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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 distance between the slit and the biprism and between the biprims and the screen 50 cm each. The angle of the biprism is `179^(@)` and its refractive index is `1.5`. If the distance between successive bright fringes is `0.0135 cm`, then the wavelenghts of lights isA. `5893xx10^(-10) cm`B. `5898xx10^(-8) cm`C. `5898xx10^(-8) m `D. `2946xx10^(-8) cm` |
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Answer» Correct Answer - B Here, `alpha+179+ alpha=180^(@)` `:. Alpha=0.5^(@)=(0.5xx pi)/(180) "rad"` Thus, `d= 2u (mu-1)alpha` `=2xx0.5(1.5-1)(0.5 xxpi)/(180)` `=0.004361m` `:. Lambda=(d)/(D) beta=(0.004361xx0.00135xx10^(-2))/(1)` `=5893Å=5893xx10^(-8) cm` |
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| 52. |
In a biprism experiment, the wavelenght of red light used is 6850Å and that of violet light used is 4050Å. The value of n for which the `(n+2)^(th)` bright band for violet light would correponded to the `n^(th)` bright band for red light for the same setting isA. 3B. 2C. `2.5`D. 4 |
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Answer» `x_((n+2)bv)=x_(nbr)` `(D)/(d)(n+2)lamda_(v)=(D)/(d)nlamda_(r)` `:.(n+2)lamda_(v)=nlamda_(r)` `4050(n+2)=6750n` `n(6750-4050)=8100` `:.n=(8100)/(2700)=3` |
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| 53. |
In a biprims experiment, the slite is illuminated by red light of wavelenght 6400. A.U. and the crosswire in the eypeice is adjusted to be at the centre of the `3^(rd)` bright band . By using blue light it is found that the `4^(th)` bright band is at the centre of the croswiere, find teh wavelength of blue light.A. 4800 A.U.B. 4500 A.U.C. 5800 A.U.D. 4600 A.U. |
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Answer» Correct Answer - A For interence by thin films, the order of thickness of film is approximately equal to wavelenght of light , `i.e.,~~1(t=10000Å)` |
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| 54. |
The disntance of `n^(th)` brigth band from central band is give byA. `((2n-1)lambda D)/(d)`B. `(n lambda D)/(d)`C. `(n lambda d)/(D)`D. `((2n) lambda D)/(d)` |
| Answer» Correct Answer - B | |
| 55. |
The cause of hearing of sound, produced in one room, in a nearby room isA. interference of sound wavesB. reflection of sounds wavesC. absorption of sound wavesD. diffraction of sound waves |
| Answer» Correct Answer - D | |
| 56. |
If the phase difference between the two light waves arriving at a point is (2n-1) `pi` radians. The condition for the minimumintensity of light at that point will beA. `I_("min") =(a_(1)-a_(2))^(2)`B. `I_("min") =a_(1)-a_(2)`C. `I_("min") =(a_(1)+a_(2))^(2)`D. `I_("min") =a_(1)^(2)-a_(2)^(2)` |
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Answer» Correct Answer - A `I= R^(2)` `=a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2) cos (alpha_(1)-alpha_(2))` `:. I_("min") =a_(1)^(2)a_(2)^(2)=-2a_(1)+a_(2)` `=(a_(1)-a_(2))^(2)` |
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| 57. |
In the experiment of interference of light, the optical paths at a point differ by `73.5` times the wavelenth of blue light of `4400Å` . The nature of the illumination at the same of point using red light of wavelength `6000Å` will beA. brightB. darkC. neither bright nor barkD. none of these |
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Answer» Correct Answer - C Path diff. `n_(b) lambda_(b)=n_(1) lambda_(r)` `=73.5xx4.4xx10^(-7)` `=(323.4xx10^(-7))/(6xx10^(-7)) =53.9` Since, path diff. `=53.9 lambda`, which is not equal to integral multiple of wavelenght or odd multiple of haf of wavelenght thus, given point is neither, bright nor dark. |
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| 58. |
When the interence of light take place, then theA. mass is conservedB. energy is conservedC. intensity at all the points is sameD. intensify at all the points is zero |
| Answer» Correct Answer - B | |
| 59. |
