correct Answer - a | |
| 52. |
A 20 D lens is used as a magnifier. Where should the object be placed to obtain maximum angular magnification? (Given, D=25 cm).A. 4.17 cmB. 4.7 cmC. 4.07 cmD. None of these |
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Answer» Correct Answer - A Focal length of the lens, `f=(1)/(20)m=(100)/(20)cm = 5 cm` Maximum angular magnification is obtained when final image is formed at D. Hence, `(1)/(-25)-(1)/(-u_(o))=(1)/(5)rArr u_(o)=4.17 cm` |
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| 53. |
The magnifying power of a telescope is `9.` When it is adjusted for parallel rays the distance between the objective and eyepiece is `20cm`. The focal lengths of lenses areA. `18 cm , 2 cm`B. `11 cm, 9 cm`C. `10 cm, 10 cm`D. `15 cm, 5 cm` |
| Answer» Correct Answer - a | |
| 54. |
The magnifying power of a telescope is `9.` When it is adjusted for parallel rays the distance between the objective and eyepiece is `20cm`. The focal lengths of lenses areA. 10 cm, 10 cmB. 15 cm, 5 cmC. 18 cm, 2 cmD. 11 cm, 9 cm |
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Answer» Correct Answer - C Here, `M = (f_0)/(f_e) = 9 :. f_0 = 9 f_e` …..(i) When adjusted for parallel rays, the distance between the objective anf eye piece is `f_0 + f_e = 20 cm` Using (i), `9 f_e + f_e = 20` or `f_e = 2 cm` From (i), `f_0 = 9 xx 2 cm = 18 cm`. |
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| 55. |
The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.A. `31.16`B. `32.16`C. `33.16`D. `34.16` |
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Answer» Correct Answer - A For eye-piece, it is given that `f_(e )=6 cm, v_(e )=-24 cm` `therefore " " (1)/(-24)-(1)/(-u_(e ))=(1)/(6)` `therefore " " u_(e )=4.8 cm` When, L = 9.8 cm, then `v_(o)=(9.8-4.8)cm = 5.0 cm` `therefore " " (1)/(5.0)-(1)/(-u_(o))=(1)/(1.0)rArr u_(o)=1.25 cm` or Magnifying power, `M = (v_(o))/(u_(o))(D)/(u_(e ))=((5.0)/(1.25))((25.0)/(4.8))=20.83` When, L = 11.8 cm, then `v_(o)=(11.8-4.8)cm = 7.0 cm` `therefore " " (1)/(7.0)-(1)/(-u_(o))=(1)/(1.0)` or `u_(o)=1.17 cm` `therefore " " M=(v_(o))/(u_(o)).(D)/(f_(e ))=((7.0)/(1.17))((25.0)/(4.8))=31.16` Therefore, range of magnifying power is from 20.83 to 31.16. |
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| 56. |
A concave mirror of focal length `f` (in air) is immersed in water `(mu=4//3)`. The focal length of the mirror in water will beA. fB. `(4)/(3)f`C. `(3)/(4) f`D. `(7)/(3) f` |
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Answer» Correct Answer - A On immersing a mirror in water, focal length of the mirror remains unchanged as medium does not affect the focal length of a mirror. |
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| 57. |
A small telescope has an objective lens of focal length `140 cm` and eye piece of focal length `5.0 cm`. What is the magnifying power of telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e. when the image is at infinity) (b) the final image is formed at the least distance of distinct vision `(25 cm)`.A. `28,33.6`B. `33.6,28`C. Both (a) and (b)D. None of these |
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Answer» Correct Answer - A i. When the telescope is i9n normal adjustment, the magnifying power is given by `M=(f_(o))/(|f_(e )|)=+(140)/(5)=28` ii. When the final image is formed at the least distance of distinct vision, then M is given by `M = (f_(o))/(|f_(e )|)(1+(f_(e ))/(D))=(140)/(5)(1+(5.0)/(25))=33.6` |
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| 58. |
The ratio of thickness of plates of two transparent medium A and B is 6 : 4. If light takes equal time in passing through them, then refractive index of A with respect to B will beA. 1.4B. 1.5C. 1.75D. 1.33 |
| Answer» Correct Answer - b | |
| 59. |
