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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. |
A beam of light consisting of wavelengths 6000 `Å` and 4500 `Å` is used in a YDSE with D = 1 m and d = 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide. |
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Answer» `beta_(1) = (lambda_(1) D)/(d) = (6000 xx 10^(-10) xx 1)/(10^(-3)) = 0.6 mm` `beta_(2) = (lambda 2 D )/(d) = 0.45 mm` Let `n_(1) ^(th)` maxima of `lambda_(1)` and `n_(2) ^(th)` maxima of `lambda_(2)` coincide at a position y. Then, `y = n_(1) P_(1) = n_(2) P_(2) = LCM` of `beta_(1)` and `beta_(2)` `implies y = LCM` of `0.6 mm` and `0.45mm` or `y = 1.8 mm` At this point,`3^("rd")` maxima for `6000 Å` and `4^("th")` maxima for `4500 Å` coincide. |
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| 52. |
In a YDSE arrangement, incident yellow light is composed of two wavelength `5890 Å` and `5895 Å`. Distance between the slits is 1 mm and the screen is placed 1 m away. Order upto which fringes can be seen on the screen will be :A. 384B. 486C. 512D. 589 |
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Answer» Correct Answer - B |
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| 53. |
The fringe width in YDSE is `2.4 xx 10^(-4)m`, when red light of wavelength `6400 Å` is used. By how much will it change, if blue light of wavelength `4000 Å` is used ?A. `9 xx10^(-4)m`B. `0.9 xx10^(-4)m`C. `4.5 xx10^(-4)m`D. `0.45 xx10^(-4)m` |
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Answer» Correct Answer - B Here, `beta_(1)=2.4 xx10^(-4)m,` ` lambda_(1)=6400 Ã…, lambda_(2)=4000 Ã…` ` :. (beta_(2))/(beta_(1))=(lambda_(2))/(lambda_(1))=(4000)/(6400)=(5)/(8)` or `beta_(2)=(5)/(8)xx beta_(1)=(5)/(8)xx2.4xx10^(-4)=1.5xx10^(-4)m` Decrease in fringe width `triangle beta=beta_(1)-beta_(2)=(2.4-1.5)xx10^(-4)=0.9xx10^(-4)m` |
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| 54. |
Light of wavelength `lambda = 5890 Å` fall on a double-slit arrangement having separation `d = 0.2 mm`. A thin lens of focal length `f = 1 m` is placed near the slits. The linear separation of fringes on a screen placed in the focal plane of the lens isA. 3 mmB. 4 mmC. 2 mmD. 1 mm |
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Answer» Correct Answer - a. `beta = (Delta lambda)/(d) = (f lambda)/(d) = (1 xx 4890 xx 10^(-10))/(0.2 xx 10^(-3))` `= 0.29 xx 10^(-2) m = 2.9 mm ~~ 3 mm` |
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| 55. |
In YDSE performed with wavelength `lambda = 5890 Å` the angular fringe width is `0.40^(@)`. What is the angular fringe width if the entire set-up is immersed in water? |
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Answer» Angular fringe width is given by `beta theta = (lambda)/(d)` `beta_(theta)^("air") = (lambda_("air"))/(d) , beta_(theta)^("water") = (lambda_("water"))/(d)` `(beta_(theta)^("air"))/(beta_(theta)^("water")) = (lambda_("air"))/(lambda_("water")) = (n_("air"))/(n_("water")) = (3)/(4)` `:. beta_(theta)^("water") = (3)/(4) beta_(theta)^(beta) = 0.40^(@) xx (3)/(4) = 0.30^(@)` |
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| 56. |
A parallel beam of sodium light of wavelength `6000 Å` is incident on a thin glass plate of `mu=1.5`, such that the angle of refraction in the plate is `60^(@)`. The smallest thickness of the plate which will make it appear dark by reflected light isA. 3926 Ã…B. 4353 Ã…C. 1396 Ã…D. 1921 Ã… |
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Answer» Correct Answer - A The condition for minimum thickness corresponding to a dark band is, `2mu t" cos"r=lambda` ` :. t=(lambda)/(2mu"cos"r)=(5890xx10^(-10))/(2xx1.5xx"cos"60^(@))` ` " "=3926xx10^(-10)m=3926Ã…` |
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| 57. |
A thin mica sheet of thickness `4xx10^(-6)` m and refractive index `(mu=1.5)` is introduced in the path of the light from upper slit. The wavelength of the wave used is `5000 Å` The central bright maximum will shiftA. 4 fringes upwardB. 2 fringes downwardC. 10 fringes upwardD. None of these |
