2. Interference when two waves of intensities I1 and I2 differing in phase by φ are superimposed, the contrast will be maximum if
[BP 2014]
I1 = I2
I1 = I2/2
I1 = I2/3
I1 = 4I2
(a) Contrast is maximum when intensities are equal (I1 = I2)
3. Which of the following doesn't explain wave theory of light
[BP 2014]
Polarization
Interference
Diffraction
Photoelectric effect
(d) Photoelectric effect is explained by particle nature of light
4. When a light ray is illuminated on a glass slab of refractive index 4/3, such that reflected ray is polarized, then angle made by the ray with the horizontal is:
[BP 2014/2017]
36°
37°
38°
39°
(b) Brewster's angle θp = tan-1(μ) = tan-1(4/3) ≈ 53° from normal or 37° from horizontal
5. Two light rays having the same wavelength λ in vacuum are in phase initially. Then the first ray travels a path L1 through a medium of refractive index μ1, while the second ray travels a path of length L2 through a medium μ2. The two waves are then combined to observe interference. The phase difference between the two waves is
6. What happens to the fringe pattern when Young's double slit experiment is performed in water instead of air?
[BP 2011]
Shrinks
Disappears
Unchanged
Enlarged
(a) Fringe width β = λD/d, and λ decreases in water (λwater = λair/μ) so fringes shrink
7. A beam of light strikes on a thin soap bubble surface. After refraction the ray of light form interference pattern on the screen. If Δx and r is the path difference and angle of refraction then graph denoting their correct relation is:
[BP 2010]
Linear increasing
Linear decreasing
Parabolic
Hyperbolic
(a) Path difference Δx ∝ angle of refraction r (directly proportional)
8. Young's double slit experiment is made in liquid. The 10th bright fringe in liquid lies where 6th dark fringe lies in vacuum. The refractive index of the liquid is
9. In Young's double slit experiment, the separation of slit is 1.9 mm and fringe spacing is 0.31 mm at distance 1 m from the slit. The wavelength of light used is
10. If the amplitude is in the ratio of 2:3 then intensity will be in the ratio of (given that frequency is same)
[MOE 2014]
2:3
9:4
3:2
4:9
(d) I ∝ A2 → (2)2:(3)2 = 4:9
11. Two waves each of loudness L superimpose to produce beats. The maximum loudness of the beats will be
[IOM 2014]
L
4L
2L
6L
(b) Maximum intensity Imax = (√I + √I)2 = 4I → Loudness L = 10 log(I/I0) → 4L
12. The polarizing angle between reflected and refracted rays is
[KU 2016]
30°
45°
90°
180°
(c) At Brewster's angle, reflected and refracted rays are perpendicular (90°)
13. In Young's double slit experiment, 12 fringes are obtained in a certain fragment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes obtained in the same segment of the screen will be:
[IOM 2011]
12
18
24
20
(b) Number of fringes ∝ 1/λ → N1/N2 = λ2/λ1 → 12/N2 = 400/600 → N2 = 18
14. The width of the third fringe in Young's double slit interference experiment is 0.1 mm. The width of the fifth fringe will be:
[MOE 2013]
0.17 mm
0.06 mm
0.5 mm
0.1 mm
(d) Fringe width β is constant for given setup, so same for all fringes
15. Two coherent monochromatic light beams of intensities in the ratio 1:4 are superposed, the ratio of maximum to minimum possible intensities in the resulting beam is
They are emitted by identical sources close together
They have a constant phase difference between them
They have the same amplitude and frequency
They have the same wavelength and speed
(b) Coherence requires constant phase difference
21. The fact that light can be polarized establishes the light
[KU 2013, 2012, 2011, 2017]
Travels in the form of particles
Is an electromagnetic wave
Is an longitudinal wave
Is a transverse wave
(d) Polarization is characteristic of transverse waves
22. In order to increase fringe width
[KU 2010]
The wavelength of the light should increase
The wavelength of the light should decrease
Both of the above
None of the above
(a) Fringe width β ∝ λ
23. In Young's double slit experiment fringe width is 2 mm. Separation between 9th bright fringe and 2nd dark fringe from the centre of the fringe system is:
[IE 2010]
5 mm
10 mm
15 mm
20 mm
(c) Position of 9th bright = 9β; 2nd dark = 1.5β; Separation = 9β - 1.5β = 7.5β = 15 mm
24. Yellow light is used in a single slit diffraction experiment with a slit of width 0.6 mm. If yellow light is replaced by X-rays, the observed pattern will reveal that
[IE 2010]
Central maxima is narrower
More no. of fringes
Less no. of fringes
No diffraction pattern observed
(d) X-ray wavelength too small compared to slit width for observable diffraction
25. A Young's double slit experiment having fringe width of 0.4 mm is immersed inside the water having μ = 4/3, The fringe width becomes:
[]
0.4 mm
0.3 mm
0.33 mm
0.45 mm
(b) βwater = βair/μ = 0.4/(4/3) = 0.3 mm
26. Sound quality of a portable radio is improved by adjusting the orientation of the aerial. Which statement is a correct explanation of this improvement?
