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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.
| 601. |
In YDSE, light of wavelength `lamda = 5000 Å` is used, which emerges in phase from two a slits distance `d = 3 xx 10^(-7) m` apart. A transparent sheet of thickness `t = 1.5 xx 106(-7) m`, refractive indes `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. `(D(mu - 1)t)/(2d)`B. `(2D(mu - 1)t)/(d)`C. `(D(mu + 1)t)/(d)`D. `(D(mu - 1)t)/(d)` |
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Answer» Correct Answer - d The path difference intorduced due to introduction of transparent sheet is given by `Delta x = (m - 1)t`. If the central maxima occupies position of nth fringe, then `(mu - 1)t = n lambda = d sin theta` `implies sin theta ((mu-1) t)/(d) = ((1.17 - 1) xx 1.5 xx 10^(-7))/(3 xx 10^(-7))` `=0.085` Hence, the angular position of central maxima is `theta = sin^(-1) (0.085)= 4.88^(@)` For small angles, `sin theta = theta = tan theta` `implies tan theta = (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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| 602. |
Three waves of equal frequency having amplitudes `10mum`, `4mum`, `7mum` arrive at a given point with successive phase difference of `pi//2`, the amplitude of the resulting wave in `mum` is given byA. (a) `4`B. (b) `5`C. (c) `6`D. (d) `7` |
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Answer» Correct Answer - B The amplitudes of the waves are `a_1=10mum`, `a_2=4mum` and `a_3=7mum` and the phase difference between 1st and 2nd wave is `pi/2` and that between 2nd and 3rd wave is `pi/2`. Therefore, phase difference between 1st and 3rd is `pi`. Combining 1st with 3rd, their resultant amplitude is given by `A_1^2=a_1^2+a_3^2+2a_1a_3cos varphi` or `A_1=sqrt(10^2+7^2+2xx10xx7cos pi)` `=sqrt(100+49-140)` `=sqrt9=3mum` in the direction of first. Now combining this with 2nd wave we have, the resultant amplitude `A^2=A_1^2+a_2^2+2A_1a_2cospi/2` or `A=sqrt(3^2+4^2+2xx3xx4cos 90^@)=sqrt(9+16)` `=5mum` |
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| 603. |
Two coherent point sources `S_1` and `S_2` are separated by a small distance `d` as shown. The fringes obtained on the screen will be A. pointsB. straight linesC. semi-circlesD. concentric circle |
| Answer» Correct Answer - D | |
| 604. |
Assertion : Intensity pattern of interference and diffraction are not same. Reason : When there are few sources of light, then the result is usually called interference but if there is a large number fo them, the word diffraction is more often used.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. |
| Answer» Correct Answer - B | |
| 605. |
Diffraction and interference of light refers to -A. quantum nature of lightB. wave nature of lightC. tranverse natre of lightD. electromengnetic nature of light |
| Answer» Correct Answer - B | |
| 606. |
Which of the following phenomena support the wave theory of light? 1. Scattering 2.Interference 3. Diffraction 4. Velocity of light in a denser medium is less than the velocity of light in the rare mediumA. 1, 2 and 3B. 1, 2 and 4C. 2, 3 and 4D. 1, 3 and 4 |
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Answer» Correct Answer - C Wave theory of light explains the phenomenon of interference diffraction and velocity of light in a denser medium or rarer medium but this theory fails to explain the scattering of light |
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| 607. |
The phenomenon of polarisation of electromagnetic wave proves that the electromagnetic waves areA. (a) LongitudinalB. (b) transverseC. (c) neither longitudinal nor transverseD. (d) None of these |
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Answer» Correct Answer - B Longitudinal waves cannot be polarised. |
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| 608. |
A parallel beam of white of white light falls on a thin film whose refractive index is 1.33. If angle of incidence is `52^(@)` then thickness of the film for the reflected light to be coloured yellow `(lambda=6000Å)` most intensively must be :A. `14 (2n + 1)m`B. `1.4(2n + 1)m`C. `0.14(2n + 1)m`D. `142(2n+1)m` |
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Answer» Correct Answer - C |
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| 609. |
