In daylight, the diameter of the iris is about 3 mm.
PART A
What is the angle between two distant point objects that can just be resolved? Assume the light's wavelength in air is 600 nm and the index of refraction inside the eye is 1.33.
PART B
What is the distance on the retina, 24 mm from the front of the eye, between the images of these two objects?
a) Angular resolution (in radians) = 1.22 (lambda)/(N D)
N is the index of refraction inside the eye. Lambda is the wavelength.
b) Multiply tha angle (in radians) from (a) by the distance to the retina, 24 mm, which is the eye lens's focal length.
PART A
To determine the angle between two point objects that can just be resolved, we can use the Rayleigh criterion. According to this criterion, two point objects can be resolved if the central maximum of one object's diffraction pattern falls on the first minimum of the other object's diffraction pattern.
The formula for the minimum resolvable angle (θ) is given by:
θ = 1.22 * (λ / D)
where λ is the wavelength of light and D is the diameter of the iris.
Given that the wavelength of light (λ) is 600 nm and the diameter of the iris (D) is 3 mm, we can substitute these values into the equation to calculate the angle (θ).
θ = 1.22 * (600 nm / 3 mm)
= 1.22 * (600 * 10^(-9) m / 3 * 10^(-3) m)
= 1.22 * (2 * 10^(-4))
= 0.0244 radians
Therefore, the angle between two distant point objects that can just be resolved is approximately 0.0244 radians.
PART B
To determine the distance on the retina between the images of the two objects, we can use the formula:
Distance on retina = (distance from the front of the eye) * (angular size of the object)
The angular size can be calculated using the formula:
Angular size = (object size / object distance)
Given that the objects are distant point objects, their sizes can be considered negligible.
Given that the distance from the front of the eye is 24 mm and the calculated angle between the objects is 0.0244 radians, we can substitute these values into the formula to find the distance on the retina.
Distance on retina = 24 mm * 0.0244 radians
= 0.5856 mm
Therefore, the distance on the retina, 24 mm from the front of the eye, between the images of these two objects is approximately 0.5856 mm.