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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.
| 451. |
A polaroid is placed at `45^(@)` to an incoming light of intensity `l_(0)` Now, the intensity of light passing through polaroid after polarisation would beA. `l_(0)`B. `l_(0)//2`C. `l_(0)//4`D. zero |
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Answer» Correct Answer - B `I=I_(0) cos^(2) theta=I_(0) cos^(2)45^(@)=I_(0)/2` |
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| 452. |
The angle of polarisation for any medium is `60^(@)` , what will be critical angle for thisA. `sin^(-1)sqrt(3)`B. `tan^(-1)sqrt(3)`C. `cos^(-1)sqrt(3)`D. `sin^(-1)1/sqrt(3)` |
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Answer» Correct Answer - D By using `mu= tan theta_(p)` `rArrmu=tan60^(@)=sqrt(3)` Also `C=sin^(-1)(1/mu)rArrC=sin^(-1)(1/sqrt(3))` |
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| 453. |
The angle of polarisation for any medium is `60^@`. What will be critical angle for this?A. (a) `sin^-1sqrt3`B. (b) `tan^-1sqrt3`C. (c) `cos^-1sqrt3`D. (d) `sin^-1(1)/(sqrt3)` |
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Answer» Correct Answer - D By using `mu=tan theta_pimpliesmu=tan 60=sqrt3`, also `C=sin^-1(1/mu)impliesC=sin^-1(1/sqrt3)` |
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| 454. |
If a wave can be polarised, it must beA. (a) an electromagnetic waveB. (b) a longitudinal waveC. (c) a progressive waveD. (d) a stationary wave |
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Answer» Correct Answer - D Transverse waves can be polarised. These include all electromagnetic waves such as light, radio waves and X-rays. On the other had longitudinal waves such as sound can not be polarised. |
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| 455. |
Two polaroid `A` and `B` are set in crossed positions. A third polaroid `C` is placed between the two making `/_theta` with the pass axis of first polaroid. Write the expression for intensity of light transmitted from second polaroid. In what orientations will be transmitted intensity be (i) minimum (ii) maximum ? |
| Answer» Correct Answer - Refer to `NCERT` | |
| 456. |
Polaroid sheets are often used for making sun glasses. This is because polaroid glass.A. (a) cut off glareB. (b) absorb more light than coloured glassesC. (c) are light-weightD. (d) remove the polarisation of direct sunlight |
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Answer» Correct Answer - A `tantheta=(mu_2)/(mu_1)=(1)/(sin C_2)xx(sin C_1)/(1)=(sin C_1)/(sin C_2)` |
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| 457. |
Which effect provides direct experimental evidence that light is a transverse, rather than a logitudinal wave motion?A. (a) Light can be diffractedB. (b) Two coherent light waves can be made to interfere.C. (c) The intensity of light from a point source falls off inversely as the square of the distance from the source.D. (d) Light can be polarised. |
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Answer» Correct Answer - D Sound, a logitudinal wave, can be diffracted, can be made to interfere, whose intensity falls off inversely as the square of distance from the source, but can not be polarised. Light a transverse wave, can be diffracted, can be made to interfere, whose intensity falls of inversely as the square of distance from the source, can also be polarised. |
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| 458. |
In YDSE, the source placed symmetrically with respect to the slit is now moved parallel to the plane of the slits it is closer to the upper slit, as shown thenA. the fringe width will increase and fringe pattern will shift downB. the fringe width will remain same but fringe pattern will shift upC. the fringe width will decrease and fringe pattern will shift down.D. the fringe width will remain same but fringe pattern will shift down. |
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Answer» Correct Answer - D Fringe width remains unchanged `beta=(lamdad)/(D)` But central bright fringe (zero path difference shift downwards therefore whole fringe pattern shifts downwards). |
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| 459. |
