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ID: 1556619 • Letter: S
Question
Show Metadata Provide my evaluation of this Provide feedback messages or contribute to the course Prepare a printable document discussion about this resource Make notes and annotations about Set a bookmark for this resource this resource PHY131s17, Homework 9: Limits of resolution of the eye The limit to the eye's acuity is actually related to diffraction by the pupil. What is the angle between two just resolvable points of light for a 2.00-mm-diameter pupil, assuming an average wavelength of 560 nm? Submit Answer Tries 0/20 Take your result to be the practical limit for the eye. What is the greatest possible distance a car can be from you ifyou can resolve its two headlights, given they are 1.20 m apart? Submit Answer Tries 0/20 What is the distance between two just-resolvable points held at an arm's length (0.900 m) from your eye? (Discuss how does your answer to compare to details you normally observe in everyday circumstances?) Submit Answer Tries 0/20 Preferences on what is marked as NEW Threaded View Chronological View SortingLEiltering options Export? Mark NEW posts no longer new For some reason, I couldn't actually figure out how to get the first answer in degrees and instead found the answer in radians. To find the answer in radians, ignore the sin part of theExplanation / Answer
(a)
From the Raley criterian formula
theta = 1.22 lmada /D
= 1.22(560 * 10^-9)/2.0 * 10^-3 m
= 0.0003416 rad
(b)
The distance between two objects seperated by a angle theta
s = r theta
r = s / theta
= 1.20 m / 0.0003416 rad
= 3.51 * 103 m
(c)
s= r theta
=0.9 * 0.0003416 rad
=0.00030744 m
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(a)
the condition for consturive interference
2nt = ( m+ 1/2) lamda
for m = 0
t= lamda/ 4n
=651/4(1.36)
=119.66 nm
lamda = 2nt /m
= 2( 1.51) (119.66 nm)/1
=361.4 nm