Interference was observed in interference chamber when air was present, now the chamber is evacuated and if the same light is used, a careful observer will seeA. no interferenceB. interference with brights bandsC. interference with dark bandsD. interference in which breadth of the fringer will be slighty increased |
| Answer» Correct Answer - D | |
| 60. |
Thomas Young demonstrated interference of light inA. 1801B. 1901C. 1897D. 1857 |
| Answer» Correct Answer - A | |
| 61. |
Interference of light was first of all experimentally observed byA. NewtonB. HuygenC. YoungeD. Fresnel |
| Answer» Correct Answer - C | |
| 62. |
In a double experiment, the distance between the slit is 1 mm and screen is 25 cm away from the slits. The wavelength of light is `6000Å`. The width of the fringe on the screen isA. `0.15 mm`B. `2.24 mm`C. `0.30` mmD. `0.12 mm` |
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Answer» Correct Answer - A `d=1 mm, D=25 cm, lambda 6000A^(@), beta = ?` `beta =(D)/(d) lambda` `=(25xx10^(-2)xx6xx10^(-7))/(1xx10^(-3))` `=150xx10^(-6) m` `0.15 m` |
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| 63. |
In double slits experiment, for light of which colour the fringe width will be minimumA. violetB. RedC. GreenD. Yellow |
| Answer» Correct Answer - A | |
| 64. |
A coherent light is incident on two paralle slits `S_(1) and S_(2)`. At a point `P_(1)` the frings will be dark if the phase difference between the rays coming from `S_(1) and S_(2)` isA. `n pi ` radiansB. `(n+0.5) pi` radiansC. `(2n+0.5) pi` radiansD. (`2n+1) pi` radians |
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Answer» Correct Answer - D Fow destructive interference, phase difference should be odd multiple of `pi`. |
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| 65. |
In order to avoid overlapping of interference frings and their obliteration (destruction) the slit width must beA. greater than fringe widthB. greater than distance between two slitsC. equal to distance between two slitsD. just less than fringe width |
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Answer» Correct Answer - D If the slite width is large then it consist of large number of point sources. Each parth gives an interference pattern which ovelaps and the fringes gets oblierated. In order. To avoid. This the slite width must be smaller than fringe width. |
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| 66. |
The tip of needle does not give a sharp image. This is due toA. interferenceB. diffractionC. polarisationD. refraction |
| Answer» Correct Answer - B | |
| 67. |
A point p is situated `90.50cm and 90.58` cm away from two coherent sources. The nature of illumination of the point p of the wavelength of light is 400Å is,A. brightB. darkC. neither bright nor barkD. noen of these |
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Answer» Correct Answer - A Path diff. `=9.58-90.50= 0.08 cm ` Now, `n=("Path diff")/("Wavelength")=(0.08xx10^(-2))/(4xx10^(-7))=2000` The pathh diff. `= 2000 lambda`= inegral multiple of wavelenght, point up appears as bright. |
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| 68. |
The path difference between the two indentical waves arrivings at a point is `100.5 lambda` . If the path difference is `44 mu m`, then wavelength of light will beA. `4375Å`B. `4000Å`C. `4738Å`D. `4873Å` |
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Answer» Correct Answer - A Path diff. `100.5Å=44xx10^(-6)` Thus, `lambda=(44xx10^(-6))/(100)=4378Å` |
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| 69. |
For the points of brighness due to two indentical light waves interfering at a point, which of the following is true ?A. crest or through of one wave concidies with the crest or through of another waveB. path difference `=n lambda`C. path oflight is maximumD. all of these |
| Answer» Correct Answer - D | |
| 70. |
Which of the following waves are diffracted by an obstacle of zise 1 cmA. Light wavesB. Sound wavesC. Ultrasonic wavesD. X-rays |
| Answer» Correct Answer - C | |
| 71. |
Which of the following is conserved, when light waves inteffere ?A. FrequencyB. AmplitudeC. ForceD. Momentum |