Four polaroids are so placed that the transmission-axis of each is inclined at an angle of `30^(@)` from the axis of the previous polaroid in the same direction. If unpolarised light-beam of intensity `I_(0)` falls on the first, polaroid, then what will be the intensity of the light emerging from the last polaroid? |
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Answer» Correct Answer - 0.21 `I_(0)` |
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| 610. |
A beam of natural light falls on a system of 5 polaroids, which are arranged in succession such that the pass axis of each polaroid is turned through `60^(@)` with respect to the precending one. The friction of the incident light intensity that passes through the system is :A. `(1)/(64)`B. `(1)/(32)`C. `(1)/(512)`D. `(1)/(128)` |
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Answer» Correct Answer - C |
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| 611. |
The polaroids `P_(1), P_(2) & P_(3)` are arranged coaxially. The angle between `P_(1)` and `P_(2)` is `37^(@)`. The angle between `P_(2)` and `P_(3)` is, if intensity of emerging light is a quarter of intensity of unploarized lightA. `theta = cos^(-1) ((5)/(4))`B. `theta = cos^(-1) ((4)/(5))`C. `theta = cos^(-1) ((4)/(5sqrt(2)))`D. `theta = cos^(-1)((5)/(4sqrt(2)))` |
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Answer» Correct Answer - D `I^(1) = (I)/(2), cos^(2) theta_(1) cos^(2)theta_(2)` |
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| 612. |
Fig. shows a surface XY separating two transparent media, medium 1 and medium 2. Lines ab and cd represent wavefronts of a light wave travelling in medium 1 and incident on XY. Line ef and gh represent wavefront of the light wave in medium 2 after rafraciton. Light travel as aA. parallel beam in each mediumB. convergent beam in each mediumC. divergent beam in each mediumD. divergent beam in one medium and convergent beam in the other medium |
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Answer» Correct Answer - a Wavefronts are parallel in both media. Therefore, light which is perpendicular to the wavefront travels as a parallel beam in each medium. |
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| 613. |
Monochromatic light of walelength 400nm and 560nm are incident simultaneously and normally on double slits apparatus whose slit sepation is 0.1 mm and screen distance is 1m. Distance between areas of total darkness will beA. 4 mmB. 5.6 mmC. 14 mmD. 25 mm |
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Answer» Correct Answer - d At the area of total darkness, minima will occur for both the wavelengths. `:. ((2n - 1))/(2) lambda_(1) = ((2m + 1))/(2) lambda_(2)` `implies (2n - 1) lambda_(1) = (2 m+ 1) lambda_(2)` or ` ((2n + 1))/((2 m + 1)) = (560)/(400) = (7)/(5)` or `10 n = 14 m + 2` By inspection: For `m= 2, n = 3`. For `m = 7, n = 10`. The distance between then will be the distance between such point, i.e., `Delta s = (D lambda_(1))/(d) {((2n_(2) + 1) - (2n_(1) + 1))/(2)}` Putting `n_(2) = 10` and `n_(1) - 3` and solving, we get `Delta x = 28` mm. |
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| 614. |
Assertion: Out of radio waves and microwaves, the former undergoes more diffraction. Reason: Radio waves have greater frequency compared to microwaves.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 The waves which consist longer wavelength have more diffraction. Since ratio waves have greater wavelength than microwaves, hence, radio waves undergo more diffraction than microwaves. |
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| 615. |
Two nicols are so oriented that the maximum amount of light is transmitted .To what fraction of its maximum value is the intensity of the transimitted light reduced when the analyser is rotated through (i)` 45^(@) (ii)90^(@) (iii)180^(@)` |
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Answer» Correct Answer - `(i) 50% (ii) "zero" (iii)100 %` |
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| 616. |
To deminstrate the phenimenon of interference, we require two sources which emit radiationA. nearly the same frequencyB. the same frequencyC. different wavelengthD. the same frequency and having a definite phase relationship |
