InterviewSolution
Saved Bookmarks
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.
| 251. |
A narrow monochromatic beam of light of intensity 1 is incident on a glass plate as shown in figure Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects `25%` of the light incident on it and transmits intensities in the interference pattern formed by two beams obtained after one reflection at each plate. |
|
Answer» Correct Answer - `[49 : 1]` |
|
| 252. |
Two coherent sources of intensity ratio `1:4` produce an interference pattern. The fringe visibility will beA. (a) `1`B. (b) `0.8`C. (c) `0.4`D. (d) `0.6` |
|
Answer» Correct Answer - B `I_1/I_2=1/4impliesI_1=k` and `I_2=4k` `:.` Fringe visibility `V=(2sqrt(I_1I_2))/((I_1+I_2))=(2sqrt(kxx4k))/((kxx4k))=0.8` |
|
| 253. |
Statement I: For the situation shown in figure two identecal coherent light sources produce interference pattern on the screen. The intensity of minima nearest to `S_(1)` is not exactly zero. Statement II: Minimum intensity is zero, when interfering waves have same intensity at the location of superposition.A. A is true and R is true and R is the correct explanation of A.B. A and R are true but R is not correct explanation of AC. A is true, R is falseD. A is false, R is true. |
|
Answer» Correct Answer - A At the location of minima, the two waves have different intensities hence minimum intensity is not equal to zero. |
|
| 254. |
Two coherent sources of intensities `I_1` and `I_2` produce an interference pattern. The maximum intensity in the interference pattern will beA. `I_(1)+I_(2)`B. `I_(1)^(2)+I_(2)^(2)`C. `(I_(1)+I_(2))^(2)`D. `(sqrt(I_(2))+sqrt(I_(2)))^(2)` |
|
Answer» Correct Answer - D `I_(1)propa_(1)^(2),propa_(2)^(2)` `I_("resultant")=I_(1)+I_(2)+2sqrt(I_(1))sqrt(I_(2))cosphi` `I_(max)=(sqrt(I_(1))+sqrt(I_(2)))^(2)` When `cosphi=1,phi=0,2pi` |
|
| 255. |
A plane monochromatic light falls normally on a diaphragm with two narrow slits separated by a distance d = 2.5 mm. A fringe pattern is formed on the screen placed at D = 100 cm behind the diaphragm. If one of the slits is covered by a glass plate of thickness `10mu m`, then distance by which these fringes will be shifted will be :A. 2 mmB. 3 mmC. 4 mmD. 5 mm |
|
Answer» Correct Answer - C |
|
| 256. |
A screen is at distance `D = 80` cm form a diaphragm having two narrow slits `S_(1)` and `S_(2)` which are `d = 2` mm apart. Slit `S_(1)` is covered by a transparent sheet of thickness `t_(1) = 2.5 mu m` slit `S_(2)` is covered by another sheet of thikness `t_(2) = 1.25 mu m` as shown if Fig. 2.52. Both sheets are made of same material having refractive index `mu = 1.40` Water is filled in the space between diaphragm and screen. Amondichromatic light beam of wavelength `lambda = 5000 Å` is incident normally on the diaphragm. Assuming intensity of beam to be uniform, calculate ratio of intensity of C to maximum intensity of interference pattern obtained on the screen `(mu_(w) = 4//3)` |
|
Answer» Correct Answer - `[(3)/(4)]` |
|
| 257. |
The amplitude of polarised light transmitted through a polariser is A. The amplitude of unpolarised light incident on it isA. `A//2`B. `A//sqrt(2)`C. `2A`D. `sqrt(2)A` |
|
Answer» Correct Answer - D `(A_(0))/(sqrt(2)) = A` |
|
| 258. |
An unpolarised light is incident on a plate of refractive index `sqrt(3)` and the reflected light is found to be completely plane polarised. The angles of incidence and refraction are respectivelyA. `60^(@) , 30^(@)`B. `30^(@) , 60^(@)`C. `Sin^(-1) ((1)/(sqrt(3))), 45^(@)`D. `Tan^(-1) ((sqrt(3))/(2)) , 30^(@)` |
|
Answer» Correct Answer - A `mu = tan i_(p) , r = 90^(@) - i_(p)` |
|
| 259. |
A beam of light strikes a glass plate at an angle of incident `60^(@)` and reflected light is completely polarised than the refractive index of the plate is-A. `1.5`B. `sqrt(3)`C. `sqrt(2)`D. `(3)/(2)` |
| Answer» Correct Answer - B | |
| 260. |
