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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.
| 501. |
Fringes are obtained with the help of a biprism in the focal plane of an eyepiece distant 1 m from the slits. A convex lens produces images of the lit in two position between biprism and eyepiece. The distances between two images of the slit in two position are `4.05xx10^(-3)m` and `2.9xx10^(-3)m` respectively. Calculate the distance between the slits. |
| Answer» `d=sqrt(d_(1)d_(2))=sqrt(4.05xx10^(-3)xx2.9xx10^(-3))=3.43xx10^(-3)m` | |
| 502. |
Two point white dots are 1mm apart on a black paper. They are viewed by eye of pupil diameter 3mm. Approximately, what is the maximum distance at which these dits can be resolved by the eye? [Take wavelelngth of light =500nm]A. `3m`B. `6m`C. `1m`D. `5` |
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Answer» Correct Answer - 4 we know `(y)/(D)ge1.22(lambda)/(d)` `rArr Dle(yd)/(1.22lambda)` `:.D=(10^(-1)xx3xx10^(-3))/(1.22xx5xx10^(-7))` `rArr D_(max)=5m` |
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| 503. |
The intensity variation in the interference pattern obtained with the help of two coherent source is `5%` of the average intensity find out the ratio of intensties of two sources. |
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Answer» `(I_(max))/(I_(min))=(105)/(95)=(21)/(19)implies((a_(1)+a_(2))^(2))/((a_(1)+a_(2))^(2))=(21)/(19)` `implies(a_(1)+a_(2))/(a_(1)-a_(2))=sqrt((21)/(19))=1.05impliesa_(1)+a_(2)=1.05a_(1)-1.05a` `0.05a_(1)=2.05a_(2)implies(a_(1))/(a_(2))=(2.05)/(0.05)=(41)/(1)` |
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| 504. |
In Fig. 16.13, at what angle `theta` above the horizon should the sun be situated so that its light reflected from the surface of still water of the pond be totally polarised? Given: refractive index of water `mu=1.327 " and " tan 53^(@)=1.327`. |
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Answer» Correct Answer - `37 ^(@)` |
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| 505. |
Light waves spreading from two sources produces steady inteference only if they haveA. congruenceB. coherenceC. same intensityD. same amplitude |
| Answer» Correct Answer - B | |
| 506. |
For different independent waves are represented by a) `Y_(1)=a_(1)sin omega_(1)t` , b) `Y_(2)=a_(2) sin omega_(2)t` c) `Y_(3)=a_(3) sin omega_(3)t` , d) `Y_(4)=a_(4) sin(omega_(4)t+(pi)/(3))` The sustained interference is possible due toA. `a & c`B. `a & d`C. `c & d`D. not possible with any combination |
| Answer» Correct Answer - D | |
| 507. |
Two coherent sources of equal intensity produce maximum intensity of `100` units at a point. If the intensity of one of the sources is reduced by 36% reducing its width, then the intensity of light at the same point will beA. (a) 90B. (b) 89C. (c) 67D. (d) 81 |
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Answer» Correct Answer - D Intensity of each source `=I_0=(100)/(4)=25unit` If the intensity of one of the source is reduced by 36% then `I_1=25unit` and `I_2=25-25xx(36)/(100)=16(unit)` Hence resultant inensity at the same point will be `I=I_1+I_2+2sqrt(I_1I_2)=25+16+2sqrt(25xx16)=81unit` |
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| 508. |
Two waves of equal amplitude and frequency interfere each other. The ratio of intensity when the two waves arrive in phase to that when they arrive `90^@` out of phase isA. (a) `1:1`B. (b) `sqrt2:1`C. (c) `2:1`D. (d) `4:1` |
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Answer» Correct Answer - C Resultant intensity `I=4I_0cos^2(varphi//2)` `impliesI_1/I_2=(cos^2(varphi_1//2))/(cos^2(varphi_2//2))=(cos^20)/(cos^2(90//2))=2/1` |
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| 509. |
Two sources of waves are called coherent ifA. (a) Both have the same amplitude of vibrationsB. (b) Both produce waves of the same wavelengthC. (c) Both produce waves of the same wavelength having constant phase differenceD. (d) Both produce waves having the same velocity |
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Answer» Correct Answer - C Two coherent sources must have a constant phase difference otherwise they can not produce interference. |