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Answer» Correct Answer - A On introducing mica sheet, fringe shift `=beta/lambda(mu-1)t` where, t is thickness of sheet So, shift `beta/((5000xx10^(-10)))xx(1.5-1)xx(4xx10^(-6))=4beta` So, central bright maximum will shift 4 fringes upward |
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| 58. |
The interference fringes for sodium light `(lambda = 5890 Å)` in a double slit experiment have an angular width of `0.2^(@)`. For what wavelength will width be `10%` greater.A. `5892Å`B. `4000Å`C. `8000Å`D. `6479Å` |
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Answer» Correct Answer - A |
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| 59. |
A double slit arrangement produces fringes for light `lambda=5890Å` which are `0.2^@` apart. If the whole arrangement is fully dipped in a liquid of `mu=4/3`, the angular fringes separation isA. (a) `0.15^@`B. (b) `0.30^@`C. (c) `0.10^@`D. (d) `0.25^@` |
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Answer» Correct Answer - A New wavelength is `3/4` times the old wavelength. So, the new angular separation is `3/4xx0.2^@` or `0.15^@`. |
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| 60. |
A thin mica sheet of thickness `2xx10^-6m` and refractive index `(mu=1.5)` is introduced in the path of the first wave. The wavelength of the wave used is `5000Å`. The central bright maximum will shiftA. two fringes upwardB. two fringes downwardC. ten fringes upwardD. None of these |
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Answer» Correct Answer - A On introducing mica sheet, fringe shift `=beta/lambda(mu-1)t` where, t is thickness of sheet So, shift `=beta/((5000xx10^(-10)))xx(1.5-1)xx(2xx10^(-6))=2beta` So, central bright maxima will shift two fringes upward. |
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| 61. |
A thin mica sheet of thickness `2xx10^-6m` and refractive index `(mu=1.5)` is introduced in the path of the first wave. The wavelength of the wave used is `5000Å`. The central bright maximum will shiftA. 1B. 2C. 5D. 10 |
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Answer» Correct Answer - D |
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| 62. |
A thin mica sheet of thickness `2xx10^-6m` and refractive index `(mu=1.5)` is introduced in the path of the first wave. The wavelength of the wave used is `5000Å`. The central bright maximum will shiftA. (a) 2 fringes upwardB. (b) 2 fringes downwardC. (c) 10 fringes upwardD. (d) None of these |
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Answer» Correct Answer - A Shift `=beta/lambda(mu-1)t` `=(beta)/((5000xx10^(-10)))xx(1.5)xx2xx10^-6=2beta`, i.e., 2 fringes upwards. |
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| 63. |
A plate of thickness t made of a material of refractive index `mu` is placed in front of one of the slits in a double slit experiment. (a) Find the changes in he optical path due to introduction of the plate. (b) Wht should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero ? Wavelength of the light used is `lamda`. Neglect any absorption of light in the plate.A. `(mu-1)(lambda)/(2)`B. `(mu-1)lambda`C. `(lambda)/(2(mu-1))`D. `(lambda)/((mu-1))` |
| Answer» Correct Answer - C | |
| 64. |
A plate of thickness t made of a material of refractive index `mu` is placed in front of one of the slits in a double slit experiment. (a) Find the changes in he optical path due to introduction of the plate. (b) Wht should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero ? Wavelength of the light used is `lamda`. Neglect any absorption of light in the plate.A. `(mu - 1) (lambda)/(2)`B. `(mu - 1) lambda`C. `(lambda)/(2(mu - 1))`D. `(lambda)/((mu - 1))` |
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Answer» Correct Answer - c Intensity at the center will be zero if path difference is `lambda // 2`. Thus is, `(mu - 1) t = (lambda)/(2)` or `t = (lambda)/(2(mu - 1)` |