When light is incident from air to glass at an angle `57^@`, the reflected beam is completely polarised. If the same beam is incident from water to glass, the angle of incidence at which reflected beam is completely polarised will beA. (a) `theta=57^@`B. (b) `thetagt57^@`C. (c) `thetalt57^@`D. (d) cannot be determined |
|
Answer» Correct Answer - C `_wmu_glt_amu_g` and `tan theta=mu` |
|
| 261. |
In case of linearly polarised light, the magnitude of the electric field vectorA. (a) does not change with timeB. (b) varies periodically with timeC. (c) increases and decreases linearly with timeD. (d) is parallel to the direction of propagation |
|
Answer» Correct Answer - B The magnitude of electric field vector varies periodically with time because it is the form of electromagnetic wave. |
|
| 262. |
At what angle of incidence will the light reflected from glass`(mu=1.5)` be completely polarisedA. `72.8^(@)`B. `51.6^(@)`C. `40.3^(@)`D. `56.3^(@)` |
|
Answer» Correct Answer - D Using, `"tan "i_(p)=mu=1.5" or "i_(p)="tan"^(-1)(1.5)=56.3^(@)`. |
|
| 263. |
In case of linearly polarised light, the magnitude of the electric field vectorA. is parallel to the direction of propagationB. does not change with timeC. increases linearly with timeD. varies perodically with time |
|
Answer» Correct Answer - D In the case of linearly polarised light the magnitude of the electric field vector varies periodically with time. |
|
| 264. |
A leaf which contains only green pigments, is illuminated by a laser light of wavelength `0.6328mum`. It would appear to beA. (a) brownB. (b) redC. (c) blackD. (d) green |
|
Answer» Correct Answer - C If an object reflects the colour of light incident on it, it will appear with that colour but if object absorbs the colour of light, it will appear to be back. Since the given wavelength does not belong to green, so it will be absorbed by the leaf and hence, it will appear to be black. |
|
| 265. |
The surface of srystals can be studied usingA. diffraction of visible lightB. diffraction of x-raysC. interference of sound wavesD. refraction of radio waves |
| Answer» Correct Answer - B | |
| 266. |
The penetration of light into the region of geometrical shadow is calledA. Diffraction of lightB. Polarisation of lightC. Interference of lightD. Rectilinear propagation of light |
| Answer» Correct Answer - D | |
| 267. |
The penetration of light into the region of geometrical shadow is calledA. (a) PolarisationB. (b) InterferenceC. (c) DiffractionD. (d) Refraction |
|
Answer» Correct Answer - C It is caused due to turning of light around corners. |
|
| 268. |
The diffraction bands observed in the case of straight edge producing diffraction effects areA. equally spaced like the interference bands but with less contrastB. unequally spaced with increasing width as we move away from the edge of geometric shadowC. unequally spaced with decreasing width as we move away from the edge of geometric shadowD. equally spaced like the interference bands but with more contrast |
| Answer» Correct Answer - C | |
| 269. |
Resolving power of a telescope increases withA. Increase in focal length of eye pieceB. Increase in focal length of objectiveC. Increase in aperture of eye pieceD. Increase in aperture of objective |
| Answer» Correct Answer - D | |
| 270. |
Assertion : All bright interference bands have same intensity. Reason : Because all bands do not receive same light from two sources.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 - C All bright interference bands have same intensity but all bands do not receive same light from two sources. |
|
| 271. |
Assertion : If the light from an ordinary source passes through a polaroid sheet, its intensity is reduced by half. Reason : The electric vectors associated with the light wave along the direction of the aligned molecules get absorbed by polaroid.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 - A A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors associated with the propagating light wave along the direction of the aligned molecules get absorbed. Thus if the light from an ordinary source passes through a polaroid sheet, its intensity get reduced by half. |
