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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.
| 551. |
When light from two sources (say slits `S_(1)` and `S_(2)`) interfere, they form alternate dark and bright fringes. Bright fringe is formed at all point where the path difference is an odd multiple of half wavelength. At the condition of equal amplitudes, `A_(1) = A_(2) = a`, the maximum intensity will be `4 a^(2)` and the visibility improves, The resultant intensity can also be indicated with phase factor as `I = 2 a^(2) cos^(2) (phi // 2)`.Using this passage, answer the following questions. At point having a path difference of `(lambda)/(4)`, the intensity |
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Answer» Correct Answer - b For a path difference of `lambda // 4`, the phase difference will be `phi = (2 pi)/(lambda) (lambda)/(4) = (pi)/(2)` `:.` intensity `= 2a^(2) cos^(2) ((phi)/(4)) = a^(2)` So, choice (b) and the other are incorrect. |
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| 552. |
When light from two sources (say slits `S_(1)` and `S_(2)`) interfere, they form alternate dark and bright fringes. Bright fringe is formed at all point where the path difference is an odd multiple of half wavelength. At the condition of equal amplitudes, `A_(1) = A_(2) = a`, the maximum intensity will be `4 a^(2)` and the visibility improves, The resultant intensity can also be indicated with phase factor as `I = 2 a^(2) cos^(2) (phi // 2)`.Using this passage, answer the following questions. If the path difference between the slits `S_(1)` and `S_(2)` is `(lambda)/(2)` the central fringe will have an intensity of |
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Answer» Correct Answer - a Path difference to the central fringe position is `lambda // 2`. Since it is an old multiple of `(lambda // 2)` the fringe formed is dark. So, choice (a) is correct and the other are wrong. |
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| 553. |
Pick out the correct statementsA. diffraction is exhibited by all electromagnetic waves but not by mechanical wavesB. diffraction cannot be observed with a plane polarized lightC. the limit of resolution of a microscope decreases with increases in the wavelength of light usedD. the width of central amximum in the diffraction pattern due to single slit increases as wavelength increases |
| Answer» Correct Answer - D | |
| 554. |
Assertion : If we look clearly at the shadow cast by an opaque object, close to the region of geometrical shadow, alternate dark and bright regions can be seen. Reason : This happens due to the phenomenon of interference.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 the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C This happens due to the phenomenon of diffraction. It is a general characteristic exhibited by all types of waves, sound waves, light waves, water waves or matter waves. |
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| 555. |
A we move away from the edge into the geometrical shadow of a straight edge, the intensity of illuminationA. DecreasesB. IncreasesC. Remains unchangedD. Increase and decreases |
| Answer» Correct Answer - A | |
| 556. |
A lens of focal length f gives diffraction pattern of F raunhoffer type of a slit having width a. If wavelength of light is `lambda`, the distance of first dark band and next bright band from axis is given byA. `(a)/(lambda)f`B. `(lambda)/(a)f`C. `(lambda)/(a f)`D. `a lambda f` |
| Answer» Correct Answer - B | |
| 557. |
Neutron diffraction pattern is used to determineA. Density of solidsB. Atomic number of elementsC. Crystal structure of solidD. Refractive index of liquid |
| Answer» Correct Answer - C | |
| 558. |
Plane parallel wavefornts encounter the interference one medium and another as shown below The wave speed is different in the two media. what will happen to the distance between wavefront and the direction of travel wavefronts as the waves enter the second media?A. the distance between wavefronts remains unchanged but the directionof wavefornt changes.B. the distance between wavefronts and the direction of wavefront both remains unchangedC. the distance between wavefronts and the direction of wavefront both changed.D. the distance between wavefrontschanged but the directionof wavefornt unchanges. |
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Answer» Correct Answer - C because wave length and direction of progation changes. |