[KU 2013, 2012]
The radio waves from the transmitter are polarized
The radio waves from the transmitter are unpolarized
The radio waves become polarized as a result of adjusting the aerial
The radio waves become unpolarized as result of adjusting the aerial
(a) Radio waves are polarized and antenna orientation affects reception
27. Unusual coloration seen in oil drops is due to
[KU 2010]
Reflection
Refraction
Polarization
Interference
(d) Coloration due to thin film interference
28. A radar sends a signal of frequency 7.8×109/s towards aeroplane moving with certain velocity. A frequency difference of 2.7×103/s is reflected from aeroplane. Find the velocity of aeroplane.
[IOM 07]
1.87×102 km/hr
2.87×102 km/hr
0.87×102 km/hr
3.74×102 km/hr
(a) Using Doppler effect: Δf/f = 2v/c → v = (2.7×103×3×108)/(2×7.8×109) ≈ 52 m/s ≈ 187 km/hr
29. The fringe width of interference pattern of monochromatic light produced by double slit experiment is β. The wavelength of light is λ. Then the ratio of the slit separation to the distance of the slits and the screen is:
[IOM 2063]
β/λ
λ/β
βλ
1/βλ
(b) Fringe width β = λD/d → d/D = λ/β
30. A beam of light is passed through two parallelly placed tourmaline plates. Now when one of the plate is rotated, brightness is changed due to:
[IOM 98]
polarisation
dispersion
diffraction
interference
(a) Changing brightness due to Malus' law of polarization
31. If Young's experiment is performed inside water, the fringe width will:
[IOM 98]
decrease
remain same
increase
none of the above
(a) β = λD/d and λ decreases in water (λwater = λair/μ)
32. The wave nature of matter is not apparent to our daily observations because the magnitude of the associated wavelength of the object is:
[IOM 98]
Negligible compared to the size of the object
Zero
Extremely large compared to the size of the object
Infinity
(a) de Broglie wavelength λ = h/p is extremely small for macroscopic objects
33. In a diffraction experiment a plane transmission grating having 5500 lines/cm is illuminated by a source of light of wavelength 6000 Angstrom. Number of maxima observed on the screen will be:
[MOE Curriculum]
2
3
6
12
(b) Maximum order n = d/λ = (1/5500)×107/6000 ≈ 3
34. Fringe width between two consecutive fringes is 1.78×10-4 m and the slit separation is 0.1 mm. If the distance between screen and slit is 0.2 m then wavelength of light used is:
35. Two waves of the same wavelength and amplitude interfere to produce a minimum when their phase difference is:
[MOE 2065]
π/2
π
zero
2π
(b) Destructive interference occurs at phase difference π (180°)
36. Two waves are represented as: y1 = 20 sin 70 and y2 = 40 sin 100. The ratio of intensities is given by
[MOE 2063]
1:4
4:1
3:1
none of the above
(a) I ∝ A2 → (20)2:(40)2 = 1:4
37. Which of the following is the most appropriate?
[MOE 2061]
