A beam of light of wavelength 420 nm, is used to obtain interference fringes in
a Young’s double-slit experiment. Find the distance of the third dark fringe on the
screen from the central maxima. Take the separation between the slits as 4.2 mm
and the distance between the screen and plane of the slits as 1.4 m.
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Answer:
Distance of the third dark fringe = 350 × 10⁻⁶ m.
Explanation:
Given:
- Wavelength of light = 420 nm = 420 × 10⁻⁹m
- Separation between slits = 4.2 mm = 4.2 × 10⁻³m
- Distance between screen and slit = 1.4 m
To Find:
- Distance of the third dark fringe on the screen
Solution:
The position of the nth dark fringe from the central maxima is given by,
where D = distance of the screen from the slit
d = separation between the slit
λ = wavelength of the light
n = 1, 2, 3..
Substituting the given data we get,
Hence the distance of the third dark fringe is 350 × 10⁻⁶ m.
TheFairyTale:
Well explained! :D
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