Answer:
D = 3,246 10⁻³ mm
Explanation:
We can treat this exercise as a diffraction problem with a circular opening. The expression that describes diffraction is
d sin θ= m λ
In our case d is the width of the pupil 3.93 mm and we will use m = 1
sin θ = λ / d
Since these angles are very small, you can approximate the sin θ=θ
θ = lam / d
This expression is exact for a linear slit, for the case of circular slits must be multiplied by a constant
θ = 1.22 λ /d
Now let's analyze the wavelength, as it is within a medium with refractive index its value changes
λ ₙ = λ ₀/ n
Let's replace
θ = 1.22 λ / (n d)
Let's calculate
θ = 1.22 550 10⁻⁹ / (1,336 3.93 10⁻³)
θ = 1,278 10⁻⁴ rad
This is the smallest angle that can be in the retina, let's use trigonometry to find the distance (y), we must use half the angle for the right triangle
Let's not forget that all angles are in radians
tan θ/2 = y / L
y = L tan θ / 2
y = 25.4 tan (1.278 10⁻⁴/2) = 25.4 0.639 10⁻⁴
y = 1.623 10⁻³ mm
The spot size twice this value
D = 2y
D = 3,246 10⁻³ mm
The smallest diameter spot the eye can produce on the retina if the pupil diameter is 3.93 mm is mathematically given as
x = 3246 10^{-3} mm
Question Parameter(s):
Assume light with a wavelength of λ = 550 nm
The distance from the pupil to the retina is 25.4 mm.
Generally, the equation for the is mathematically given as
∅ = lam / d
Therefore
∅ = 1.22 λ / (n d)
∅ = 1.22 550 10^{-4} / (1,336 3.93 10^{-3})
∅= 1,278 10^{-4} rad
In conclusion,
y = L tan θ / 2
y = 25.4 tan (1.278 10{-4}/2)
y= 25.4 0.639 10^{-4}
y = 1.623 10^{-3} mm
Since, spot size twice the value
x = 2y
x = 3246 10^{-3} mm
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