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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.
| 201. |
In Fraunhofer diffraction experiment, L is the distance between screen and the obstacle, b is the size of obstacle and `lambda` is wavelength of incident light. The general condition for the applicability of Fraunhofer diffraction is :A. `(b^(2))/(Llambda)gtgt1`B. `(b^(2))/(Llambda)=1`C. `(b^(2))/(Llambda)ltlt1`D. `(b^(2))/(Llambda)ne1` |
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Answer» Correct Answer - D |
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| 202. |
Light of wavelength `5000Å` falls on a plane reflecting surface. What are the wavelength and frequency of reflected light ? For what angle of incidence is the reflected ray normal to the incident ray ? |
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Answer» wavelength of incident `Å = 5000 xx10^(-10)` m Speed of light `C=3xx10^(8)` m Freuency of incident light is given by the relation , `V=( C)/(lamda)` `=(3xx10^(8))/(5000xx10^(-10))=6xx10^(14)` Hz the wavelength and frequency of indent light is the same as that of refected ray. Hence the wavelrngth of reflected light is 5000 `Å` and its frequancy is `6xx10^(14)` Hz . When reflected , angle r is `90^(@)` Accoding to the law of reflection , the angle of incidence is alwayes equal to the angle of reflection , Hence we can write the sum as : angle i+ angle r =90 angle i + angle i =90 angle `=(90)/(2) = 45^(@)` therefore , the angle of incldence for the given consition is `45^(@)` |
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| 203. |
In double slit experiment using light of wavelength `600 nm`, the angular width of a fringe formed on a distant screen is `0.1^(@)`. What is the spacing between the two slits ?A. `3.44xx10^(-4)m`B. `1.54xx10^(-4)m`C. `1.54xx10^(-3)m`D. `1.44xx10^(-3)m` |
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Answer» Correct Answer - A Angular width fo fringe, `theta=0.1^(@)=0.1xx(pi)/(180)"rad" =(3.14)/(1800)"rad" ` ` :. d=(lambda)/(theta)=(600xx10^(-9))/(3.14//1800)=3.44xx10^(-4) m` |
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| 204. |
In double slit experiment using light of wavelength `600 nm`, the angular width of a fringe formed on a distant screen is `0.1^(@)`. What is the spacing between the two slits ? |
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Answer» Wavelength of light used, `lambda=6000 nm =600 xx10^(-9)` Angular width of fringe, `theta=0.1^(@)=0.1xx(pi)/(180)=(3.14)/(1800)"rad"` Angular width of a fringe is related to slit spacing (d) as : `theta=(lambda)/(d)` `d=(lambda)/(theta)` `=(600xx10^(-9))/((3.14)/(1800))=3.44xx10^(-4)m` Therefore, the spacing between the slits is `3.44xx10^(-4)m.` |
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| 205. |
Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light ? For what angle of incidence is the reflected ray normal to the incident ray? |
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Answer» Given `lambda=5000 Å =5xx10^(-7)m` The wavelenght and frequency of reflected light remains the same. `therefore` Wavelength of reflected light, `lambda=5000Å` Frequency of reflected light `upsilon=(c)/(lamda)=(3xx10^(8))/(5xx10^(-7))=6xx10^(14)Hz` The reflected ray is normal to incident if angle of incidence i=45 |
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| 206. |
A beam of light of wavelength 590 nm is focussed by a converging lens of diameter 10.0 cm at a distnce of 20 cm from it. find the diameter of he disc image formed.A. `5.76xx10^(-4)cm`B. `7.51xx10^(-4)cm`C. `8.72xx10^(-4)cm`D. `9.80 xx10^(-4)cm` |
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Answer» Correct Answer - A Given wavelength `lambda=590 nm =590xx10^(-9)m` Diameter of lens `=10xx10^(-2)m` `:.` Radius of lens `=5xx10^(-2)m` Focal length of lens `=20xx10^(-2)m` Now, the diameter of image disc `D=2R=2(1.22(lambdaD)/b)=((1.22xx590xx10^(-9)xx20xx10^(-2))/(5xx10^(-2)))xx2` `~=5.76xx10^(-4)cm` |
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| 207. |
Two wavelength of sodium light `590 nm` and `596 nm` are used, in turn, to study the diffraction taking placed at a single slit of aperture `2 xx 10^(-4)m`. The distance between the slit and screen is `1.5 m`. Calculate the separation between the positions of first maxima of diffraction pattern obtained in the two cases. |
