58 + Interview Questions in INTERFERENCE-AND-DIFFRACTION-OF-LIGHT IN GENERAL AWARENESS Page 1 InterviewSolution

1.	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. `3xx10^(10)m`B. `3.35xx10^(11)m`C. `1.95xx10^(11)m`D. `4.32xx10^(10)m`
Answer» Correct Answer - C Limit of resolution of the telescope, `alpha=(1.22lambda)/(a)=(d)/(x) rArr d=(1.22lambda x)/(a)` `=(1.22xx5xx10^(-7)xx8xx10^(16))/(0.25)=1.95xx10^(11)m`

Discussion

2.	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. 5 mmB. 2.5 mmC. 1.25 mmD. 1.0 mm
Answer» Correct Answer - B Distance between first and sixth minima `x=(5lambdaD)/(d)` `therefore d=(5lambdaD)/(x)=(5xx5000xx10^(-10)xx0.5)/(0.5xx10^(-3))` `d=2.5xx10^(-3)m=2.5mm`

Discussion

3.	A parallel beam of light of wavelength `6000Å` gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is :A. `6xx10^(-3)` radB. `1.8xx10^(-3)` radC. `3xx10^(-3)` radD. `2xx10^(-3)` rad
Answer» Correct Answer - D We have d `sin theta =n lambda` `0.3xx10^(-3) xx theta=6000xx10^(-10)` The angular position ` theta=2xx10^(-3)` rad

Discussion

4.	A narrow slit of width 2 mm is illuminated by monochromatic light fo wavelength 500nm. The distance between the first minima on either side on a screen at a distance of 1 m isA. 5 mmB. 0.5 mmC. 1 mmD. 10 mm
Answer» Correct Answer - B Distance `=(2lambda)/(b)xxd=(2xx5xx10^(-4))/(2)xx1000=0.5mm`

Discussion

5.	In double slit experiment , the distance between two slits is `0.6mm` and these are illuminated with light of wavelength `4800 Å`. The angular width of dark fringe on the screen at a distance 120 cm from slits will beA. `8xx10^(-4)` radB. `6xx10^(-4)` radC. `4xx10^(-4)` radD. `16xx10^(-4)` rad
Answer» Correct Answer - A Given, `d=0.6mm=0.6xx10^(-3)m` `lambda=4800"Å"=4.8xx10^(-7)m,n=1` Angular fringe width of first minima `(2x)/(D)=2(2n-1)(lambda)/(2d)=(2xx1-1)(lambda)/(d)` `therefore (2x)/(D)=((2-1)xx4.8xx10^(-7))/(0.6xx10^(-3))=8xx10^(-4)` rad

Discussion

6.	A parallel beam of light of intensity I is incident on a glass plate. `25%` of light is reflected in any reflection by upper surface and `50%` of light is reflected by any reflection from lower surface. Rest is refracted The ratio of maximum to minimum intensity in interference region of reflected rays is A. `(((1)/(2)+sqrt((3)/(8)))/((1)/(2)-sqrt((3)/(8))))^(2)`B. `(((1)/(4)+sqrt((3)/(8)))/((1)/(2)-sqrt((3)/(8))))^(2)`C. `(5)/(8)`D. `(8)/(5)`
Answer» Correct Answer - A The intensity of light reflected from upper surface is `I_(0)=I_(0)xx25%=I_(0)xx(25)/(100)=(I_(0))/(4)` The intensity of transmitted light from upper surface is `I=I_(0)-(I_(0))/(4)=(3I_(0))/(4)` `therefore` The intensity of reflected light from lower surface is `I_(2)=(3I_(0))/(4)xx(50)/(100)=(3I_(0))/(8)` `therefore (I_("max"))/(I_("min")) =((sqrt(I_(1))+sqrt(I_(2)))^(2))/((sqrt(I_(1))-sqrt(I_(2)))^(2))=(((1)/(2)+sqrt((3)/(8)))^(2))/(((1)/(2)-sqrt((3)/(8)))^(2))`

Discussion

7.	A thick glass slab `(mu= 1.5)` is to be viewed in reflected white light. It is proposed to coat the slab with a thin layer of a material having refractive index 1.3 so that the wavelength 6000 Å os suppressed. Find the minimum thickness of the coating required.
Answer» Ray-1 and ray-2 both are reflected from denser medium. Hence, they are in phase. `:. 2mut = lambda/2` for minima or ` t=lambda/(4mu) = 600/(4xx1.3)` ` = 1154Å` .

