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Question
A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen that is 40 cm away. What is the distance between the two first-order minima?
Solution 1
Data: λ = 550 nm = 550 × 10-9 m,
a = 0.4 mm = 4 × 10-4 m
D = 40 cm = 40 × 10-2 m
`"y"_"md" = "m" (lambda"D")/"a"`
∴ `"y"_"1d" = 1 (lambda"D")/"a"` and
`2"y"_"1d" = (2 lambda"D")/"a"`
`= (2 xx 550 xx 10^-9 xx 40 xx 10^-2)/(4 xx 10^-4)`m
= 2 × 550 × 10-6
= 1100 × 10-6
= 1.100 × 10-3 m
= 1.10 mm
Solution 2
Given:
λ = 550 nm = 550 × 10-9 m, a = 0.4 mm = 0.4 × 10-3 m
D = 40 cm = 40 × 10-2 m
To find: Distance between two first order minima
Formula: Width of central maxima, Wc = `(2λ"D")/"a"`
Calculation:
From formula,
Wc = `(2 xx 550 xx 10^-9 xx 40 xx 10^-2)/(0.4 xx 10^-3)`
= 2 × 550 × 10-6
= 1100 × 10-6
= 1.100 × 10-3 m
= 1.10 mm
Distance between two first-order minima = Width of central maxima = 1.10 mm
Distance between two first-order minima is 1.10 mm.
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