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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?
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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_1 d = 1 (lambda D)/a` and
`2 y_1 d = (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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