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| 65. |
In Young’s double slit experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate having same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength` 5400overset(@)A`. It is found that the point P on the screen where the central maximum fell before the glass plates were inserted, now has ` 3/ 4` the original intensity. It is also observed that what used to be `5^(th)` maximum earlier, lies below the point P while the`6_(th)` minimum lies above P. Calculate the thick- ness of glass plates. Absorption of light by glass plate may be neglected. |
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Answer» Correct Answer - `9.3mu"m"` |
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| 66. |
A double slit arrangement produces interference fringes for sodium light `(lambda=589nm)` that have an angular separation of `3.50xx10^-3` radian. For what wavelength would the angular separation be 10% greater?A. (a) `527nm`B. (b) `648nm`C. (c) `722nm`D. (d) `449nm` |
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Answer» Correct Answer - B Angular separation `=lambda/d` For angular separation to be 10% greater `lambda` should be 10% greater. New wavelength is `(589+(589)/(10))nm` or `(589+58.9)nm`, i.e., `647.9nm`, i.e., 648nm` |
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| 67. |
In Young’s double slit experiment a monochromatic light of wavelength` lambda` from a distant point source is inci- dent upon the two identical slits.The interference pattern is viewed on a distant screen. Intensity at a point P is equal to the intensity due to individual slits (equal to` I_(0)`). A thin piece of glass of thickness t and refractive index `mu` is placed in front of the slit which is at larger distance from point P, perpendicular to the light path. Assume no absorption of light energy by the glass. (a) Write intensity at point P as a function of t. (b) Write all values of t for which the intensity at P is minimum. |
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Answer» Correct Answer - (a)`4I_(0)cos^(2)((pi)/(3)+(pi)/(lambda)(mu-1)t)` (b)`t=((2n-1)lambda)/(2(mu-1))-(lambda)/(3(mu-1))"wheren=1,2,3...` |
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| 68. |
In Young’s double slit experiment a transparent sheet of thickness t and refractive index `mu` is placed in front of one of the slits and the central fringe moves away from the central line. It was found that when temperature was raised by` Deltatheta`the central fringe was back on the central line (at C). It is known that temperature coefficient of linear expan- sion of the material of the transparent sheet is`alpha`. A young scientist modeled that the refractive index of the material changes with temperature as `Deltamu"= – gama Deltatheta`. Find `Deltatheta` in terms of other given quantities. D and d are given and have usual meaning. |
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Answer» Correct Answer - `Deltatheta=[(mu-1)/((mu-1)alpha-gamma)]` |
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| 69. |
The figure represents two identical slits in a Young’s double slit experiment. The width of each slit is b and distance between the centres of the two slits is d. Consider a point P on the screen that is close to centre of the screen. `Delta x_(t)` represents the optical path difference to point P from the top edges of the two slits and `Delta x_(b)` represents the optical path difference to P from the bottom edges of the two slits. Find` Delta x_(b) – Delta x_(t)`. It is given that `D gtgt d.` |
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Answer» Correct Answer - `(db)/(D)` |
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| 70. |
Interference fringes are produced by a double slit arrangement and a piece of plane parallel glass of refractive index 1.5 is interposed in one of the interfering beam. If the fringes are displaced through 30 fringe widths for light of wavelength `6xx 10^(-5) cm`, find the thickness of the plate. |