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| 559. |
Calculate the specific rotation of sugar solution from the following data: length of the polaimeter tueb=0.21 concentration of sugar solution =`80kgm^(-3)` angle of rotation =`10^(@)` |
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Answer» Correct Answer - `0.01rad m^(-1)kg^(-1)m^(3)` |
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| 560. |
A quartz plate is a half wve plate for light whose wave length. Is `lambda`. Assuming that the variations is the indices of refraction with wavelength can be neglected, how would this behave with repsect to light of wavelengt `lamda=2lamda`. |
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Answer» Correct Answer - Plane will behave as a quarter wave plate for wavelength `lambda` |
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| 561. |
In YDSE if a slab whose refractive index can be varied is placed in front of one of the slits. Then, the variation of resultant intensity at mid-point of screen with `mu` will be best represented by `(mu is greater than or equal to 1) |
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Answer» `Delta x= (mu-1)t`, , For `mu=1, Delta x=0` `:. I = "maximum" = I_(0)` , As `mu` increases path difference `Deltax` also increases. , or `Delta x = 0 to (lambda)/(2)`, intensity will decrease from `I_(0)` to zero. Then for `Delta x= (lambda)/(2) " to " lambda`, intensity will increase from zero to `I_(0)`. Hence option 3 is correct. |
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| 562. |
A light wave travels through three transparent materials of equal thickness rank in order from the highest to lowest the indices of refraction `n_(1),n_(2)` and `n_(3)`.A. `n_(3)gtn_(1)gtn_(2)`B. `n_(1)ltn_(3)ltn_(2)`C. `n_(3)ltn_(1)ltn_(2)`D. `n_(2)gtn_(3)gtn_(1)` |
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Answer» Correct Answer - A `nprop(1)/(lamda)` |
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| 563. |
In YDSE if a slab whose refractive index can be varied is placed in front of one of the slits. Then, the variation of resultant intensity at mid-point of screen with `mu` will be best represented by `(mu is greater than or equal to 1)A. B. C. D. |
| Answer» Correct Answer - C | |
| 564. |
Two coherent monochormatic light source are located at two vertices of an equilateral trangle. If the intensity due to each of the source independently is `1Wm^(-2)` at the third vertex. The resultant intensity due to both the sources at that point (i.e at the third vertex) is (in `Wm^(-2)` )A. zeroB. `sqrt(2)`C. 2D. 4 |
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Answer» Correct Answer - D at the third vertex path diff = 0, hence intensity is max. |
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| 565. |
By an anodizing process, a transparent film of aluminium oxide of thickness `t = 250` nm and index of refraction `n_(2) = 1.80 ` is deposited on a sheet of polished aluminium What is the color of utenslis made from this sheet with observer in white light? Assume normal incidence of the light. |
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Answer» We must find which colours in the visible region, having vacuum wavelengths from 400 nm (violet) to 700 nm (red), will interfere constructively and which destructively. From the problem, with `n_(1) = 1` (air) maximum constructive interference occurs for `lambda = (2(n_(2)//n_(1))t)/(m + (1)/(2)) = ((900 nm))/(m + (1)/(2)` `(m = 0, 2,...` Only the value corresponding to `m = 1`, that is, `lambda_(1) = 450 nm` (violet) is the visible range. we infer that the red-orange-yellow end of the spectrum will be strongly reflected, while the violet-blue end will be greatly diminished in intensity as compared with the illumination white light. |
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| 566. |
Two monochromatic and coherent point sources of light are placed at a certain distance from each other in the horizontal plane. The locus of all thos points in the horizontal plane which have constructive interference will be-A. a hyperbolaB. family of hyperbolaC. family of straight lineD. family of parabolas |
| Answer» Correct Answer - B | |
| 567. |
As a soap bubble evaporates, it appears black just before it breaks. Explain this phenomenon in terms of the phase changes that occur on reflection form the two surface of the soap film. |