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Answer» Correct Answer - 4 Sustained interference is possible with coherent source only. |
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| 617. |
The refractive index of glass with respect to water is `1.125` . If the speed of light in water is `2.25 xx 10^(8) ms^(-1)` , what is the speed of light in glass ? |
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Answer» Correct Answer - `2.0 xx 10^(8) ms^(-1)` |
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| 618. |
The intensity ratio of two waves is `9:1`. If they produce interference, the ratio of maximum to minimum intensity will beA. `4:1`B. `2:1`C. `9:1`D. `3:2` |
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Answer» Correct Answer - A `(I_(max))/(I_(min)) = ((sqrt(I_(1)) + sqrt(I_(2)))^(2))/((sqrt(I_(1)) - sqrt(I_(2)))^(2))` |
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| 619. |
Consider interference between two sources of intensity I and 4I. Find out resultant intensity where phase difference is (i). `pi//4` (ii). `pi` (iii). `4pi` |
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Answer» Correct Answer - (i). 7.8I (ii). I (iii). 9I (i). `I_("result")=I+4I+2xxsqrt(I)xxsqrt(4I)cos((pi)/(4))` `=5I+2sqrt(2)I=7.8I` (ii). `I_("result")=I+4I+2sqrt(I)xxsqrt(4I)xxcospi=5I-4I=I` (iii). `I_("result")=I+4I+2sqrt(I)xxsqrt(4I)cos4pi=5I+4I=9I` |
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| 620. |
Two beam of light having intensities I and 4I interfere 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 resultant intensities at A and B is : `(2001 , 2M)`A. 3IB. 4IC. 5ID. 6I |
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Answer» Correct Answer - D |
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| 621. |
A thin film of specific material can be used to decrease the intensity of reflected light.There is destructive interference fo waves from uppe rand lower surface of the fil,m.These films are called non-reflecting coatings.The process of coating the lens of or surface with non-reflectig film is called bloomingas shown in fig. Magnesium fluoride`(MgF_(2))` is genertally used as anti-reflectioncoating.If refrative index of `MgF_(2)` is `1.35` then minimum thickness of film required is `(Take lambda=550nm)`A. `112.4mm`B. `78.2mm`C. `99.6mm`D. None |
| Answer» Correct Answer - C | |
| 622. |
Two waves of same intensity produce interference. If intensity at maximum is 4I, then intensity at the minimum will be: |
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Answer» Correct Answer - B |
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| 623. |
A thin film of specific material can be used to decrease the intensity of reflected light.There is destructive interference fo waves from uppe rand lower surface of the fil,m.These films are called non-reflecting coatings.The process of coating the lens of or surface with non-reflectig film is called bloomingas shown in fig. Choose the correct statementsA. Both `1` and `2` suffer change of `pi` upon relfectionB. Onlu `1` suffers a phase change of `pi` upon reflectionC. Only` 2 `suffers a phase chagne of `pi` upon reflection.D. Neither `1` and`2` suffer a phase change upon reflection. |
| Answer» Correct Answer - A | |
| 624. |
Intensities of the two waves of light are I and `4I`. The maximum intensity of the resultant wave their superposition isA. (a) `5I`B. (b) `9I`C. (c) `16I`D. (d) `25I` |
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Answer» Correct Answer - B `I_(max)=I_1+I_2+2sqrt(I_1I_2)` So, `I_(max)=I+4I+2sqrt(I.4I)=9I` |
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| 625. |
Two waves of intensity I undergo interference. The maximum intensity obtained isA. (a) `I//2`B. (b) `I`C. (c) `2I`D. (d) `4I` |
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Answer» Correct Answer - D For maximum intensity `varphi=0^@` `:. I=I_1+I_2+2sqrt(I_1I_2)cos varphi` `=I+I+2sqrt(I.I)cos0^@=4I` |
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| 626. |
If two light waves having same frequency have intensity ratio `4:1` and they interfere, the ratio of maximum to minimum intensity in the pattern will beA. (a) `9:1`B. (b) `3:1`C. (c) `25:9`D. (d) `16:25` |