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| 510. |
Two light sources are said to be coherent if they are obtained fromA. (a) Two independent point sources emitting light of the same wavelengthB. (b) A single point sourceC. (c) A wide sourceD. (d) Two ordinary bulbs emitting light of different wavelengths |
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Answer» Correct Answer - B When two sources are obtained from a straight source, the wavefront is divided into two parts. These two wavefronts acts as if they emanated from two sources having a fixed phase relationship. |
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| 511. |
A point source (A) is kept on the axis of a hemi- spherical paperweight made of glass of refractive index`mu = ( 3 )/( 2)` . The distance of the point source from the centre (O) of the sphere is R where R is radius of the hemisphere. Use paraxial approximations for answering following questions (a) Find the change in radius of curvature of the wave- fronts just after they enter the glass at O. (b) Find the radius of curvature of the wavefronts at point P just outside the glass. |
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Answer» Correct Answer - (a)`(R)/(2)` (b)10R |
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| 512. |
A polariser in used toA. Reduce intensity of lightB. increases intensity of lightC. produce polarised lightD. analyse polarised light |
| Answer» Correct Answer - C | |
| 513. |
The limit of resolution of an optical instrument arises on account ofA. ReflectionB. DiffractionC. PolarisationD. Interference |
| Answer» Correct Answer - B | |
| 514. |
The angle of diffraction of the second order maximum of wavelength `5xx10^(-5)` cm is `30^(@)` in the case of a plane transmission grating. How many lines are there in 1 cm of the grating surface. |
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Answer» Correct Answer - 1000 |
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| 515. |
Assertion: Standard optical diffraction cannot be used for discrimination between different X-ray wavelengths. Reason: The grating spacing is not of the order of X-ray wavelengths.A. (a) If both the assertion and reason are true and reason explains the assertion assertion.B. (b) If both assertion and reason are true but reason does not explain the assertionC. (c) If assertion is true but reason is false.D. (d) If assertion is false but reason is true. |
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Answer» Correct Answer - A X-rays are electromagnetic radiations whose wavelengths are of the order of `1Å(=10^(-10)m)`. A standard optical diffraction grating cannot be used to discriminate between different wavelenghts in the X-ray wavelength range. For `lambda=1Å(=0.1nm)` and `d=3000nm`, for example, the first order maximum occurs at `theta=sin^-1((mlambda)/(d))=sin^-1` ((1)(0.1))/(3000)=0.0019^(@)` This is too close to the central maximum to be practicle. A grating with `d~~lambda` is desirable, but since X-ray wavelength are about equal to atomatic diameter, such gratings cannot be constructed directly. Hence, option (a) is true. |
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| 516. |
Assertion : The clouds in the sky generally appear to be whitish. Reason : Diffraction due to clouds is efficient in equal measures its all wavelengths.A. (a) If both the assertion and reason are true and reason explains the assertion assertion.B. (b) If both assertion and reason are true but reason does not explain the assertionC. (c) If assertion is true but reason is false.D. (d) If assertion is false but reason is true. |
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Answer» Correct Answer - C A cloud is a large collection of very tiny droplets of water or ice crystals. The droplets are so small and light that they can float in air. Clouds appear white because they reflect the ligth of sun. Light is made up of colours of rainbow and when they are added up together, white colour is obtained and clouds reflect all the colours by exactly the same amount so they generally look white. |
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| 517. |