| Answer» As the soap bubble becomes very thin, the thickness of the bubble approaches zero, Since light reflecting off the front of the soap surface is phase- shifted `180^(@)` and light reflecting off the back of the soap film is phase-shifted `0^(@)`, the reflected light meets the conditions for a minimum. Thus, the soap film appears blacks. | |
| 568. |
A circular planar wire loop is dipped in a soap solution and after taking it out, held with its plane vertical in air. Assuming thickness of film at the top very small, as sunlight falls on the soap film, & observer receive reflected light.A. the top portion appears dark while th efirst colour to be observed as one moves down is red.B. the top portion appears violet while the first colour to be observed as one moves down is indigo.C. the top portion appears dark whilt the first colour to be observed as one move down is violet.D. the top portion appears dark while the first colour to be observed as one move down depends on the refractive index of the soap solution. |
| Answer» Correct Answer - C | |
| 569. |
Soap bubble appears coloured due to the phenomenon ofA. interferenceB. diffractionC. dispersionD. reflection |
| Answer» Correct Answer - A | |
| 570. |
Two beams of ligth having intensities I and 4I interface to produce a fringe pattern on a screen. The phase difference between the beams is `(pi)/(2)` at point A and `pi` at point B. Then the difference between the resultant intensities at A and B isA. 2IB. 4IC. 5ID. 7I |
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Answer» Correct Answer - B Here, `I_(1) = I, I_(2) = 4I, theta_(1) = pi//2, theta _(2) = pi` Resultant intensity `I_(theta_(1)) = I_(1)+I_(2) + 2sqrt(I_(1)I_(2))cos theta_(1)` `= I+4I+2sqrt(Ixx4I)cos pi//2 = 5I` Resultant intensity `I_(theta_(1)) = I_(1)+I_(2)+2sqrt(I_(1)I_(2))cos theta_(2)` `= I+4I+2sqrt(1xx4I)cos pi` `= 5I-4I=I :. I_(theta_(1))-I_(theta_(2)) = 5I - I = 4I` |
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| 571. |
Two beams of ligth having intensities I and 4I interface to produce a fringe pattern on a screen. The phase difference between the beams is `(pi)/(2)` at point A and `pi` at point B. Then the difference between the resultant intensities at A and B isA. `2 I`B. `4 I`C. `5 I`D. `7 I` |
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Answer» Correct Answer - b `I (phi) = I_(1) + I_(2) + 2 sqrt(I_(1) I_(2)) cos phi` Here, `I_(1) = I` and `I_(2) = 4 I` At point A, `phi = (pi)/(2)` `:. I_(A) = I + 4 I = 5 I` At point B, `phi = pi` `:. I_(B) = I + 4 I - 4 I = I` `:. I_(A) - I_(B) = 4 I` |
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| 572. |
Two beams of ligth having intensities I and 4I interface to produce a fringe pattern on a screen. The phase difference between the beams is `(pi)/(2)` at point A and `pi` at point B. Then the difference between the resultant intensities at A and B isA. `2I`B. `4I`C. `5I`D. `9I` |
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Answer» Correct Answer - B `DeltaI=(I_(1)+I_(2)+2sqrt(I_(1)I_(2))cosphi_(1))-(I_(1)+I_(2)+2sqrt(I_(1)I_(2))cosphi_(2))` `=2sqrt(I_(1)I_(2))(cosphi_(1)-cosphi_(2))=2xx2I("cos"phi/2-cospi)=4I` |
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| 573. |
A circular planar wire loop is dipped in a soap solution and after taking it out, held with its plane vertical in air assuming thickness of film at the very small as sunlight falls on te soap film & observer receives reflected light.A. Th top portion appears dark while the first colour to be observed as one moves down is red.B. The top portion appears violet while the first colour to be observed as one moves down in indigoC. The top portion appears dark while the first colour to be observed as one move down in violet.D. the top portion appears dark while the first to be observed as one move down depends on the refractive index of the soap solution. |
| Answer» Correct Answer - C | |
| 574. |
Two beams of ligth having intensities I and 4I interface to produce a fringe pattern on a screen. The phase difference between the beams is `(pi)/(2)` at point A and `pi` at point B. Then the difference between the resultant intensities at A and B isA. 2 IB. 4 IC. 5 ID. 7 I |
