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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.
| 1. |
The slit width, when a light of wavelenght 6500Å is incident on a slit, if first minima for red light is at `30^(@)` isA. `1xx10^(-6) cm`B. `5.2xx10^(-6) cm`C. `1.3xx10^(-6) cm`D. `2.6xx10^(-6) cm` |
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Answer» Correct Answer - C `lamda=6500A^(@),n=1,a=?theta=30^(@)` For `n^(th)` mininma, `sintheta=nlamda` `:.a=(nlamda)/(sintheta)` `=(1xx6.5xx10^(-7))/(sin30)` `1.3xx10^(-6)m` |
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| 2. |
Two stars distance two light yers are just resolved by a telescope. The wavelenght of light used in is `0.25`m. If the wavelenght of light used in `5000Å` then the minimum distance between the stars isA. `1.22xx10^(11)m`B. `22.44 xx10^(11)m`C. `3.66xx10^(11)m`D. `4.88xx10^(11)m` |
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Answer» Correct Answer - D `(d)/(x)=(1.22lamda)/(D)` `:.d=(1.22xx5xx10^(-7)xx2xx10^(16))/(0.25)=4.88xx10^(10)m` |
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| 3. |
Two stars are situated at a distance of 8 light years from the earth. These are to be just resolved by a telescope of diameter 0.25 m. If the wavelength of light used is 5000 Å, then the distance between the stars must beA. `3 xx10^(10) m`B. `3.35 xx10^(11) m`C. `1.35 xx10^(11) m`D. `4.32 xx10^(10) m` |
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Answer» Correct Answer - C `d=(1.22lamda xx)/(D)=(1.22xx5xx10^(-7)xx8xx10^(16))/(0.25)` `=1.95xx10^(11)m` |
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| 4. |
The distance between the first and the sixth minima in the diffraction pattern of a single slit is 0.5 mm. The screen is 0.5 m away from the slit. If the wavelength of light used is 5000 Å. Then the slit width will beA. `2.5mm`B. `5 mm`C. `1.25 mm`D. `1 mm` |
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Answer» Correct Answer - A `x_(6)-x_(1)=(5Dlamda)/(a)` `:.a=(5Dlamda)/(x_(6)xxx_(1))=(5xx0.5xx5xx10^(-7))/(0.5xx10^(-3))` `=25xx10^(-4)m` |
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| 5. |
A double slit experiment is performed with light of wavelength `500nm`. A thin film of thickness `2mum` and refractive index `1.5` is introduced in the path of the upper beam. The location of the central maximum willA. remains unshiftedB. shift downward by nearly two fringesC. shift upwards by nearly two fringesD. shift downward by 10 fringes |
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Answer» Correct Answer - C `lamda=500nm,t=2mum,mu=1.5,N=?` `N=((mu-1)t)/(lamda)` `=((1.5-1)xx2xx10^(-6))/(5xx10^(-7))` N=2 shifts upward |
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| 6. |
Lights of wavelengths `lambda_(1)=4500 Å, lambda_(2)=6000 Å` are sent through a double slit arrangement simultaneously. ThenA. no interference pattern will be formedB. the thrid bright fringe of `lambda_(1)` will coincide with the fourth bright fringe of `lambda_(2)`C. the thrid bright fringe of `lambda_(2)` will coincide with the fourth bright fringe of `lambda_(1)`D. the fringes of wavelenths `lambda_(1)` will bw wider than the fringes of wavelenghts `lambda_(2)` |
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Answer» Correct Answer - C `x_(n)=(D)/(d)n_(1) lambda_(1)=(D)/(d) n_(2) lambda_(2)` `:. (n_(1))/(n_(2))=(lambda_(2))/(lambda_(1))` `:. n_(1)=(lambda_(2))/(lambda_(1)) n_(2)=(6000)/(45000)n_(2)=(4)/(3) n_(2)` `:. 3n_(1)=4n_(2)` Thus `3^(rd)` bright ringe of `lambda_(2)` will coinide with `4^(th)` bright fringe of `lambda_(1)` |
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| 7. |
In double slit experiment, the angular width of the fringes is `0.20^@` for the sodium light `(lambda=5890Å)`. In order to increase the angular width of the fringes by `10%`, the necessary change in the wavelength isA. increased of `589 Å`B. decreases of `589 Å`C. increases of `6479 Å`D. 0 |