Discussion

8.	An oil film covers the surface of a small pond. The refractive index of the oil is greater than that of water. At one point on the film, the film has the smallest non-zero thickness for which there will be destructive interference in the reflected light when infrared radiation with wavelength 800 nm is incident normal to the film. When this film is viewed at normal incidence at this same point, for what visible wavelengths, if any, will there be constructive interference? (Visible light has wavelengths between 400nm and 700 nm )
Answer» Ray-1 is reflected from denser medium and ray-2 from denser medium. `:. Deltax = 2mut = lambda = 800nm` for destruction interference. ` mut = 400nm ` For constructive interference , ` Deltax = 2mut = (2n-1) lambda/2` ` :. lambda = ((4mut)/(2n-1))= ((1600)/(2n-1))` For `n=1, lambda=1600nm ` For `n=2, lambda = 533nm` For `n=3, lambda = 320 nm` The only wavelength lying in the given range is 533 nm.

Discussion

9.	Determine what happens to the double slits interference pattern if one of the slits is covered with a thin, transparent film whose thickness is `lambda/(2(mu-1))`, where `lambda` is the wavelength of the incident light and mu is the index of refraction of the film.
Answer» Path difference produced by slab, ` Deltax = (mu-1)t = lambda/2` `lambda/2` path difference is equivalent to `180^@` phase difference. Hence, maxima and minima interchange their positions.

Discussion

10.	A possible means for making an airplane invisible to radar is to coat the plane with an anti reflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is `mu=1.5`. How thick is the oil film? Refractive index of the material of airplane wings is greater than the refractive index of polymer.
Answer» `2mut = lambda//2` This is the condition for destructive interference . `:. t= lambda/(4mu) = (3.0)/ (4xx 1.5) = 0.5 cm`

Discussion

11.	Interference fringes are produced on a screen by using two light sources of intensities / and 9/. The phase difference between the beams `pi/2` is at point P and `pi`at point Q on the screen. The difference between the resultant intensities at point P and Q isA. 2IB. 4IC. 6ID. 8I
Answer» Correct Answer - C According to question, resultant intensity of interferring wave is given by `I_(P)=I_(1)+I_(2)+2sqrt(I_(1)I_(2)cos phi)` For `phi=(pi)/(2), I_(P)=I+9I=10I` Again at point Q, resultant intensity is given by `I_(Q)=I_(Q)=I_(1)+I_(2)+2sqrt(I_(1)I_(2)cos phi)` For `phi=pi, I_(Q)=I+9I+(-2sqrt(9(I)^(2)))` `=10I-6I=4I` Now, difference between the resultant intensity is given by `Delta I=I_(P)-I_(Q)=10I-4I=6I`

Discussion

12.	Light waves from two coherent sources having intensities I and 2I cross each other at a point with a phase difference of `60^(@)` . The intensity at the point will beA. `4.414 I`B. `5.455 I`C. `4I`D. `6.441I`
Answer» Correct Answer - A Here, `I_(1)=I, I_(2)=2I and phi =60^(@)` We know that, the amplitude A of resultant wave is `A=sqrt(A_(1)^(2)+A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos phi)` `A^(2)=A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos phi` We also know that, `I prop A^(2)` `therefore` Required intensity, `I=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cos phi` `=I+2I+2sqrt(I xx 2I) cos 60^(@)` `=4.414I`

Discussion

13.	nth bright fringe if red light `(lambda_(1)=7500" Å")` coincides with (n +1)th bright fringe of green light `(lambda_(2)=6000" Å")` The value of n, isA. 4B. 5C. 3D. 2
Answer» Correct Answer - A Since, the two bright fringes coincide `therefore nlambda_(1)=(n+1)lambda_(2)` `or (n+1)/(n)=(lambda_(1))/(lambda_(2))=(7500)/(60000)=(5)/(4)` `or 1+(1)/(n)=(5)/(4) rArr n=4`

Discussion

14.	A fringe width of a certain interference pattern is `beta=0.002` cm What is the distance of 5th dark fringe centre?A. `1xx10^(-2)cm`B. `11xx10^(-2)cm`C. `13xx10^(-3)cm`D. `3.28xx10^(6)cm`
Answer» Correct Answer - C The distance of 7th dark fringe from centre is given by `x_(n)=(2n+1)(lambda l)/(2d)` as n=6 for 7 th dark fringe So, `x_(7)=(13)/(2)(lambdaD)/(d) " as " (lambdaD)/(d)=beta` So, `x_(7)=((13)/(2))beta=(13)/(2)xx0.002=13xx10^(-3)cm`

Discussion

15.	In an interference pattern the position of zeroth order maxima is 4.8 mm from a certain point P on the screen. The fringe width is 0.2 mm. The position of second maxima from point P isA. 5.1 mmB. 5 mmC. 40 mmD. 5.2 mm
Answer» Correct Answer - A The distance between zeroth order maxima and second order minima is `y_(1)=(beta)/(2)+beta=(3)/(2)beta=(3)/(20xx0.2mm=0.3mm` `therefore` The distance of second maxima from point P is `y=(4.8+0.3)mm=5.1 mm `