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Answer» Path difference due to the glass slab `Deltax=(mu-1)t` Thirty fringes are displaced due to the slab Hence `Deltax= 30lambda` `rArr (mu-1)t=30lambda` `:. T=(30lambda)/(mu-1)=(30xx6xx10^(-5))/(1.5-1)=3.6xx10^(-3)cm` |
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| 71. |
In Young’s double-slit experiment, the separation between two slits is`" d = 0.32 mm "`and the wavelength of light used is` lambda = 5000overset(@)A`. Find the number of maxima in the angular range`-sin^(-1)(0.6)lethetalesin^(-1)(0.6)` |
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Answer» Correct Answer - 769 |
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| 72. |
Coherent light of wavelength` lambda" = 500 nm"` is sent through two narrow parallel slits in a large vertical wall. The two slits are 5 `mu`m apart. In front of the wall there is a semi cylindrical screen with its hori- zontal axis at the line running on the wall parallel to the slits and midway between them. Radius of the cylin- drical screen is `"R = 2.0 m."` Find the vertical height of the second order interference maxima from the centre (O) of the screen. |
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Answer» Correct Answer - 0.4m |
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| 73. |
Assertion: To observe diffraction of light the size of obstacle/aperture should be of the order of `10^-7m`. Reason: `10^-7m` is the order of wavelength of visible light.A. (a) If both assertion and reason are true and reason is the correct explanation of assertion.B. (b) If both assertion and reason are true but reason is not the correct explanation of assertion.C. (c) If assertion is true but reason is false.D. (d) If assertion and reason both are false. |
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Answer» Correct Answer - A For diffraction to occur, the size of an obstacle/ aperture is comparable to the wavelength of light wave. The order of wavelength of light wave is `10^-7m` so diffraction occurs. |
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| 74. |
The angle between pass axis of polarizer and analyser is `45^(@)`. The percentage of polarized light passing through analyser isA. `75%`B. `25%`C. `50%`D. `100%` |
| Answer» Correct Answer - B | |
| 75. |
When a transparent parallel plate of uniform thickness `t` and refractive index `mu` is interposed normally in the path of a beam of light, the optical path isA. `mu t`B. `mu//t`C. `(mu-1)t`D. None of these |
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Answer» Correct Answer - A |
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| 76. |
In a YDSE, fringes are produced by monochromatic light of wavelength 5450Å. A thin plate of glass of refractive index 1.5 is placed normally in the path of one of the intefering beams and the central bright band of the fringe system is found to move into the position previoysly occupied by the third band from the centre. select the correct alternatieA. The thickness of the plate is `3.27xx10^(-6)m`B. If the separation between the sources is `1000Å` , the angular position of first maxima is 0.01 radC. The thickness of the plate is `1.59xx10^(-6)m`D. A maxima is still obtained at the centre of screen |
| Answer» Correct Answer - A::D | |
| 77. |
When a transparent parallel plate of uniform thickness `t` and refractive index `mu` is interposed normally in the path of a beam of light, the optical path isA. `(mu-1)t`B. `(mu+1)t`C. `mu t`D. `(mu)/(t)` |
| Answer» Correct Answer - A | |
| 78. |
In young’s double slit experiment (with identical slits) the intensity of a maxima is I. P is a point on the screen where` 10^(th)` maxima is formed with light of wave- length `lambda = 6000 overset(@)A`. Find the intensity at point P if the entire experimental set up is submerged in water of refractive index `mu = (4) /( 3)` . Assume that intensity due to individual slits remains unchanged after the system is dipped in water. |
