YDSE Problems I
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Q. In YDSE d=2mm D=2m wavelength=500nm.Intensity of 2 slits are I and 9I then find intensity at y=1/6mm.
Q.
In a Young’s double-slit experiment, the path difference, at a certain point on the screen between two interfering waves is of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to
Q. The relative permittivity of distilled water is 81. The velocity of light in it will be :
(Given μr=1)
(Given μr=1)
- 4.33×107 m/s
- 3.33×107 m/s
- 5.33×107 m/s
- 2.33×107 m/s
Q. In an interference experiment, the ratio of amplitudes of coherent waves is a1a2=13. The ratio of maximum and minimum intensities of fringes will be
- 4
- 2
- 9
- 18
Q. 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?
Q. White light is incident normally on a glass plate (in air) of thickness 500 nm and refractive index of 1.5. The wavelength (in nm) in the visible region (400 nm-700 nm) that is strongly reflected by the plate is
- 450
- 600
- 400
- 500
Q. In the Young's double slit experiment, the distance between the slits varies in time as d(t)=d0+a0 sinωt; where d0, ω and a0 are constants. The difference between the largest fringe width and the smallest fringe width, obtained over time, is given as :
- 2λD(d0)(d20−a20)
- λDd20a0
- λDd0+a0
- 2λDa0(d20−a20)
Q.
In an interference arrangement similar to Youngs double slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150 m. The intensity I(θ)is measured as a function of q, where q is defined as shown. If I0 is maximum intensity, then I(θ) for 0≤θ≤90° is given by