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I am having difficulty with this question. I don't know what to do. Thanks for any help

Suppose you are standing on a straight highway and watching a car moving away from you at 20.0 m/s. The air is perfectly clear, and after 11 minutes you see only one taillight. If the diameter of your pupil is 7.00 mm and the index of refraction of your eye is 1.33, what is the distance between the taillights.

Taillights are often red so you can assume that λ=700 nm.

(Hint: Assume the eye is acting like a single slit with a circular aperture. You will also need to take into consideration the fact that the wavelength of light changes inside your eye since the index of refraction is different. Please show all steps.)

When light is projected through a very small hole, the light emerging from the other end of the hole undergoes both diffraction and interference. The resulting pattern very much resembles a “bull’s eye” and the resulting pattern follows the same equation as that produced by a single slit, where the nodes are predicted by the equation

n*lambda = d*sin(theta).

This means that as light passes through any small aperture, the resulting light waves will interfere with one another generating a pattern of alternating nodes and antinodes.

At first glance this might seem to be an interesting but inconsequential result of the wave nature of light, but if fact leads us to a very important principle of nature!

Suppose for a moment, that instead of having a single source pass light through our small aperture, instead we pass the light from two different, but closely spaced light sources through the aperture. What would happen?

The light from each source would, of course, generate its own diffraction and interference pattern and if the two sources are very closely spaced, these two interference pattern would overlap! If the first node of the first pattern corresponds to the central antinode of the second pattern it would be impossible to tell that these two patterns were different and, in fact, it would be impossible to tell that there were two discrete sources! As a direct consequence of this, there is a real, practical limit to how closely spaced two discrete objects can be and yet still be detectable as separate objects. This limit is called the diffraction limit.

The equation describing the diffraction limit comes directly from the equation describing the interference pattern of a single slit or aperture.

n*wavelength = d*sin(theta)

Two modifications of this equation are made to determine the diffraction limit. The variable n is replaced by n = 1.22 where this number is the value for n where the first node overlaps the central antinode of the adjacent diffraction pattern so as to make it impossible to resolve the images of the two light sources. In addition, the sin(theta) can be replaced by just theta since the angular separation is so small that the small angle approximation almost always applies here.

Therefore, the final equation for the diffraction limit is

1.22*wavelength = d*sin(theta) = d*theta = d*(x/L)

1.22*wavelength = d*x/L

At first the diffraction limit might always seem to be a disadvantage, but sometimes we can use this wave characteristic to our advantage.

Pictures printed in a magazine are made up of very tiny dots which cannot be seen by the casual observer due to the diffraction limit. In that way we can mix small dots on a piece of paper with the knowledge that due to the wave nature of light, the individual dots will blend together as their nodes and antinodes overlap.

In this case the wavelength in the equation will need to be wavelength of the light within the eye. This can be found by dividing the wavelength of the light in air by the index of refraction of the eye:

wavelength = 700nm/1.33 = 526nm = 5.26x10^(-7)m

d = aperture of the iris = 7.0mm = 7.0x10^(-3)m

D = the distance to the car = 20m/s*11min*60sec/min = 13,200m

x = the answer to the question - the distance between the headlights

Putting this all together:

1.22*5.26x10^(-7)=7.0x10^(-3)*x/13,200

Solving for x: x=1.21m - the distance between the two headlights!

Physics

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.