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Answer» Correct Answer - `(I)/(4)` |
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| 79. |
Calculate the minimum thickness of a soap bubble film `(mu=1.33)`that results in constructive interference in the reflected light if the film is illuminated with light whose wavelength in free space is `lambda=600nm`. |
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Answer» For constructive interference in case of soap film, `2mut = (n-1/2)lambda (when, n=1,2,3,…..)` For minimum thickness `t, n=1 or 2mut=lambda//2` The minimum thickness of a soap bubble, `t=lambda/(4mu) = (600)/(4xx1.33)=112.78 nm` |
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| 80. |
Two slits are one millimeter apart and the screen is placed one meter away. What should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single pattern. |
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Answer» We have `a theta=lambda (or) theta= (lambda)/(a)` (a = width of each slit) `10 (lambda)/(d)= 2(lambda)/(a)` `:. a= (d)/(5) = (1)/(5)= 0.2 mm` |
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| 81. |
To observe diffraction, the size of the obstacleA. should be `lambda//2`, where `lambda` is the wavelength.B. should be of the order of wavelength.C. has no relation to wavelength.D. should be much larger than the wavelength. |
| Answer» Correct Answer - B | |
| 82. |
Two parallel slits `0.6mm` apart are illuminated by light source of wavelength `6000Å`. The distance between two consecutive dark fringe on a screen `1m` away from the slits isA. (a) `1mm`B. (b) `0.01mm`C. (c) `0.1m`D. (d) `10m` |
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Answer» Correct Answer - A Distance between two consecutive dark fringes `(lambdaD)/(d)=(6000xx10^(-10)xx1)/(0.6xx10^-3)` `=1xx10^-3m=1mm` |
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| 83. |
When a thin transparent plate of Refractive Index `1.5` is introduced in one of the interfearing beam produces 20 fringes shift. If it is replaced by refractive index `1.7` , the number of fringes that undergo displacement isA. 23B. 14C. 28D. 7 |
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Answer» Correct Answer - B Shift `= (beta)/(lambda) (mu - 1)t` |
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| 84. |
What should be the width of each slit to obtain `10` maxima of the double slit interference pattern within the central maximum of single slit diffraction pattern ? (NCERT Solved example) |
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Answer» We want `atheta=lambda,theta=(lambda)/(a)` `10 (lambda)/(d)=2(lambda)/(a)a=(d)/(5)=0.2 nm` Notice that the wavelength of light and distance of the screen do not enter in the calculation of a. |
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| 85. |
If the distance between the first maxima and fifth minima of a double-slit pattern is 7 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, then wavelength of the light used isA. 600 nmB. 525 nmC. 467 nmD. 420 nm |
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Answer» Correct Answer - A For minima `y_(n)=(2n-1)(lamdaD)/(2d)` where n is the no. of minimum For maxima `y_(n)=(nlamdaD)/(d)` `y_(5^(th)"dard")=(9lamdaD)/(2d):y_(1^(@)" maxima")=(lamdaD)/(d)` By equation `y_(5^(@)" min")-y_(1^(@)" max")=7xx10^(-3)` `(9lamdaD)/(2d)-(lamdaD)/(2d)=7xx10^(-3)implieslamda=(7xx10^(-3)xx15xx10^(-5)xx2)/(7xx50xx10^(-2))` `thereforelamda=600nm` |
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| 86. |
On introducing a thin film in the path of one of the two interfering beam, the central fringe will shift by one fringe width. If `mu=1.5,` the thickness of the film is (wavelength of monochromatic light is `lambda`)A. `4lambda`B. `3lambda`C. `2lambda`D. `lambda` |
| Answer» Correct Answer - C | |
| 87. |
