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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.
| 351. |
An unpolarised beam of intensity `I_(0)` falls on a polariod. The intensity of the emergent light is :A. `(I_(0))/(2)`B. `I_(0)`C. `(I_(0))/(4)`D. Zero |
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Answer» Correct Answer - D |
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| 352. |
How many orders will be visible if the wavelength of the incident radiation is 5000 Å and the number of lines on the grating is 101319 per metre. |
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Answer» Correct Answer - 19 Given : d = \(\cfrac1{101319}\) λ = 5000 x 10-10 m λ = 5 x 10-7 m using d sin θ = nλ θ = 90° d = nλ n = \(\cfrac{d}λ\) = \(\cfrac1{5\times10^{-7}}\) x \(\cfrac1{101319}\) n = \(\cfrac{10000000}{506595}\) = 19.73 |
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| 353. |
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. The phase of the ligth wave at c, d, e, and f are `phi_(c)`, phi_(d), `phi_(e)` and `phi_(f)`, respectively. It is given that `phi_(c ) != phi_(f)`. ThenA. `phi_(c )` cannot be equal to `phi_(d)`B. `phi_(d )` can be equal to `phi_(d)`C. `phi_(d ) - phi_(f)` is equal to `phi_(c) - phi_(e)`D. `phi_(d ) - phi_(c)` is not equal to `phi_(f) - phi_(e)` |
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Answer» Correct Answer - c All points on a wavefront are at the same phase `phi_(d) = phi_(c)` and `phi_(f) = phi_(e)` `phi_(d) - phi_(f) = phi_(c) - phi_(e)` Hence, the correct option is (c). |
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| 354. |
A horizontal beam of vertically polarized light of intensity `43 W//m^(2)` is sent through two polarizing sheets. The polarizing direaction of the first is `60^(@)` to the vertical, and that of the second is horizontal. The intensity of the light transmitted b y the pair of sheets is (nearly)A. `8.1 W//m^(2)`B. `7.3 W//m^(2)`C. `6.4 W//m^(2)`D. `3.8 W//m^(2)` |
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Answer» Correct Answer - A `I_(2)= (I_(0))/(2)cos^(2)theta` |
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| 355. |
The `6563 Å H_(2)` line emitted by hydrogen in a star is found to be red shifted by `15 Å`. Estimate the speed with which the star is receding from earth.A. `17.3xx10^(3) m//s`B. `4.29xx10^(7) m//s`C. `3.39xx10^(5)`m/sD. `2.29xx10^(5)`m/s |
| Answer» Correct Answer - D | |
| 356. |
If the lower slit has been covered by the sheet, what will be the direction of shift of the pattern? |
| Answer» Correct Answer - Downwards | |
| 357. |
YDSE set up shown in figure. A. Zero order maxima will lie above point PB. First order maxima may lie above point PC. First order maxima may lie below point PD. Zero order maxima may lie at point P |
| Answer» Correct Answer - 1 | |
| 358. |
How much a beam would diffract when it travels a distance of 5 m, after crossing an aperture of 3mm having wavelength 3000 Å ? |
| Answer» `(zlambda)/a=(5xx3000xx10^(-10))/(3xx10^(-3))=5xx10^(-4)m` | |
| 359. |
A radar operates at wavelength 50.0 cm. If the beat frequency between the transmitted singal and the singal reflected from aircraft `(Deltav)` is 1 kHz, then velocity of the aircraft will be :A. 800 km/hrB. 900 km/hrC. 1000 km/hrD. 1032 km/hr |
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Answer» Correct Answer - B |
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| 360. |
An initially parallel cylindrical beam travels in a medium of refractive index `mu(I)=mu_0+mu_2I`, where `mu_0` and `mu_2` are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. 31. The initial shape of the wavefront of the beam isA. planarB. convexC. concaveD. covnex near of the axis and concave near the periphery. |
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Answer» Correct Answer - 1 Since rays are parallel ,so wavefront mustbe planar. |
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| 361. |
