Light of wavelength 550 nm falls normally on a slit of width 22.0 10 5 cm. The angular position of the second minima from the central maximum will be (in radians) :
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Hey mate,
◆ Answer- π/6 rad
◆ Explaination-
# Given-
λ = 550 nm = 5.5×10^-7 m
d = 22×10^-5 cm = 22×10^-7 m
# Solution-
Let θ be the angular position of second minima from central maxima-
sinθ = 2λ / d
sinθ = (2 × 5.5×10^-7) / (2.2×10^-7)
sinθ = 1/2
Taking sin inverse-
θ = 30°
θ = π/6 rad
Therefore, angular position of second minima from central maxima is π/6 rad.
Hope this is helpful...
◆ Answer- π/6 rad
◆ Explaination-
# Given-
λ = 550 nm = 5.5×10^-7 m
d = 22×10^-5 cm = 22×10^-7 m
# Solution-
Let θ be the angular position of second minima from central maxima-
sinθ = 2λ / d
sinθ = (2 × 5.5×10^-7) / (2.2×10^-7)
sinθ = 1/2
Taking sin inverse-
θ = 30°
θ = π/6 rad
Therefore, angular position of second minima from central maxima is π/6 rad.
Hope this is helpful...
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