Discussion

16.	In a YDSE experiment, d = 1mm, `lambda`= 6000Å and D= 1m. The minimum distance between two points on screen having 75% intensity of the maximum intensity will beA. 0.50 mmB. 0.40 mmC. 0.30 mmD. 0.20 mm
Answer» Correct Answer - D `3/4 I_(max) = I_(max) cos^2 (phi/2)` ` :. phi/2 = pi/12` and `(5pi)/6` `:. phi= pi/6` and `(5pi)/3 = ((2pi)/lambda) (Deltax) = ((2pi)/lambda) (yd)/D` `:. y_1= (lambdaD)/(12d) = ((6000 xx 10^-10)(1))/((12)(10^-3))` `= 0.05 xx (10^-3) m = 0.05mm` `y_2 =5 ((lambdaD)/(12d))= 5xx 0.05 mm ` =0.25mm `Deltay = y_2-y_1 = 0.2mm` .

Discussion

17.	Coherent light with wavelength 600nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first order bright fringe is at 4.94 mm from the centre of the central bright fringe. For what wavelength of light will the first order dark fringe be observed at this same point on the screen?
Answer» `y=omega_1 = omega_2/2` `:. (lambda_1D)/d = (lambda_2D)/(2d)` `rArr lambda_2 = 2lambda_1 = 1200nm` .

Discussion

18.	The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of 4th dark fringe after a glass sheet of refractive index 1.5 is introduced. The thickness of glass sheet would beA. `4.8 mu m`B. `4.2 mu m`C. `5.4 mu m `D. `3.0 mu m `
Answer» Correct Answer - B Shift `=((mu-1)tD)/d = 3.5, omega= (3.5lambdaD)/d` `t=(3.5lambda)/(mu-1)` `=((3.5)(6000 xx 10^-10))/(1.5 -1)` =`4.2 xx 10^(-6)m` =4.2 `mu`m.

Discussion

19.	Two very narrow slits are spaced `1.80 mu`m apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light of `lambda = 550nm ?` (Hint : The angle `theta` is not small).
Answer» `d sin theta_1 = lambda/2 ` ` sin theta_1 = lambda/(2d) = (550 xx 10^(-9))/(2 xx 1.8 xx 10^-6)` ` :. theta_1 = 8.78^@ ` ` y_1/D = tan theta_1` ` :. Y_1 = D tan theta_1 ` `= 35 tan 8.78^@` = 5.41 cm `d sin theta_2 = (3lambda)/2` ` :. sin theta_2 = (3lambda)/(3d) = (3 xx 550 xx 10^-9)/(2xx 1.8 xx 10^-6)` ` :. theta_2 = 27.27^@` ` :. y_2 = D tan theta_2` ` = 35tan 27.27^@` = 18cm ` Deltay = y_2 - y_1 = 12.6 cm.

Discussion

20.	Two slits spaced 0.450nm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500nm?
Answer» The required distance= one fringe width `=omega = (lambdaD)/d` ` = ((500 xx 10^(-9))(0.75))/(0.45 xx (10^-3)) ` ` = 8.33 xx 10^-4` = 0.83mm.

Discussion

21.	In a double slit experiment, the distance between the slits is 5.0 mm and the slits are 1.0m from the screen. Two interference patterns can be seen on the screen one due to light with wavelength 480nm, and the other due to light with wavelength 600nm. What is the separation on the screen between the third order bright fringes of the two intergerence patterns?
Answer» Distance `= (3lambda_2D)/d - (3lambda_1D)/d)` = (3(lambda_2-lambda_1)D)/d ` ` =( 3 xx (600 - 480) xx 10^(-9) xx 1.0)/(5.0 xx 10^-3)` ` = 7.2 xx 10^(-5) m ` = 0.072mm .

Discussion

22.	With two slits spaced 0.2 mm apart and a screen at a distance of 1 m, the third bright fringe is found to be at 7.5 mm from the central fringe. The wavelength of light used isA. 400nmB. 500nmC. 550nmD. 600nm
Answer» Correct Answer - B `y= (3lambdaD)/d` `:. lambda = (yd)/(3D) = ((7.5 xx 10^-3)(0.2 xx 10^-3))/((3)(1))` ` =500xx 10^-9 m = 500nm `

Discussion

23.	Red light of wavelength 625 nm is incident normally on a optical diffraction grating with `2xx10^(5)` lines/m. Including central principal maxima, how many maxima may be observed on a screen which is far from the grating?A. 15B. 17C. 8D. 16
Answer» Correct Answer - B For principal maxima in grating spectra `(sin theta)/(N)= n lambda` where, n=(1, 2, 3) is the order of principal maximum and `theta` is the angle of diffraction. `n=(1)/(lambdaN)=(1)/(6.25xx10^(-7)xx2xx10^(5))=8` `therefore` Number of maxima `=2n+1=2xx8+1=17`

Discussion

24.	Consider sunlight incident on a slit of width `10^(4) Å` . The image seen through the slit shallA. be a fine sharp slit white in colour at the centreB. a bright slit white at the centre diffusing to zero intensity at the edgesC. a bright slit white at the centre diffusing to regions of different coloursD. only be a diffused slit white in colour
Answer» Correct Answer - A Here, the width of the slit is `10^(4)` Å, i.e. 10000 Å. The wavelength of (visible) sunlight varies from 4000 Å to 8000 Å. As width of slit ` a gt lambda` (wavelength of light), therefore no diffraction occurs. The image seen through the slit shall be a fine sharp slit white in colour at the centre.

