Answer:
A) Therefore laser1 has the maximum closest to the central maximum
B) Δₓ = 0.8
Explanation:
A) The expression for the constructive interference of a double slit is
d sin θ = m λ
let's use trigonometry to find the angle
tan θ = y / L
in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we subjugate
d y / L = m λ
y = m λ L / d
let's apply this equation for each case
a) Lares 1 has a wavelength λ₁ = d / 20, the screen is at L = 6.00 m
they ask us for the first axiom m = 1,
let's calculate
y₁ = 1 (d / 20) 6.00 / d
y₁ = 0.3
Laser 2, λ₂ = d / 15
λ₂ = 1 (d / 15) 6.00 / d
λ₂ = 0.4
Therefore laser1 has the maximum closest to the central maximum
b) let's find the distance of each requested value
second maximum m = 2 of laser 1
yi '= 2 (d / 20) 6 / d
y1 '= 0.6
3rd minimum of laser 2
the expression for destructive interference is
d sinθ = (m + 1/2) lam
y = (m ) λ L / d
in this case m = 3
let's calculate
y2 '= (3+0.5) (d / 15) 6 / d
y2 '=21/15
They ask us for the dalt of these interference
Δₓ = y3 -y2'
Δₓ = 21/15 - 0.6
Δₓ = 0.8