Respuesta :
Answer:
d = 0.247 mm
Explanation:
given,
λ = 633 nm
distance from the hole to the screen = L = 4 m
width of the central maximum = 2.5 cm
2 y = 0.025 m
y = 0.0125 m
For circular aperture
[tex]sin \theta = 1.22\dfrac{\lambda}{d}[/tex]
using small angle approximation
[tex]\theta = \dfrac{y}{D}[/tex]
now,
[tex]\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}[/tex]
[tex]y = 1.22\dfrac{\lambda\ D}{d}[/tex]
[tex]d = 1.22\dfrac{\lambda\ D}{y}[/tex]
[tex]d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}[/tex]
d =0.247 x 10⁻³ m
d = 0.247 mm
the diameter of the hole is equal to 0.247 mm