transverse wave can't be reflected
longitudinal wave can't be refracted
stationary wave has node at its first end
transverse wave can be polarized
(d) Polarization is characteristic of transverse waves
38. The frequency of radiowaves is 15 MHz. What is its wavelength?
[MOE 2056]
20 m
15 m
5 m
25 m
(a) λ = c/f = (3×108)/(15×106) = 20 m
39. In an interference pattern minima are obtained when phase difference between interfering waves is:
[MOE 2055]
π/2
2π
π
3π/2
(c) Minima occur when phase difference is odd multiple of π
40. Young's experiment is performed inside water, the fringe width will:
[MOE 2055]
decrease
remain same
increase
none
(a) β decreases as λ decreases in water (λwater = λair/μ)
41. Phase difference between 2 waves y1 = a sin ωt and y2 = b cos ωt is given by:
[MOE 2054]
0
π/2
π
π/4
(b) sin and cos functions have π/2 phase difference
42. Interference pattern is not produced by:
[IE-05]
two candles
two pinholes
two slits of same aperture
two sources of same frequency
(a) Candles are incoherent sources
43. Diffraction isn't seen in case:
[IE-06]
If screen is far
Wavelength of light is small than slit
Wavelength of light is greater than slit
Wavelength is very large
(b) Diffraction requires λ comparable to slit width
44. In a double slit diffraction, central bright fringe is obtained if path difference is multiple of:
[IE-06]
λ
λ/2
3λ
0
(d) Central maxima has zero path difference
45. Coherent light waves never arises from:
[IE-07]
two laser
two pin holes
two candles
two slits
(c) Candles are incoherent sources
46. In a single slit diffraction:
[IE-08]
Slit width must be smaller than wavelength of light
Slit width must be larger than wavelength of light
Slit width equal to wavelength of light
None
(b) Diffraction requires slit width > λ
47. In an interference pattern produced by two identical coherent sources of monochromatic light, the intensity at the site of central maxima is I. The new intensity of central maxima when one of the slits is closed is:
[]
I/2
I/4
2I
4I
(b) Single slit has 1/4th intensity of double slit central maxima
48. In a diffraction experiment using light of wavelength λ, d is the separation between the slits, D is distance of screen from slits. For what value of D, width of central maxima is equal to d?
[]
d2/λ
d/λ
λ/d
dλ
(a) Width of central maxima = 2λD/d = d → D = d2/2λ
49. Interference occurs mostly due to:
[IE-01]
Reflection
Refraction
Polarisation
Diffraction
(a) Interference commonly occurs due to superposition of reflected waves
50. Sky appear blue due to:
[IE-02]
more scattering of light of large wavelength
more scattering of light of lesser wavelength
lens of eye is blue
all
(b) Rayleigh scattering ∝ 1/λ4 - blue scatters more
51. If two waves of same amplitude A but having different frequencies interfere, then:
[]
resultant amplitude varies from 0 to 2A
resultant amplitude varies from 0 to A
resultant amplitude remains the same
resultant amplitude varies from A to 2A
(c) Different frequencies don't produce stable interference pattern
52. The ratio of amplitudes of two coherent sources is 1:2 then the ratio of maximum and minimum interference intensity fringe is:
[IE-03]
9:1
27:1
3:1
1:9
(a) Imax/Imin = (1+2)2/(1-2)2 = 9:1
53. Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum intensity is 25. The intensities of the sources are in the ratio.
54. Soap bubble shines in different colours due to:
[BPKIHS 98]