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Answer» Correct Answer - A The angular freingwdith is, `theta=(lambda)/(d) and theta_(0)=(lambda_(0))/(d) :. (lambda_(0))/(lambda)=(theta_(0))/(theta)=(10)/(100)=0.1` `:. lambda_(0)=0.1lambda= 0.1 xx5890=598 Å` |
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| 8. |
A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?A. the angular width of the central maximum of the diffraction pattern will increasesB. the angular width of the central maximum of the diffraction pattern will decreaseC. the angular width of the central maximum will decreaseD. diffraction pattern is not observed on the screen in the case of electrons |
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Answer» Correct Answer - B ` theta=(lambda)/(d)` According to de Broglie `lambda=(h)/(mv)` `:.` If velocity of electron is increases its wavelengths decreases and therefore, its angular bandwith also decreases. |
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| 9. |
In a biprism expeirment , the distance between the slit and the biprism is 20 cm and the distance between the biprism and focal and plane of the epyepiece is 80 cm. Whith the separation between two virtual images of the slit seperation between two virtual images of the slit is 0.4 cm, an interfence pattern is obtained with a light of wavelength `5500Å` .The distance between the third and eighth bands on the same side of the central bright band isA. `1.68 mm`B. `3.68 mm`C. `2.68mm`D. `0.68` mm |
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Answer» Correct Answer - D `x_(8b)-x_(3b)=8beta-3beta=5beta=5(Dlamda)/(d)` `=(5xx1xx5.5xx10^(-7))/(4xx10^(-3))=(275xx10^(-5))/(4)` =0.68 mm |
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| 10. |
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. `6000Å`B. `5000Å`C. `3000Å`D. `45000Å` |
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Answer» Correct Answer - A `beta_(1)-beta_(2)=(D_(1))/(d) lambda-(D_(2))/(d) lambda` `=(lambda)/(d)(D_(1)-D_(2))` `=(lambda)/(d)(D_(1)-D_(2))` `=(lambda)/(d)(D_(1)-D_(1)+0.05)` `=(lambda)/(d)xx0..05` `:. lambda=(( beta_(1)-beta_(2))d)/(0.05)=(3xx10^(-5)xx10^(-3))/(0.05)` `=6000 Å` |
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| 11. |
Consider interference between waves form two sources of intensities `I&4I`.Find intensities at point where the phase difference is `pi`A. `I=I_(2)`B. `I=2I_(2)`C. `I=3I_(2)`D. `I=0` |
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Answer» Correct Answer - A `I= sqrt(I_(1)I_(2)) 2 cos (alpha_(1)-alpha_(2))` `=I_(4)+4I_(1)+ sqrt(I_(1)4I_(1)) 2 cos pi` `=I_(1)` |
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| 12. |
If an interference pattern have maximum and minimum intensities in `36:1` ratio, then what will be the ratio of amplitudes?A. `5:7`B. `7:4`C. `4:7`D. `7:5` |
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Answer» Correct Answer - D `(I_(max))/(I_(min))=(36)/(16),(a_(1))/(a_(2))=?` `(I_(max))/(I_(min))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))` `(36)/(1)=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))` on solving it `:.(a_(1))/(a_(2))=(7)/(5)` |
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| 13. |
Two waves having amplitudes in the ratio `5:1` produce interference. The ratio of the maximum to minimum intesnity isA. `25:1`B. `6:4`C. `9:4`D. `3:2` |
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Answer» Correct Answer - C `(a_(1))/(a_(2))=(5)/(1) :. a_(1)=5 a_(1)` `(I_("max"))/(I_("min"))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))=((5a_(2)+a_(2))^(2))/((5a_(2)-a_(2)))=(9)/(4)` |
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| 14. |
Two light waves of amplitudes `A_(1)` and `A_(2)` superimpose with each other such that `A_(1) gt A_(2)`. The difference between maximum and minimum amplitudes isA. `A_(1)`B. `2A_(2)`C. `2A_(1)`D. `A_(2)` |
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Answer» Correct Answer - B `A_("max")-A_("min")= A_(1)+A_(2)-A_(1)+A+A_(2)` `=2A_(2)` |
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| 15. |