(a) For the telescope described what is the separation between the objective lens and eye piece ? (b) If this telescope is used to view a `100 m` tall tower `3 km` away, what is the height of the image of the tower formed by the objective lens ? (c ) What is the height of the final image of the tower if it is formed at 25 cm` ? |
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Answer» (a) Here, in normal adjustment, Separation between objective lens and eye lens `= f_0 + f_e = 140 + 5 = 145 cm` (b) Angle subtended by tower `100 m` tall at `3 km` `prop = (100)/(3 xx 1000) = (1)/(30) radian` If `h` is the height of image formed by the objective, then `prop = (h)/(f_0)=(h)/(140)` `:. (h)/(140) = (1)/(30)` or `h = (140)/(30) cm = 4.7 cm` ( c) Magnification of final image `= 4.7 xx 6 = 28.2 cm`. |
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| 60. |
A small telescope has an objective lens of focal length `140 cm` and eye piece of focal length `5.0 cm`. What is the magnifying power of telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e. when the image is at infinity) (b) the final image is formed at the least distance of distinct vision `(25 cm)`. |
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Answer» Here, `f_0 = 140 cm, f_e = 5.0 cm` Magnifying power = ? (a) In normal adjustment, Magnifying power `= (f_0)/(-f_e) = (140)/(-5) = - 28` (b) When final image is at the least distance of distinct vision, Magnifying power `= (-f_0)/(f_e) (1 + (f_e)/(d)) = (140)/(-5) (1 + (5)/(25)) = -33.6`. |
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| 61. |
The refractive indices of water and glass with respect to air are 1.2 and 1.5 respectively. The refractive index of glass with respect to water isA. 0.6B. 0.8C. 1.25D. 1.75 |
| Answer» Correct Answer - c | |
| 62. |
The refractive indices of water and glass with respect to air are 1.2 and 1.5 respectively. The refractive index of glass with respect to water isA. `2.6/1.5`B. `1.5/2.6`C. `1.3/1.5`D. `1.5/1.3` |
| Answer» Correct Answer - d | |
| 63. |
If refractive indices of glass and water with respect to air are `3//2` and `4//3` respectively, what is the refractive index of glass with respect to water ? |
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Answer» Here, `._(g)mu^(a) = (3)/(2)` and `._(w)mu^(a) = (4)/(3)` `._(g)mu^(w) = (._gmu^(a))/(._wmu^a) = (3//2)/(4//3) = (9)/(8)`. |
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| 64. |
The focal length of objective and eye lens of a microscope are `4 cm` and `8 cm` respectively. If the least distance of distinct vision is `24 cm` and object distance is `4.5 cm` from the objective lens, then the magnifying power of the microscope will beA. 18B. 32C. 64D. 20 |
| Answer» Correct Answer - b | |
| 65. |
An astronomical telescope arranged for normal adjustment has a magnification of 6. If the length of the telescope is 35cm, then the focal lengths of objective and euye piece respectivley =, areA. 30cm,5cmB. 5cm,30cmC. 40cm,5cmD. 30cm,6cm |
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Answer» Correct Answer - A In normal adjustment,the object and the final image both are at infinity and the separation between the objective and the eye pieces si `f_(u)+f_(e)` Thus , `f_(u)+f_(e)=35cm…..(i)` Also, the magnifying power of the telescope in normal adjustment is `m=-(f_(u))/(f_(e))=-6 ("give")` `f_(u)=6f_(e) .....(ii)` On solving Eqs. (i) and (ii) we get `f_(v)=30cm and f_(e)=5cm` |
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| 66. |
Magnification of a compound microscope is 30. Focal length of eye`-` piece is `5 cm` and the image is formed at a distance of distinct vision of `25 cm`. The magnificatio of the objective lens isA. 6B. 5C. `7.5`D. 10 |
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Answer» Correct Answer - B Here, `m = 30, f_e = 5 cm, d = 25 cm` As `m_e = 1 + (d)/(f_e) = 1 + (25)/(5) = 6` As `m = m_e xx m_0 :. 30 = 6 xx m_0` `m_0 = (30)/(6) = 5`. |