|
| 272. |
Diffraction of light isA. the bending of light at the surface of separation when it travels from rarer medium of denser mediumB. the bending of light at the surface of separation when it travels from denser medium to rarer mediumC. encroachment of light into the geometrical shadow of the obstancle placed in its pathD. emergence of a light ray grazing the surface of separation when it travels denser to rarer medium |
| Answer» Correct Answer - C | |
| 273. |
In YDSE set up (see fig.), the light sources executes SHM between P and Q according to the equation `x = A sin omega t`, S being the mean position. Assume `d rarr 0` and `A lt lt L`. `omega` is small enough to neglect Doppler effect. If the sources were stationary at S, intenstiy at O would be `I_(0)` Read the paragraph carefully and answer the following questions. When the source comes toward the point Q,A. the bright fringe will be less bright.B. the dark fringe will no longer remain darkC. the fringe width will increaseD. none of these |
|
Answer» Correct Answer - d Intensity of the fringes will increase. |
|
| 274. |
STATEMENT-1 In `YDSE` centela fringe is always a bright fringe. STATEMENT-2 If path difference at central fringe is zero then it will be a bright fringe.A. Statement-1 is true Statemetnt-2 is True,Statement -2 is a correct explanation for statement -1.B. Statement-1 is true Statemetnt-2 is True,Statement -2 is NOT a correct explanation for statement -1.C. Statement -1 is true,statement -2 is falseD. Statement -1 is False ,statement -2 is True. |
| Answer» Correct Answer - D | |
| 275. |
In the arrangement shown in Fig., slits `S_(1)` and `S_(4)`are having a variable separation Z. Point O on the screen is at the common perpendicular bisector of `S_(1) S_(2)` and `S_(3) S_(4)`. The minimum value of Z for which the intensity at O is zero isA. `(2 lambda D)/(d)`B. `(lambda D)/(2d)`C. `(lambda D)/(3d)`D. `(lambda D)/(d)` |
|
Answer» Correct Answer - d Intensity at O is zero if `delta//2 = pi//2`. That is, `delta = (2 pi)/(beta) (Z)/(2) = pi implies Z = beta =(lambda D)/(d)` |
|
| 276. |
In a arrangement shown in the fig.the distance D is latge compared to the separation d between this slits ,find: The position of first bright fringe for the minimum value of `d`isA. `d//2`belowB. `d` aboveC. `3d//2`belowD. `3d//2`above |
| Answer» Correct Answer - C | |
| 277. |
In a arrangement shown in the fig.the distance D is latge compared to the separation d between this slits ,find: The fringe width isA. `(3Dlambda)/(4d)`B. `(3Dlambda)/(2d)`C. `(Dlambda)/(d)`D. `(2Dlambda)/(d)` |
| Answer» Correct Answer - B | |
| 278. |
In a arrangement shown in the fig.the distance D is latge compared to the separation d between this slits ,find: The minimum value of d for which there is a dark fringe at the point `O`.A. `sqrt(lambdaD//2)`B. `sqrt(Dlambda//3)`C. `sqrt(lambdaD)`D. cannot be calcualted |
| Answer» Correct Answer - B | |
| 279. |
A block o plastic having a thin air cavity (whose thickenss is comparable to wavelength of light waves) is shown in fig. The thickness of air cavity (which can be considered as air wedge for interference pattern) is varying linearly from one end to other as shown. A broad beam of monochromatic light is incident normally from the top of the plastic box. Some ligth lis reflected back above and below the cavity .The plastic layers than wavelength of incident light . An observer when looking down from top sees an interference pattern consisting of eight dark fringes and seven bright fringe along the wedge. Take wavelength of incident light in air as `lambda_(0)` and refractive index of plastic as `mu` Assume that the thickness of the ends of air cavity are such that formation of fringe takes place there. Determine the distance of 4th dark fringe from the left end of air cavity.A. `(2 L)/(6) + lambda_(0)`B. `L_(1) +(3 L)/(4)`C. `(4 L)/(7)`D. `(5 L)/(7)` |
|
Answer» Correct Answer - c For 4th dark fringe from left end, `2mu t = (n - 3) lambda_(0)` `2 mu (L_(2) + (L_(1) - L_(2))/(L) x) = 2 mu L_(1) - 3 lambda_(0)` `x = (4L)/(7)` |
