Answer:
a) size of the bright spot is proportional to the hollow size
b) as the size of the hole increases, the circular point decreases.
Explanation:
a) In this case the diameter of the hole is much greater than the wavelength, as the size of the hole is many orders greater than the wavelength we are in the part of geometric optics,
Consequently the size of the bright spot is proportional to the hollow size.
Consequently the size increases
b) in this case the hole diameter d = 100 10⁻⁶m and the wavelength that for the green color is lam = 500 nm = 5 10⁻⁷ m
We see that angles are very small so the wavelength of the office is greater than the wavelength, but you can observe the effects of diffraction
d sin θ = 1.22 m λ
the numerical constant appears by solving the equation in polar coorθdinates, because the hole is circular
the first zero occurs for m = 1
sin θ = 1.22 λ / d
In these experiments the angles are small
sin θ = θ
we substitute
θ = 1.22 λ/ d
θ = 1.22 500 10⁻⁹ / 100 10⁻⁶
θ = 6.1 10⁻³
without the hole diameter increases by 20%
d’ = 1.2 d
we substitute
θ'= 1.22 λ / d'
θ’ = 1.22 λ /1.2 d
θ‘= 1.22 λ /d [tex]\frac{1}{1.22}[/tex]
θ ’= θ 0.83
θ ’= 6.1 10⁻³ 0.83
θ' = 5 10⁻³ rad
Therefore, the answer is that as the size of the hole increases, the circular point decreases.