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| 67. |
In a compound microscope magnification will be large, if the focal length of the eye piece isA. LargeB. SmallerC. Equal to that of objectiveD. Less than that of objective |
| Answer» Correct Answer - b | |
| 68. |
A beam of light consisting of red, green and blue colours is incident on a right angled prism, fig. 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 ` |
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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 crictical 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 = 1.424`, `sin C = (1)/(mu) = (1)/(1.424) = 0.702, C = 44.6^@` `:.` green ray will also 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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| 69. |
Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle 30degree at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is A. 28B. 30C. 32D. 34 |
| Answer» Correct Answer - B | |
| 70. |
For having large magnification power of a compound microsopeA. length of the microscope tube must be smallB. focal length of objective lens and eye piece should be largeC. focal length of the objective lens and eye piece should be smallD. focal length of eye piece must be smaller than the focal length of objective lens |
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Answer» Correct Answer - D In order to increase the magnifying power of a compound microscope, the object should have large focal length and eye piece should have small focal length. |
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| 71. |
A beam of light consisting of red, green, and blue colors is incident on a right-angled prism. 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, repectively. The prism will A. Separate part of the blue colour from the red and greenB. Separate part of the red colour from the greem and blue coloursC. Separate all the three colours from one anotherD. None of these |
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Answer» Correct Answer - B The critical angles `[C=sin^(-1)(1)/(n)]` for need (n=1.39), green (n=1.44) and blue (n=1.47) lights are `46^(@) , 44^(@) and 43^(@)` respectively. All colous will strike the hypotenuse face at `45^(@)`. Hence green and blue rays will totally reflected while red rays will get through. |
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| 72. |
A beam of light consisting of red, green, and blue colors is incident on a right-angled prism. 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, repectively. The prism will A. separate part of the rod colour from the green and blue colours.B. separate all the three colours from the other two colors.C. not separate even partially any colour from the other two colors.D. separate part of the rod colour from the green and blue colours. |
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Answer» Correct Answer - A Critical angle `sinC=(1)/(mu)` so for`C_(R)gt45^(@),C_(G)lt45^(@),C_(B)lt45^(@)` So total internal reflection will be for`G&B`. |
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| 73. |
Critical angle of glass is `theta_(1)` and that of water is `theta_(2)`. The critical angle for water and glass surface would be `(mu_(g)=3//2, mu_(w)=4//3)`A. less than `theta_(2)`B. between `theta_(1)` and `theta_(2)`C. greater than `theta_(2)`D. less than `theta_(1)` |
| Answer» Correct Answer - C | |
| 74. |
A small angled prism of refractive index 1.4 is combined with another small angled prism of refractive index 1.6 to produce disperison without deviation. If the angle of first prism is `6(@)`, then the angle of the second prism isA. `8^(@)`B. `6^(@)`C. `4^(@)`D. `2^(@)` |