|
| 280. |
Monochromatic light of wavelength `lambda_1` travelling in medium of refractive index `n_1` enters a denser medium of refractive index `n_2`. The wavelength in the second medium isA. (a) `lambda_1(n_1/n_2)`B. (b) `lambda(n_2)/(n_1)`C. (c) `lambda_1`D. (d) `lambda((n_2-n_1)/(n_1))` |
|
Answer» Correct Answer - A `n_1=(c)/(v_1)=(vlambda)/(vlambda_1)=(lambda)/(lambda_1)` `n_2=(c)/(v_2)=(vlambda)/(vlambda_2)=(lambda)/(lambda_2)` Now, `n_1/n_2=(lambda_2)/(lambda_1)` or `lambda_2=(n_1/n_2)lambda_1` |
|
| 281. |
A beam of light propagating in medium A with index of refraction `n(A)` passes across an interface into medium B with index of refraction `n(B)`. If `v(A)` and `v(B)` are the speeds of light in A and B respectively. Then which of the following is true?A. (a) `v(A)gtv(B)` and `n(A)gtn(B)`B. (b) `v(A)gtv(B)` and `n(A)ltn(B)`C. (c) `v(A) lt v(B)` and `n(A)gtn(B)`D. (d) `v(A) lt v(B)` and `n(A) lt n(B)` |
|
Answer» Correct Answer - B Light travels faster in rarer medium. This alone decides is favour of option (b). |
|
| 282. |
Two waves of light in air have the same wavelength and are intially in phase. They then travel through plastic layers with thickness of `L_(1) = 3.5 mm` and `L_(2) = 5.0 mm` and indices of refraction `n_(1) = 1.7` and `n_(2) = 1.25` as shown in figure. The rays later arrive at a common point. The longest wavelength of light for which constructive interference occurs at the point is A. `0.8 mu m`B. `1.2 mu m`C. `1.7 mu m`D. `2.9 mu m` |
|
Answer» Correct Answer - b `Delta x = `path difference between two light waves `= [n_(1) L_(1) + L_(2) - L_(1)] - [n_(2) L_(2)]` `Delta phi =` phase difference between two waves For longest wavelengths, `Delta phi` is the smallest. For constructive interference, `Delta phi = 2 pi` `I_(max) = n_(1) L_(1) - n_(2) L_(2) + (L_(2) - L_(1))` `(1.7) (3.5 xx 10^(-6)) - (1.25) (5.0 xx 10^(-6))` `+ (5.0 - 3.5) xx 10^(-6)` ` = (5.95 - 6.25 + 1.5) xx 10^(-6) = 1.2 5 10^(-6)m` `= 1.2 mu m` |
|
| 283. |
This interference film is used to measure the thickness of slides, paper, etc. The arrangement is as shown in fig. For the sake of clarity, the two strips are shown thick. Consider the wedge formed in between strips 1 and 2. If the interference pattern because of the two waves reflected from wedge surface is observed, then from the observed data we can compute thickness of paper, refractive index of the medium filled in wedge, number of bonds formed, etc. Considre the strips to be thick as compared to wavelength of ligth and light is incident normally. Neglect the effect due to reflection from top surface of strip 1 and bottom surface of strip 2. Take `L = 5 cm` and `lambda_(air) = 40 nm`. For data in question53 determine the distance of point B from the 20th dark band. Counting of dark point has to start from the contact point.A. 4 cmB. 3 cmC. 5 cmD. 2cm |
|
Answer» Correct Answer - a For 20th dark band, `x = 20 xx 5 xx 10^(-4) m = 1 m` So, required separation `= S - 1 = 4 cm` |
|
| 284. |
Light is incident at an angle `phi` with the normal to a plane containing two slits of separation d. Select the expression that correctly describes the positions of the interference maxima in terms of the incoming angle `phi` and outgoing angle `theta`. A. `sin phi = sin theta = (m + (1)/(2)) (lambda)/(d)`B. `d sin theta = m lambda`C. `sin phi - sin theta = (m + 1) (lambda)/(d)`D. `sin phi + sin theta = m (lambda)/(d)` |
|
Answer» Correct Answer - d Path difference `= d sin phi + d sin theta` For maxima, `Delta x = m lambda` `implies sin phi + sin theta = (m lambda)/(d)` |
|
| 285. |
In a double-slit experiment, two parallel slits are illuminated first bylight of wavelength 400 nm and then by light of unknown wavelength. The fourth-order dark fringe resulting from the known wavelength of light falls in the same place on the screen as the second-order bright fringe from the unknown wavelength. The value of unknown wavelength of light isA. 900 nmB. 700 nmC. 300 nmD. none of these |