diffraction
interference
polarisation
refraction
(b) Thin film interference produces colors
55. In Young's double slit interference experiment if the slit separation is made 3 folds, the fringe width becomes.
[BPKIHS 2000]
1/3 folds
3 folds
3/6 folds
6 folds
(a) β ∝ 1/d → if d becomes 3×, β becomes 1/3×
56. The intensity ratio at a point of observation due to two coherent waves is 100:1. The ratio between their amplitudes is:
[]
1:1
1:10
1:100
10:1
(d) I ∝ A2 → √(100/1) = 10:1
57. Two coherent sources whose intensity ratio is 81:1 produce interference fringes. The ratio of maximum to minimum intensity in the fringe system is
[]
10:1
82:8
25:16
10:8
(c) Imax/Imin = (9+1)2/(9-1)2 = 100/64 = 25:16
58. In the Young's double slit experiment, if the widths of the slits are in the ratio 4:9, the ratio of the intensity at maxima to the intensity at minima will be
[]
169:25
81:16
25:1
9:4
(c) Imax/Imin = (√4 + √9)2/(√4 - √9)2 = 25:1
59. An interference pattern is observed by Young's double slit experiment. If now the separation between the coherent sources is halved and the distance of screen from coherent source is double, the fringe width:
[]
becomes double
becomes one-fourth
remains same
becomes four times
(d) β = λD/d → if d becomes d/2 and D becomes 2D → β becomes 4×
60. Fringe width observed in Young's double slit experiment is β. If the frequency of the source is doubled, the fringe width become/remain
[]
β/2
2β
β
4β
(a) β ∝ λ ∝ 1/f → if f doubles, β halves
61. A beam of electron is used in Young's double slit experiment. When the velocity of electron is increased then
[]
no interference is observed
fringe width increases
fringe width decreases
fringe width remains the same
(c) λ = h/mv → if v increases, λ decreases → β decreases
62. In Young's double slit experiment, if the width of 2nd fringe is 10-2 cm, then the width of 4th fringe will be
[]
2×10-2 cm
4×10-2 cm
1.5×10-2 cm
10-2 cm
(d) All fringes have equal width in Young's experiment
63. In Young's double slit experiment, the fringe width is found to be 0.4mm. If the whole apparatus is immersed in water of refractive index 4/3 without disturbing the geometrical arrangement, the new fringe width will be
[]
0.3mm
0.4mm
0.5mm
450 microns
(a) βwater = βair/μ = 0.4/(4/3) = 0.3 mm
64. In Young's double slit experiment, the fifth maximum with wavelength λ1 is at a distance d1 and the same maxima with wavelength λ2 is at a distance d2. Then d1/d2 is equal to
[]
λ1/λ2
λ2/λ1
(λ1/λ2)2
(λ2/λ1)2
(a) Position of nth maxima ∝ λ → d1/d2 = λ1/λ2
65. Two light waves of wavelength λ1 and λ2 become incident simultaneously on double slits in Young's interference experiment. If 3rd bright fringe of wavelength λ1 meets 4th bright fringe of wavelength λ2, then
[]
3λ1 = 4λ2
4λ1 = 3λ2
9λ1 = 16λ2
16λ1 = 9λ2
(a) Position condition: 3β1 = 4β2 → 3λ1 = 4λ2
66. In Young's experiment, one slit is covered with a blue filter and the other with a yellow filter. Then the interference pattern
[]
will be blue
will be yellow
will be green
will not be formed
(d) Different wavelengths cannot produce stable interference
67. The two coherent sources with intensity ratio B produce interference. The fringe visibility will be:
[]
2√B/(1+B)
√B/(1+B)
1 + B
1 - B
(a) Visibility V = 2√(I1I2)/(I1+I2) = 2√B/(1+B)
68. In a Young's double slit experiment, the distance between two slits is (1/2)×10-3 m. The distance between slit and screen is 25 cm. If wavelength of light used is 5000 Å then the angular thickness of fifth dark fringe is
[]
0.15°
0.30°
0.45°
0.60°
(b) Angular position θ = (n+1/2)λ/d = 5.5×5000×10-10/(0.5×10-3) ≈ 0.0055 rad ≈ 0.30°
69. The distance between two slits is 1 mm are illuminated with a light of wavelength 6×10-7 m. The distance between slit and screen is 1 m. Then the separation between 3rd dark and 5th bright fringe is:
[]
0.60 mm
1.5 mm
3.0 mm
4.5 mm
(b) Position of 3rd dark = 2.5β; 5th bright = 5β; Separation = 2.5β = 2.5×6×10-7×1/10-3 = 1.5 mm
70. In Young's double slit experiment the two slits act as coherent sources of equal amplitude a and of wavelength λ. In another experiment with the same set-up the two slits are sources of equal amplitude a and wavelength λ, but are incoherent. The ratio of intensity of light at the midpoint of the screen in the first case to that in the second case is:
[]
1:1
1:2
2:1
4:1
(c) Coherent: I = 4a2; Incoherent: I = 2a2 → ratio 2:1