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Answer» Correct Answer - b We know that the resultant amplitude of two interfering waves is given by `R^(2) = a^(2) + B^(2) + 2a b cos phi`, where R is the amplitude of resultant wave, a is the amplitude of one wave, b is the amplitude of second wave, and `phi` as the phase difference between the two wave at a point. Also `I prop ("amplitude")^(2)` `:. I prop I_(1) + I_(2) + 2sqrt I_(1) sqrt I_(2) cos phi` Applying Eq. (i) when phase diffenence is `pi // 2` `I_(pi // 2) prop I + 4 I` `implies I_(pi // 2) prop 5 I` Again, applying Eq. (i) when phase difference is `pi` `I_(pi) prop I + 4 I + 2 sqrt I sqrt( 4I) cos pi` `:. I_(pi) prop I implies I_(pi//2) - I_(pi) prop 4 I` Correct option is (b) |
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| 575. |
A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate as shown in Figure. The observed interference fringes from this combination shall be A. straightB. circularC. equally spacedD. having fringe spacing which increases as we go outward |
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Answer» Correct Answer - a Locus of equal thickness are lines running parallel to the axis of the cylineder. Hence straight fringe will be observed. |
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| 576. |
A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate with the curved surface downwards. Monochromatic light is incident normally from the top. The observed interference fringes from the combination do not follow on of the following statements.A. The fringes are straight and parallel to the length of the plece.B. The line of contact of the cylindrical glass pice and the glass plate appears dark.C. the fringe spacing increases as we go outwards.D. The fringes are formed due to the interference of light rays reflected from the curved surface of the cylindrical piece and te top surface of the glass plate. |
| Answer» Correct Answer - C | |
| 577. |
Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm. If yellow light is replaced by X-rays, then the observed pattern will reveal,A. that the central maximum is narrowerB. more number of fringesC. less number of fringesD. no diffraction pattern |
| Answer» Correct Answer - D | |
| 578. |
Assertion: The pattern and position of fringes always remain same even after the introduction of transparent medium in a path of one of the slit. Reason: The central fringe is bright or dark is depend upon the initial phase difference between the two coherence sources.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 - C If a transparent medium of thickness t and refractive index `mu` is introduced in the path of one of the slits, then effective path in air is increased by an amount `(mu-1)` due to introduction of plate. Therefore, the zeroth fringe shifts to a new position where the two optical paths are equal. In such case fringe width remains unchanged. The central fringe is bright or dark depends upon the initial phase difference between the two coherent sources. |
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| 579. |
Assertion (A) : The maximum intensity in interference pattern is four times the intensity in interference pattern is four times the intensity due to each slit of equal width. Reason (R ) : Intensity is directly proportinal to square of amplitude.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 - B Let a be the amplitude of waves from each slit. Intensity is directly proportional to square of amplitude i.e, `Ipropa^2`. Due to each slit `I_1propa^2` When destructive interference occurs, then resultant amplitude `=a-a=0` `:.` Minimum intensity, `I_(min)=0` When constructive interference occurs, the resultant amplitude `=a+a=2a` `:.` Maximum intensity, `I_(max)prop(2a)^2` Hence, `I_(max)prop4a^2` or `I_(max)=4` times the intensity due to each slit. |
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| 580. |
Sound waves shown more diffraction as compare to light rays-A. wavelength of sound waves is more as compare to light rysB. wavelength of light rays is more as compare to sound wavesC. wavelength of sound waves and light rays is sameD. none of these |
| Answer» Correct Answer - A | |
| 581. |
Assertion : No diffraction is produced in sound waves near a very small opening. Reason : For diffraction to take place, the aperture of opening should be of same order as wavelength of waves.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 The diffraction of sound is only possible when the size of opening should be of the same order as its wavelength and the wavelength of sound is of the order of `1.0m`, hence, for a very small opening no diffraction is produced in sound waves. |
