Answer:
A) geometric optics, B) geometric optics , c) geometric optics ,
e) geometric optics, f) geometric optics
Explanation:
For this exercise we must use the condition for interference and diffraction so that these phenomena are relevant the wavelength must be comparable to the gap spacing
λ> = a
Lam when the spacing is much greater than the wavelength, the description of geometric optics is more and more exact
let's analyze each situation
a) let's find the wavelength
v = λ f
λ= v / f
λ= 343/1000
λ = 0.343 m
0.343 << 1m
therefore the description of the geometric optics of
b) red light passes through the pupil of the eye
red light has a wavelength of 700 num or more, the lojo pupil has a maximum of 8 me
λ = 700 10⁻⁹ m = 7 10⁻⁷ m
a = 8 mm 10⁻³
longitudinal is much less therefore the geometric optics is correct
c) luz azul lam = 450 nm = 450 10⁻⁹ m = 4.5 10⁻⁷ m
again the wavelength is much less than the diameter of the pupil, for which the description with the optics is generally sufficient
d) a radio A transmits up to a maximum of f = 1400 Khz = 1,400 10⁶ Hz
let's find the wavelength
c = λf
λ = c / f
λ= 3 108 / 1,400 106
λ= 2.14 102 m
in this case the wavelength is greater than the width of the gate, so the description of diffraction should be used to explain the phenomenon
e) X-rays have wavelength lam = 10-10 m
the separation of the elbow bones is of the order of a few millimeters, for local the wavelength is much less than the separation, therefore with the relations of geometric optics it is sufficient
