Question

A. Calculate the diffraction limit of the human eye, assuming a wide-open pupil so that your eye acts like a lens with diameter 0.8 centimeter, for visible light of 500-nanometer wavelength.
Express your answer using two significant figures.

B. How does this compare to the diffraction limit of a 10-meter telescope?
Express your answer using two significant figures.

C. Now remember that humans have two eyes that are approximately 7 centimeters apart. Estimate the diffraction limit for human vision, assuming that your "optical interferometer" is just as good as one eyeball as large as the separation of two regular eyeballs.
Express your answer using two significant figures.

Answer 1

A. The Dawes limit tells us that the resolving power is equal to 11.6 / d, where d is the diameter of the eye’s pupil in units of centimeters. The eye's pupil can dialate to approximately 7 mm, or 0.7 cm. So 11.6 / .7 = 16.5 arc seconds, or about a quarter arc minute ~ 17 arc seconds

Although, the standard answer for what people can really see is about 1 arc minute. 

B. It is considered as linear, so given a 10 meter telescope (10,000 mm):

10000 / 7 = 1428 times better for the 10 meter scope ~ 1400 times better (in 2 significant figures)

C. For a 7 cm interferometer, that is just similar to a 7 cm scope. Therefore we would expect 

11.6 / 7 = 1.65 arc seconds ~ 1.7 arc seconds

This value is what we typically can get from a 7 cm scope. 

Answer 2

Answer:

a) 16 arc seconds

b) 1250

c)1.785 arc seconds

Explanation:

Given data:

lens diameter = 0.8 cm

wavelength 500 nm

a) the diffraction of the eye is given as

$= 2.5\times 10^5 \frac{\lmbda}{D}$ arc seconds

$= 2.5\times 10^5 \frac{5\times 10^{-7}}{8\times 10^{-3}}$ arc seconds

= 16 arc seconds

b) we know that

$\frac{DIffraction\ limit\ of\ eye}{diffraction\ limit\ of\telescope}$

$= \frac{2.5\times 10^5(\frac{\lambda}{D_{eye}})}\frac{2.5\times 10^5(\frac{\lambda}{D_{telescope}})}$

$\frac{\theta_{eye}}{\theta_{telescope}} = \frac{10}{8\times 10^{-3}} = 1250$

c) $\theta_{eye} = 2.5\times 10^{5} \frac{5\times 10^{-7}}{7\times 10^{-2}}$$\theta_{eye} = 1.78\ arc\ second$