To make the central fringe at the center`O` , mica sheet of refractive index `1.5` is introduced Choose the corect statement. A. The thickness of sheet is `(d^(2))/(D)` in front of `S_(1)`B. The thickness of sheet is `(d^(2))/(D)` in front of `S_(2)`C. The thickness of sheet is `(d^(2))/(2D)` in front of `S_(1)`D. The thickness of sheet is `(d^(2))/(2D)` in front of `S_(1)` |
| Answer» Correct Answer - A | |
| 460. |
it the two waves represented dy `y_(1)=4cos omegat` and `y_(2)=3 cos(omegat+pi//3)` interfere at a point, then the amplitude of the resulting wave will be aboutA. 7B. 5C. 6D. `3.5` |
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Answer» Correct Answer - C Given, `phi=pi/3` `A=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cosphi)` `=sqrt(16+9+2(4)(3)cos 60^(@))approx sqrt(36)=6` |
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| 461. |
If two waves represented by `y_(1)=4sinomegat` and `y_(2)=3sin(omegat+(pi)/(3))` interfere at a point find out the amplitude of the resulting waveA. 7B. 6C. 5D. `3.5` |
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Answer» Correct Answer - B Here phase difference `phi=pi//3` So, amplitude of resulting wave `A=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2) cos phi)` `=sqrt(4^(2)+3^(2)+2xx4xx3 cos pi//3)approx 6` |
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| 462. |
The colour are characterized by which of following character of light-A. FrequencyB. AmplitudeC. WavelengthD. velocity |
| Answer» Correct Answer - A | |
| 463. |
Phenonmenon of interfernece is observed-A. only for light wavesB. only for sound wavesC. for both sound and light wavesD. none of above |
| Answer» Correct Answer - C | |
| 464. |
Whose fringe width will be larger, the one for red light or the one for yellow light, all other things be the same? |
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Answer» we know that, fringe width, `beta=(lambdaD)/d` Now, for same value of D and d, `B prop lambda` Now, `lambda_(red) gtlambda_(yellow)` Hence, `beta_(red) gt beta_(yellow)` Hence, fringe width will be greater when red coloured light is used. |
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| 465. |
Whose fringe width will be more, the one for red light or the one for yellow light, all other things be the same ? |
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Answer» We know that `beta=(lambdaD)/d` Now for same value of D and d `beta propv` Now `lambda_(red) gt lambda_(Yellow)` Hence `beta_(red) gt beta_(yellow)` Hence, fringe width will be greater when red coloured light is used. |
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| 466. |
In a YDSE green light of wavelength 500 nm is used. Where will be the second bright fringe be formed for a set up in which separation between slits is 4 mm and the screen is placed 1 m from the slits? |
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Answer» `y_(2)=(2lambdaD)/d=(2xx500xx10^(-9)xx1)/(4xx10^(-3))=0.25 mm` Note: While reporting our answer we should report `y_(2)=pm0.25mm` as fringe will be formed both above and below central bright fringe. |
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| 467. |
Two radio transmitters radiating in phase are located at point A and B, 250 m apart. The ratio wave have frequency of 3MHz. A radio receiver is moved out from point B along a line BC (perpendicular to AB). The distance from B beyond which the detector does not detect any minima si : A. 200 mB. 400 mC. 600 mD. 500 m |
| Answer» Correct Answer - C | |
| 468. |
Red light of wavelength `750 nm` enters a glass plate of refractive index `1.5`. If velocity of light in vacuum is `3 xx 10^(8) m//s`, calculate velocity, wavelength and frequency of light in glass. |
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Answer» Correct Answer - (i) `4 xx 10^(14)` Hz (ii) `2 xx 10^(8) ms^(-1)` (iii) 500 nm |
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| 469. |
An electromagnetic wave of wavelength `lamda_(0)` (in vacuum) passes from P towards Q crossing three difference media of refractive index `mu,2mu` and `3mu` respectively as shown in figure `phi_(P)` and `phi_(Q)` be the phase of the wave at points P and Q. Find the phase difference `phi_(Q)-phi_(P)`. [take `mu=1`] |