For constructive interference to take place between two monochromatic light waves of wavelength `lambda` , the path difference should beA. `((2n-1)lambda)/4`B. `2nlambda`C. `((2n+1)lambda)/2`D. `nlambda` |
| Answer» Correct Answer - C | |
| 518. |
The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit isA. (a) `1:4:9`B. (b) `1:2:3`C. (c) `1:(4)/(9pi^2):(4)/(25pi^2)`D. (d) `1:(1)/(pi^2):(9)/(pi^2)` |
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Answer» Correct Answer - C `I=I_0[(sin alpha)/(alpha)]^2`, where `alpha=varphi/2` For `n^(th)` secondary maxima `dsintheta=((2n+1)/(2))lambda` `impliesalpha=varphi/2=pi/2[dsintheta]=((2n+1)/(2))pi` `:. I=I_0[(sin((2n+1)/(2))pi)/(((2n+1)/(n))pi)]^2=(I_0)/({((2n+1))/(2)pi}^2)` So `I_0:I_1:I_2=I_0:(4)/(9pi^2)I_0:(4)/(25pi^2)I_0` `=1:(4)/(9pi^2):(4)/(25pi^2)` |
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| 519. |
Answer the following questions : (a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the piture on our TV screen. Suggest a possible expanation. (b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffractions and interference patterns. What is the justification of this principle ? |
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Answer» Weak rader singals sent by a low fluing aircrrfit with the TV singnals received by the antenna As a result , the Tv signals may get distored Hence when a low fying aircroft passes overhead , we sometimes notice a silght shaking of the picture on our TV screen . (b) the principle o dlinear superpostion of wave displacement is essential toour understanding of intersity disibutions and interference pattens this is because superpostion follows from the linear character of a differential equation that governs wave motion If `y_(1) and y_(2)` are the the solutions of the second order wave equation then an linear combination of `y_(1)and y_(2)` will also be the solution of the wave equation . |
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| 520. |
What is the shape of the wavelength in case Light diverging from a point source. |
| Answer» It is spherical wavefront. | |
| 521. |
A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit isA. `0`B. `pi//2`C. `pi`D. `2 pi` |
| Answer» Correct Answer - D | |
| 522. |
The coherent point sources `S_(1)` and `S_(2)` vibrating in same phase emit light of wavelength `lambda`. The separation between the sources is `2lambda`. Consider a line passingh through `S_(2)` and perpendicular to the line `S_(1)S_(2)`. What is the smallest distance from `S_(2)` where a minimum of intensity occurs due to interference of waves from the two sources?A. `(7lambda)/(12)`B. `(15lambda)/(4)`C. `(lambda)/(2)`D. `(3lambda)/(4)` |
| Answer» Correct Answer - A | |
| 523. |
Two coherent point sources `S_(1)` and `S_(2)` vibrating in phase emit light of wavelength `lambda`. The separation between the sources is `2lambda`. Consider a line passing through `S_(2)` and perpendicular to line `S_(1) S_(2)`. Find the position of farthest and nearest minima.. |
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Answer» `Delta x_(min) = (2n-1) (lambda)/(2)` The farthest minima has path difference `lambda//2` while nearest minima has path difference `(3//2)lambda`. For the nearest minima. `S_(1)P-S_(2)P= (3)/(2)lambda`, (as maximum path difference is `2lambda`] `implies sqrt((2lambda)^(2)+D^(2)) - D=(3)/(2)lambda implies (2lambda)^(2)+D^(2)= ((3)/(2)lambda+ D)^(2)` `implies 4lambda^(2)+D^(2) = (9)/(4)lambda^(2)+D^(2)xx2xx(3)/(2)lambda xxD` `implies 3D = 4lambda- (9lambda)/(4) = (7lambda)/(4) implies D = (7)/(12) lambda` For the farthest minima, `S_(1)P-S_(2)P = (lambda)/(2)` `implies sqrt(4lambda^(2)+D^(2))-D = (lambda)/(2)` `implies 4 lambda^(2)+D^(2)= (lambda^(2))/(4)+D^(2)+Dlambda implies D= 4lambda-lambda//4 = (15lambda)/(4)` |
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| 524. |
Two coherent point sources `S_(1)` and `S_(2)` vibrating in phase emit light of wavelength `lambda`. The separation between the sources is `2lambda`. Consider a line passing through `S_(1)` and perpendicular to line `S_(1) S_(2)`. Find the position of farthest and nearest minima.. |