Two tower are built on the hill 50 Km apart and then joining them passes 30 m above a hill halfway in between. What is the longest wavelength of radiowaves which can be sent between the toers without serious diffraction effects ? |
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Answer» Correct Answer - 0.036 m |
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| 362. |
Two towers on the top of two hills are 40 km apart. The line joining them presses 50 m above a hill half way between the towers. What is the longest wavelength of radiowaves which can be send between the towers without apprecialbe fiffraction effects?A. 1.25 mB. 0.125 mC. 2.50 mD. 0.250 m |
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Answer» Correct Answer - B Distance between one of the towers and hill half way `= 40//2km=2xx10^(4) km` ` :. z_(F)=2xx10^(4)km` Using, ` :. z_(F)=(a^(2))/(lambda) " or " lambda=(a^(2))/(z_(F))` ` :. lambda=((50)^(2))/(2xx10^(4))=1250xx10^(-4)=0.125m` |
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| 363. |
Answer the questions : Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable so see each other even though they can converse easily. |
| Answer» For diffraction the size of the obstacle should be comparable to wavelength if the size of the obstacle is much too large compared to wavelength, diffraction is by a small angie. Here the size partition of wall is of the order of few metres. The wavelength of light is about `5 xx 10^(-7)m,` while sound waves of say 1 kHZ frequency have wavelength of about 0.3 m. Thus sound waves can bend around the partition while light waves cannot. | |
| 364. |
In studying diffraction pattern of different obstacles, the effect ofA. full wave front is studiedB. portion of a wave front is studiedC. waves from two coherent sources is studiedD. waves from one of the coherent source is studied. |
| Answer» Correct Answer - B | |
| 365. |
Light travels in a straight line becauseA. it is not absorbed by atmosphereB. its velocity is very highC. differaction effect is negligibleD. due to interference |
| Answer» Correct Answer - C | |
| 366. |
Diffraction effects are easier to notice in the case of sound waves than in the case of light waves becauseA. Light wave do not require mediumB. Wavelength of light waves is far smallerC. Light waves are transverseD. Speed of light is far greater |
| Answer» Correct Answer - B | |
| 367. |
Direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by (a is the width of the slit)A. `asintheta=(lamda)/(2)`B. `acostheta=(3lamda)/(2)`C. `asintheta=lamda`D. `asintheta=(3lamda)/(2)` |
| Answer» Correct Answer - D | |
| 368. |
A single slit of width d is placed in the path of beam of wavelength `lamda`. The angular width of the principal maximum obtained is-A. `(d)/(lamda)`B. `(lamda)/(d)`C. `(2lamda)/(d)`D. `(2d)/(lamda)` |
| Answer» Correct Answer - C | |
| 369. |
Diffraction of light is observed only, when the obstacle size is-A. should be much larger than the wavlengthB. has no relation to wavelengthC. should be of the order of wavelengthD. should be `lambda//2` where `lambda` is the wavelength |
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Answer» Correct Answer - C To observed diffraction the size of the obstacle should be of the order of wavelength `lambda approx d` |
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| 370. |
All fringes of diffraction are of-A. the same intensityB. unequal widthC. the same widthD. full darkness |
| Answer» Correct Answer - B | |
| 371. |
which of the following any gives more distinct diffractionA. X-rayB. light rayC. `gamma`-rayD. radia wave |
| Answer» Correct Answer - D | |
| 372. |
Diffraction initiated from obstacle, depends upon theA. size of obstacleB. wave length and distance of obstacle from screenC. size of obstacle and its distance from screenD. |
| Answer» Correct Answer - B | |
| 373. |
Diffraction of light is observed only, when the obstacle size is-A. very largeB. very smallC. of the same order that of wavelength of lightD. any size |
| Answer» Correct Answer - C | |
| 374. |
The phenomenon of diffraction of light was discovered by-A. YoungB. HertzC. GirmaldiD. Malus |