In a biprism experimentn the distancce virtua images of the slit is `2.5 mm` and the eypiece is a distance of 1 m from the slits. If the fringe width is `0.3` mm then the frequency of sources of lights isA. `4xx10^(14)Hz`B. `2xx1^(15)Hz`C. `4xx10^(12)Hz`D. `2xx1^(12)Hz` |
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Answer» Correct Answer - A `n=(V)/(lambda)=(v)/(d)(D)/(beta) " " ( :. lambda=d beta//D)` `=(3xx10^(8)xx1)/(2.5xx10^(-3) xx0.3xx10^(-3))=4xx10^(14)Hz` |
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| 16. |
In the double-slit experiment, the distance of the second dark fringe from the central line are `3 mm`. The distance of the fourth bright fringe from the central line isA. 6 mmB. 8 mmC. 12mmD. 16 mm |
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Answer» Correct Answer - B `x_(n("bright"))= n beta " " :. x_(4b)= 4 beta` and `x_(n("dark"))=(2n-1)b//2 " ":. x_(2b)=1.5 beta` `:.x_(4)= (4)/(1.5)xx(4xx3)/(1.5)=8 mm` |
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| 17. |
Two coherent monochromatic light beams of intensities 4/ and 9/ are superimosed the maxmum and minimum possible intenties in the resulting beam areA. 3 I and 2 IB. 9 I and 5 IC. 16 I and 3 ID. 25 I and I |
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Answer» Correct Answer - D `I=I_(1)+I_(2)+ sqrt(I_(1)I_(2))2 cos (alpha_(1) alpha_(2))` `:. I_("max")= 4I + 9 I + sqrt(4 I xx 9 I) 2 cos theta ` `= 12 I + 12 I =25 I` `and I_("min") 4 I + 9 I + sqrt( 4 I xx 9 I) 2 cos pi =I` |
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| 18. |
In young double slit experiment the ratio of intentsities of bright and dark bands is 16 which meansA. the ratio of their amplitudes is 5B. intensites of individual source are 25 and 9 unit respectivelyC. the ratio of their amplitudes is 4D. intensities of individual sources are 4 and 3 units respectively |
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Answer» Correct Answer - B `(I_("max"))/(I_("min")) =((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))=(16)/(1)` `=:. (a_(1))/(a_(2))=(5)/(3)` But `I propa^(2)` `:. (I_(1))/( I_(2))=(a_(1)^(2))/(a_(2)^(2))=(25)/(9)` |
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| 19. |
In the experiment of interfernece, p is the number of bright bands for a light of wavelength `lamda_(1)`. If the source of light is replaced by `lamda_(2)` then the number of bright bands will beA. `p lambda_(2)//lambda_(1)`B. `p lambda_(1)//lambda_(2)`C. `p lambda_(1)`D. `p lambda_(2)` |
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Answer» Correct Answer - B `n_(1)lambda_(1)=n_(2)lambda_(2)` `n_(2)=(n_(1)lambda_(1))/(lambda_(2))=(p lambda_(1))/(lambda_(2))` |
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| 20. |
The distance between any two successive dark bands is given byA. ` beta =(D)/(d) lambda`B. ` beta =(lambda)/(d)`C. `beta=(d lambda)/(D)`D. `beta =D,d` |
| Answer» Correct Answer - A | |
| 21. |
In the interference pattern, energy isA. created at the position of maximaB. destroyged at the position of minimaC. conserved but is redistributedD. not conserved |
| Answer» Correct Answer - C | |
| 22. |
Light energy hasA. particle like propertiesB. wave like propertiesC. sometimes particle like properties and sometimes wave like propertiesD. neither wave nor particle nature |
| Answer» Correct Answer - C | |
| 23. |
Wave nature of light follows becauseA. light travels in a straight lineB. light exhibites the phenomena of reflection and refractionC. light exhibits the phenomenon of interferenceD. light cause the phenomenon of photelectric effect |
| Answer» Correct Answer - C | |
| 24. |
For interrference of light, sources must have,A. same freqencyB. frequencies very does to each other widely different freqeuncieC. widely different frequenciesD. none of these |
| Answer» Correct Answer - A | |
| 25. |
The transverse nature of light is shown byA. interference of lightB. refraction of lightC. polarisation of lightD. dispersion of light |
| Answer» Correct Answer - C | |
| 26. |
Colours thin films are due toA. dispersion of lightB. interference of lightC. absorption of lightD. scattering of light |