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Answer» (a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of the two slits (b) Given that: Wavelength of the light beam, `lambda_(1)=590 nm=5.9xx10^(-7)m` Wavelength of another light beam, `lambda-(2)=596nm=5.96xx10^(-7)m` Distance the slits from the screen `=D=1.5m` Distance between the two slits `alpha=2xx10^(-4)m` For the first secondary maxima, `sintheta=(3lambda_(1))/(2alpha)=(chi_(1))/(D)` `chi_(1)=(3lambda_(1)D)/(2alpha) and chi_(2)=(3lambda_(2)D)/(2alpha)` `therefore` Separation between the positions of first secondary maxima of two sodium lines `chi_(1)-chi_(2)(3D)/(2alpha)(lambda_(2)-lambda_(1))` `=(3xx1.5)/(2xx2xx10^(-4))(5.96xx10^(-7)-5.9xx10^(-7))` `=6.75xx10^(-5)m` |
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| 208. |
A source S is kept directly behind the slit `S_(1)` in a double slit apparatus. What will be the phase difference at P, if a liquid of refraction index `mu`is filled : (wavelength of light in aire is `lambda` due to the source, assume `lgtgtd,Dgtgtd`). (a) Between the screen and the slits. (b) Between the slits & the source S. In this case, find the minimum distance between the points on the screen where the intensity is half the maximum intensity on the screen. |
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Answer» Correct Answer - `[(a)Deltaphi=((1)/(l)+(mu)/(D))(pid^(2))/(lambda);(b) Deltaphi=((mu)/(l)+(1)/(D))(pid^(2))/(lambda);D_("min")=(beta)/(2)=(lambdaD)/(2d)]` |
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| 209. |
In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by `5xx10^-2m` towards the slits the change in fringe width is `3xx10^-5m`. If separation between the slits is `10^-3m`, the wavelength of light usedA. (a) `6000Å`B. (b) `5000Å`C. (c) `3000Å`D. (d) `4500Å` |
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Answer» Correct Answer - A `beta=(lambdaD)/(d)impliesbetapropD` `impliesbeta_1/beta_2=D_1/D_2implies(beta_1-beta_2)/(beta_2)=(D_1-D_2)/(D_2)implies(Deltabeta)/(DeltaD)=beta_2/D_2=lambda_2/d_2` `=lambda_2=(3xx10^-5)/(5xx10^(-20))xx10^-3=6xx10^-7m=6000Å` |
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| 210. |
In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by m `5 xx 10^(-2)m` towards the slits, the change in fringe width is m `3 xx 10^(-5)m` . If separation between the slits is `10^(-3)m` , the wavelength of light used isA. `4000 Å`B. `6000 Å`C. `5890 Å`D. `8000 Å` |
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Answer» Correct Answer - B |
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| 211. |
For what distance is ray optics a good approximation when the aperture is `2 mm` wide and wavelength is `600 nm` ? |
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Answer» Ray optics is a good approximately upto Fresnel distance only. `Z_(F)=a^(2)/lambda=((2xx10^(-3))^(2))/((600xx10^(-9)))=6.7m` Hence, the ray optics hold good upto distance of 6.7 m. |
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| 212. |
In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by m `5 xx 10^(-2)m` towards the slits, the change in fringe width is m `3 xx 10^(-5)m` . If separation between the slits is `10^(-3)m` , the wavelength of light used is |
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Answer» Correct Answer - `[6000Å]` |
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| 213. |
For what distance is ray optics a good approximation when the aperture is 4 mm wide and the wavelength is 500 nm?A. 22 mB. 32 mC. 42 mD. 52 m |
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Answer» Correct Answer - B Fresnel distance, `z_(F)=(a^(2))/(lambda)=((4xx10^(-3))^(2))/(500 xx 10^(-9))` ` :. z_(F)=32m` |
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| 214. |
What is the shape of the wavelength in case Light emerging out of a convex lens when a point source is placed at its focus. |
| Answer» It is plane wavefront. | |
| 215. |
Red light is generally used to observe diffraction pattern from single slit. If blue light is used instead of red light, then diffraction pattern.A. Will be more clearB. Will contractC. Will expandD. Will not be visualized |
| Answer» Correct Answer - B | |
| 216. |
For what distance is ray optics a good approximation when the aperture is `3 mm` wide and wavelength is `500 nm` ? (NCERT Solved Example)A. 32mB. 69mC. 16 mD. 8m |
| Answer» Correct Answer - A | |
| 217. |
A monochromatic beam of light is used for the formation of fringes on a screen by illuminating the twoslits in the Young,s double slit interfrence experiment. When a thin film of mic is interposed in the path of one of the interfering beamsA. the fringe-width increasesB. the fringe-width decreasesC. the fringe pattern disappearsD. fringe-width remains the same but the pattern shifts |