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Answer» Correct Answer - A `(I_(max))/(I_(min))=((sqrt(I_1/I_2)+1)/(sqrt(I_1/I_2)-1))^2=((sqrt(4/1)+1)/(sqrt(4/1)-1))^2=9/1` |
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| 627. |
Two coherent monochromatic light beams of intensities I and `4I` are superposed. The maximum and minimum possible intensities in the resulting beam areA. (a) `5I` and `I`B. (b) `5I` and `3I`C. (c) `9I` and `I`D. (d) `9I` and `3I` |
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Answer» Correct Answer - C `I_(max)=(sqrt(I_1)+sqrt(I_2))^2=(sqrtI+sqrt(4I))^2=9I` `I_(min)=(sqrt(I_1)-sqrt(I_2))^2=(sqrtI-sqrt(4I))^2=I` |
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| 628. |
Wavefront is the locus of all points, where the particles of the medium vibrate with the sameA. phseB. amplitudeC. frequencyD. period |
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Answer» Correct Answer - A On the wavefront, all the points are in the same phases |
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| 629. |
Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam areA. 5I and 3IB. 9I and 3IC. 4I and ID. 9I and I |
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Answer» Correct Answer - D We know that That maximum intensity `I_(max)=(sqrt(I_(1))+sqrt(I_(2)))^(2).....(i)` The minimum intensity `I_(min)=(sqrt(I_(1))-sqrt(I_(2)))^(2) ....(ii)` So, the ratio of the maximum and minimum intensities is `I_(max)/I_(min)=(sqrt(I_(1))+sqrt(I_(2)))^(2)/((sqrt(I_(1))-sqrt(I_(2)))^(2))=(sqrt(4I)+sqrt(I))^(2)/((sqrt(4I)-sqrt(I))^(2))` `rArrI_(max)/I_(min)=((3sqrt(I))/(sqrt(I)))^(2)=9/1=9:1` Hence, the maximum and minimum intensities intensible in the resulting pattern are 9I and I |
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| 630. |
Two doherent sources of intensity ratio `alpha` interfere in interference pattern `(I_(max)-I_(min))/(I_(max)+I_(min))` is equal toA. `(2alpha)/(1+alpha)`B. `(2sqrt(alpha))/(1+alpha)`C. `(2alpha)/(1+sqrt(alpha)`D. `(1+alpha)/(2alpha)` |
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Answer» Correct Answer - B `(I_(max)-I_(min))/(I_(max)+I_(min))=((a_(1)+a_(2))^(2)-(a_(1)-a_(2))^(2))/((a_(1)+a_(2))^(2)+(a_(1)-a_(2))^(2))` `[becauseI_(max)=(a_(1)+a_(2))^(2),I_(min)=(a_(1)-a_(2))^(2)` where a=amplitude] `=(4a_(1)a_(2))/(2(a_(1)^(2)+a_(2)^(2)))=(2a_(1)a_(2))/((a_(1)^(2)+a_(2)^(2)))` Now, dividing the numerator and denominator by `a_(1)a_(2)` we get `(I_(max)-I_(min))/(I_(max)+I_(min))=2/((a_(1)/a_(2)+a_(2)/a_(1)))(because I_(1)/I_(2)=alpharArra_(1)/a_(2)=sqrt(alpha))` `rArr(I_(max)-I_(min))/(I_(max)+I_(min))=2/((sqrt(alpha+1/sqrt(alpha))))=(2sqrt(alpha))/((alpha+1))` |
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| 631. |
Two coherent sources of intensity ratio `beta` interfere, then `(I_(max)-I_(min))/(I_(max)+I_(min))` isA. `(beta)/(1+beta)`B. `(2sqrt(beta))/(1+beta)`C. `(2sqrt(beta))/(1+sqrt(beta))`D. `(2beta)/(1+sqrt(beta))` |
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Answer» Correct Answer - B `(I_(1))/(I_(2)) = beta , I_(max) = (sqrt(I_(1)) + sqrt(I_(2)))^(2) I_(min) = (sqrt(I_(1)) - sqrt(I_(2)))^(2)` |
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| 632. |
The interference pattern is obtained with two coherent light sources of intensity ration n. In the interference pattern, the ratio `(I_(max)-I_(min))/(I_(max)+I_(min))` will beA. `sqrt(n)/(n+1)`B. `(2sqrt(n))/(n+1)`C. `sqrt(n)/((n+1)^(2))`D. `(2sqrt(n))/((n+1)^(2))` |
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Answer» Correct Answer - B It is given that, `I_(2)/I_(1)=nrArrI_(2)=nI_(1)` `:.` Ratio of intensities is given by `(I_(max)-I_(min))/(I_(max)+I_(min))=((sqrtI_(2)+sqrt(I_(1)))^(2)-(sqrt(I_(2)-I_(1)))^(2))/((sqrt(I_(1))+sqrt(I_(2)))^(2)+(sqrt(I_(2))-sqrt(I_(1)))^(2))` `=((sqrt(I_(2)/I_(1))+1)^(2)-(sqrt(I_(2)/I_(1))-1)^(2))/((sqrt(I_(2)/I_(1))+1)^(2)+(sqrt(I_(2)/I_(1))-1)^(2))` `=((sqrt(n)+1)^(2)-(sqrt(n)-1)^(2))/((sqrt(n)+1)^(2)+(sqrt(n)-1)^(2))=(2sqrt(n))/(n+1)` |
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| 633. |
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference patten, the ratio `(I_(max)-I_(min))/(I_(max)+I_(min))` will beA. (a) `(sqrtn)/((n+1)^2)`B. (b) `(2sqrtn)/((n+1)^2)`C. (c) `(sqrtn)/(n+1)`D. (d) `(2sqrtn)/(n+1)` |