Discussion

25.	Two identical radiators have a separation of `d=lambda//4` where `lambda` is the wavelength of the waves emitted by either source. The initial phase difference between the sources is `lambda//4`. Then the intensity on the screen at a distant point situated at an angle `theta=30^@` from the radiators is (here `I_0` is intensity at that point due to one radiator alone)A. `I_(0)`B. `2I_(0)`C. `3I_(0)`D. `4I_(0)`
Answer» Correct Answer - B The intensity at a point on screen is given by `I=4I_(0)cos^(2)(phi//2)` When `phi` is the phase difference. In this problem `phi` arises (i) due to initial phase differnce of `(pi)/(4)` and (ii) due to path difference for the observation point situated at `theta=30^(@)`. Thus, `phi=(pi)/(4)+(2pi)/(lambda)(d sin theta)=(pi)/(4)+(2pi)/(lambda) xx (lambda)/(4) (sin30^(@))=(pi)/(4)+(pi)/(4)=(pi)/(2)` Hence, `(pi)/(2)=(pi)/(4) and I=4I_(0) cos^(2)((pi)/(4))=2I_(0)`

Discussion

26.	When a compact disc is illuminated by a source of white light, coloured lines are observed. This is due toA. dispersionB. diffractionC. interferenceD. refraction
Answer» Correct Answer - B The fine rulings, each `0.5mu m` wide, on a compact disc function as a diffraction grating. When a small source of white light illuminates a disc, the light is diffracted from the rulings.

Discussion

27.	Light of wavelength `lamda` is incident on a slit of width d. the resulting diffraction pattern is observed on a screen at a distance D. the linear width of the principal maximum is equal to the width of the slit if D equalsA. `(d^(2))/(2lambda)`B. `(d)/(lambda)`C. `(2lambda^(2))/(d)`D. `(2lambda)/(d)`
Answer» Correct Answer - A The linear width of central principal maximum `=(2lambdaD)/(d)` If it is equal to width of slit (d), then `(2lambdaD)/(d)=d rArr D=(d^(2))/(2lambda)`

Discussion

28.	A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.0 mm wide and the resulting diffraction pattern is observed on a screen 2m away. What is the distance between the first dark fringe on either side of the central bright fringe?
Answer» For the diffraction at a single slit, the position of minima is given by ` d sin theta = nlambda ` For small value of theta, `sin theta~~theta=y/D` `:. dy/d=lambda or y=D/d lambda` Substituting the values ,we have `y=(2xx6xx10^(-7))/(1xx10^(-3))` `=1.2xx10^(-3)m` `=1.2 mm`. `:.` Distance between first minima on either side of central side of central maxima=`2y=2.4mm.`

Discussion

29.	A single slit of width a is illuminated by violet light of wavelength `400nm` and the width of the diffraction pattern is measured as y. When half of the slit width is covered and illuminated by yellow light of wavelength `600nm`, the width of the diffraction pattern is
Answer» Correct Answer - C Given, `lambda_(1)=4000 nm, lambda_(2)=600 nm, d_(1)=d and d_(2)=(d)/(2)` In single slit diffraction experiment, width of central maxima `y=(2lambdaD)/(d) rArr (y_(1))/(y_(2))=(lambda_(1))/(lambda_(2))xx(d_(2))/(d_(1))` `therefore (y_(1))/(y_(2))=(400)/(600)xx((d)/(2))/(d)=(1)/(3)` So, `y_(2)=3y_(1)=3y`

Discussion

30.	A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the secondary maximum in the pattern for red light of wavelength 6500 Å ?A. 4400 ÅB. 4100 ÅC. 4642.8 ÅD. 9100 Å
Answer» Correct Answer - C Distance of nth secondary maxima from central maxima `x=((2n+1)lambdaD)/(2a)` For red light, `x=((4+1)D)/(2a)xx6500` For unknown of light, `x=((6+1)D)/(2a)xx lambda` According to the question, `5xx6500=7xx lambda` Wavelength of light `lambda=(5)/(7)xx6500=4642.8` Å