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Answer» `Delta x_(min) = (2n - 1) (lambda)/(2)` The farthest mimima has path difference `lambda//2` while the nearest minima has path difference `(3//2)lambda`. For the nearest minima, `S_(1) P - S_(2) P = (3)/(2) lambda` [as maximum path difference is `2lambda`] `implies sqrt((2lambda)^(2) + D^(2)) - D = (3)/(2) lambda` `implies (2lambda)^(2) + D^(2) = ((3)/(2) lambda + D)^(2)` `implies 4lambda^(2) + D^(2) = (9)/(4) lambda^(2) + D^(2) xx 2 xx (3)/(2) lambda xx D` `implies 3D = 4lambda - (9lambda)/(4) = (7lambda)/(4) implies D = (7)/(12) lambda` For the farthest minima, `S_(1)P - S_(2)P = (lambda)/(2)` `impliessqrt (4 lambda^(2) + D^(2)) - D = (lambda)/(2)` `4lambda^(2) + D^(2) = (lambda^2)/(4) + D^(2) + D lambda implies D = 4 lambda - lambda//4 = (15 lambda)/(4)`. |
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| 525. |
Two coherent narrow slits emitting light of wavelength `lambda` in the same phase are placed parallel to each other at a small separation of `2lambda`. The light is collected on a screen S which is placed at a distance D(`gtgt lambda`) from the slit `S_1` as shown in figure. Find the finite distance x such that the intensity at P is equal to intensity O.. |
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Answer» Correct Answer - `[sqrt(3)D]` |
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| 526. |
Electrimagnetic waves are transverse is nature is evident byA. PolarizationB. interferenceC. refelctionD. diffraction |
| Answer» Correct Answer - 1 | |
| 527. |
When the angle of incidence on a material is `60^(@)`, the reflected light is completely polarised. The velocity of the refracted ray inside the materials is (in m//sec^(-1))A. `3xx10^(8)`B. `((3)/(sqrt(2)))xx10^(8)`C. `sqrt(3)xx10^(8)`D. `0.5xx10^(8)` |
| Answer» Correct Answer - C | |
| 528. |
In YDSE, let A and B be two slits. Films of thickness `t_(A)` and `t_(B)` and refractive `mu_(A)` and `mu_(B)` are placed in front of A and B, respectively. If `mu_(A) t_(A) = mu_(A) t_(B)`, then the central maxima willA. not shiftB. shift toward AC. shift toward BD. (b) if `t_(B) lt t_(A)` and (C ) if `t_(B) lt t_(A)` |
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Answer» Correct Answer - d `Delta x = (mu_(A) - 1) t_(A) - (mu_(B) - 1) t_(B)` `= mu_(A) t_(A) - mu_(B) t_(B) - t_(B) + t_(B)` `= t_(B) - t_(A)` If `Delta x gt 0`, the fringe pattern will shift upward. If `Delta x lt 0`, the fringe pattern will shift downward. |
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| 529. |
In YDSE, let A and B be two slits. Films of thickness `t_(A)` and `t_(B)` and refractive `mu_(A)` and `mu_(B)` are placed in front of A and B, respectively. If `mu_(A) t_(A) = mu_(A) t_(B)`, then the central maxima willA. Not shiftB. Shift towards `S_(2)` irrespective of amounts of `t_(1)` and `t_(2)`C. Shift towards `S_(2)` irrespective of amounts of `t_(1)` and `t_(2)`D. Shift towards `S_(1)` if `t_(2)gt t_(1)` and towards `S_(2)` if `t_(2)lt t_(1)` |
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Answer» Correct Answer - D Shift `= (mu -1) (tbeta)/(lambda)` |
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| 530. |
In the `YDSE` shown the two slits are covered with thin sheets having thickness `t` & `2t` and refractive index `2mu` and `mu` .Find the position (y) of central maxima. A. zeroB. `(tD)/(d)`C. `(-tD)/(d)`D. none of these |
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Answer» Correct Answer - B At the central maxima `Deltax=0` But upward shift `=((2mu-1)tD)/(d)` and Downward shift `=((mu-1)2tD)/(d)` So net shift `y=(tD)/(d)[2mu-1-2mu+2]impliesy=(tD)/(d)` |
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| 531. |
In a YDSE two thin transparent sheet are used in front of the slits `S_(1)`&`S_(2)` having `mu_(1)=1.6`and `mu_(2)=1.4` respectively.If both sheets have thickness`t` ,the central maximum is observed at a distance of `5mm`from center`O`.Now the sheets are replaced by another two sheets each of refractive index `(mu_(1)-mu_(2))/(2)` but having thickness `t_(1)&t_(2)=(t_(1)-t_(2))/(2)`.Now central maximum is observed at distance of `8mm` from center`O` on the same side as before.Find the thickness `t_(1)` (in`mu`m) [Given:d=1mm,D=1m] |
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Answer» Correct Answer - [33] |