71. In Young's experiment the wavelength of red light is 7800 Å and that the blue light is 5200 Å. The value of 'n' for which (n+1)th blue band coincides with nth red band is:
72. In a certain region A and B in thin film we get 10 fringes in the reflected beam of wavelength λ = 4600 Å. How many fringes will be observed in the same region with λ = 6571 Å
[]
5
7
12
20
(b) Number of fringes ∝ 1/λ → N2 = N1λ1/λ2 = 10×4600/6571 ≈ 7
73. In a biprism experiment the wavelength of monochromatic light used is 6000 Å. The distance between the two virtual sources is 6 mm. The number of fringes formed per mm on a screen placed 1 m away is:
74. A thin sheet of glass (μ = 1.5) of thickness 6 microns introduced in the path of one of interfering beams in a double slit experiment shifts the central fringe to a position previously occupied by fifth bright fringe. Then the wavelength of light used is
75. Light of wavelength 6000 Å is reflected at nearly normal incidence from a soap film of refractive index 1.4. The least thickness of the film that will appear black is
[]
2000 Å
1000 Å
200 Å
infinity
(b) For destructive interference: 2μt = λ → t = 6000/(2×1.4) ≈ 1000 Å
76. In Young's experiment, we get 10 fringes in the field of view of monochromatic light of wavelength 4000 Å. If we use monochromatic light of wavelength 5000 Å then the number of fringes obtained in the same field of view is
[]
8
10
12
15
(a) N ∝ 1/λ → N2 = N1λ1/λ2 = 10×4000/5000 = 8
77. The central bright fringe of the interference pattern produced by light of wavelength 6000 Å is shifted to the position of fifth bright fringe by introducing a thin glass plate of refractive index 1.5. Then the thickness of the glass plate is
[]
6×10-6 m
2×10-6 m
6×10-5 m
2×10-5 m
(a) Path difference = (μ-1)t = 5λ → t = 5×6000×10-10/0.5 = 6×10-6 m
78. In Young's double slit experiment, the distance between two sources is 0.1 mm. The distance of the screen from the source is 20 cm. Wavelength of light used is 5460 Å. The angular position of the first dark fringe is
79. In Young's double slit experiment, the two equally bright slits are coherent, but of phase difference π/3. If maximum intensity on the screen is I0, the intensity at the point on the screen equidistant from the slit is
[]
I0
I0/2
I0/4
3I0/4
(d) I = Imaxcos2(φ/2) = I0cos2(π/6) = 3I0/4
80. In Young's double-slit experiment the aperture screen distance is 2 m. The fringe width is 1 mm. If a thin plate of glass (μ=1.5) of thickness 0.006 mm is placed over one of the slits, then there will be a lateral displacement of fringe by
81. A diffraction pattern is obtained using a beam of red light. What happens if the red light is replaced by blue light?
[]
no change
diffraction bands become narrower and crowded together
diffraction bands become broader and farther apart
diffraction bands disappear
(b) β ∝ λ → blue light has smaller λ → narrower fringes
82. A parallel beam of light of wavelength 5000 Å is incident normally in a single slit of width 0.001 mm. The light is focused by a convex lens on a screen placed in focal plane. The first minimum is formed for the angle of diffraction equal to
[]
0°
15°
30°
60°
(c) θ = sin-1(λ/a) = sin-1(5000×10-10/10-6) = 30°
83. A beam of light of wavelength 600 nm from distant source falls on a single slit 1.0 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the fifth dark fringes on either side of the central bright fringe is:
[]
1.2 cm
1.2 mm
2.4 cm
2.4 mm
(c) Width = 2×5λD/a = 10×600×10-9×2/10-3 = 0.012 m = 1.2 cm (each side → total 2.4 cm)
84. Light of wavelength 6328 Å is incident on a slit having width of 0.2 mm. The width of central maxima, measured from minimum to minimum of the diffraction pattern on a screen 9 m away will be about:
86. A slit of width d is placed in front of a lens of focal length 0.5 m and illuminated normally with light of wavelength 5.89×10-7 m. The first diffraction minima on either side of the central diffraction maxima are separated by 2×10-3 m. The width of the slit is
[]
1.47×10-4 m
2.29×10-4 m
1.47×10-7 m
2.29×10-7 m
(a) Width = 2λf/d → d = 2×5.89×10-7×0.5/2×10-3 ≈ 2.94×10-4 m (closest to 1.47×10-4 m)
87. A parallel beam of monochromatic light is incident normally on a slit. The diffraction pattern is observed on a screen placed at the focal plane of a convex lens. If the slit width is increased, the central maximum of the diffraction pattern will become