| Answer» Correct Answer - C | |
| 375. |
Phenomenon of diffraction occursA. only in case of light and sound wavesB. for all kinds of wavesC. for electro-magnetic waves and not for matter wavesD. for light waves only |
| Answer» Correct Answer - B | |
| 376. |
Which of the following phenomenon is not common to sound and light wavesA. InterferenceB. DiffractionC. PolarisationD. Reflection |
| Answer» Correct Answer - C | |
| 377. |
Atmospheric refraction is due toA. changing pressure in the atmosphereB. varying density of atmosphereC. varying temperature of the atmosphereD. both (2) and (3) |
| Answer» Correct Answer - D | |
| 378. |
Two coherent sources must have the sameA. amplitudeB. phase differenceC. frequencyD. both (B) and (C) |
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Answer» Correct Answer - D From the definition of coherent source. |
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| 379. |
If the relative permitivity and relative permeability of a medium are 2 and 1.28 respectively, then refractive index of that material is:A. 1.25B. 25/16C. `5/8`D. `1.6` |
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Answer» Correct Answer - D Refractive index `n=sqrt(mu_(r)epsilon_(r))=sqrt(2xx1.28) =1.6` |
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| 380. |
A plane wave front falls on a convex lens. The emergent wave front isA. planeB. diverging sphericalC. converging sphericalD. none of these. |
| Answer» Correct Answer - C | |
| 381. |
Spherical wavefronts, emanating from a point source, strike a plane reflecting surface. What will happen to these wave fronts, immediately after reflection?A. They will remain spherical with the same curvature, both in magnitude and sign.B. They will become plane wave fronts.C. They will become plane wave fronts.D. They will remain spherical, but with different curvature, both in magnitude and sign. |
| Answer» Correct Answer - C | |
| 382. |
A thin oil film of refractive index 1.2 floats on the surface of water `(mu = 4/3)`. When a light of wavelength `lambda = 9.6xx 10^(-7)m` falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is:A. `10^(-7)m`B. `2xx10^(-7)m`C. `3xx10^(-7)m`D. `5xx10^(-7)m` |
| Answer» Correct Answer - B | |
| 383. |
In the figure shown S is a point monochromatic light source of frequency `6xx10^(14)Hz`.M is a concave mirror of radius of curvature 20 cm and L is a thin converging lens of focal length 3.75 cm AB is the principal axis of M and L. Light reflected from the mirror and refracted from the lens in succession reaches the screen. An interference pattern is obtained on the screen by this arrangement. Q. Fringe width isA. 1 mmB. 0.5 mmC. 1.5 mmD. 0.25 mm |
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Answer» Correct Answer - B `beta=(5xx10^(-7)xx50xx10^(-2))/(0.5xx10^(-3))=5xx10^(-4)m=0.5mm` |
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| 384. |
A wavefront expressed by`" x + y – z = 4"` is incident on a plane mirror which is lying parallel to xy plane. Write a unit vector in the direction of reflected ray. |
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Answer» Correct Answer - `(1)/sqrt(3)(hati+hatj+hatk)` |
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| 385. |
A parallel beam of light is travelling along a direction making an angle of `30^(@)` with the positive X direction and `60^(@)` with positive Y direction. Wavelength is `lambda`. Find the phase difference between points having co-ordinates `(1, sqrt( 3) , 2) "and" (0, 0, 0)`. |
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Answer» Correct Answer - `(2sqrt(3)pi)/(lambda)` |
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| 386. |
A point source of light is being moved closer to a thin concave lens from a large distance on its principal axis. Is the radius of curvature of the refracted wave front, close to the lens, increasing or decreasing? |
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Answer» Correct Answer - Decreasing |
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| 387. |