| Answer» Correct Answer - B | |
| 27. |
Fringes are produced with monochromatic light of wavelengths `4.45xx10^(-5)` cm. A thin glass plate of R.I.5 in then normally placedin one of the paths of interfering waves and the central bright band of the fringe system is found to move into the position, previously occupied by the thrie bright band from the system. The thickness of glass plate will beA. `32.7xx 10^(-4) cm`B. `32.7xx 10^(-5) cm`C. `16.7xx 10^(-5) cm`D. `12.5xx 10^(-5) cm` |
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Answer» Correct Answer - B `Delta beta=(beta)/(lambda) (mu-1)t` `3 beta=( beta)/(5.45xx10^(-7)) (1.5-1)t` |
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| 28. |
Band with for red light of wavelength `6400Å` us 0.32 mm. If red lilght is replaced by blue light of wavelength `4800Å`, then the change in bandwidth will beA. 4800ÅB. 6000ÅC. 6400ÅD. 5600Å |
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Answer» Correct Answer - B `beta_(1)=03.322 mm lambda_(1)6400A^(@) lambda_(2) 4800A^(@)` `beta_(1)-beta_(2)= ?` `beta prop lambda` `:. (beta_(2))/(beta_(1))=(lambda_(2))/(lambda_(1))` `beta_(2)=(4800)/(6400)xx0.32=0.24 mm` Now, `beta_(1)-beta_(2)=0.32-0.24 mm ` |
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| 29. |
A wavelength of light `5600Å` produces 60 fringes. What will be the number of fringes produced at same distance if wavelength of light used is `4800Å`?A. 51B. 70C. 60D. 45 |
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Answer» Correct Answer - B `lambda_(1)5600, n_(1) 60 n_(2) = ?` For `lambda_(2)=4800Å` For same path difference, `n_(1) lambda_(1)= n_(2) lambda_(2)` `:.n_(2)=(n_(1) lambda_(1))/(lambda_(2)) =(60xx5600)/(4800) =70` |
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| 30. |
Light and radio wavs both are electromagnetic in nature. Radiowaves undergo appericable diffraction around buildings, where as light waves do not. This is becauseA. wavelength of radiowavesis very small as comparedto the size of the buildingB. wavelengths of radio waves is in the region of 200 to 400 mC. light waves are transverse where as radiowaves are longitudinalD. light waves travel faster than radiowaves |
| Answer» Correct Answer - B | |
| 31. |
Both light and soucnd waves produced diffraction. It is more difficult to bserve the diffraction with light waves becauseA. light wave do not equired mediumB. wavelength of light waves is far smallerC. light waves are transverseD. speed of light is far greater |
| Answer» Correct Answer - B | |
| 32. |
The main difference in diffraction and inteference isA. in diffraction, all the maxima are of decreasing intensity whereas in interference all the maxima are of equal intensityB. in diffraction, all the maxima are of equal intensity whereas in interference all the maixa are of decreasing intensityC. in diffraction, all the maxima are of decreasing intensity whereas in interference all the maxima are of decreasing intensityD. in diffraction, all the maxima are of equal intensity whereas in interference all the maxima are of equal intensity |
| Answer» Correct Answer - A | |
| 33. |
The main difference in diffraction and interference is thatA. in diffraction, the fringe width of different firnges are not equal whereas in interference the fringes widths are equalB. it cannot be observed with white lightC. unlike diffraction the interference fringesareD. nono of these |
| Answer» Correct Answer - A | |
| 34. |
Radio telescope is used to seeA. stars and to measure diametersB. distance stars and planetsC. high frequency wavesD. sun and to measure its temeprature |
| Answer» Correct Answer - B | |
| 35. |
The path difference produced by two waves is 3.75 `mu`m and the wavelength is 5000 Å. The point isA. uncertainB. darkC. partially brightD. bright |
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Answer» Correct Answer - B Path difference `= n lambda` `:.n=("Path diff")/(lambda)` `=(3.75xx10^(-6))/(5xx10^(-7)) =(37.5)/(5)=7.5` Path difference is equal to odd multiple of hal of wagvelength given is dark. |