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| 532. |
In YDSE, both slits produce equal intensities on the screen. A 100% transparent thin film is placed in front of one of the slits. Now, the intensity on the centre becomes 75% of the previous intensity. The wavelength of light is 6000Å and refractive index of glass is 1.5. Thus, minimum thickness of the glass slab isA. `0.2 mu m`B. `1.0 mu m`C. `1.4 mu m`D. `1.6 mu m` |
| Answer» Correct Answer - D | |
| 533. |
White light is incident normally on a glass plate of thickness `0.50 xx 10^(-6)` and index of refraction 1.50. Which wavelengths in the visible region `(400 nm - 700 nm)` are strongly reflected by the plate? |
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Answer» The light of wavelength `lambda` is strongly reflected if `2 mu d = (n + (1)/(2)) lambda` Where n is a non-negative integer. Here `2 mu d = 2 xx 1.50 xx 0.5 xx 10^-6 m = 1.5 xx 10^-6 m`. Putting `lambda = 400` nm in Eq.(i), we get `1.5 xx 10^-6 m = (n + (1)/(2)) (400 xx 10^-(6) m)` or ` n = 3.25` Putting `lambda = 700 nm` in Eq. (i), we get `1.5 xx 10^(-6) m = (n + (1)/(2)) (700 xx 10^(-9) m)` or `n = 1.66` Thus, within 400 nm to 700 nm, the integer n can take the values 2 and 3. Putting these values of n in (i), the wavelengths become ` lambda = (4 mu d)/(2 n + 1) = 600 nm and 429 nm` Thus, light of wavelenghts 429 nm and 600 nm are strongly reflected. |
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| 534. |
White light is incident normally on a glass plate (in air) of thickness `500 nm` and refractive index of `1.5`. The wavelength (in `nm`) in the visible region `(400 nm-700nm)` that is strongly reflected by the plate is:A. `450`B. `600`C. `400`D. `500` |
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Answer» Correct Answer - B For strong reflection. `2mut=(lambda)/(2),(3lambda)/(2),(5lambda)/(2).....rArr lambda=4mut,(4mut)/(3),(4mut)/(5),(4mut)/(7).....` `rArr 3000 nm. 1000 nm, 600 nm, 333 nm.` `rArr` only option is `600 nm`. |
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| 535. |
Light wavelength 650 nm is incident normally upon a glass plate the glass plate rests on top of second plate so that hey touch at one end and are separated by 0.0325 mm at the other end as shown in the figure. Which range of values contains the horizontal separation between adjacent bright fringes?A. 1.1 mm to 1.4 mmB. 1.4 mm to 2.8 mmC. 2.8 mm to 4.2 mmD. 4.2 mm to 5.6 mm |
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Answer» Correct Answer - A `d=` distance between the adjacent bright fringe `d=(Llamda)/(2D)implies((120xx10^(-3))(650xx10^(-9)))/(2xx(0.325xx10^(-3)))` `d=1.2mm` |
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| 536. |
As shown in figure waves with identical wavelengths and amplitudes and which are intially in phase travel through difference media. Ray 1 travels through air and Ray 2 through a transparent medium for equal length L, in four different situations. Rays reach a common point on the screen. The number of wavelengths in length L is `N_(2)` for Ray 2 and `N_(1)` for Ray 1. In the following table, values of `N_(1)` and `N_(2)` are given for all four situations. The order or the situations according to the intensity of the light at the common point in descending order is Situations 1 2 3 4 `N_(1)` 2.75 1.80 3.00 3.25 `N_(2)` 2.75 2.80 3.25 4.00 A. `I_(3)=I_(4)gtI_(2)gtI_(1)`B. `I_(1)gtI_(3)=I_(4)gtI_(2)`C. `I_(1)gtI_(2)gtI_(3)gtI_(4)`D. `I_(2)gtI_(3)=I_(4)gtI_(1)` |
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Answer» Correct Answer - D Path difference `Deltax_(1)=(lamda)/(2),Deltax_(2)=lamda,Deltax_(3)=(lamda)/(4),Deltax_(4)=(3lamda)/(4)` Therefore `I_(2)gtI_(3)=I_(4)I_(1)` |
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| 537. |
Two identical coherent sources are placed on a diameter of a circle of radius R at separation x `(lt lt R)` symmetrical about the center of the circle. The sources emit identical wavelength `lambda` each. The number of points on the circle of maximum intensity is `(x = 5 lambda)`A. 20B. 22C. 24D. 26 |