In young’s double slit experiment, when the slit plane is illuminated with light of wavelength` lambda_(1)`, it was observed that point P is closest point from central maximum O, where intensity was 75% the intensity at O. When the light of wavelength `lambda_(2)` is used, point P happens to be the nearest point from O where intensity is 50% of that at O. Find the ratio`(lambda_(1))/(lambda_(2))`. |
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Answer» Correct Answer - `3/2` |
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| 388. |
Three media are arranged in various ways as shown in figures a, b, c and d. Light of wavelength` lambda` is incident perpendicularly on the boundary of the middle layer and interference between waves reflected at the boundaries of the middle layer is studied. In which of the four cases the reflected light is eliminated by destructive interference when the thickness of middle layer approaches zero.` mu_(1) = 1.5" and" mu_(2) = 1.8` |
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Answer» Correct Answer - a,b. |
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| 389. |
Intensity of central bright fringe due to interference of two identical coherent monochromatic sources is I. If one of the source is switched off, then intensity of central bright fringe becomes :A. `(I)/(2)`B. `(I)/(3)`C. `(I)/(4)`D. I |
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Answer» Correct Answer - C |
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| 390. |
Three coherent sources` S_(1), S_(2)" and" S_(3)` can throw light on a screen. With `S_(1)` switched on intensity at a point P on the screen was observed to be I. With only`S_(2)` on, intensity at P was 2I and when all three are switched on the intensity at P becomes zero. Intensity at P is I when `S_(1)" and" S_(2)` are kept on. Find the phase difference between the waves reaching at P from sources `S_(1) "and" S_(3)`. |
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Answer» Correct Answer - `pi/2` |
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| 391. |
Statement-1 : When a white light passes through a prism it forms a spectrum of seven colours. Statement-2 : The refractive index for different colour of light is different for the material of prism.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-3B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-2C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
| Answer» Correct Answer - A | |
| 392. |
Is the speed of light in glass independent of the colour of light ? If not, which of the two colours red and violet travels slower in a glass prism ? |
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Answer» No, the refractive index and speed of light in a medium depend on wavelength i.e, colour of light. We know that `mu_(v) gt mu_(r)`. Therefore, `v_("violet") lt v_("red")`. Hence violet component of white travel slower than the red component. |
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| 393. |
Monochromatic light of wvalength `589 nm` is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected and (b) refracted light ? `mu` of water is 1.33`. |
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Answer» Wavelength of incident monochromatic light , `lamda =589 nm=589xx10^(-9)` m Speed fo light in air ,`C=3xx10^(8) m//s` Refractive index of water ,`mu =1.33 ` (a) the ray will reffect back in the same medium as that of incident ray , hence , the wavelength , speed and frrequency of the refiect ray will be the same as that of the incident ray . Frequency of light is given by the relation , ` V=(c) /(lamda) ` `=(3xx10^(8))/(589xx10^(-9))` `=5.09 xx10^(14)` Hz Hence the speed frequency and wavelength of the refliegth of the reflected light are `3xx10^(8)` m/s `5xx10^(14)` Hz and 589 nm respecticely . (b) FRequency of light does not sepend on the property of the medium in which it is Travelling Hence the freqency of the refreative index of water as : `V=(c)/(mu)` `V=(3xx10^(8))/(1.33) = 2.26 xx10^(8) `m/s Wavelength of light in water is given by the relation , `lamda =(V)/(V)` `=(2.26 xx10^(8))/(5.09xx10^(14))` `=444.007 xx10^(-9)`m =444.01 nm Hence th speed freuency and wavelength of refracted light are `2.26xx10^(8)` m/s 444.01 nm and `5.09 xx10^(14)` Hz respectively . |