| Answer» Correct Answer - A | |
| 72. |
The reagion or locus of all dark points produced due to interference of two identical light waves isA. bright band of bright fringsB. dark bond or dark fringC. band widthD. spectrum |
| Answer» Correct Answer - B | |
| 73. |
The centre of the diffraction pattern in Fraunhofer diffraction is alwaysA. balckB. brightC. sometimesdark sometimes brightD. dark for short wave lenghts and brights long wave lengths |
| Answer» Correct Answer - B | |
| 74. |
The condition for observing Fraunhofer diffraction from a single slit is that the light wavefront incident on the slit should beA. shpericalB. cylindericalC. planeD. elliptical |
| Answer» Correct Answer - C | |
| 75. |
For different independent waves are represented by a) `Y_(1)=a_(1)sin omega_(1)t` , b) `Y_(2)=a_(2) sin omega_(2)t` c) `Y_(3)=a_(3) sin omega_(3)t` , d) `Y_(4)=a_(4) sin(omega_(4)t+(pi)/(3))` The sustained interference is possible due toA. (i) and (iii)B. `(i) and (iv)`C. `(iii) and (iv)`D. not possible at all |
| Answer» Correct Answer - D | |
| 76. |
Interfeence phenomenon can take placeA. in all wavesB. in tranverse waves onlyC. in longitudinal wavesD. in standing waves only |
| Answer» Correct Answer - A | |
| 77. |
One important similarity between sound and light is that bothA. can pass throug even in the absence of any mediumB. are transverse wavesC. travel at the same speed in airD. can show interference effect |
| Answer» Correct Answer - D | |
| 78. |
Which of the following undergoe maximum diffraction ?A. `alpha`- raysB. `gamma`- raysC. Radio wavesD. Light waves |
| Answer» Correct Answer - C | |
| 79. |
the penomenon of nonuniform distribution of energy due to the superposition of two light waves isA. reflection of lightB. interference of lightC. diffraction of lightD. polarisation of light |
| Answer» Correct Answer - B | |
| 80. |
If two light waves each of amplitude a travelling a travelling through a medium arrive at a point in opposite phase then the resultant amplitude R at that points isA. `R=0`B. `R=a`C. `R=2a`D. `R=4a` |
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Answer» Correct Answer - A `a_(1)= a_(2)= a and alpha_(1)-alpha_(2)= pi, R=?` `R= sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1) cos (alpha_(1)-alpha_(2)))` `=sqrt(a^(2)+a^(2)+ 2 a_(1)a_(1) cos pi)` `R =0` |
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| 81. |
Intensity of light at point is directly proportional toA. amplitude of waveB. square of amplitude of waveC. cube of amplitude of waveD. wavelength of wave |
| Answer» Correct Answer - B | |
| 82. |
Phenomenon of bending of waves withoug a changed in medium is calledA. reflectionB. rerfractionC. interferenceD. diffraction |
| Answer» Correct Answer - D | |
| 83. |
Two waves of equal amplitude and frequency interfere each other. The ratio of intensity when the two waves arrive in phase to that when they arrive `90^@` out of phase isA. `2:1`B. `1:1`C. `sqrt(2):1`D. `4:1` |
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Answer» Correct Answer - A For same phase, phas diff =0 `:. I_("max")=I_(2)+I_(2) +sqrt(I_(1)I_(2)) 2 cos (alpha_(1)-alpha_(2))` `I+I+2I=4I " "...(i)` For phae diff. `=90^(@)` `I_("min")=I_(1)I_(2)+sqrt(I_(1)I_(2)) 2 cos 90` `=2I` Thus, `(I_("max"))/(I_("min"))=(4I)/(2I)=(2)/(I) " "...(ii)` |
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| 84. |
The phenomenon of producing oalternate points of maximum and minimum intensity due to the superposition of two light waves isA. refraction of lightB. reflection of lightC. interference of lightD. diffraction of light |
| Answer» Correct Answer - C | |
| 85. |
A source of light is said to be monochromatic if it will emit light waves ofA. same wavelengthsB. differerent wavelengthsC. decreasing wavelenghtsD. increasing wavelngths |
| Answer» Correct Answer - A | |
| 86. |
A monochromatic visible light, consiSt ofA. a single ray of lightB. light of a single wavelengthC. light consistaing of many wavelength with a single colourD. light of a single wavelengths with all the colours of the spectrum of white light |