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Answer» Correct Answer - D Let `I_1/I_2=n/1` `(I_(max)-I_(min))/(I_(max)+I_(min))=((sqrtI_1+sqrtI_2)^2-(sqrtI_1-sqrtI_2)^2)/((sqrtI_1+sqrtI_2)^2(sqrtI_1-sqrtI_2)^2)` `=(4sqrt(I_1I_2))/(2(I_1+I_2))` Dividing numberator and denomenator by `I_2`, required ratio `=(2sqrt(I_1//I_2))/((I_1/I_2+1))=(2sqrtn)/(n+1)` |
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| 634. |
If the ratio of the intensity of two coherent sources is 4 then the visibility `[(I_(max)-I_(min))//(I_(max)+I_(min))]` of the fringes isA. `4`B. `4//5`C. `3//5`D. `9` |
| Answer» Correct Answer - B | |
| 635. |
Assertion Two coherent sources transmit waves of equal intensity `I_(0)` Resultant intensity at a point where path difference is `lambda/3` is also `I_(0)`. Reason In interference resultant intensity at any point is the average intensity of two individual intensities. |
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Answer» Correct Answer - C `phi=(2pi)/lambda.Deltax=(2pi)/lambdaxxlambda/3=(2pi)/3` and `I=4I_(0)cos^(2)(phi/2)=4I_(0)xx1/4=I_(0)` |
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| 636. |
Two waves of same frequency and same amplitude from two monochromatic sources are allowed to superpose at a certain point. If in one case the phase difference is 0 and in other case it is `pi//2` then the ratio of the intensities in the two cases will beA. `1:1`B. `2:1`C. `4:1`D. None of the above |
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Answer» Correct Answer - B Where phase difference is 0, intensity is maximum where, `phi=pi/2 or phi/2=pi/4` the intensity will be `I=I_(max)"cos"^(2)phi/2=I_(max)/2` `:.` The desired ratio is `2:1` |
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| 637. |
Light of wavelength `5000xx10^(-10)m` is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of 5 mm from the central maximum on a screen placed at a distance of 3m from the slit. Then the width of the slits isA. `3cm`B. `0.3cm`C. `0.03cm`D. `0.01cm` |
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Answer» Correct Answer - C `a sin theta = n lambda, a (Y)/(D)= n lambda a = (n lambda D)/(Y)` |
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| 638. |
Two slits at a distance of `1mm` are illuminated by a light of wavelength `6.5xx10^-7m`. The interference fringes are observed on a screen placed at a distance of `1m`. The distance between third dark fringe and fifth bright fringe will beA. `0.65 mm`B. `1.63 mm`C. `3.25 mm`D. `4.88 mm` |
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Answer» Correct Answer - B `DeltaY=(5lambdaD)/d-(5lambdaD)/(2d)=(5lambdaD)/(2d)` `=(5xx6.5xx10^(-7)xx1)/(2xx10^(-3))m=1.63mm` |
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| 639. |
It is found that what waves of same intensity from two coherent sources superpose at a certain point, then the resultant intensity is equal to the intensity of one wave only. This means that the phase difference between the two waves at that point isA. zeroB. `pi/3`C. `(2pi)/3`D. `pi` |
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Answer» Correct Answer - C Intensity, `I=4I_(0)"cos"^(2)phi/2` `because I=I_(0)` `:. phi=(2pi)/3` |
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| 640. |
Light of wavelength `5000Å` is incident normally on a slit. First minimum of diffraction pattern is formed at a distance of 5 mm from the central maximum. If slit width is 0.2 mm, then distance between slit and screen will be :A. 1 mB. 1.5 mC. 2.0 mD. 2.5 m |
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Answer» Correct Answer - B |
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| 641. |
A parallel beam of monochromatic light of wavelength `5000Å` is incident normally on a single narrow slit of width `0.001mm`. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the angle of diffraction equal toA. `0^(@)`B. `15^(@)`C. `30^(@)`D. `60^(@)` |
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Answer» Correct Answer - C In single narrow slit, for first minima, `d sin theta=lambda rArr sin theta=lambda/d` `theta=sin^(-1)((5000xx10^(-10))/(0.001xx10^(-3)))=30^(@)` |
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| 642. |
