Answer:
The path-length difference is [tex]dsin\theta=1.08*10^{-3}mm[/tex]
The angle is [tex]\theta = 0.5157^o[/tex]
Explanation:
From the question we are told that
The distance of separation is d = 0.12 mm = [tex]0.12*10^{-3} m[/tex]
The distance from the screen is D = 0.63 m
The wavelength is [tex]\lambda = 540nm = 540 *10^{-9}m[/tex]
From the question we can deduce that the the two maxima's are at the m=0 and m=2
Now the path difference for this second maxima is mathematically represented as
[tex]d sin \theta = m \lambda[/tex]
Where d[tex]dsin\theta[/tex] is the path length difference
Substituting values
[tex]dsin \theta = 2 * 540*10^{-9}[/tex]
[tex]dsin\theta = 1.08*10^{-6}m[/tex]
converting to mm
[tex]dsin\theta = 1.08*10^{-6} * 1000 mm[/tex]
[tex]dsin\theta=1.08*10^{-3}mm[/tex]
To obtain the angle we make [tex]\theta[/tex] the subject
[tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]
Substituting values
[tex]\theta = sin ^{-1} [\frac{1.08*10^{-6}}{0.12*10^-3} ][/tex]
[tex]\theta = 0.5157^o[/tex]
The Pathlength difference between the waves at second maximum on the screen is; 1.08 × 10^(-6) m
We are given;
Distance between two slits; d = 0.12 mm = 0.12 × 10^(-3) m
Distance of slit from screen; D = 0.6 m
We want to find the path length at second maxima m = 2
λ = 540 nm = 540 × 10^(-9) m
Formula for Pathlength is;
dsin θ = mλ
Where mλ is the Pathlength difference.
Since at m = 0, the pathlength is zero,
Thus;
Pathlength difference = (2 × 540 × 10^(-9)) - 0
Pathlength difference = 1.08 × 10^(-6) m
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