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Rayleigh criterion and circular apertures-- check my work?
Think of the pupil of your eye as a circular aperture 5.00 mm in diameter. Assume you are viewing light of wavelength 550 nm, to which your eyes are maximally sensitive.
a) What is the minimum angular separation at which you can distinguish two stars?
b) What is the maximum distance at which you can distinguish the two headlights of a car mounted 1.50 m apart?
\ThetaR = sin-1(1.22\lambda/d) and sin\Theta = 1.22\lambda/d, where d is the diameter of the aperture and theta is the angle from the central axis through the opening to the first diffraction minimum. \ThetaR is the minimum resolvable angular separation in radians.
small angle approximations
I'm pretty sure my answer to (a) is right. Due to the small angle approximation, \ThetaR = 1.22*5.5e-7/5e-3 = 1.34e-4 radians * (180/pi) = 7.69e-3 degrees.
I'm less sure about (b). Call the distance between the headlights s and the orthogonal distance from one headlight to your eye l. For small angles, sin\Theta is approximately \Theta is approximately tan\Theta = s/l. Using the answer obtained in (a), s/l = 1.34e-4 radians. l = 1.5/1.34e-4 = 1.12e4 m = 11.2 km. I don't know what mistake I would've made, but that seems awfully far.
Let me know what you think! Thanks :)
Homework Statement
Think of the pupil of your eye as a circular aperture 5.00 mm in diameter. Assume you are viewing light of wavelength 550 nm, to which your eyes are maximally sensitive.
a) What is the minimum angular separation at which you can distinguish two stars?
b) What is the maximum distance at which you can distinguish the two headlights of a car mounted 1.50 m apart?
Homework Equations
\ThetaR = sin-1(1.22\lambda/d) and sin\Theta = 1.22\lambda/d, where d is the diameter of the aperture and theta is the angle from the central axis through the opening to the first diffraction minimum. \ThetaR is the minimum resolvable angular separation in radians.
small angle approximations
The Attempt at a Solution
I'm pretty sure my answer to (a) is right. Due to the small angle approximation, \ThetaR = 1.22*5.5e-7/5e-3 = 1.34e-4 radians * (180/pi) = 7.69e-3 degrees.
I'm less sure about (b). Call the distance between the headlights s and the orthogonal distance from one headlight to your eye l. For small angles, sin\Theta is approximately \Theta is approximately tan\Theta = s/l. Using the answer obtained in (a), s/l = 1.34e-4 radians. l = 1.5/1.34e-4 = 1.12e4 m = 11.2 km. I don't know what mistake I would've made, but that seems awfully far.
Let me know what you think! Thanks :)