|
Answer» Correct Answer - b `y_(n) = (n + (1)/(2)) (D lambda)/(d) , n = 1, 2, 3` `(3 + (1)/(2)) (D lambda_(1))/(d) = 2 xx (D lambda_(2))/(d)` `lambda_(2) = (7)/(4) xx 400 = 700` nm |
|
| 286. |
In the YDSE, the monochromatic source of wavelength `lambda` is placed at a distance `d/2` from the central axis (as shown in the figure), where d is the separation between the two slits `S_1 and S_2` . (a)Find the position of the central maxima. (b) Find the order of interference formed at O. (c)Now, S is placed on centre dotted line. Find the minimum thickness of the film of refractive indes `mu =1.5` to be placed in front of `S_2` so that intensity at O becomes `3/4`th of the maximum intensity. (Take `lambda=6000Å, d = 6mm`.) |
|
Answer» Correct Answer - [(a) 4 mm above 0; (b) 20; (c ) 2000 `Å`] |
|
| 287. |
In YDSE, the amplitude of intensity variation fo the two sources is found to be `5%` of the average intensity. The ratio of the intensities of two interfering sources isA. 2564B. 1089C. 1681D. 869 |
|
Answer» Correct Answer - c `(I_(max))/(I_(min)) = ((sqrt I_(1) + sqrt I_(2))^(2))/((sqrt I_(1) - sqrt I_(2))^(2))` `= ((1 + sqrt((I_(2))/(I_(1))))/(1 - sqrt((I_(2))/(I_(1)))))^(2)` For given information, `I_(max) = (105)/(100) I` `I_(min) = ((95)/(100)) I` where `I + I_(1) + I_(2)` is average intensity `:. (I_(max))/(I_(min)) = (105)/(95) = ((1 + sqrt((I_(2))/(I)))/(1 - sqrt((I_(2))/(I_(1)))))^(2)` `implies sqrt((I_(2))/(I_(1))) = 0.0244 = gt (I_(2))/(I_(1)) = 0.0059` `implies (I_(1))/(I_(2)) = 1681`. |
|
| 288. |
Absolute refractive indices of glass and water are `3//2` and `4//3`. The ratio of velocity of light in glass and water will be |
|
Answer» Correct Answer - `8 :9` |
|
| 289. |
A ray of light of frequency `5 xx 10^(14)` Hz is passed through a liquid . The wavelength of light measured inside the liquid is found to be 450 nm . Calculate (i) wavelength of light in vacuum (ii) refractive index of liquid (iii) velocity of light in the liquid . Take velocity of light in vacuum as ` 3 xx 10^(8) ms^(-1)` |
|
Answer» Correct Answer - (i) 600 nm (ii) 1.33 (iii) `2.25 xx 10^(8) ms^(-1)` |
|
| 290. |
Which of the following statement is correctA. if two polariods are placed at `90^(@)` to each other then , the transmitted intensity is zero.B. if two polariods are placed at `90^(@)` to each other and one more polaroid is placed between these two (bisecting the angle between them), then intensity of light will be half.C. polarized light has angle of electric field along the propagation of light.D. All of them |
|
Answer» Correct Answer - A No part of light is transmitted through to polaroids kept at `90^(@)` to each other. If one more polaroid is inserted between the two then, first polaroid will rotate the light by `90^(@)` second polaroid which is kept `45^(@)` angle with the first polaroid will rotate it again by `90^(@)` polarized light has angle of electric field perpendicular to the direction of motion. |
|
| 291. |
A light wave is travelling along - Y direction. For this light wave, which of the following equations of planes may represent its wavefront.A. x = 6B. `x + y + 2z = 8`C. y = 8D. z = 2 |
|
Answer» Correct Answer - A |
|
| 292. |
Light waves travel in vacuum along the y-axis. Which of the following may represent the wavefront?A. x = constantB. y = constantC. z = constantD. `x + y + z` = constant |
|
Answer» Correct Answer - b Velocity of light is perpendicular to the wavefront. |
|
| 293. |
Figure shows wavefront P Passing through two systems A and B, and emerging as Q and then as R. Then systems A and B could, respectively, be A. a prism and a convergent lensB. a convergent lens and a prismC. a divergent lens and a prismD. a convergent lens and a divergent lens |
|
Answer» Correct Answer - b P to Q: convergence increasing, Q to R: changing. |
|
| 294. |
A wave front AB passing through a system C emerges as DE. The system C could be A. a slitB. a biprismC. a prismD. a glass slab |
|
