contestada

a double-slit experiment has slit spacing 0.035 mm, slit-to-screen distance 1.2 m, and wavelength 490 nm. what's the phase difference between two waves arriving at a point 0.52 cm from the center line of the screen?

Respuesta :

We can calculate by using Young's double-slit experiment equation that the phase difference between two waves arriving at a point 0.52 cm from the center of the screen is 0.62π .

Young's double-slit experiment experiment was based on the hypothesis that if light were wave-like in nature, then it should behave in a manner similar to ripples or waves on a pond of water. Where two opposing water waves meet, they should react in a specific manner to either reinforce or destroy each other. So as we have from Young's double-slit experiment,

σ / λ = ΔΦ / 2π

σ = dsinθ

(dsinθ) / λ = ΔΦ / 2π

ΔΦ = (dsinθ)2π/λ

= (0.000035) x sinθ x 2π / 490 x 10^-9

sinθ = 0.0052/1.2

putting the values we get

ΔΦ = (9.5246x 10^-7) / (490  x 10^-9)

ΔΦ =  1.94 = 0.62π

Hence the phase difference is 0.62π .

To learn more about the Young's double-slit experiment , visit here

https://brainly.com/question/17167388

#SPJ4

Otras preguntas