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Answer» Correct Answer - D Shift `= (mu -1) (tbeta)/(lambda)` |
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| 218. |
When a compact disc is illuminated by a source of white light, coloured lines are observed. This is due toA. DispersionB. DiffractionC. InterfernceD. Refraction |
| Answer» Correct Answer - B | |
| 219. |
Light is incident on a polarizer with intensity `I_0`. A second prism called analyzer is kept at a angle of `15^@`, from the first polarizer then the intensity of final emergent light will be A. (a) `0.5I_0`B. (b) `0.3I_0`C. (c) `0.46I_0`D. (d) `0.83I_0` |
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Answer» Correct Answer - C `I=I_0/2cos^2(15^@)` `I=I_0/4*2cos^2(15^@)` `I=I_0/4[1+cos(30^@)]` `I=I_0/4*[1+sqrt3/2]` `I=I_0/8[2+sqrt3]` `I=0.46I_0` |
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| 220. |
Assertion (A) : When tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the center of the shadow of the obstacle. Reason (R ) : Destructive interference occurs at the centre of the shadow.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 As the waves diffracted from the edges of circular obstacle, placed in the path of light interfere constructively at the centre of the shadow resulting in the formation of bright spot. |
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| 221. |
Assertion: Coloured spectrum is seen when we look through a muslim cloth. Reason: It is due to the diffraction of white light on passing through fine slits.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 It is quite clear that the coloured spectrum is seen due to diffraction of white light on passing through fine slits made by fine threads in the muslin cloth. |
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| 222. |
When an unpolarised light of inensity `I_0` is incident on a polarizing sheet, the intensity of the light which does not get transmittted isA. zeroB. `I_(0)`C. `(I)/(2)I_(0)`D. `(1)/(4)I_(0)` |
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Answer» Correct Answer - 3 `I=I_(0) cos^(2)theta` Intensity of polarized light =`(I_(0))/(2)` `:.` Intensity of untransmitted light `=I_(0)-(I_(0))/(2)=(I_(0))/(2)` |
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| 223. |
When an unpolarised light of inensity `I_0` is incident on a polarizing sheet, the intensity of the light which does not get transmittted isA. (a) ZeroB. (b) `I_0`C. (c) `1/2I_0`D. (d) `1/4I_0` |
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Answer» Correct Answer - C If an unpolarised light is converted into plane polarised light by passing through a polaroid, its intensity becomes half. |
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| 224. |
A light has amplitude A and angle between analyser and polariser is `60^@`. Light is reflected by analyser has amplitudeA. (a) `Asqrt2`B. (b) `A//sqrt2`C. (c) `sqrt3A//2`D. (d) `A//2` |
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Answer» Correct Answer - D The amplitude will be `A cos 60^@=A//2` |
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| 225. |
A light has amplitude A and angle between analyser and polariser is `60^(@)`. Light is reflected by analyser has amplitudeA. `Asqrt(2)`B. `A//sqrt(2)`C. `sqrt(3)A//2`D. `A//2` |
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Answer» Correct Answer - D The amplitude will be `A cos 60^(@)=A//2` |
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| 226. |
A polariser and an analyser are oriented so that maximum light is transmitted what will be the intensity of outcoming light when analyer is rotated through `60^(@)`. |
| Answer» According to malus law `I=I_(0)cos^(2)theta=I_(0)cos^(2)60^(2)60^(@)=I_(0)[(1)/(2)]^(2)=(I_(0))/(4)` | |
| 227. |
When a polariser and analyser have their axes inclined to one another at `30^(@)` , the amount of light transmitted is 5 SI units. What is the maximum intensity of light transmitted and at what angle between the two? |
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Answer» Correct Answer - 6.67 SI units, `theta=0^(@)` or `+-180^(@)` |
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| 228. |
A polariser and an analyser are oriented so that the maximum amount of lights is transmmitted. Fraction of its maximum value is the intensity of the transmitted through reduced when the analyzer is rotated through (intensity of incident light `= I_(o)`) , a) `30^(@)` , b) `45^(@)` , c) `60^(@)`A. `0.375 I_(0), 0.25I_(0), 0.125 I_(0)`B. `0.25 I_(0), 0.375 I_(0), 0.125 I_(0)`C. `0.125 I_(0), 0.25 I_(0), 0.0375 I_(0)`D. `0.125 I_(0), 0.375 I_(0) , 0.25 I_(0)` |