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Answer» Correct Answer - C We know about the thin prism, `delta_(m)=(mu-1)A` In given condition, `A_(1)(mu_(1)-1)A_(2)(mu_(2)-1)` Given, `mu_(1)=1.4, mu_(2)=1.6 and A_(1)=6^(@)` `"Now", 6(1.4-1)=A_(2)(1.6-1)` `A_(2)=(6xx0.4)/(0.6) Rightarrow A_(2)=(2.4)/(0.6)` `A_(2)=4^(@)` Hence, the angle of the second prism is `4^(@)`. |
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| 75. |
Two lenses of power `+ 10 D and - 5 D` are placed in contact, (i) Calculate the focal length of the combination (ii) where should an object be held from the combination so as to obtain a virtual image of magnification `2` ? |
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Answer» Correct Answer - (i) 20 cm (ii) `-10 cm` (i) Here, `P_1 = + 10 D, P_2 = -5 D` `P = P_1 + P_2 = + 10 - 5 = 5 D` `F = (100)/(P) cm = (100)/(5) = 20 cm` (ii) `m = + 2`, as image is virtual From `m = (f)/(u + f)` `2 = (20)/(u + 20)` `2 = (20)/(u + 20)` `2u + 40 = 20` `u = (20 - 40)/(2) = -10 cm`. |
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| 76. |
A small air bubble in a glass sphere of radius `2 cm` appears to be `1 cm` from the surface when looked at, along a diameter. If the refractive index of glass is `1.5`, find the true position of the air bubble. . |
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Answer» Correct Answer - `1.2 cm` Here, `mu_1 = 1, mu_2 = 1.5, R = -2 cm` Incident ray `OA` in glass is refracted in air, along `AB`, and appears to come from `I` `u = PO = ? B = PI = -1 cm` As refractive occurs from denser to rarer medium, `:. -(mu_2)/(u) + (mu_1)/(v) = (mu_1 - mu_2)/( R)` `-(1.5)/(u)+(1)/(-1) = (1 - 1.5)/(-2) = (1)/(4)` `(1.5)/(u) = - 1- (1)/(4) = -(5)/(4)` `u = (-4 xx 1.5)/(5) = -1.2 cm` The air bubble `O` lies at `1.2 cm` from the refracting surface within the sphere. |
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| 77. |
What is the ratio of velocities of two light waves travelling in vacuum and having wavelength `4000Å and 8000Å` ? |
| Answer» The ratio is one. This is because in vacuum, all colours travel with the same velocity. | |
| 78. |
Assertion `:` Withing a glass slab, a double convex air bubble is formed. This air bubble behaves like a converging lens. Reason`:` Refrative index of air is more than the refractive index of glass.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
| Answer» Correct Answer - d | |
| 79. |
refractive indices of glass for blue, red and yellow colours are `mu_b,mu_r and mu_y`. Write them in decreasing order of values. |
| Answer» Values of `mu` in decreasing order are `mu_b,mu_y,mu_r`. | |
| 80. |
If `mu_(v)=1.530 and mu_(R)=1.5145`, then disperisve power of a crown glass isA. 0.0164B. 0.00701C. 0.0132D. 0.032 |
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Answer» Correct Answer - A Here, `mu_(V)=1.5230 and mu_(R)=1.5145` Therefore, means refractive index. `mu=(1.523+1.5145)/(2)=1.5187` Now, disperisive power, `omega` is given by `omega=(mu_(V)-mu_(R))/(mu-1)=(1.5230-1.5145)/(1.51787-1)=0.0164` |
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| 81. |
The relative luminosity of wavelength `600 nm` is 0.6. Find the radiant flux of `600 nm` needed to produce the same brightness sensation as produced by `120 W` of radiant flux at `555 nm`A. `50 W`B. `72 W`C. `120 xx (0.6)^(2)W`D. `200 W` |
| Answer» Correct Answer - D | |
| 82. |
Considering normal incident of ray, find equivalent refractive index of combination of two slabs shown in figure. A. `1.8`B. `1.43`C. 2D. none of these |
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Answer» (b) The equivalent refractive index of combination of slab for normal incidence is `mu=(Sigma t_(l))/(Sigma (t_(l))/(mu_(l)))` In this case` mu=(t_(1)+t_(2))/((t_(1))/(mu_(1))+(t_(2))/(mu_(2)))=(10+15)/(10/(4/3)+15/(3/2))` `=25/(7.5+10)=25/(17.5=1.43` |
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| 83. |