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| 36. |
The distance of `n^(th)` minima from the centre of interference pattern isA. `x_(n)=(D)/(d)n lambda`B. `x_(n)=(D)/(d) (2n-1) (lambda)/(2)`C. `x_(n)=Dd`D. `(2n-1) (lambda)/(2)` |
| Answer» Correct Answer - B | |
| 37. |
The optical path difference between the two light waves arriving at a point on the screen isA. `(xd)/(D)`B. `(xD)/(d)`C. `(d)/(xD)`D. `(dD)/(x)` |
| Answer» Correct Answer - A | |
| 38. |
Polarization oflight takes place due to many process. Wchih of the following will not cause polarization ?A. reflactionB. double refractionC. scatteringD. diffraction |
| Answer» Correct Answer - D | |
| 39. |
For the sustained interference of light, the necessary condition is that the two sources shouldA. be narrowB. be very close to each otherC. emit light waves continuouslyD. all of these |
| Answer» Correct Answer - D | |
| 40. |
Two sources of light are said to be equally bright if they will emit waves of sameA. velocityB. wavelengthsC. amplitudeD. phase |
| Answer» Correct Answer - C | |
| 41. |
If path difference between two interfering waves is zero, then the point will beA. darkB. brightC. as it isD. either dark of bright |
| Answer» Correct Answer - B | |
| 42. |
When destrutive interference obtained, the phase difference isA. `0, 2pi, 4pi`,...B. `pi, 3pi, 5pi,....`C. `(pi)/(2),(3pi)/(2),(5pi)/(2),.....`D. `(pi)/(4),(pi)/(2),(3pi)/(4),pi,....` |
| Answer» Correct Answer - B | |
| 43. |
To obtain a sustained iterference pattern, we requie two sourece which emit radiation ofA. the same frquencyB. nearly the same frequencyC. the same frequency and have a defined phase relationshipD. different wavelenghths |
| Answer» Correct Answer - C | |
| 44. |
For steady interference, two two sources of light must beA. coherentB. monochoromaticC. equally brightD. all of these |
| Answer» Correct Answer - D | |
| 45. |
When interference of light place, the intensity of lightA. is maximum at same points and at rest of the points is minimumB. increases at some point and remains the same at rest of the pointsC. decreases at same and remains same ate rest of the pointsD. does not change at all |
| Answer» Correct Answer - A | |
| 46. |
The interference patter in which the positions of maximum and minimum intensity of light remain fixed isA. constructive interferenceB. destructive interferenceC. sustained interferenceD. noen of these |
| Answer» Correct Answer - C | |
| 47. |
The locus of all bright points produced due to the superpostion of two indentical light waves isA. bright fringsB. dark fringeC. band widthD. spectrum |
| Answer» Correct Answer - A | |
| 48. |
Let `I_(1) and I_(2)` be the intensity of two light waves of amplitudes `a_(1) and a_(2)` respectively . The resultant ampltiudes I at a point due to the superposition of two light waves isA. `I=I_(1)^(2)+I_(2)^(1)`B. `I=((I_(1)+I_(2)))/((I_(1)-I_(2)))`C. `I=I_(1)I^(2)+I_(2)^(2)+2sqrt(I_(1) I_(2) cos (alpha_(1)-alpha_(2)))`D. `I=I_(2)+I_(2)+sqrt(I_(1)I_(2)) 2 cos (a_(1)-a_(2))` |
| Answer» Correct Answer - D | |
| 49. |
Two waves having the intensities in the ratio of `9 : 1` produce interference. The ratio of maximum to minimum intensity is equal toA. `10:8`B. `9:1`C. `4:1`D. `2:1` |
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Answer» Correct Answer - C `(I_(1))/(I_(2))=(a_(1)^(2))/(a_(2)^(2))=(9)/(1) " " :. (a_(1))/(a_(2))=(3)/(1)` `(I_("max"))/(I_("min"))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))=((a_(1)+3a_(1))^(2))/((a_(1)-3a_(1))^(2))=(4)/(1)` |
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| 50. |
In a biprism experimentn the the band width is found to be increased by 25% of the initial. If the distance between sliteand eyepeice is increased by 20 cm then the initial distance between slit and epepiece isA. 80 cmB. 90 cmC. 70 cmD. 60 cm |
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Answer» Correct Answer - A `beta_(2)=1.25 beta_(1) and D_(2)=D_(1)+0.2` `rArr (D_(2))/(d) lambda=1.25 D_(1)` `D_(1)=(0.2)/(0.25)=0.8 mm =80 cm` |
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