Discussion

31.	Three waves of equal frequency having amplitudes `10mum`, `4mum`, `7mum` arrive at a given point with successive phase difference of `pi//2`, the amplitude of the resulting wave in `mum` is given byA. 4B. 5C. 6D. 7
Answer» Correct Answer - B The amplitude of the waves are `a_(1)=10mum,a_(2)=4mu m and a_(3)=7mu m` and phase difference between Ist and IInd wave is `(pi)/(2)` and that between Iind and IIIrd wave is `(pi)/(2)`. Then, phase difference between Ist and IIIrd is `pi`. Combining Ist with IIIrd, their resultant amplitude is given by `A_(1)^(2)=a_(1)^(2)+a_(3)^(2)=2a_(1)a_(3) cos phi` `or A_(1)=sqrt(10^(2)+7^(2)+2xx10xx7 cos pi)` `=sqrt(100+49-140)=sqrt(9)=3 mu m` Now, combining this with IInd wave, we have the resultant amplitude `A^(2)=A_(1)^(2)+a_(2)^(2)+2A_(1)a_(2) "cos"(pi)/(2)` `or A=sqrt(3^(2)+4^(2)+2xx3xx4xxcos90^(@))` `=sqrt(9+16)=sqrt(25)=5mu m`

Discussion

32.	In the phenomenon of interference,A. sources must be coherentB. amplitudes must be sameC. wavelengths must be sameD. intensities may be different
Answer» Correct Answer - A::C::D

Discussion

33.	A plane wave front of wavelength `lamda` is incident on a single slite of width b. What is the angular width for secondary maximum ?A. `(lambda)/(2b)`B. `(lambda)/(b)`C. `(2lambda)/(b)`D. `(b)/(lambda)`
Answer» Correct Answer - B `beta=(lambdaD)/(d) and theta=(beta)/(D)` `therefore theta=(lambda)/(d)` `therefore` Width of single slit = b So, `theta=(lambda)/(b)`

Discussion

34.	If the ratio of amplitude of two waves is `4:3`, then the ratio of maximum and minimum intensity isA. `16:18`B. `18:16`C. `49:1`D. `1:49`
Answer» Correct Answer - C In the interference pattern, `I_("max")=(a+b)^(2)and I_("min")=(a-b)^(2)` `(I_("max"))/(I_("min"))=((a+b)^(2))/((a-b)^(2))=((4+3)^(2))/((4-3)^(2))` `((7)/(1))^(2)=(49)/(1)=49:1`

Discussion

35.	If the ratio of amplitude of wave is`2 : 1`, then the ratio of maximum and minimum intensity isA. `9:1`B. `1:9`C. `4:1`D. `1:4`
Answer» Correct Answer - A The ratio of maximum to minimum intensity is given as `(I_("max"))/(I_("min"))=((a+b)^(2))/((a-b)^(2))` `=((2+1)^(2))/((2-1)^(2))=(9)/(1)`

Discussion

36.	If white light is used in a biprism experiment, thenA. fringe pattern disappersB. all fringes will be colouredC. central fringe will be white others will be colouredD. central fringe will be dark
Answer» Correct Answer - C When white light is used in a biprism experiment central spot will be white, while the surrounding fringes will be coloured.

Discussion

37.	How will the diffraction pattern of single slit change when yellow light is replaced by blue light? The fringe will beA. widerB. narrowerC. brighterD. fainter
Answer» Correct Answer - B Frings width `prop` wavelength of light. Since, wavelength of blue colour is less than yellow colour. Therefore, fringe will become narrower.

Discussion

38.	In an interference experiment, the spacing between successive maxima or minima is (Where the symbols have their usual meanings)A. `lambda d//D`B. `lambda D//lambda`C. `d D//lambda`D. `lambda d//4D`
Answer» Correct Answer - B In an interference experiment the spacing between successive maxima and minima is called the fringe width and is given by `beta=D lambda//d`

Discussion

39.	Light propagates 2 cm distance in glass of refractive index 1.5 in time `t_(0)`. In the same time `t_(0)`, light propagates a distance of 2.25 cm in medium. The refractive index of the medium isA. `4//3`B. `3//2`C. `8//3`D. None of these
Answer» Correct Answer - A For a given time, optical path remains constant. `therefore mu_(1)x_(1)=mu_(2)x_(2)` `rArr 1.5xx2=mu_(2)xx2.25` `rArr mu_(2)=(1.5xx2)/(2.25)=(4)/(3)`

Discussion

40.	In the YDSE apparatus shown in figure, `dltltD` and `d = 6lambda`. Find (a) total number of maximas and minimas on the screen . (b) two y-coordinates corresponding to third order maxima.
Answer» (a) In the given set up, `dltltD` Therefore, we can apply ` Deltax =d sin theta ` `(Deltax)_min = 0` at `theta = 0^@` and `(Deltax)_max = 6lambda` at` theta = 90^@` Therefore, total number of maximas are eleven corresponding to, `Deltax = 0, +-lambda, +-2lambda, +-3lambda, +-4lambda and +-5lambda` (b) Third order minima lies at `Deltax = +-3lambda` `:. d sin theta = +-3lambda or 6 lambda sin theta = +-3 lambda ` or sin theta = +- 1/2` or `theta = +-(30^@)` Now, `y/D = tan theta` and `y= D tan theta` `:. y = +-D tan30^(@)` or `y=+- D/sqrt3`.