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Answer» Correct Answer - C Optical path difference between (OPD) P & Q (O.P.D) `=2.25lamda_(0)xx1+(3.5lamda_(0))xx2+3lamda_(0)xx3=18.25lamda_(0)` and phase difference `Deltaphi=(2pi)/(lamda_(0))xxDeltax=(pi)/(2)` |
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| 470. |
In a standard `YDSE` apparatus a time film `(mu=1.5,t=2.1mum)` is placed in front of upper slit.How far above or below the center point of the screen are two nearest maximum located? Take `D=1m,d=1mm,lambda=4500Å`.(Symbols have usual meaning).A. `1.5mm`B. `0.6mm`C. `0.15mm`D. `0.3mm` |
| Answer» Correct Answer - C,D | |
| 471. |
In a YDSE apparatus, we use white light then :A. the fringe next totthe central will be redB. the central fringe will be white.C. the fringe next to the central will be violet.D. there will not be a completely dark fringe. |
| Answer» Correct Answer - B,C,D | |
| 472. |
To observe a stationary interference pattern formed by two light waves, it is not necessary that they must haveA. the same frequencyB. same amplitudeC. a constant phase differenceD. the same intensity. |
| Answer» Correct Answer - B,D | |
| 473. |
The intensity of interference waves in an interference pattern is same as `I_(0)`. The resultant intensity at the point of observation will beA. `I = 2I_(0) [1+cos phi]`B. `I= I_(0)[1+cos phi]`C. `I= ([1+cos phi])/(I_(0))`D. `I= ([1+cos phi])/(2I_(0))` |
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Answer» Correct Answer - D `I = I_(1)+I_(2)+2sqrt(I_(1))sqrt(I_(2)) cos phi` `I = I_(0)+I_(0)+2I_(0)cos phi` `= 2I_(0) (1+cos phi)` |
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| 474. |
For the sustained interference of light, the necessary condition is that the two sources shouldA. have constant phase difference onlyB. be narrowC. be close to each otherD. of same amplitude with constant phase difference |
| Answer» Correct Answer - D | |
| 475. |
To observe a stationary interference pattern formed by two light waves, which of the following is not a necessary condition:A. Same frequencyB. Same amplitudeC. Constant phase differenceD. Same intensity |
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Answer» Correct Answer - A |
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| 476. |
Refractive index of a thin soap film of a uniform thickness is 1.34. Find the smallest thickness of the film that gives in interference maximum in the reflected light when light of wavelength `5360 Å` fall at normal incidence. |
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Answer» Condition for maximum interference in the reflected light, in case of thin film interference, can be expressed as `2 mu t cos r = (2 n - 1) lamda//2, n = 1, 2,…` [for plane parallel films] where `mu` is the refractive index of film relative ot the surrounding: `t` is the thickness of film , and r is the angle of refraction. For normal incidence, `r = 0` `:. 2mu t = (2n - 1) lambda//2, n = 1, 2, 3 ,...` For minimum thickness, `n = 1` `:. 2 mu t_(min) = lambda//2` or `t_(min) = (lambda)/(4 mu)` Given: `lambda = 5360 Å` and `mu = 1.34` `:. t_(min) = (5360)/(4 xx 1.34)` or `t_(min) = 1000`. |
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| 477. |
Consider a case of thin film interference as shown.Thickness of film is equal to wavelength of light in `mu_(2)` A. Reflected light will be maxima if `mu_(1)ltmu_(2)ltu_(3)`B. Reflected light will be maxima if `mu_(1)ltmu_(2)gtu_(3)`C. Transmitted light will be maxima if `mu_(1)ltmu_(2)gtu_(3)`D. Transmitted light will be maxima if `mu_(1)gtmu_(2)gtu_(3)` |
| Answer» Correct Answer - A::C | |
| 478. |
Which of the following is not an essential condition for interference?A. The two interfering waves must propagation in almost the same directionB. The waves must propagation in almost the same directionC. The amplitudes of the two waves must be equalD. The two interfering beams of light must originate from the same source |
| Answer» Correct Answer - C | |
| 479. |
Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation when blue- green light of wavelength 500 nm is used ? |