A radio transmitting station operating at a frequency of 120 MHz has two identical antennas that radiate in phase. Antenna B is 9 m to the right of antenna A. consider point P at a horizontal distance x to the right of antenna A as shown in figure. The value of x and order for which the constructive interference will occur at point P are A. `x = 14.95 m, n = 1`B. `x = 5.6 m, n = 2`C. `x = 1.65 m, n = 3`D. `x = 0 m, n = 3.6` |
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Answer» Correct Answer - a.,b.,c Path difference, `delta = BP - AP` `sqrt(x^(2) + 9^(2)) - x = n lambda` `implies x^(2) + 9^(2) - n^(2) lambda^(2) + x^(2) + 2 n lambda x` `implies x = (9^(2) - n^(2) lambda^(2))/(2 n lambda)` `lambda = ( c) /(v) = (3 xx 10^(8))/(120 xx 10^(6)) = 2.5 m` `n = 1, x = 14.95 m, n = 3, x = 1.6 m,` `n = 2, x = 5.6 m, n = 4` not possible |
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| 88. |
A small transparent slab containing material of `mu=1.5` is placed along `AS_(2)`(figure). What will be the distance from O of the principle maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab ?A. 0.19 D and -0.33 DB. 0.19 D and -0.55 DC. 0.33 D and -0.65 DD. 0.33 D and -0.75 D |
| Answer» Correct Answer - A | |
| 89. |
Two slits, 4 mm apart, are illuminated by light of wavelength `6000 Å`. What will be the fringe width on a screen placed 2 m from the slitsA. 0.12 mmB. 0.3 mmC. 3.0mmD. 4.0mm |
| Answer» Correct Answer - B | |
| 90. |
A screen is placed `50 cm` from a single slit, which is illuminated with `6000 Å` light. If the distance between the first and third minima in the diffraction pattern is `3.00 mm`, what is the width of the slit ?A. `1 xx 10^(-4)m`B. `2 xx 10^(-4)m`C. `0.5 xx 10^(-4)m`D. `4 xx 10^(-4)m` |
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Answer» Correct Answer - B The position of `n^(th)` minima in the diffraction pattern is `x_(n)=(n D lambda)/(d)` ` :. x_(3)-x_(1)=(3-1)(D lambda)/(d)=(2D lambda)/(d)` or `d=(2D lambda)/(x_(3)-x_(1))=(2xx0.50xx6000xx10^(-10))/(3xx10^(-3))=2xx10^(-4)m` |
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| 91. |
A screen is placed at `50cm` from a single slit, which is illuminated with `600nm` light. If separation between the first and third minima in the diffraction pattern is `3.0mm`, then width of the slit is:A. (a) `0.4mm`B. (b) `0.1mm`C. (c) `0.3mm`D. (d) `0.2mm` |
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Answer» Correct Answer - D `y_3-y_1=(3lambdax)/(alpha)-(lambdax)/(alpha)` `implies3xx10^-3=(2xx600xx10^-9xx0.5)/(alpha)` `=a=0.2nm` |
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| 92. |
Plane microwaves are incident on a long slit having a width of 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at `theta=30^@`.A. `2.5cm`B. `5cm`C. `7.5cm`D. `10cm` |
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Answer» Correct Answer - A `a sin theta = n lambda` |
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| 93. |
A slit of width d is placed in front of a l ens of focal length 0.5m and is illuminated normally with light of wavelength `5.89xx10^(-7)m`. The first diffraction minima on either side of the central diffraction maximum are separated by `2xx10^(-3)m` . The width d of the slit is `"_________"`m.A. `1.47xx10^(-4)m`B. `2.94xx10^(-4)m`C. `1.47xx10^(-7)m`D. `2.92xx10^(-7)m` |
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Answer» Correct Answer - B `w = 2y= (2lambda D)/(a) = (2f lambda)/(a)` |
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| 94. |
A two-slit inteference, experiment uses coherent light of wavelength `5 xx 10^(-7)m`. Intensity in the interference pattern for the following points are `I_(1) , I_(2) , I_(3)`, and `I_(4)`, respectively 1. A point that is close to one slit than the other by `5 xx 10^(-7)m`. 2. A point where the light waves received from the two slits are out of phase by `(4 pi)/(3)` rad. 3. A point that is closer to one slit than the other by `7.5 xx 10^(-7) m`. 4. A point where the light waves received by the two slits are out of phase by `(pi)/(36)` rad. Then which of following statements is/are correct?A. `I_(1) gt I_(4) gt I_(2) gt I_(3)`B. `I_(1) gt I_(4) gt I_(2) gt I_(3)`C. `4 I_(2) = I_(4)`D. `I_(3) = 0` |
