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प्रश्न
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
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उत्तर
Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 × 10−2 m
In case of constructive interference, we have the relation for the distance between the two fringes as:
u = `nlambda D/d`
Where,
n = Order of fringes = 4
λ = Wavelength of light used
∴ `lambda = (ud)/(nD)`
= `(1.2 xx 10^(-2) xx 0.28 xx 10^(-3))/(4xx1.4)`
= 6 × 10−7
= 600 nm
Hence, the wavelength of the light is 600 nm.
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संबंधित प्रश्न
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