Discussion

41.	A small aperture is illuminated with a parallel beam of `lambda = 628 nm`. The emergent beam has an anglur divergence of `2^(@)`. The size of the aperture isA. 180 mB. 18 `mu`mC. 1.8 mD. 0.18 m
Answer» Correct Answer - B From diffraction at a single slit, Size of aperture, `a=(lambda)/(sin theta)` `or a=(3141.59xx10^(-10))/(sin1^(@))=18xx10^(-6)m=18mu m`

Discussion

42.	Two coherent sources are 0.3 mm apart. They are 0.9m away from the screen. The second dark fringe is at a distance of 0.3cm from the centre. Find the distance of fourth bright fringe from the centre. Also, find the wavelength of light used.
Answer» Correct Answer - A::B::C Given, ` d= 0.3 xx 10^(-3)m, D=0.9m ` ` (3lambdaD)/(2d) = 0.3 xx 10^(-2)cm` (the distance of second dark fringe) ` :. (lambdaD)/d = (0.3 xx 10^(-2))(2/3)` `= 0.2 xx 10^(-2)m = 0.2cm ` (i) Distance of fourth bright fringe from centre =`(4lambdaD)/d = 0.8cm (ii) `lambda = (d/D) (0.2 xx 10^-2)m = ((0.3 xx (10^-3))/0.9) (0.2 xx 10^-2))` `= 6.67 xx 10^(-7)m`.

Discussion

43.	The condition for diffraction of mth order minima isA. `d sin theta_(m)=m lambda, m=1,2,3, …`B. `d sin theta_(m)=(m lambda)/(2), m=1,2, 3, …`C. `d sin theta_(m)=(m+1)(lambda)/(2), m=1,2,3,…`D. `d sin theta_(m)=(m-1)(lambda)/(2), m=1, 2, 3,…`
Answer» Correct Answer - A For obtaining mth secondary minima at a point on screen, path difference between the diffracted waves `Delta=d sin theta_(m)= pm mlambda` where, m = 1, 2, 3, …..

Discussion

44.	Calculate the resolving power of a telescope when light of wavelength `540 nm` is used. Diameter of objective lens is `6 cm`.A. `9.1 xx10^(-4)"rad"^(-1)`B. `0.1xx10^(-4)"rad"^(-1)`C. `5xx10^(-4)"rad"^(-1)`D. `6xx10^(-4)"rad"^(-1)`
Answer» Correct Answer - B Given `D=6cm=6xx10^(-2)m` `lambda=540nm=540xx10^(-9)m` Resolving power of the telescope `RP=(D)/(1.22lambda)=(6xx10^(-2))/(1.22xx540xx10^(-9))` `=(6)/(122xx54)xx10^(8)` rad In terms of `"rad"^(-1)` `RP^(-1)=(1)/(RP)=(122xx54)/(6xx10^(8))"rad"^(-1)` `=0.1 xx 10^(-4)"rad"^(-1)`

Discussion

45.	If the aperature of a telescope is decreased resolving power willA. increaseB. decreaseC. remain sameD. zero
Answer» Correct Answer - B The resolving power `=(a)/(1.22 lambda)` were, a = aperture of telescope `therefore` Resolving power `prop` aperture of telescope If resolving power is decreased, the resolving power will decrease.

Discussion

46.	Interference was observed in interference chamber when air was present, now the chamber is evacuated and if the same light is used, a careful observer will seeA. interference in which width of the fringe will be slightly increasedB. Interference with bright bondC. Interference with dark bondD. All of the above
Answer» Correct Answer - A The refreactive iondex of air is slightly more than 1. When chamber is evacuated, refractive index decreases and hence the wavelength increases and fringe width also increases.