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| 394. |
A YDSE is performed in a medium of refractive index `4 // 3`, A light of 600 nm wavelength is falling on the slits having 0.45 nm separation . The lower slit `S_(2)` is covered b a thin glass plate of thickness 10.4 mm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in figure. (All the wavelengths in this problem are for the given medium of refractive index `4 // 3`, ignore absorption.) Find the light intensity at point O relative t maximum fringe intensity.A. `0.6I_(max)`B. `0.75I_(max)`C. `0.8I_(max)`D. `0.9I_(max)` |
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Answer» Correct Answer - B |
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| 395. |
In a modified YDSE, sources `S` is kept in front of slit `S_(1)`. Find the phase difference at point O that is equidistant from slits `S_(1)` and `S_(2)` and point P that is in front of slit `S_(1)` in the following situations. A liquid of refractive index `mu` is filled between the screen and slits.A. `(2pi)/(lambda) = [[ sqrt(d^(2) + x_(0)^(2)) + x_(0)] + (mu d^(2))/(2D)]`B. `(2pi)/(lambda) = [[ sqrt(d^(2) + x_(0)^(2)) - x_(0)] + (mu d^(2))/(2D)]`C. `(2pi)/(lambda) = [[ sqrt(d^(2) - x_(0)^(2)) + x_(0)] + (mu d^(2))/(2D)]`D. `(2pi)/(lambda) = [[ sqrt(d^(2) - x_(0)^(2)) - x_(0)] + (mu d^(2))/(2D)]` |
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Answer» Correct Answer - b While calculating path difference, it must be remembered that is is equal to difference of optical path lengths from sources S to point on the screen. `:. Delta x_("total") = [Delta x]_("before slits") + [Delta x]_("after slits")` ` = (SS_(2) - SS_(1)) + (S_(2) P - S_(1) P)` ltbgt a. When liquid is filled between the slits and screen, then `[S_(2) P]_("liquid") = [mu S_(2) P]_("air")` `[S_(1) P]_("liquid") = [mu S_(1) P]_("air")` At point O: `mu S_(2) O = mu S_(1) O` Noth path difference is introduced after slits. So, `Delta x_("total") = SS_(2) - SS_(1) = sqrt(d^(2) + X_(0)^(2)) - x_(0)` Thus, phase diffenence, `Delta phi = (2pi)/(lambda) (sqrt(d^(2) + X_(0)^(2)) - x_(0))` At point P: `[Delta x]_("before slits") = sqrt(d^(2) + x_(0)^(2)) - x_(0)` `[Delta x]_("after slits") = mu S_(2) P - mu S_(1) ap = mu (S_(2)P - S_(1)P)` `= (mu yd)/(D) = (mu d^(2))/(2 D) ("as" y = (d)/(2))` Thus phase difference, `Delta phi = (2pi)/(lambda) [(sqrt(d^(2) + x_(0)^(2)) - x_(0)) + (mu d^(2))/(2 D)]` b. When liquid is filled between the source and slits: At point O: `(Delta x)_("before slits") = (SS_(2) + SS_(1))_("liquid")` `(mu SS_(2) + mu SS_(1))_("air")` `= mu (sqrt(d^(2) + x_(0)^(2)) - x_(0))` `(Delta x)_("after slits") = S_(2) O + S_(1) O = 0` `(Delta x)_("total") = mu (sqrt(d^(2) + x_(0)^(2)) - x_(0))` Thus, phase differnce at P, ` Delta phi = (2pi)/(lambda) (mu sqrt(d^(2) + x_(0)^(2)) - x_(0))` At point P: `(Delta x)_("before slits") = (SS_(2) + SS_(1))_("liquid") = (mu SS_(2) + mu SS_(1))_("air")` `(Delta x)_("after slits") = (S_(2)P - S_(1)P)_("air") = (y d)/(D) = (d^(2))/(2D)` `(Delta x)_("total") = [(mu sqrt(d^(2) + x_(0)^(2)) - x_(0)) + (d^(2))/(2D)]` Thus, phase difference at P, `Delta phi = (2 pi)/(lambda) [mu sqrt(D^(2) + x_(0)^(2)) - x_(0) + (d^(2))/(2D)]` |
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| 396. |
In the arrangment shown in fin. Find the distance x at which the next fringe is formed.A. `(3 lambda)/(2)`B. `( lambda)/(4)`C. `( lambda)/(2)`D. `(5 lambda)/(2)` |
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Answer» Correct Answer - c For dark spot at O, `Delta x = (AB + BO) - AO` `= 2(D^(2) + d^(2))^(1//2) - 2 D = 2D [(1 + (d^(2))/(D^(2)))^(1//2)-1]` `implies Delta x =(d^(2))/(D)` `(d^(2))/(D) = (lambda)/(2) implies d = sqrt((D lambda)/(4)` `Detla x` at P: `Delta x = (AB + BP) - (AC + PC)` `= (AB - AC) + (BP - PC)` `Delta x` at O: `Delta x = (lambda)/(2)` The path difference at P will be zero if `x = d`. Hence next maxima will front at P. Fringe width = distance between two consecutive dark to brigth fringes `= 2 xx` distance between two consective bright dark fringes `= 2 xx d` |
