What is the spatial resolution of the eye for two red lights separated by 5cm?

[captainjack2000]

Homework Statement

Two red lights are separated by 5cm, perpendicular to your line of sight. How far away from them are you if they can just be resolved with the naked eye?

Homework Equations

angular resolution =sin(theta)=1.22lambda / D
spatial resolution = 1.22(focal length * lambda)/D

The Attempt at a Solution

the angular resolution of the eye is about 1 min of arc = 1/60th of a degree
the wavelength of red light is about 650nm
In the equation above D is diameter of the eye's pupil - approximately 5mm?
spatial resolution=5cm

I am not quite sure how to put all this information together?
using the information I have;
5cm=1.22(focal length *650nm)/5mm
which gives the focal length of 315 m

 

[mgb_phys]

You seem to have mixed up multiple ways of solving this.

If you are given 1' as the resolution then it's simply a triangle - 1' vertex and a base 5cm wide = find how long it it. (hint an angle this small sin(angle) = angle ,in raidans)

If you are assuming this is the diffraction limit of the eye then you need the diameter of the pupil and the wavelength to get the angle. Then the same as above.

Either way the focal length doesn't come into it

 

[captainjack2000]

We are only given the separation distance of the red lights but I think it is assumed that we know the wavelength of the light and the angular resolution of the eye.

Since the angle is so small and sin(theta)=theta can it also be assumed that tan(theta)=theta? If so the vector triangle would give theta=5cm/distance where theta is 1/60th of a degree=2.9E-4 degrees
distance =171.89m

Am I still doing this wrongly?
When do you use the equation resolution=1.22lambda/D?

thanks

 

[mgb_phys]

Yes, 1 arc-min is 290 micro-rad, so the distance is 0.05m / 290e-6 = 170m
With angles this small tan=sin=theta

Or if you want to use the diffraction limit, (remember 1.22 is for the half angle)
= 2.44 wavelength / d = 2.44 * 600e-9 / 0.005 = 0.00029 rad = 290 microrad = same answer

 

[captainjack2000]

thank you that makes much more sense now