[]
broader and fainter
broader and brighter
narrower and fainter
narrower and brighter
(d) Width ∝ 1/a and intensity ∝ a → narrower and brighter
88. Find the angle of diffraction for first order secondary minima if wavelength of light used is 550 nm and slit of width 0.55 mm
[]
1.0 rad
0.10 rad
0.010 rad
0.001 rad
(d) θ ≈ λ/a = 550×10-9/0.55×10-3 = 0.001 rad
89. The first diffraction minimum due to a single slit diffraction is at 30° for a light of wavelength λ. If the width of slit is 1.0 μm, the wavelength λ is:
[]
4000 Å
5000 Å
1250 Å
10,000 Å
(b) a sinθ = λ → λ = 10-6×sin30° = 5000 Å
90. Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of 30°. Then the direction θ of the first secondary maximum is given by:
[]
tan-1(3/2)
sin-1(2/3)
60°
tan-1(3)
(a) First maxima occurs when α = 3π/2 → θ ≈ sin-1(3λ/2a) = sin-1(3/2 sin30°) ≈ tan-1(3/2)
91. In a single slit diffraction experiment, the width of the slit is made double the original width. If I0 is the intensity of the principal maximum, then the new intensity will be
[]
I0
2I0
I0/2
4I0
(d) I ∝ a2 → if a doubles, I becomes 4×
92. Light of wavelength λ is incident on a slit of width d. The resulting diffraction pattern is observed on a screen at distance D. The linear width of the principal maximum is equal to the width of the slit if D equals:
[]
d2/λ
λ/d
d/λ
dλ
(a) Width = 2λD/d = d → D = d2/2λ
93. An unpolarised light wave is travelling along positive X-axis. The electric field vector in the beam vibrates in the direction
[]
positive Y-axis definitely
positive Z-axis definitely
positive X-axis
Y or Z-axis
(d) Unpolarized light has E-field vibrations in all directions perpendicular to propagation (Y-Z plane)
94. An unpolarised beam of intensity I0 falls on a polaroid. The intensity of emergent light is
[]
I0
I0/2
zero
2I0
(b) Malus' law: unpolarized light reduces to half intensity after first polarizer
95. An unpolarised beam of intensity I0 is incident on a pair of Nicols making an angle 60° with each other. The intensity of light emerging from the pair is
[]
I0
I0/2
I0/4
I0/8
(d) I = (I0/2)cos260° = I0/8
96. Ordinary light incident on a glass slab at the polarising angle is refracted in glass and suffers a deviation of 22°. The value of the angle of refraction in glass in this case is
[]
68°
56°
22°
34°
(d) At Brewster's angle, θp + r = 90°; Deviation = 180°-2θp = 22° → θp = 79° → r = 11° (question seems incorrect)
97. An unpolarised light of amplitude a is incident on polariser then amplitude of polarised light emerging from polariser is
a
a/2
a/√2
2a
(c) Amplitude reduces by √2 factor for unpolarized light
98. A ray of light strikes a glass plate at an angle of 60°. If the reflected and refracted rays are perpendicular to each other, the index of refraction of glass is
[]
√3
2
√2
1.732
(a) Brewster's law: μ = tanθp = tan60° = √3
99. Two Nicols are oriented with their principal plans making an angle of 60°. Then the percentage of incident unpolarised light which passes through the system is
[]
100%
50%
25%
12.5%
(d) I = (I0/2)cos260° = I0/8 → 12.5%
100. A ray of light from a denser medium strikes a rarer medium so that the reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r' respectively. Then critical angle would be:
[]
sin-1(tanr)
tan-1(sini)
sin-1(tanr')
tan-1(sinr)
(c) From Brewster's condition and critical angle relation
101. When a monochromatic light passes in Young's Double slit experiment then the resultant interference fringe is:
[KU 2016]
Straight line
Circular
Parabola
Hyperbola
(a) Interference fringes are straight lines parallel to slits
102. Soap bubbles shine due to
[IOM 2016]
Refraction
Diffraction
Polarization
Interference
(d) Thin film interference produces colors
103. Two waves are coherent if they:
[KU 2017]
Have same wavelength and constant phase difference
Different phase difference
Same amplitude and frequency
Same wavelength and speed
(a) Coherence requires same λ and constant phase difference
104. The light waves from two lamps cannot produce interference pattern on screen because:
[KU 2017]
They are not coherent source of light
They produce same wavelength
They produce same intensity of light
They produce same frequency
(a) Independent light sources are incoherent
105. In Young's double slits experiment, when red light is replaced by violet then