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Answer» Correct Answer - a `I = I_(0) cos^(2).(xd sin theta)/(lambda)` (1) Path difference, `delta = d sin theta = 5 xx 10^(-7) m,` `I_(1) = I_(0) cos^(2).(pi xx 5 xx 10^(-7))/((5 xx 10^(-7) m)) = I_(0) cos^(2) (pi) = I_(0)` (2) `I_(2) = I_(0) cos^(2) ((4pi)/(2 xx 3) "rad")` `implies I_(0) cos^(2) ((4 pi)/(6)) = 0.25 I_(0)` (3) `I_(3) = I_(0) cos^(2) ((pi xx 7.5 xx 10^(-7)m)/(5 xx 10^(-7)m))` `=I_(0) cos^(2) (1.5 pi "rad") = I_(0) cos^(2) (270^(@)) = 0` (4) `I_(4) = I_(0) cos^(2) ((pi)/(2 xx 3)) "rad" = 0.75 I_(0)`. |
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| 95. |
In a double slit interference pattern, the first maxima for infrared light would beA. at the same place as the first maxima for green lightB. closer to the centre than the first maxima for green lightC. farther from the centre than the first maxima for green lightD. infrared light does not produce an interference pattern |
| Answer» Correct Answer - C | |
| 96. |
In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other then in the interference pattern.A. the intensity of both the maxima and minima increasesB. the intensity of the maxima increases and the minima has zero intensityC. the intensity of the maxima decreases and that of minima increasesD. the intensity of the maxima decreases and the minima has zero |
| Answer» Correct Answer - A | |
| 97. |
A double slit is illuminated by light of wave length 6000 Å. The slit are 0.1 cm apart and the screen is placed one metre away. Calculate. (i). The angular position of the `10^(th)` maximum in radian and (ii). Separation of the two adjacent minimal. |
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Answer» `lamda=6000Å=6xx10^(-7)m,d=0.1cm=1xx10^(-3)m,D=1m,n=10` Angular position `theta_(n)=(nlamda)/(d)=(10xx6xx10^(-7))/(10^(-3))=6xx10^(-3)rad` (ii). Separation between two adjacent minima =fringe width `beta` `beta=(lamdaD)/(d)=(6xx10^(-7)xx1)/(1xx10^(-3))=6xx10^(-4)m=0.6mm` |
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| 98. |
A screen is placed 50cm from a single slit, which is illuminated with 6000Å light. If distance between the first and third minima in the diffraction pattern is 3.0 mm, what is the width of the slit?A. `0.1mm`B. `0.2mm`C. `0.4mm`D. `0.8mm` |
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Answer» Correct Answer - B Position of first minima on a single slit diffraction pattern is given by `d sin theta = n lambda` For small value of `theta, sin theta ~~ theta = (y)/(D)` `:. (y.a)/(D) = n lambda or y= (n lambda D)/(a)` `:.` Distance between third order minima and first order minima will bew `Delta y = y_(3)-y_(1) = ((3-1)(lambda D))/(a) = (2lambda D)/(a)` |
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| 99. |
A double slit of separation 1.5 mm is illuminated by white light (between 4000 and 8000 `Å`). On a screen 120 cm away colored interference pattern is formed. If a pinhole is made on this screen at a distance of 3.0 mm from the central white fringe, some wavelengths will be absent in the transimitted light. Find the second longest wavelength (in `Å`) which will be absent in the transmitted light. |
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Answer» Correct Answer - 5769 `Å` In a double-slit interference pattern, the distance x of a dark fringe from the central achormatic fringe is `x = (D)/(2d) (2n + 1 ) (lambda)/(2) , n = 0, 1, 2,…` `lambda = (75000)/(2n+ 1) Å` (where `n = 0, 1, 2,…)` Thus, in the range 4000 - 8000 `Å`, the absent wavelength are 6818, 5769, 5800, 4421. |
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| 100. |
Light of wavelength `520nm` passing through a double slit,produced interference pattern of relative intensity versus deflection angle `theta` as shown in the figure.find the separation of between the slits. |
| Answer» Correct Answer - `1.98xx10^(-2)mm` | |