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Answer» Fringe spacing `=(Dlamda)/(d)=(1xx5xx10^(-7))/(1xx10^(-3))m` `=5xx10^(-4)m=0.5mm` |
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| 480. |
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 central bright bandC. interference with central dark bandD. interference in which breadth of the fringe will be slightly increased. |
| Answer» Correct Answer - B | |
| 481. |
A slit of width a is illuminated by white light. For red light `(lambda=6500Å)`, the first minima is obtained at `theta=30^@`. Then the value of a will beA. (a) `3250Å`B. (b) `6.5xx10^-4mm`C. (c) `1.24` micronsD. (d) `2.6xx10^-4cm` |
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Answer» Correct Answer - C For first minima `theta=lambda/a` or `a=lambda/theta` `:. a=(6500xx10^-8xx6)/(pi)` (As `30^@=pi/6` radian) `=1.24xx10^-4cm=1.24` microns |
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| 482. |
A slit of width d is illuminated by red light of wavelength `6500Å` For what value of d will the first maximum fall atangle of diffractive of `30^(@)` |
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Answer» For first secondary maximum of the diffraction pattern, `d isn theta =(3lambda)/2` `:. D=(3lambda)/(2 sin theta)=(3xx6,500xx10^(-10)m)/(2xx sin 30^(@))=1.95 xx 10^(-6)m` |
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| 483. |
In a single slit diffraction experiment, the width of the slit is made double its original width. Then the central maximum of the diffraction pattern will becomeA. narrower and fainterB. narrower and brighterC. broader and fainterD. broader and brighter |
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Answer» Correct Answer - B The angular width of the central maximum is `2lambda//a` where a is the width of the slit. If the value of a is doubled, the angular width of the central maximum decreases to half its earlier value. This implies that the central maximum becomes much sharper. Furthermore if a is doubled, the intensity of the central maximum becomes four times. Thus the central maximum becomes much sharper and brighter. |
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| 484. |
Statement - I : Diffraction of sound waves is evident in daily experince than that of light waves Statement- II : The wave length of sound waves is comparitively than that of light waves.A. Statement-I is true and Statement-II is true and Statement-II is the correct explanation of Statement-I.B. Statement-I asnd Statement-II are true but Statement-II is not the correct explanation of Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
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Answer» Correct Answer - A Condition for differaction is width of the slit `d le lambda` |
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| 485. |
To ensure almost `100%` transmittivity, photographic lenses are often coated with a thin layer of dielectric maerial, like `MgF_(2)(mu=1.38)` . The minimum thickness of the film to be used so that at the centre of visible spectrum `(lambda = 5500 Å)` there is maximum transmission.A. `1000Å`B. `2000Å`C. `4000Å`D. 500Å` |
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Answer» Correct Answer - A for interfrence at thin films `(2mu cosd r) = (lambda)/(2)` Here `i~~r~~0` `:. T = (lambda)/(4mu)` |
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| 486. |
A long horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and the after reflection is seen on a screen 1 m away from the slit. If the mirror reflects only 64% of the light falling on it, the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen isA. `8:1`B. `3:1`C. `81:1`D. `9:1` |
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Answer» Correct Answer - c Intensity of direct ray `= I_(0) = kA_(0)^(2)` Intensity of reflected ray `= (64)/(100) I_(0) = k ((8 A_(0))/(10))^(2)` `:. (I_(max))/(I_(min)) = ((A_(0) + 0.8 A_(0))^(2))/((A_(0) - 0.8 A_(0))^(2)) = ((1.85)/(0.8))^(2) = (81)/(1)` |
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| 487. |