A luminous point object is moving along the principal axis of a concave mirror of focal length 12cm towards it. When its distance from the mirror is 20cm its velocity is `4cm//s`. The velocity of the image in `cm//s` at that instant isA. `6`towards the mirrorB. `6` away form the mirrroC. `9`away form the mirrorD. `9`towards the mirror. |
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Answer» Correct Answer - C `(1)/(v)+(1)/(u)=(1)/(f) rArr(1)/(v)+(1)/(-20)=(1)/(-12) rArr v=-30cm` Differentiling equation`(1)/(v)+(1)/(u)=(1)/(f)`w,r.t.time `-(1)/(v^(2))(dv//dt)-(1)/(u^(2))(du//dt)=0` ` dv//dt=-((v)/(u))^(2)xxdu//dt rArr =-(30/(20))^(2) xx4=9cm//s` `dv//dt=9cm//s`a way from the mirror. |
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| 84. |
The magnification produced by the objective lens and the eye lens of a compound microscope are 25 and 6 respectively. The magnifying power of this microscope isA. 19B. 31C. 150D. `sqrt(150)` |
| Answer» Correct Answer - c | |
| 85. |
An object is present on the principal axis of a concave mirror at a distance `30cm`from it.Focal length of mirror is`20cm` If object starts moving with `2cms^(-1)`perpendicular to principal axis above the principal axis then,A. image moves with velocity`4cms^(-1)`below the principal axis.B. image moves with velocity`4cms^(-1)`above the principal axis.C. image moves with velocity`8cms^(-1)`below the principal axis.D. image moves with velocity`8cms^(-1)`above the principal axis. |
| Answer» Correct Answer - A | |
| 86. |
An object is present on the principal axis of a concave mirror at a distance `30cm`from it.Focal length of mirror is`20cm` If object starts moving with `2cms^(-1)` along principal axis towards the mirror thenA. image starts moving with`8cms^(-1)`away from the mirrorB. image starts moving with`8cms^(-1)` towards the mirrorC. image starts moving with`4cms^(-1)`towards the mirrorD. image starts moving with`4cms^(-1)`away from the mirror |
| Answer» Correct Answer - A | |
| 87. |
The plane faces of two identical planoconvex lenses each having focal length of `40 cm` are pressed against each other to form a usual convex lens. The distance from this lens, at which an object must be placed to obtain a real, inverted image with magnification one isA. 80 cmB. 40 cmC. 20 cmD. 162 cm |
| Answer» Correct Answer - b | |
| 88. |
A parallel beam of light emerges from the opposite surface of the sphere when a point source of light lies at the surface of the sphere. The refractive index of the sphere isA. `3/2`B. `5/3`C. `2`D. `5/2` |
| Answer» Correct Answer - c | |
| 89. |
The plane faces of two identical plano convex lenses, each with focal length f are pressed against each other using an optical glue to form a usual convex lens. The distance from the optical centre at which an object must be placed to obtain the image same as the size of object isA. `f/4`B. `f/2`C. `f`D. `2 f` |
| Answer» Correct Answer - c | |
| 90. |
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. 5 RB. 3 RC. 2 RD. `1.5 R` |
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Answer» Correct Answer - A Here, `mu_1 = 1, mu_2 = 1.5`, Let `PO = OQ =x` As light is travelling from Rerer to Denser medium, `:. (mu_1)/(-u) + (mu_2)/(v) = (mu_2 - mu_1)/( R)` `:. (1)/(-(-x)) + (1.5)/(x) = (1.5 - 1)/( R)` `(1)/(x) + (1.5)/(x) = (0.5)/( R)` or `(2.5)/(x) = (0.5)/( R)` or `x = 5 R`. |
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| 91. |
A mark placed on the surface of a sphere is viewed through glass from a position directly opposite. If the diameter of the sphere is `10 cm` and refractive index of glass is `1.5`, find the position of the image. |
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Answer» Correct Answer - 20 cm towards mark from the surface opposite to mark |
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| 92. |
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. 5 RB. 3 RC. 2 RD. 1.5 R |
| Answer» Correct Answer - A | |
| 93. |