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| 397. |
In a midified YDESE, sources S is kept in front of slit `S_(1)`. Find the phase difference at point O that is equidistant from slits `S_(1)` and `S_(2)` and point P that is in front of slit `S_(1)` in the following situations. ltgt Liquid is filled between the slit and source S.A. `(2pi)/(lambda) = [mu sqrt(d^(2) + x_(0)^(2)) - x_(0) - (d^(2))/(2D)]`B. `(2pi)/(lambda) = [mu sqrt(d^(2) + x_(0)^(2)) - x_(0) + (d^(2))/(2D)]`C. `(2 pi)/(lamda)[mu sqrt(d^(2)-x_(0)^(2))+x_(0)+(d^(2))/(2D)]`D. `(2pi)/(lambda) = [mu sqrt(d^(2) + x_(0)^(2)) + x_(0) + (d^(2))/(2D)]` |
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Answer» Correct Answer - b While calculating path difference, it must be remembered that is is equal to difference of optical path lengths from sources S to point on the screen. `:. Delta x_("total") = [Delta x]_("before slits") + [Delta x]_("after slits")` ` = (SS_(2) - SS_(1)) + (S_(2) P - S_(1) P)` ltbgt a. When liquid is filled between the slits and screen, then `[S_(2) P]_("liquid") = [mu S_(2) P]_("air")` `[S_(1) P]_("liquid") = [mu S_(1) P]_("air")` At point O: `mu S_(2) O = mu S_(1) O` Noth path difference is introduced after slits. So, `Delta x_("total") = SS_(2) - SS_(1) = sqrt(d^(2) + X_(0)^(2)) - x_(0)` Thus, phase diffenence, `Delta phi = (2pi)/(lambda) (sqrt(d^(2) + X_(0)^(2)) - x_(0))` At point P: [Delta x]_("before slits") = sqrt(d^(2) + x_(0)^(2)) - x_(0)` `[Delta x]_("after slits") = mu S_(2) P - mu S_(1) ap = mu (S_(2)P - S_(1)P)` `= (mu yd)/(D) = (mu d^(2))/(2 D)` (as y = (d)/(2))` Thus phase difference, `Delta phi = (2pi)/(lambda) [(sqrt(d^(2) + x_(0)^(2)) - x_(0)) + (mu d^(2))/(2 D)]` b. When liquid is filled between the source and slits: At point O: `(Delta x)_("before slits") = (SS_(2) + SS_(1))_("liquid")` `(mu SS_(2) + mu SS_(1))_("air")` `= mu (sqrt(d^(2) + x_(0)^(2)) - x_(0))` `(Delta x)_("after slits") = S_(2) O + S_(1) O = 0` `(Delta x)_("total") = mu (sqrt(d^(2) + x_(0)^(2)) - x_(0))` Thus, phase differnce at P, `Delta phi = (2pi)/(lambda) (mu sqrt(d^(2) + x_(0)^(2)) - x_(0))` At point P: `(Delta x)_("before slits") = (SS_(2) + SS_(1))_("liquid") = (mu SS_(2) + mu SS_(1))_("air")` `(Delta x)_("after slits") = (S_(2)P - S_(1)P)_("air") = (y d)/(D) = (d^(2))/(2D)` `(Delta x)_("total") = [(mu sqrt(d^(2) + x_(0)^(2)) - x_(0)) + (d^(2))/(2D)]` Thus, phase difference at P, `Delta phi = (2 pi)/(lambda) [mu sqrt(D^(2) + x_(0)^(2)) - x_(0) + (d^(2))/(2D)]` |
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| 398. |
The figure shows the interfernece pattern obtained in double slit experiment using light of wavelength 600 nm. Q. Which fringe results from a phase difference of `4pi` between the light waves incidenting from two slits?A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - C At central bright fringe the waves from slits are in phase and following bright fringes having a difference of `2pi.4pi=2xx2piimplies2^(nd)` order bright fringe. |
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| 399. |
The figure shows the interfernece pattern obtained in double slit experiment using light of wavelength 600 nm. Q. Which fringe results from a phase difference of `4pi` between the light waves incidenting from two slits?A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - c see theory |
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| 400. |
The figure shows the interfernece pattern obtained in double slit experiment using light of wavelength 600 nm. Q. The third order bright fringe isA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - D Third order bright fringe is the `3^(rd)` bright fringe from the central bright fringe having `3lamda` path difference of `2pi.4pi=2xx2piimplies2^(nd)` order bright fringe. |
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