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Answer» Correct Answer - A `I = (I_(0))/(2) cos^(2) theta` |
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| 229. |
The box of a pin hole camera, of length L, has a hole of radius a . It is assumed that when the hole is illuminated by a parallel beam of light of wavelength `lamda` the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say b_(min)) when:A. `a = (lambda^(2))/(L)and b_(min) = ((2lambda^(2))/(L))`B. `a = sqrt(lambdaL)and b_(min) = ((2lambda^(2))/(L))`C. `a = sqrt(lambda L) and b_(min) = sqrt(4 lambda L)`D. `(lambda^(2))/(L) and b_(min) = sqrt(4 lambda L)` |
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Answer» Correct Answer - C Let b is the radius of spot = geometrical spread + spread due to diffraction `b = a + (lambda L)/(a)` `(db)/(da) = 0 implies a= sqrt(lambda L)` Hence `b_(min) = sqrt(L lambda) + (lambda L)/(sqrt(L lambda))` |
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| 230. |
In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength `lambda`), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate isA. (a) `2lambda`B. (b) `(2lambda)/(3)`C. (c) `lambda/3`D. (d) `lambda` |
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Answer» Correct Answer - A According to given condition `(mu-1)t=nlambda` for minimum t, `n=1` So, `(mu-1)t_(min)=lambda` `t_(min)=(lambda)/(mu-1)=(lambda)/(1.5-1)=2lambda` |
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| 231. |
In a biprism experiement, by using light of wavelength `5000Å`, `5mm` wide fringes are obtained on a screen `1.0m` away from the coherent sources. The separation between the two coherent sources isA. (a) `1.0mm`B. (b) `0.1mm`C. (c) `0.05mm`D. (d) `0.01mm` |
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Answer» Correct Answer - B `d=(Dlambda)/(beta)=(1xx5xx10^-7)/(5xx10^-3)=10^-4m=0.1mm`. |
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| 232. |
In a biprism experiement, by using light of wavelength `5000Å`, `5mm` wide fringes are obtained on a screen `1.0m` away from the coherent sources. The separation between the two coherent sources isA. 1.0 mmB. 0.1 mmC. 0.05 mmD. 0.01 m |
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Answer» Correct Answer - B |
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| 233. |
When two light waves meet at a palceA. their displacement add upB. their intensities add upC. both will add upD. Energy becomes zero |
| Answer» Correct Answer - B | |
| 234. |
A perons with a normal near point (25cm) uses a compound microscope consisting of an objective lens of focal length 8.0 mm and an eye-lense of focal length 2.5 cm. A small objective placed at a distance of 9.0 mm in front of the objective lens produces an image, which is then magnified by the eye-lens to produce a virtual image at the near point. Assume that the eye is placed close to the eye-piece. The magnifying power of the microscope isA. 34B. 8C. 11D. 88 |
| Answer» Correct Answer - D | |
| 235. |
A perons with a normal near point (25cm) uses a compound microscope consisting of an objective lens of focal length 8.0 mm and an eye-lense of focal length 2.5 cm. A small objective placed at a distance of 9.0 mm in front of the objective lens produces an image, which is then magnified by the eye-lens to produce a virtual image at the near point. Assume that the eye is placed close to the eye-piece. The distance between the objective and the eye-lens isA. 7.20 cmB. 9.47 cmC. 2.27 cmD. 4.93 cm |
| Answer» Correct Answer - B | |
| 236. |
The idea of secondary wavelets for the propagation of a wave was first given byA. NewtonB. HuygensC. MaxwellD. Fresnel |
| Answer» Correct Answer - B | |
| 237. |
Two light waves superimposing at the mid-point of the screen are coming from coherent sources of light with phase difference 3p rad. Their amplitudes are 1 cm each. The resultant amplitude at the given point will be.A. 5 cmB. 3 cmC. 2 cmD. zero |
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Answer» Correct Answer - D Resultant amplitude, `A=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos phi)` Here, `A_(1)=A_(2)=1cm,phi=3pi`rad ` :. A=sqrt(I^(2)+I^(2)+2xx1xx1xxcos3pi)=sqrt(2+2xx(-1))=0` |
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| 238. |
