Question

Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.3 m from the slits. The center of the 3rd order bright band on the screen is separated from the central maximum by 0.90 cm. a) What is the frequency of the light, the slit separation, and the angle of the 3rd order bright band

Answer 1

Answer:

a) f = 4.76 10¹⁴ Hz, b) d = 2.73 10⁻⁴ m, c) θ = 6.923 10⁻³ rad

Explanation:

a) In this problem the frequency of light is asked, let's use the relationship between the speed of the wave, its wavelength and its frequency

           c = λ f

           f = c /λ

           f = $\frac{3 \ 10^8}{630 \ 10^{-9}}$

           f = 4.76 10¹⁴ Hz

b) slit separation (d)

the expression for the constructive interference of the double-slit experiment is

          d sin  θ = m λ

let's use trigonometry

          tan θ = y / L

          tan θ = $\frac{sin \theta}{cos \theta}$

in general the angles are small, so we can approximate

          tan θ = sin θ

          tan θ = y/L

we substitute

          d y / L = m λ

          d = m L λ / y

we calculate

          d = 3  1.3  630 10⁻⁹ /0.90 10⁻²

          d = 2.73 10⁻⁴ m

c) the angle

           tan θ = y / L

           θ = tan⁻¹ y / L

           θ = tan⁻¹ 0.9 10⁻² / 1.3

           θ = tan⁻¹ 6,923 10⁻³

let's find the angle in radians

           θ = 6.923 10⁻³ rad