A parallel beam of light of wavelength 600nm is incident normally on a slit of width 'd'. If the distance between slits and screen is 0.8m and the distance of second order maximum from the centre of the screen is 15mm, calculate the width of the slit.
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Hello dear,
● Answer-
6.4×10^-5 m
● Explaination-
# Given-
D = 0.8 m
x = 15 mm = 1.5×10^-2 m
n = 2
λ = 600 nm = 6×10^-7 m
# Solution-
For constructive interference,
x / D = nλ / d
d = nλD / x
d = 2 × 6×10^-7 × 0.8 / 1.5×10^-2
d = 6.4×10^-5 m
Therefore, slit width is 6.4×10^-5 m.
Hope this is helpful...
● Answer-
6.4×10^-5 m
● Explaination-
# Given-
D = 0.8 m
x = 15 mm = 1.5×10^-2 m
n = 2
λ = 600 nm = 6×10^-7 m
# Solution-
For constructive interference,
x / D = nλ / d
d = nλD / x
d = 2 × 6×10^-7 × 0.8 / 1.5×10^-2
d = 6.4×10^-5 m
Therefore, slit width is 6.4×10^-5 m.
Hope this is helpful...
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