Two light waves arrives at a certain point on a screen. The waves have the same wavelength. At the arrival point, their amplitudes and phase differences are : (I) `2a_(0),6a_(0)" and "pi " rad"` (II)`3a_(0),5a_(0)" and "pi " and"` (III) 9a_(0), 7a_(0), " and "3pi " rad"` (IV) 2a_(0), 2a_(0) " and "0`. The pair/s which has greatest intensity is/are :A. IB. IIC. II,IIID. I, IV |
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Answer» Correct Answer - B |
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| 239. |
Consider interference between waves form two sources of intensities `I&4I`.Find intensities at point where the phase difference is `pi`A. `I`B. `5I`C. `4I`D. `3I` |
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Answer» Answer: `I=R^(2)=a_(1)^(2)+a_(2)^(2)_2a_(1)a_(2)cosdelta` `=I+4I+4Icospi` `I=5I-4I=I` |
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| 240. |
Assertion: A narrow pulse of light is sent through a medium. The pulse will retain its shape as it travels through the medium. Reason: A narrow pulse is made of harmonic waves with a large range of wavelengths.A. (a) If both assertion and reason are true and reason is the correct explanation of assertion.B. (b) If both assertion and reason are true but reason is not the correct explanation of assertion.C. (c) If assertion is true but reason is false.D. (d) If assertion and reason both are false. |
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Answer» Correct Answer - D A narrow pulse is made of harmonic waves with a large range of wavelength. As speed of propagation is different for different wavelengths, the pulse cannot retain its shape while travelling through the medium. |
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| 241. |
Assertion: When a light wave travels from a rarer to a denser medium, it loses speed. The reduction in speed imply a reduction in energy carried by the light wave. Reason: The energy of a wave is proportional to velocity of wave.A. (a) If both assertion and reason are true and reason is the correct explanation of assertion.B. (b) If both assertion and reason are true but reason is not the correct explanation of assertion.C. (c) If assertion is true but reason is false.D. (d) If assertion and reason both are false. |
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Answer» Correct Answer - D When a light wave travel from a rarer to a denser medium it loses speed, but energy carried by the wave does not depend on its speed. Instead, it depends on the amplitude of wave. |
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| 242. |
Assertion: When a light wave travels from a rarer to a denser medium, it loses speed. The reduction in speed imply a reduction in energy carried by the light wave. Reason: The energy of a wave is proportional to velocity of wave.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. |
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Answer» Correct Answer - D When a light wave travel from a rarer to a denser medium, its speed decreases but energy carried by the wave remains constant. It depends on the amplitude of wave. The frequency also remains constant. |
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| 243. |
Statement I: In calculating the disturbance produced by a pair of superimposed incoherent wave trians, you can add their intensities. Statement II: `I_(1) + I_(2) + 2 sqrt(I_(1) I_(2)) cos delta`. The average value of `cos delta = 0` for incoherent waves.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. |
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Answer» Correct Answer - A Average value of `(cos delta=underset(t)overset(t=T)intcos(phi_1-phi_2)dt)/T=0` Here `phi_1` and `phi_2` are constantly randomly, fluctuating phases of the two wave trains and integral is taken over a long time (relative to periods of the individual waves). |
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| 244. |
Distance between the slit shown in figure is `d = 20 lambda`, where `lambda` is the wavelength of light used. Find the angle `theta` where a. central maxima (where path difference is zero) is obtained, and b. third-order maxima is obtained. |
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Answer» Ray 1 has a longer path than that of ray 2 by a distance d `sin 45^(@)` before reaching the slits. Afterwords ray 2 has a path longer than ray 1 by a distance `d sin theta` The net path difference is therefore, `d sin theta -d sin 45^(@)` (i) Central maximum is obtained, where not path difference is zero or `d sin theta-d sin 45^(@) =0 or theta=45^(@)` (ii) Third order maxima is obtained, where net path difference is `3lambda` or `d sin theta -d sin 45^(@)=3lambda` `:. sin theta = sin 45^(@)+(3lambda)/d` Putting `d=20 lambda` we have `sin theta=sin 45^(@)+(3lambda)/(20 lambda) or theta ~~59^(@)` |
