How Far Away Are Car Taillights When They Merge Due to Diffraction?

  • Thread starter Thread starter Dart82
  • Start date Start date
  • Tags Tags
    Diffraction Power
Click For Summary
SUMMARY

The discussion focuses on calculating the distance at which car taillights appear to merge due to diffraction effects. Using the Rayleigh criterion for resolution, the participant applied the formula θ_min = 1.22(λ/D) with λ as the wavelength of red light (660 nm) and D as the diameter of the eye's pupil (7 mm). The calculation yielded a distance of approximately 15,385 meters, which the participant questioned as incorrect. The confusion arose from the application of the Pythagorean theorem in the context of the problem.

PREREQUISITES
  • Understanding of diffraction and the Rayleigh criterion
  • Basic knowledge of optics, specifically light wavelength and refractive index
  • Familiarity with algebraic manipulation and solving equations
  • Concepts of geometry, particularly the Pythagorean theorem
NEXT STEPS
  • Study the Rayleigh criterion in detail, including its applications in optics
  • Learn about the effects of diffraction on light and how it impacts visibility
  • Explore the relationship between wavelength, refractive index, and light behavior
  • Practice solving problems involving geometric optics and distance calculations
USEFUL FOR

Students studying physics, particularly those focusing on optics and diffraction, as well as educators looking for examples of real-world applications of optical principles.

Dart82
Messages
56
Reaction score
0

Homework Statement

Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the pupils of your eyes (average refractive index = 1.36) have diameters of about 7.0 mm. The taillights of this car are separated by a distance of 1.3 m and emit red light (wavelength = 660 nm in vacuum). How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?
[URL=http://img58.imageshack.us/my.php?image=resolvingpoweruz3.jpg][PLAIN]http://img58.imageshack.us/img58/1871/resolvingpoweruz3.th.jpg[/URL][/PLAIN]

Homework Equations


s=r(theta)
Rayleigh criterion for resolution: theta min = 1.22 (lambda/D)

The Attempt at a Solution



lambda eye = lambda vacuum/ n eye
(660x10^-9m) / 1.32 = 5x10^-7m

1.22(lambda eye/D)=s/r
1.22(5x10^-7m)/0.007m = 1.3m/r
r = 15385 meters
What am i doing wrong here??
 
Last edited:
Physics news on Phys.org
the answer should be?
 
the only thing i can think of is:
a^2 + b^2 = c^2

(15385m^2) - (.65m^2 ) = answer ...i got 15384.99 which i know isn't right
 

Similar threads

Diffraction related to Resolving Power
  • · Replies 1 ·
Replies
1
Views
3K
Resolving Power and minimum seperation
  • · Replies 5 ·
Replies
5
Views
4K
The Rayleigh Criterion and the Wavelength of Light
  • · Replies 2 ·
Replies
2
Views
4K
Solving Power Problems: Finding d & l
  • · Replies 2 ·
Replies
2
Views
4K