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Answer» Correct Answer - A Path diff at P is , `Delta = 2(x//2 cos theta) = x cos theta` For intensity to be maximum `Delta x = n lambda , x cos theta = n lambda` `(n lambda)/(x) gt 1 , x = 5 lambda , n= 1,2,3,4,5` Therefore in all for quadrants these can be 20 maximum |
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| 538. |
In a double slit experiment ,the separation between the slits is `d=0.25cm` and the distance of the screen `D=100`cm from the slits .if the wavelength of light used in `lambda=6000Å`and `I_(0)`is the intensity of the central bright fringe.the intensity at a distance `x=4xx10^(-5)`in form the central maximum is-A. `I_(0)`B. `I_(0)//2`C. `3I_(0)//4`D. `I_(0)//3` |
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Answer» Correct Answer - C path diff.`(xd)/(D) rArr` path diff.=`(4xx10^(-5)xx0.25xx10^(-2))/(1)` Path diff.=`1xx10^(-7)` phase diff.=`("path diff.")/(lambda)xx2pi rArr` phase.diff=`(1xx10^(-7))/(6xx10^(-7))xx2pi` phase diff.=`(2pi)/(6) rArr` Phase.diff,=`(pi)/(3)rArr phi=60^(@)` `I_(R)=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cos60^(@) rArr I_(R)=I+I+2Ixx(1)/(2)` `I_(R)=3I rArr I_(R)=(3I_(0))/(4)` |
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| 539. |
In a double slit experiment, the separation between the slits is d and distance of the screen from slits is D. If the wavelength of light used is `lambda` and I is theintensity of central bright fringe, then intensity at distance from central maximum isA. `I cos^(2) ((pi^(2)xd)/(lambda D))`B. `I^(2)sin^(2) ((pi xd)/(2 lambda D))`C. `I cos^(2) ((pi xd)/(lambda D))`D. `I sin^(2) ((pi xd)/(lambda D))` |
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Answer» Correct Answer - C `I = I_(0)cos^(2) ((theta)/(2))` |
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| 540. |
The slits in a double-slit interference experiment are illuminated by orange light `(lambda = 60 nm)`. A thin transparent plastic of thickness t is placed in front of one of the slits. The nunber of fringes shifting on screen is plotted versus the refractive index `mu` of the plastic in graph shown in figure. The value of t is A. 4.8 mmB. `640 mu m`C. `24 mu m`D. none of these |
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Answer» Correct Answer - c N = no. of fringes shifted shift of central maxima `= n lambda = (mu - l) t` `implies t = 24 mu m`. |
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| 541. |
A parallel beam of light in incident on a liquid surface such that the wave front makes an angle `30^(@)` with of the surface and has a width of `sqrt(3)m` , the width of the refracted beam is ____ `(._(a)mu_(L) = sqrt(3))`A. `3 m`B. `sqrt(3)m`C. `(sqrt(11))/(3)m`D. `sqrt((11)/(3))m` |
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Answer» Correct Answer - D `mu_(1) sin i = mu_(2)sin r, (W_(i))/(W_(r)) = (cos i)/(cos r)` |
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| 542. |
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 separation between 1st and 2nd dark fringes form the left end of air cavityA. `(3 L)/(7) + (2 lambda_(0))/(mu)`B. `(5 L)/(7)`C. `(4 L)/(7)`D. `(6 L)/(7)` |
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Answer» Correct Answer - d 1st dark fringe is at left end only. So, 2nd dark fringe would be at `2 mu [L_(2) + (L_(1) - L_(2))/(L) x] = (n - 1) lambda_(0) (n - 1) lambda_(0)` `x = (6L)/(7)` |
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| 543. |
In a YDSE with identical slits, the intensity of the central bright fringe is `I_(0)`. If one of the slits is covered, the intensity at the same point isA. `2 I_(0)`B. `I_(0)`C. `I_(0) //2`D. `I_(0) //4` |
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Answer» Correct Answer - d Let intensity of one slit be `I`. For maxima, `Delta phi = 0` `implies I_(0) = I + I + 2 sqrt(II) cos (0)` `implies I = (I_(0))/(4)`. |
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| 544. |
A monochromatic beam of light passes from a denser medium into a rarer medium. As a resultA. (a) Its velocity increasesB. (b) Its velocity decreasesC. (c) Its frequency decreasesD. (d) Its wavelength decreases |
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Answer» Correct Answer - A Light travels faster in rarer medium. |