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| 245. |
In YDSE, if a bichromatic light having wavelengths `lambda_(1)` and `lambda_(2)` is used, then maxima due to both lights will overlaps at a certain distance y from the central maxima. Take separation between slits as d and distance between screen and slits as D. Then the value of y will beA. `((lambda_(1) + lambda_(2))/(2D))d`B. `(lambda_(1) - lambda_(2))/(D) xx 2d`C. LCM of `(lambda_(1) D)/(d)` and `(lambda_(2) D)/(d)`D. HCF of `(lambda_(1) D)/(d)` and `(lambda_(2) D)/(d)` |
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Answer» Correct Answer - c Let `n_(1)^(th)` maxima corresponding to `lambda_(1)` be over lapping with `n_(2)^(th)` maxima corresponding to `lambda_(2)`. Then, the required distance `y = (n_(1) y_(1) D)/(d) = (n_(2) lambda_(2) D)/(d)` = LCM of `(lambda_(1) D)/(d)` and `(lambda_(2) D)/(d)` |
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| 246. |
In the adjacent diagram, CP represents a wavefront and AO & BP, the corresponding two rays. Find the condition on `theta` for constructive interference at P between the ray BP and reflected ray OP. A. `costheta=(3lamda)/(2d)`B. `costheta=(lamda)/(4d)`C. `sectheta-costheta=(lamda)/(d)`D. `sectheta-costheta=(4lamda)/(d)` |
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Answer» Correct Answer - B In `DeltaOPR,(PR)/(OP)=costhetaimpliesOP=(d)/(costheta)` in `DeltaCOPcos2theta=(OC)/(OP)` ltbr. `impliesOC=OPcos2theta=(dcos2theta)/(costheta)` So path difference `=CO+OP+(lamda)/(2)` `=(dcos2theta)/(costheta)+(d)/(costheta)+(lamda)/(2)` `=(d(2cos^(2)theta-1))/(costheta)+(d)/(costheta)+(lamda)/(2)=2dcostheta+(lamda)/(2)` Now for constructive interference at P between B P and OP, path difference `=nlamda` `implies2dcostheta+(lamda)/(2)=nlamdaimplies2dcostheta=(n-(1)/(2))lamda` `impliescostheta=((2n-1)/(4d))lamda`: For `n=1,costheta=(lamda)/(4d)` |
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| 247. |
Two wave are represented by the equations `y_(1)=asinomegat` ad `y_(2)=acosomegat` the first waveA. leads the second by `pi`B. lags the seconds by `pi`C. leads the seconds by `(pi)/(2)`D. lags the seconds by `(pi)/(2)` |
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Answer» Correct Answer - D `y_(1)=asinomegat=acos(omega-pi//2)` `y_(2)=acosomegat` |
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| 248. |
Two transparent slabs have the same thickness as shown in figure. One in made of material A of refractive index 1.5. The other is made of two materilas B and C with thickness in the ratio 1:2. The refractive index of C is 1.6. If a monochromatic parallel beam passing through the slabs has the same number of wavelengths inside both, the refractive index of B is A. 1.1B. 1.2C. 1.3D. 1.4 |
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Answer» Correct Answer - c Total number of waves `=((1.5)t)/(lambda)` `("Total number of waves") = (("optical path length")/("wavelength"))` For B and C: Total number of waves `=(n_(0) ((t)/(3)))/(lambda) + (1.6((2t)/(3)))/(lambda)` Equating (i) and (ii), we get `n_(0) = 1.3` |
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| 249. |
The ratio of intensities of two waves is `9 : 1` When they superimpose, the ratio of maximum to minimum intensity will become :-A. `10:8`B. `9:1`C. `4:1`D. `2:1` |
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Answer» Correct Answer - C |
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| 250. |
A diffraction grating is ruled with 600 lines per millimetre. When monochromatic light falls normally on the grating, the first order diffracted beams are observed on the far side of the grating each making an angle of `15^@` with the normal to the grating. What is the frequency of the light?A. (a) `1.2xx10^(13)Hz`B. (b) `4.7xx10^(13)Hz`C. (c) `1.9xx10^(14)Hz`D. (d) `7.0xx10^(14)Hz` |
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Answer» Correct Answer - D The angle of diffraction `theta_n` for nth order diffraction pattern of the light is given by `dsin theta_n=nlambda` where d=spacing between adjacent slits in diffraction grating `=10^-1/600m` n=1 for first-order diffracted beam. `theta_1=15^@` `implies(dsintheta_1)/(1)=(10^-3)/(600)sin15^@=4.3xx10^-7m` Thus frequency of light, `f=c/lambda=(3xx10^8)/(4.3xx10^-7)=7.0xx10^(14)Hz`. |
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