Answer» Correct Answer - c A slit would give divergent, a biprism would give double, a glass slab would give a parallel N100wavefront. Edge is downward. |
|
| 295. |
Figures shows a wave front P passing through two systems A and B, and emerging as Q and then as R. The systems A and B could, respectively, be A. (a) a prism and a convergent lensB. (b) a convergent lens and a prismC. (c) a divergent lens and a prismD. (d) a covergent lens and a divergent lens |
|
Answer» Correct Answer - D P to Q convergence increasing, Q to R direction changing. |
|
| 296. |
The similarity between the sound waves and light waves isA. (a) Both are electromagnetic wavesB. (b) Both are longitudinal wavesC. (c) Both have the same speed in a mediumD. (d) They can produce interference |
|
Answer» Correct Answer - D Sound wave and light waves both shows interference. |
|
| 297. |
A point is emitting light of wavelength `6000Å` is placed at a very small height `h` above a flat reflecting surface `MN` as shown in the figure.The intensity of the reflected light is `36%` of the incident intensity.Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance `D` from it. If the intensity at pbe maximum then the minimum distance through which the reflecting suface `MN` should be displaced so that at `P` again becomes a mixture.?A. `3xx10^(-10)m`B. `6xx10^(-10)m`C. `1.5xx10^(-10)m`D. `12xx10^(-10)m` |
| Answer» Correct Answer - A | |
| 298. |
A point sources `S` emitting light of wavelength `600nm` is placed at a very small height `h` above the flat reflecting surface `AB`(see figure).The intensity of the reflected light is`36%` of the intensity.interference firnges are observed on a screen placed parallel to the reflecting surface a very large distance `D` from it. (A)What is the shape of the interference fringes on the screen? (B)Calculate the ratio of the minimum to the maximum to the maximum intensities in the interference fringes fromed near the point `P` (shown in the figure) (c) if the intenstities at point `P` corresponds to a maximum,calculate the minimum distance through which the reflecting surface `AB` should be shifted so that the intensity at `P` again becomes maximum.A. circleB. squareC. parbolaD. straight line |
| Answer» Correct Answer - A | |
| 299. |
A point sources `S` emitting light of wavelength `600nm` is placed at a very small height `h` above the flat reflecting surface `AB`(see figure).The intensity of the reflected light is`36%` of the intensity.interference firnges are observed on a screen placed parallel to the reflecting surface a very large distance `D` from it. (A)What is the shape of the interference fringes on the screen? (B)Calculate the ratio of the minimum to the maximum to the maximum intensities in the interference fringes fromed near the point `P` (shown in the figure) (c) if the intenstities at point `P` corresponds to a maximum,calculate the minimum distance through which the reflecting surface `AB` should be shifted so that the intensity at `P` again becomes maximum. |
|
Answer» Correct Answer - (i). Circular (ii). `(1)/(16)` (iii). 300 nm. (i). S is a point source, fringes formed will be circular (ii). Ratio of minimum and maximum intensities intensity of light direct from source `I_(1)=I_(0)`(say) Intensity after reflection `I_(2)=0.36I_(0)` `therefore(I_(min))/(I_(max))=((sqrt(I_(1))sqrt(I_(2)))/(sqrt(1))+sqrt(I_(2)))^(2)=((0.4)/(1.6))^(2)=(1)/(16)` (iii). Shift of AB for same intensity if intensity at P correspionds to maximum it means that constructive interference occurs at P. `therefore` Path difference between direct waves from S and reflected waves, from reflector AB, is `nlamda` Let AB is shifted by x (towards P or away from P) `therefore` Additional path difference introduced =2x For minimum value of `x_(1),n=1` `therefore` Path difference `=1xxlamda=600nm` `2x=600nmimpliesx=300nm` |
|
| 300. |
In a grating experiment red line of wavelength `7000xx10^(-10)`m in the third order coincides with violet line in the fifth order. What is the wavelength of the voilet line? |
|
Answer» Correct Answer - `4200xx10^(-10)`m |
|