Question

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.

Which laser has its first maximum closer to the central maximum?
What is the distance Image for Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas las between the first maxima (on the same side of the central maximum) of the two patterns?
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas las= ______ m
What is the distance Deltay_max-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Deltay_max-min = ______ m

Answer 1

A) Therefore laser1 has the maximum closest to the central maximum

B) Δₓ = 0.8

Step-by-step 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