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| 582. |
White light reflected from a soap film (Refractive index `=1.5` ) has a maxima at 600 mm and a minima at 450 nm at with no minimum in between. Then, the thickness of the film isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C We know for maxima `2mut cos r=((2n+1)lambda_(1))/2` for minima `2mut cos r=(n+1)lambda_(2)` Given here, `lambda_(1)=600nm and lambda_(2)=450nm` Dividing Eq (i) by (ii) and substituting values of `lambda_(1)` and `lambda_(2)` we get `((2n+1)lambda_(1))/(2(n+1)lambda_(2))=1` `90m+90=60n+60` `30 n =30 rArr n=1` Using one of the equations we get In given condition `2mut = (2n+1) lambda/2` `2xx1.5xxt=(2xx1+1)600/2` `3t=(3xx600)/2` `rArr t=300nm =3xx10^(-6)m=3units` |
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| 583. |
Light waves can be polarised as they areA. transverseB. of high frequencyC. longitudinalD. reflected |
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Answer» Correct Answer - A Only transverse waves can be polarised |
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| 584. |
Which of the following cannot be polarised?A. Radio wavesB. Ultraviolet raysC. Infrared raysD. Untrasonic waves |
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Answer» Correct Answer - D Ultrasonic waves are longitudinal waves |
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| 585. |
Which of the following cannot be polarised?A. RadiowavesB. ultraviolet raysC. Infrared raysD. Ultrasonic waves |
| Answer» Correct Answer - D | |
| 586. |
Waves that cannot be polarised areA. LongitudinalB. TransverseC. ElectromagneticD. Light |
| Answer» Correct Answer - A | |
| 587. |
Which of the following cannot be polarised?A. Ultraviolet raysB. Ultrasonic wavesC. X-raysD. Radiowaves |
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Answer» Correct Answer - A |
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| 588. |
Through which character we can distiguish the light waves from sound wavesA. InterferenceB. refractionC. polarisationD. reflection |
| Answer» Correct Answer - C | |
| 589. |
Through which character we can distiguish the light waves from sound wavesA. InterferenceB. reflectionC. RefractionD. Polarisation |
| Answer» Correct Answer - D | |
| 590. |
The critical angle of a transparent crystal is `60^(@)`. Then its polarizing angle isA. `theta = tan^(-1)((2)/(sqrt(3)))`B. `theta = sin^(-1)(sqrt(2))`C. `theta = cos^(-1)((1)/(sqrt(2)))`D. `theta = cot^(-1)(sqrt(2))` |
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Answer» Correct Answer - A `mu = (1)/(sin c) and mu = tan theta` |
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| 591. |
In figure, if a parallel beam of white light is incident on the plane of the slit, then the distance of the nearest white spot on the screen form O is [assume `d lt lt D, lambda lt lt D`] |
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Answer» Correct Answer - d The nearest white spot will be at P, the central maxima. `y = (2d)/(3) - (d)/(2) = (d)/(6)` |
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| 592. |
In figure, if a parallel beam of white light is incident on the plane of the slit, then the distance of the nearest white spot on the screen form O is [assume `d lt lt D, lambda lt lt D`] A. d/4B. d/2C. d/3D. d/6 |
| Answer» Correct Answer - D | |
| 593. |
Laser light is considered to be coherent because it consists ofA. many wavelengthsB. uncoordinated wavelengthsC. coordinated waves of exactly the same wavelengthD. divergent beams |
| Answer» Correct Answer - C | |
| 594. |
Calculate the width of central maxima, if light of 9000Å incidents upon a slit of width 1.5 mm. Screen is kept 150 cm away from the slit.A. 1.8 mmB. 1.30 mmC. 0.9 mmD. 0.9 cm |
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Answer» Correct Answer - A Width of central maxima `=(2Dlambda)/d=(2xx1.5xx9xx10^(-7))/(1.5xx10^(-3)) =18xx10^(-4) m=1.8 mm` |
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| 595. |
Light of wavelength `lambda` is incident on a slit of width d and distance between screen and slit is D. Then width of maxima and width of slit will be equal if D is ––A. `(2lambda^(2))/d`B. `d^(2)/(2lambda)`C. `d/lambda`D. `(2lambda)/d` |
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Answer» Correct Answer - B In diffraction fringe width `beta=(lambdaD)/d` when `d/2=beta" then"d/2=(lambdaD)/drArrD=d^(2)/(2lambda)` |