| Answer» Correct Answer - B | |
| 87. |
If the ratio of amplitude of two waves is `4:3`, then the ratio of maximum and minimum intensity isA. `16:9`B. `9:16`C. `1:49`D. `49:1` |
| Answer» Correct Answer - D | |
| 88. |
Monochromatic light hasA. different wavelenghtsB. same wavelengthsC. unidirectionalD. coherent wavelengths |
| Answer» Correct Answer - B | |
| 89. |
Crystalline structure of solids can be studied by using the method ofA. diffractionB. interferenceC. polarisationD. refraction |
| Answer» Correct Answer - A | |
| 90. |
Interference fringes are obtained due to the interference of wave from two coherent sources of light with amplitudes `a_(1)` and `a_(2)(a_(1)=2a_(2))`. The ratio of the maximum and minimum intensities of light in the interference pattern isA. `2:1`B. `4:9`C. `9:1`D. `9:4` |
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Answer» Correct Answer - C `(I_("max"))/(I_("min"))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))=((2a_(2)+a_(2))^(2))/((2a_(2)-a_(2)))=(9)/(1)` |
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| 91. |
If two waves are not coherent, then it obtainedA. steady intereferenceB. no interferenceC. diffused interferenceD. diminished interference |
| Answer» Correct Answer - C | |
| 92. |
Which colour should be used to increases the resolving power of a microscopeA. violetB. redC. yellowD. green |
| Answer» Correct Answer - A | |
| 93. |
Resolving power of a telescope can be increased byA. increasing diameter of the objectiveof the telescopeB. decreasing diameter of the objectiveof the telescopeC. increasin thewavelengths of lightD. none of these |
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Answer» Correct Answer - A `R=(D)/(1.22 lambda)` `R propD` |
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| 94. |
To increse fringe width, by keeping distance between slit and screen constant, we need to ensure thatA. d increases `lambda` constantB. d increases `lambda` decreasesC. d decreases `lambda` decreasesD. d decreases `lambda` increases |
| Answer» Correct Answer - D | |
| 95. |
Resolving power of telescop can be increased byA. larger focal length of eyepieceB. small focal length of eyepieceC. large aperture of the objective lensD. small focal length of the objective lens |
| Answer» Correct Answer - C | |
| 96. |
Distance between screen and source is decreased by 25%. Then the percentage change in fringe width isA. `20%`B. `31%`C. `75%`D. `25%` |
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Answer» Correct Answer - D `D_(1)-D_(12)=25%, lambda_(1)-beta_(2)=?` `D_(1)-D_(2)=(25)/(100)=(1)/(4)` `:. beta prop D` `(beta_(2))/(beta_(1))=(D_(2))/(D_(1))=(3)/(4)(D)/(D)` `beta_(2)=3(3beta_(1))/(4)` `beta_(1)-beta_(2)=beta_(1)(3 beta_(1))/(4)=(beta_(1))/(4)` % change `=((beta_(1)-beta_(1))/(beta_(1)))100=(beta_(1))/( 4xx beta_(1))xx 100=25%` |
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| 97. |
Then nth bright band of red light of wavelength `6750Å` is coincide with `(n+1)^(th)` bright band of green light of wavelength `5400Å`. Find the value of `n`.A. 5B. 4C. 3D. 6 |
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Answer» Correct Answer - B `lambda_(1)=6750Å, lambda_(2) 5400Å, n=?` `n=(lambda_(2))/(lambda_(1)-lambda_(2))` Provided `lambda_(1) gt lambda_(2)` `=(5400)/(6750-5400)=4` |
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| 98. |
The ability of an optical instrument to show the images of two adjacent point as separate is calledA. dispersive powerB. magnifying powerC. resolving powerD. none of these |
| Answer» Correct Answer - C | |
| 99. |
The optical path difference at a point on the screen which is equidistant from two sources isA. zeroB. greater than zeroC. less than zeroD. none of these |
| Answer» Correct Answer - A | |
| 100. |
The limit of resolution of an optical instrument arises on account ofA. interferenceB. diffrationC. polarisationD. none of these |
| Answer» Correct Answer - B | |