Discussion

47.	The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit isA. `1:4:9`B. `1:2:3`C. `1:(4)/(9pi^(2)):(4)/(25pi^(2))`D. `1:(4)/(pi^(2)):(9)/(pi^(2))`
Answer» Correct Answer - C The ratio of intensities of successive maxima is `1:((2)/(3pi))^(2):((2)/(5pi))^(2):((2)/(7pi))^(2)=1:(4)/(9pi^(2)):(4)/(25pi^(2))`

Discussion

48.	A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observe on screen 1 m away. It is observed that the first minimum is at a distance of `2.5 mm` from the centre of the screen. Find the width of the slit.A. 0.2 mmB. 0.3 mmC. 0.4 mmD. 0.5 mm
Answer» Correct Answer - A Distance of first minima in diffraction pattern due to a single slit is `x_(1)=(Dlambda)/(a)` It is given that `D=1m, lambda=500 mm=500xx10^(-19)m` `x_(1)=2.5mm=2.5 xx 10^(-3)m, a=?` `therefore a=(Dlambda)/(x_(1))=(1xx500xx10^(-9))/(2.5xx10^(-3)) m ` `=200xx10^(-6)m=0.2 mm`

Discussion

49.	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 is
Answer» Correct Answer - D The phase difference `(phi)` between the wavelets from the top edge and the bottom edge of the slit is `phi=(2 pi)/(lambda)(d sin theta)`, where d is the slit width. The first minima of the diffraction pattern occurs at `sin theta=(lambda)/(d)` So, `phi=(2pi)/(lambda)(d xx (lambda)/(d))=2pi`

Discussion

50.	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?
Answer» At` S_2, Deltax= 2lambda` Therefore, the minima closest to `S_2` will be corresponding to the path difference `Deltax = (3lambda)/2`. Suppose this point is P at a distance y from `S_2`.Then, `S_1P - S_2P = (3lambda)/2` `sqrt(y^2 + (S_1 S_2)^2) -y = (3lambda)/2` or ` sqrt(y^2+(2lambda)^2) = (y + (3lambda)/2)` Squaring and then solving this equation, we get `y= (7lambda)/12`.

Discussion

Explore topic-wise InterviewSolutions in .

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. `3xx10^(10)m`B. `3.35xx10^(11)m`C. `1.95xx10^(11)m`D. `4.32xx10^(10)m`

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. 5 mmB. 2.5 mmC. 1.25 mmD. 1.0 mm

A parallel beam of light of wavelength `6000Å` gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is :A. `6xx10^(-3)` radB. `1.8xx10^(-3)` radC. `3xx10^(-3)` radD. `2xx10^(-3)` rad

A narrow slit of width 2 mm is illuminated by monochromatic light fo wavelength 500nm. The distance between the first minima on either side on a screen at a distance of 1 m isA. 5 mmB. 0.5 mmC. 1 mmD. 10 mm

In double slit experiment , the distance between two slits is `0.6mm` and these are illuminated with light of wavelength `4800 Å`. The angular width of dark fringe on the screen at a distance 120 cm from slits will beA. `8xx10^(-4)` radB. `6xx10^(-4)` radC. `4xx10^(-4)` radD. `16xx10^(-4)` rad

A thick glass slab `(mu= 1.5)` is to be viewed in reflected white light. It is proposed to coat the slab with a thin layer of a material having refractive index 1.3 so that the wavelength 6000 Å os suppressed. Find the minimum thickness of the coating required.

Determine what happens to the double slits interference pattern if one of the slits is covered with a thin, transparent film whose thickness is `lambda/(2(mu-1))`, where `lambda` is the wavelength of the incident light and mu is the index of refraction of the film.

Interference fringes are produced on a screen by using two light sources of intensities / and 9/. The phase difference between the beams `pi/2` is at point P and `pi`at point Q on the screen. The difference between the resultant intensities at point P and Q isA. 2IB. 4IC. 6ID. 8I

Light waves from two coherent sources having intensities I and 2I cross each other at a point with a phase difference of `60^(@)` . The intensity at the point will beA. `4.414 I`B. `5.455 I`C. `4I`D. `6.441I`

nth bright fringe if red light `(lambda_(1)=7500" Å")` coincides with (n +1)th bright fringe of green light `(lambda_(2)=6000" Å")` The value of n, isA. 4B. 5C. 3D. 2

A fringe width of a certain interference pattern is `beta=0.002` cm What is the distance of 5th dark fringe centre?A. `1xx10^(-2)cm`B. `11xx10^(-2)cm`C. `13xx10^(-3)cm`D. `3.28xx10^(6)cm`

In an interference pattern the position of zeroth order maxima is 4.8 mm from a certain point P on the screen. The fringe width is 0.2 mm. The position of second maxima from point P isA. 5.1 mmB. 5 mmC. 40 mmD. 5.2 mm

In a YDSE experiment, d = 1mm, `lambda`= 6000Å and D= 1m. The minimum distance between two points on screen having 75% intensity of the maximum intensity will beA. 0.50 mmB. 0.40 mmC. 0.30 mmD. 0.20 mm

The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of 4th dark fringe after a glass sheet of refractive index 1.5 is introduced. The thickness of glass sheet would beA. `4.8 mu m`B. `4.2 mu m`C. `5.4 mu m `D. `3.0 mu m `

Two very narrow slits are spaced `1.80 mu`m apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light of `lambda = 550nm ?` (Hint : The angle `theta` is not small).

