The limit to the eye’s acuity is actually related to diffraction by the pupil. W

Question

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 3.00–mm–diameter pupil, assuming an average wavelength of 560 nm?

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 3.00–mm–diameter pupil, assuming an average wavelength of 560 nm? What is the distance between two just–resolvable points held at an arm’s length (0.700 m) from your eye? (Discuss how does your answer to compare to details you normally observe in everyday circumstances?)

Take your result to be the practical limit for the eye. What is the greatest possible distance a car can be from you if you can resolve its two headlights, given they are 1.20 m apart?

Explanation / Answer

Given,

D = 3 mm ; lambda = 560 nm

1)We know that

sin(theta) = 1.22 lambda/D

sin(theta) = 1.22 x 560 x 10^-9/3 x 10^-3 = 2.2773 x 10^-4

theta = sin^-1 (2.2773 x 10^-4) = 0.013 deg

Hence, theta = 0.013 deg

2)y = 1.2 m

We know that, tan(theta) = y/L

L = y/tan(theta)

L = 1.2/tan(0.013) = 5288.84 m

Hence, L = 5288.84 m

3)L = 0.7 m

Again using, tan(theta) = y/L

y = L tan(theta)

y = 0.7 x tan(0.013) = 1.59 x 10^-4 m

Hence, y = 1.59 x 10^-4 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.