Light is incident at an angle `phi`with the normal to a vertical plane containing two narrow slits (S1 and S2) at separation d. The medium to the left of slit plane is air and wavelength of the incident light is `lambda`. The medium to the right of the slit plane has refractive index `mu`. Find all values of angular position `(theta) `of a point P where we will observe constructive interference. Wavelength of incident light is `lambda`. |
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Answer» Correct Answer - `theta=sin^(-1)[(1)/(mud)(nlambda-dsinphi)];[n=0,pm1,pm2..]` |
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| 488. |
In the arrangement shown, S is a point source of monochromatic light. S1 and S2 are two slits located sym- metrically with respect to the source with separation between them` d_(1)`. Parallel to this slit plane there are two more slits (S3 and S4) at separation` d_(2)`. These slits are also symmetrically located with respect to S. A screen is at a distance `L_(2)` from this slit plane. How does the intensity on the screen change with y and `d_(1)`? |
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Answer» Correct Answer - `"Intensity" propcos^(2)((pid_(1)d_(2))/(2lambdaL_(1)))cos^(2)((pid_(2)y)/(2lambdaL_(2)))` |
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| 489. |
When two narrow slits separated by a small distance are illuminated by a light of wavelength ` 5 xx 10^(-7)` m , interference fringes of width `0.5` mm are obtained on a screen . What should be the wavelength of light source to obtain fringes `0.3` mm wide , if the distance between the screen and the slits is reduced to half of the initial value . |
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Answer» Correct Answer - `6 xx 10^(-7)`m |
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| 490. |
A very thin prism has an apex angle A and its material has refractive index` mu = 1.48` . Light is made to fall on one of the refracting faces at near normal incidence. Interference results from light reflected from the outer surface and that emerging after reflection at the inner surface. When violet light of wavelength `lambda "= 400 nm"` is used, the first construc- tive interference band is observed at a distance `"d = 3.0 cm"` from the apex of the prism. (a) Find the apex angle A. (b) If red light `(lambda" = 800 nm")` is used, at what distance from the apex will we observe the first constructive interfer- ence band. |
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Answer» Correct Answer - `(a)1.3xx10^(-4)"degree"` `(b)6 "cm"` |
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| 491. |
Three narrow slits A, B and C are illuminated by a parallel beam of light of wavelength` lambda` , P is a point on the screen exactly in front of point A. Slit plane is at a distance D from the screen`(D gtgt lambda)`. It is know that `"BP – AP ="(1)/(3)` (a) Find d in terms of D and` lambda` (b) Write the phase difference between waves reaching at P from C and A. (c) If intensity of P due to any of the three slits individu- ally is `I_(0)`, find the resultant intensity at P. |
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Answer» Correct Answer - `(a)sqrt((2Dlambda)/(3))` `(b)(8pi)/(3)` `(c)3I_(0)` |
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| 492. |
Two plane mirrors, a source S of light, emitting wavelengths of` lambda_(1) = 4000overset(@)A" and" lambda_(2) = 5600overset(@)A`and a screen are arranged as shown in figure. The angle `theta` shown is 0.05 radian and distance a and b are 1 cm and 38 cm respectively.(a) Find the fringe width of interference pattern formed on screen by the blue light. (b) Calculate the distance of first black line from central bright fringe. (c) Find the distance between two black lines which are nearest to the central bright fringe. |
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Answer» Correct Answer - (a)`80mu"m"` (b)280`mu"m"`(`c)560mu"m"` |
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| 493. |
Two point source separated by `d=5mum` emit light of wavelength `lamda=2mum` in phase A cicular wire of radius `20mum` is placed aournd the source as shown in figure.A. Points A and B are dark and points C and D are bright.B. Point A and B are bright and point C and D are darkC. Point A and C are dark and points B and D are bright.D. Points A and C are bright and points B and D are dark. |