Two slits spaced 0.450nm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500nm?

With two slits spaced 0.2 mm apart and a screen at a distance of 1 m, the third bright fringe is found to be at 7.5 mm from the central fringe. The wavelength of light used isA. 400nmB. 500nmC. 550nmD. 600nm

Red light of wavelength 625 nm is incident normally on a optical diffraction grating with `2xx10^(5)` lines/m. Including central principal maxima, how many maxima may be observed on a screen which is far from the grating?A. 15B. 17C. 8D. 16

When a compact disc is illuminated by a source of white light, coloured lines are observed. This is due toA. dispersionB. diffractionC. interferenceD. refraction

A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.0 mm wide and the resulting diffraction pattern is observed on a screen 2m away. What is the distance between the first dark fringe on either side of the central bright fringe?

A single slit of width a is illuminated by violet light of wavelength `400nm` and the width of the diffraction pattern is measured as y. When half of the slit width is covered and illuminated by yellow light of wavelength `600nm`, the width of the diffraction pattern is

A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the secondary maximum in the pattern for red light of wavelength 6500 Å ?A. 4400 ÅB. 4100 ÅC. 4642.8 ÅD. 9100 Å

Three waves of equal frequency having amplitudes `10mum`, `4mum`, `7mum` arrive at a given point with successive phase difference of `pi//2`, the amplitude of the resulting wave in `mum` is given byA. 4B. 5C. 6D. 7

In the phenomenon of interference,A. sources must be coherentB. amplitudes must be sameC. wavelengths must be sameD. intensities may be different

A plane wave front of wavelength `lamda` is incident on a single slite of width b. What is the angular width for secondary maximum ?A. `(lambda)/(2b)`B. `(lambda)/(b)`C. `(2lambda)/(b)`D. `(b)/(lambda)`

If the ratio of amplitude of two waves is `4:3`, then the ratio of maximum and minimum intensity isA. `16:18`B. `18:16`C. `49:1`D. `1:49`

If the ratio of amplitude of wave is`2 : 1`, then the ratio of maximum and minimum intensity isA. `9:1`B. `1:9`C. `4:1`D. `1:4`

If white light is used in a biprism experiment, thenA. fringe pattern disappersB. all fringes will be colouredC. central fringe will be white others will be colouredD. central fringe will be dark

How will the diffraction pattern of single slit change when yellow light is replaced by blue light? The fringe will beA. widerB. narrowerC. brighterD. fainter

In an interference experiment, the spacing between successive maxima or minima is (Where the symbols have their usual meanings)A. `lambda d//D`B. `lambda D//lambda`C. `d D//lambda`D. `lambda d//4D`

Light propagates 2 cm distance in glass of refractive index 1.5 in time `t_(0)`. In the same time `t_(0)`, light propagates a distance of 2.25 cm in medium. The refractive index of the medium isA. `4//3`B. `3//2`C. `8//3`D. None of these

In the YDSE apparatus shown in figure, `dltltD` and `d = 6lambda`. Find (a) total number of maximas and minimas on the screen . (b) two y-coordinates corresponding to third order maxima.

A small aperture is illuminated with a parallel beam of `lambda = 628 nm`. The emergent beam has an anglur divergence of `2^(@)`. The size of the aperture isA. 180 mB. 18 `mu`mC. 1.8 mD. 0.18 m

Two coherent sources are 0.3 mm apart. They are 0.9m away from the screen. The second dark fringe is at a distance of 0.3cm from the centre. Find the distance of fourth bright fringe from the centre. Also, find the wavelength of light used.

The condition for diffraction of mth order minima isA. `d sin theta_(m)=m lambda, m=1,2,3, …`B. `d sin theta_(m)=(m lambda)/(2), m=1,2, 3, …`C. `d sin theta_(m)=(m+1)(lambda)/(2), m=1,2,3,…`D. `d sin theta_(m)=(m-1)(lambda)/(2), m=1, 2, 3,…`

Calculate the resolving power of a telescope when light of wavelength `540 nm` is used. Diameter of objective lens is `6 cm`.A. `9.1 xx10^(-4)"rad"^(-1)`B. `0.1xx10^(-4)"rad"^(-1)`C. `5xx10^(-4)"rad"^(-1)`D. `6xx10^(-4)"rad"^(-1)`

If the aperature of a telescope is decreased resolving power willA. increaseB. decreaseC. remain sameD. zero

The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit isA. `1:4:9`B. `1:2:3`C. `1:(4)/(9pi^(2)):(4)/(25pi^(2))`D. `1:(4)/(pi^(2)):(9)/(pi^(2))`

A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observe on screen 1 m away. It is observed that the first minimum is at a distance of `2.5 mm` from the centre of the screen. Find the width of the slit.A. 0.2 mmB. 0.3 mmC. 0.4 mmD. 0.5 mm