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| 545. |
When a parallel beam of monochormatic light suffers refraction while going from a rarer medium into a denser medium, which of the following are correct ? a) the width of the beam decreases b) the width of the beam increases c) the refracted beam makes more angle with the interface d) the refracted beam makes less angle with the interfaceA. a, c trueB. b,d trueC. a,d trueD. b,c true |
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Answer» Correct Answer - D `(W_(i))/(W_(r)) = (cos i)/(cos r), i gt r` |
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| 546. |
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. The fringe width `beta` can be expressed asA. `beta = beta_(0) sin omega t`B. `beta = beta_(0) cos omega t`C. `beta = beta_(0) sin 2 omega t`D. none of these |
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Answer» Correct Answer - d `beta = (lambda D)/(d) =` constant |
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| 547. |
When a light wave passes from a rarer medium to a denser medium, there will be a phase change of `pi` radians. This difference brings change in the conditions for constructive and destructive interference. This phenomena also reasons the fromation of interference pattern in thin films like, oily layer, soap film, etc., but has no reason on the shifting of fringes from the central portion outward. The shift is dependent on the refractive index of the material as per the relation, `Delta y = (mu - 1) t` Thin film interference happens withA. point or spherical sourceB. broad sourceC. film thickness of the order of `10,000 Å`D. very thick transparent slabs |
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Answer» Correct Answer - b., c Broad sources provide wide angular incidence of ligth. The thickness should be small, since the path difference should be comparable with the wavelength. Thick slabs cannot bring wavelength comparable path difference. So, choices (a) and (d) are wrong. Choices (b) and (c ) are correct. |
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| 548. |
In an experiment the two slits are `0.5mm`apart and the fringes are observed to`100cm`from the plane of theslits ,The distance of the`11th`bright fringe from the ist bright fringe is `9.72mm`.calcualte the wavelengthA. `4.85xx10^(-5)cm`B. `4.85xx10^(-5)m`C. `4.86xx10^(-7)cm`D. `4.86xx10^(-5)cm` |
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Answer» Answer: Given `d=.5mm =5xx10^(-2)cm,` `X_(n)=X_(11)-X_(1)=9.72mm` `:.X_(n)=(nlambdaD)/(d) n=11-1=10` `rArr lambda=(X_(n)D)/(d) =(.972xx5xx10^(-2))/(10xx100) rArr lambda=4.86xx10^(-5)cm` |
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| 549. |
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. The fractional change in intensity of the central maximum as function of time isA. `(A sin omega t)/(L)`B. `(2 A sin omega t)/(L)`C. `(3 A sin omega t)/(L)`D. `(4 A sin omega t)/(L)` |
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Answer» Correct Answer - b `I prop (1)/(("distance")^(2))` `:. (Delta I)/(I_(0)) = (l_(2) - (L - x)^(2))/(L^(2)) ~~ (2 A sin omega t)/(L)` |
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| 550. |
When light from two sources (say slits `S_(1)` and `S_(2)`) interfere, they form alternate dark and bright fringes. Bright fringe is formed at all point where the path difference is an odd multiple of half wavelength. At the condition of equal amplitudes, `A_(1) = A_(2) = a`, the maximum intensity will be `4 a^(2)` and the visibility improves, The resultant intensity can also be indicated with phase factor as `I = 2 a^(2) cos^(2) (phi // 2)`.Using this passage, answer the following questions. If the slits `S_(1)` and `S_(2)` are arranged as shown in Fig. the ratio of internsity of fringe at P and R is |
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Answer» Correct Answer - a Since the ligth from slit `S_(1)` and `S_(2)` reach the point P with a path difference of `lambda // 2`, the point P will be dark, `I = 0`. Since path difference is zero for R, there will be a bright fringe at R. So, ratio of intensity will be zero. So choice (a) is correct. |
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