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Answer» Correct Answer - D Path difference for points A & C is O so constructive interference take place. Path difference for B & D `=5mum=(5lamda)/(2)` So distructive interference take place. |
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| 494. |
Find the maxinum intensity in case of interference of n identical waves each of intensity `I_(0)` if the interference is (a) coherent and (b) incoherent. |
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Answer» The resultant intensity is given by `I = I_(1) + I_(2) + 2 sqrt( I_(1) I_(2)) cos phi` a. The sources are said to be coherent if they have constant phase difference between them. The intensity will be maximum when `f = 2 np,` the sources are in same phase. Thus `I_(max) I_(1) + I_(2) + 2 sqrt(I_(1) I_(2)) = ( sqrt I_(1) + sqrt I_(2))^(2)` Similarly, for n indentical waves, `I_(max) = (sqrt I_(0) + sqrt I_(0) + ....)^(2) = n^(2) I_(0)` b. The incoherent sources have phase difference that varies randomly with time. Thus, `[cos phi]_(av) = 0` Hence, `I = I_(2) + I_(2)` Hence, for n idental waves, `I = I_(0) + I_(0) + ... = n I_(0)` |
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| 495. |
Two incoherent sources of light emitting light of intensity `I_(0)` and `3I_(0)` interfere in a medium. Calculate, the resultant intensity at any point. |
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Answer» Since, the two sources are incoherent, hence we do not obtain any sustained interference pattern. We observe an average intensity `I=I_(1)+2sqrt(I_(1)I_(2))cosphi` Now, `cosphi=0` in one complete cycle. `:.I=I_(1)+I_(2)` which is `4I_(0)` |
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| 496. |
In a Fresnel biprism experiment, the two positions of lens give separation between the slits as 16cm and 9cm respectively. What is the actual distance of separation?A. 12 cmB. 12.5 cmC. 13 cmD. 14 cm |
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Answer» Correct Answer - A In Fresnel biprism experiment, the actual distance of separation between the two slits, `d=sqrt(d_(1)d_(2))=sqrt(16xx9)=12cm` |
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| 497. |
In a Fresnel biprism experiment, the two positions of lens give separation between the slits as 16cm and 9cm respectively. What is the actual distance of separation?A. 10.5 cmB. 12 cmC. 13 cmD. 14 cm |
| Answer» Correct Answer - B | |
| 498. |
A parallel coherent beam of light falls on fresnel biprism of biprism of refractive index `mu` and angle `alpha` .The fringe width on a screen at a distance `D` from biprism will be (wavelength =`lambda`)A. `(lamda)/(2(mu-1)alpha)`B. `(lamdaD)/(2(mu-1)alpha)`C. `(D)/(2(mu-1)alpha)`D. none of these |
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Answer» Correct Answer - A `beta=((a+b)lamda)/(2a(mu-1)alpha)=((1+(b)/(a))lamda)/(2(mu-1)alpha)` For parallel beam, `a=infty,` so `beta=(lamda)/(2(mu-1)alpha)` |
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| 499. |
In a biprism experiment with sodium light,bands of width `0.0195cm` are observed at`100cm` from slit.On introducing a convex lens `30cm` away from the silt between biprism and screen,two images of the slit are seen`0.7cm`apart of `100cm` distance from the slit .Calcualte the wavelength of sodium light . |
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Answer» Correct Answer - `lambda=5850Å` `x_1=(n_1lambda_1D)/d` & `x_2=(n_2lambda_2D)/d` so, `x_2/x_1=(n_2lambda_2)/(n_1lambda_1)` `2.7/2=(30xxlambda_2)/(20xxlambda_1)` `lambda_2=0.9 lambda_1` `lambda_2=5400 Å` |
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| 500. |
The path difference between two interfering waves at a point on screen is 171.5 times the wavelength if the path difference is 0.01029 cm find the wavelength. |
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Answer» Path difference `=171.5lamda=(343)/(2)lamda=` odd multiple of half wavelength. It means dark finge is observed Accoding to question `0.01029=(343)/(2)lamdaimplieslamda=(0.01029xx2)/(343)=6xx10^(-5)cmimplieslamda=6000Å` |
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