a beam of light consisting of two wavelengths ,800nm and 600nm is used to obtain the interference fringes in a young's double slit experiment on a screen placed 1.4m away.if the two slits seperated by 0.28nm,calculate the least distance from the central bright maximum where the bright fringes of the two wavelenghts coincide
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D = 1.4 m
d = 0.28 mm = 0.28 × 10-3 m
λ1 = 800 × 10-9 m
λ2 = 600 × 10-9 m
let y be the common distance of the bright fringes by both wavelength.
y = n1λ1D/d = n2λ2D/d
=> n1λ1 = n2λ2
=> 800 × 10-9 × n1 = 600 × 10-9 × n2
=> 4 n1 = 3 n2
n1 ≠ 0, n2 ≠ 0
For y to be minimum since n are integers
n1 = 3 and n2 = 4
y = n1λ1D/d
=> y = 3×800 × 10-9 ×1.4/0.28 × 10-3
=> y = 1.2×10-2 m
=> y = 1.2 cm
HOPE THIS HELP..
D = 1.4 m
d = 0.28 mm = 0.28 × 10-3 m
λ1 = 800 × 10-9 m
λ2 = 600 × 10-9 m
let y be the common distance of the bright fringes by both wavelength.
y = n1λ1D/d = n2λ2D/d
=> n1λ1 = n2λ2
=> 800 × 10-9 × n1 = 600 × 10-9 × n2
=> 4 n1 = 3 n2
n1 ≠ 0, n2 ≠ 0
For y to be minimum since n are integers
n1 = 3 and n2 = 4
y = n1λ1D/d
=> y = 3×800 × 10-9 ×1.4/0.28 × 10-3
=> y = 1.2×10-2 m
=> y = 1.2 cm
HOPE THIS HELP..
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