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| 596. |
`V_(0)` and `V_(E)` represent the velocities, m0 and mE the refractive indices of ordinary and extraordinary rays for a doubly refracting crystal. ThenA. (a) `V_0geV_E`, `mu_0lemu_E` if the crystal is calciteB. (b) `V_0leV_E`, `mu_0lemu_E` if the crystal is quartzC. (c) `V_0leV_E`, `mu_0gemu_E` if the crystal is calciteD. (d) `V_0geV_E`,`mu_0gemu_E` if the crystal is quartz |
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Answer» Correct Answer - C In double refraction light rays always splits into two rays (O-ray and E-ray). O-ray has same velocity in all direction but E-ray has differenct velocity in different direction. For calcite `mu_e`lt`mu_0`impliesv_egt`v_0` `"For quartz" `mu_e`gt`mu_0impliesv_0`gt`v_e` |
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| 597. |
A light of wavelength `6000 Å` shines on two narrow slits separeted by a distance 1.0 mm and illuminates a screen at a distance 1.5 m away. When one slit is covered by a thin glass plate of refractive index 1.8 and other slit by a thin glass plate of refractive index `mu`, the central maxima shifts by 0.1 rad. Both plates have the same thickness of 0.5 mm. The value of refractive index `mu` of the glass isA. 1.4B. 1.5C. 1.6D. none of these |
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Answer» Correct Answer - c Changed in path difference for any point on screen is `|mu - 1.8|t`. For central maxima, phase difference `= 0`. Hence, `d sin theta - |mu - 1.8|t = 0` `implies d theta = |mu - 1.8|t` [`theta` is very small and is in radium] `implies |mu - 1.8| = (10^(3) xx 0.1)/(0.5 xx 10^(-3)) = 0.2` `mu = 2` or `1.6`. |
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| 598. |
A double-slit experiment is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of slits and screen is 1.33 m. The slits are illuminated by a parallel bam of light whose wavelength in air is `6830 Å`. Then the fringe width isA. (a) `6.3xx10^-4m`B. (b) `8.3xx10^-4m`C. (c) `6.3xx10^-2m`D. (d) `6.3xx10^-5m` |
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Answer» Correct Answer - A The experimental setup is in a liquid, therefore the wavelength of light will change. `lambda_(liquid)=(lambda_(air))/(mu)=(6300)/(1.33)` `=(6300xx10^(-10))/(1.33)m` Fringe width `beta=(lambda_(liquid)D)/(d)=(lambda_(air)D)/(mud)=(6300xx10^(-10))/(1.33)xx(1.33)/(10^-3)` `=6.32xx10^-4m` |
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| 599. |
A double-slit experiment is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of slits and screen is 1.33 m. The slits are illuminated by a parallel bam of light whose wavelength in air is `6830 Å`. Then the fringe width isA. `6.3 xx 10^(-4) m`B. `8.3 xx 10^(-4) m`C. `6.3 xx 10^(-2) m`D. `6.3 xx 10^(-5) m` |
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Answer» Correct Answer - a The experimental set-up is in a liquid, therefore the wavelength of light will change `lambda_("liquid") = (lambda_("air"))/(mu) = (6300)/(1.33) = (6300 xx 10^(-10))/(1.33) m` Fringe width, `beta = lambda_("liquid") D/(d) = (lambda_("air") D)/(mu d) = (6300 xx 10^(-10))/(1.33) xx (1.33)/(10^(-3))` `= 6.3 xx 10^(-4) m` |
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| 600. |
In YDSE, light of wavelength `lamda = 5000 Å` is used, which emerges in phase from two slits distance `d = 3 xx 10^(-7) m` apart. A transparent sheet of thickness `t = 1.5 xx 10^(-7) m`, refractive index `n = 1.17`, is placed over one of the slits. Where does the central maxima of the interference now appear from the center of the screen? (Find the value of y?) A. (a) `(D(mu-1)t)/(2d)`B. (b) `(2D(mu-1)t)/(d)`C. (c) `(D(mu+1)t)/(d)`D. (d) `(D(mu-1)t)/(d)` |
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Answer» Correct Answer - D The path difference introduced due to introduction of transparent sheet is given by `Deltax=(m-1)t`. If the central maxima occupies position of nth fringe, then `(mu-1)t=nlambda=dsintheta` `sin theta=((mu-1)t)/(d)=((1.17-1)xx1.5xx10^-7)/(3xx10^-7)=0.085` Hence the angular position of central maxima is `theta=sin^-1(0.085)=4.88^@` For small angles `sin theta~=theta~=tantheta` `tantheta=y/D` `y/D=((mu-1)t)/(d)` Shift of central maxima is `y=(D(mu-1)t)/(d)` This formula can be used if D is given. |
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