A parallel beam of monochromatic light of wavelength `5000Å` is incident normally on a single narrow slit of width `0.001mm`. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the angle of diffraction equal toA. (a) `0^@`B. (b) `15^@`C. (c) `30^@`D. (d) `60^@` |
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Answer» Correct Answer - C For the first minima `dsin theta=lambda` `impliessin theta=lambda/dimpliestheta=sin^-1((5000xx10^(-10))/(0.001xx10^-3))=30^@` |
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| 643. |
The first diffraction minima due to a single slit diffraction is at `theta=30^(@)` for a light of wavelength `5000Å` The width of the slit isA. `5xx10^(-5)cm`B. `10xx10^(-5)cm`C. `2.5xx10^(-5)cm`D. `1.25xx10^(5)cm` |
| Answer» Correct Answer - B | |
| 644. |
When the mica sheet of thickness `7`micros and `mu=1.6` is placed in the path of one of interfering beams in the biprism experiment then the central fringe gets at the position of seventh bright fringe.The wavelength of light used will beA. `4000Å`B. `5000Å`C. `6000Å`D. `7000Å` |
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Answer» Correct Answer - 3 `lambda=((mu-1))/(n)....(1)` Accoding to question `n=7,mu=1.6,t=7xx10^(-6)meter...(2)` From eqs (1) and(2) `lambda=6xx10^(-7)`meter |
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| 645. |
A glass plate of `1.2xx10^(-6)`m thickness is placed in the path of one of the interfering beams in a biprism arrangement using monochromatic light of wavelength 6000 Å. If the central bands shifts by a distance equal to the width of the bands, find the refractive index of glass |
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Answer» Correct Answer - 1.5 |
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| 646. |
When a mica sheet of thickness 7 microns and `mu=1.6` is placed in the path of one of interfering beams in the biprism expriment then the central fringe gets at the position of seventh bright fringe. What is the wavelength of light used? |
| Answer» `lamda=((mu-1)t)/(n)=((1.6-1)7xx10^(-6))/(7)=6xx10^(-7)`meter | |
| 647. |
In the ideal double-slit experiment, when a glass-plate(refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wave-length `lamda`), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate isA. `2lamda`B. `(2lamda)/(3)`C. `(lamda)/(3)`D. `lamda` |
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Answer» Correct Answer - A As path difference due to slab `=(mu-1)t` `implies(mu-1)t=nlamda` for munimum thickness t of plate, n should be minimum i.e. `n=1therefore(mu-1)t=lamda` `impliest=(lamda)/(mu-1)impliest=(lamda)/(1.5-1)impliest=2lamda` |
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| 648. |
In a biprism experiment , the path length of the one of the interfering beams is increased by `25 xx 10^(-4)` mm . As a result , the central bright fringe shifts to the position perviously occupied by 5th bright band . Calculate the wavelength of light used . |
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Answer» Correct Answer - 5000 Å |
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| 649. |
In a biprism experiment , the distance of the `15` th bright band from the centre of the interference pattern is 6 mm . Calculate the distance of the 25th bright band and 31st dark band . |
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Answer» Correct Answer - 10 mm , 12.2 mm Given, 15th bright β = 6 mm β = \(\frac{\lambda D}d\) 6 mm = \(\frac{15D\lambda}{d}\) \(\frac{\lambda D}d=\frac6{15}\) (i) 25th bright band β = \(\frac{n \lambda D}d\) β = \(\frac{25\lambda D}d\) (\(\because\) n = 25) \(\because \frac{\lambda D}d=\frac6{15}\) β = \(\frac{25\lambda D}d\) β = 25 x \(\frac6{15}\) β = 30/3 β = 10 mm fringe width for dark band β = (2n + 1)\(\frac{\lambda D}d\) fringe = (2 x 31 + 1) \(\frac{\lambda D}d\) β = 63\(\frac{\lambda D}d\) β = 63 x \(\frac6{15}\) β = 25.2 mm |
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| 650. |
80 gm of impure sugar when dissolved in a litre of water gives an optical rotation of `9.9^(@)` when placed in a tube of length 20 cm. If concentration of sugar solution is 75 gm/litre then specific rotation of sugar is :A. `44^(@)`B. `55^(@)`C. `66^(@)`D. `73^(@)` |
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Answer» Correct Answer - D |
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