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 `10cm,` 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 = ?` Refraction from rarer to denser medium, `-(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) , (1)/(30) , f = 40 cm` `P = 100//f = (100)/(40) = 2.5 D`. |
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| 94. |
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 m`, 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 to denser medium, therefore `(-mu_1)/(u) + (mu_2)/(v) = (mu_2 - mu_1)/( R)` `(-1)/(-60)+(1.5)/(v) = (1.5 - 1)/(25)` `(3)/(2 v) = (1)/(50) - (1)/(60) = (1)/(300)` `v = (300 xx 3)/(2) = 450 cm` As `v` is positive, image formed on the other side of the object, i.e., in refracting denser medium at `45 cm` from the pole. Power of the refracting surface, `P = (mu_2 - mu_1)/( R) = (1.5 - 1)/(0.25) = (0.5)/(0.25) = 2 D`. |
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| 95. |
A beam of light of wavelength `400 nm` is incident normally on a right angled prism as shown in Fig. It is observed that light just grazes along the surface `AC` after falling on it. If refractive index `mu` of the material of prism varies with wavelength `lamda` as `mu = 1.2 + (b)/(lamda^2)` Calculate the value of `b and mu` of prism material for `lamda = 500 nm`. Given `theta = sin^-1 (0.625)`. . |
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Answer» As the ray goes grazingly along `AC, theta` must be the critical angle. From `mu = (1)/(sin C) = (1)/(sin theta) = (1)/(0.625) = 1.6` As `mu = 1.2 + (b)/(lamda^2)` `1.6 = 1.2 + (b)/((400)^2` `b = 0.4 xx (400)^2 = 64000 nm^2` For `lamda = 500 nm`, the refractive index of the material of the prism is `mu = 1.2 + (b)/(lamda^2) = 1.2 + (64000)/((500)^2) = 1.2 + 0.256` `mu = 1.456`. |
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| 96. |
Assertion `:` There is no dispersion of light refracted through a rectangular glass slab. Reason `:` Dispersion of light is the phenomenon of splitting of a beam of white light into its constituent colours.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B 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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| 97. |
The time required for the light to pass through a glass slab ( refractive index=1.5) of thickness u mm is (c=`3xx10^(8)ms^(-1)` speed of light in free space)A. `10^(-11)`sB. `2xx10^(-11)`sC. `2xx10^(11)`sD. `2xx10^(-5)`s |
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Answer» Correct Answer - B (b) we have `n_(a)c_(a)=n_(g)c_(g)` `n_(g)/n_(a)=c_a/c_g` `3/2=(3xx10^(8))/c_(g)` `c_(g)=2xx10^(8)` Time = `"Distance"/"Speed"` `t=(4xx10^(-3))/(2xx10^(8))` `t=2xx10^(-11)s` |
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| 98. |
What is the time taken (in seconds ) to cross a glass of thickness 4 mm and `mu= 3` by lightA. `4xx10^(-11)`B. `2xx10^(-11)`C. `16xx10^(-11)`D. `8xx10^(-10)` |
| Answer» Correct Answer - a | |
| 99. |
Electromagnetic radiation of frequency `n`, wavelength `lambda`, travelling with velocity v in air, enters a glass slab of refractive index `mu`. The frequency, wavelength and velocity of light in the glass slab will be respectivelyA. `n/mu, lambda/mu, v/mu`B. `n, lambda/mu, v/mu`C. `n, lambda, v/mu`D. `n/mu, lambda/mu, v` |
| Answer» Correct Answer - b | |
| 100. |
A plane mirror placed at the origin has `hat(i)` as the normal vector to its reflecting surface. The mirror beings to translate with a velocity `hat(i) + hat(j) + hat(k)`. At the same time an object which was initially at `hat(i) + hat(j)` starts moving with a velocity `(hat(i)+hat(j)) m//s` Now choose the correct options.A. Initial position of the image will be `-hat(i)+hat(j)`B. The velocity of the image will be `hat(i)+hat(j)`C. The velocity of the imahe relative to the object will be zeroD. The velocity of the image relative to the mirror will be `-hat(k)` |
| Answer» Correct Answer - A::B::C::D | |