Question Video: Relation of Maximum Distance for Resolvability of Spatial Features to Feature Separation

The headlights of a car are 1.3 m apart. What is the maximum distance at which an eye with a pupil diameter of 0.40 cm can resolve these two headlights? Assume that the wavelength of light from the headlight is 555 nm.


Video Transcript

The headlights of a car are 1.3 meters apart. What is the maximum distance at which an eye with a pupil diameter of 0.40 centimeters can resolve these two headlights? Assume that the wavelength of light from the headlight is 555 nanometers.

Letโ€™s begin by highlighting some of the important information weโ€™ve been given. Weโ€™re told the headlights of the car are 1.3 meters apart. Weโ€™ll call that distance ๐‘‘. Weโ€™re told that the pupil of the eye that is observing the headlights has a diameter of 0.40 centimeters. Weโ€™ll call that value ๐‘‘ sub ๐‘. Weโ€™re told we can assume that the wavelength of light coming from the headlights is 555 nanometers. Weโ€™ll call that value ๐œ†.

We want to know the maximum distance from the headlights the eye can resolve the two as two distinct targets. Weโ€™ll call that maximum distance capital ๐ท. Letโ€™s begin by drawing a diagram of our scenario.

The two car headlights shine light visible to the observing eye, which is some distance capital ๐ท away. If youโ€™ve ever looked at a carโ€™s headlights very far away in real life, youโ€™ll know that there is a certain distance beyond which the two headlights seem to merge in your vision. And they appear to just be one bright source. We want to solve for that distance from the eye, capital ๐ท, just before the light from the headlights seems to merge and come from one source.

This whole question of being able to resolve distinct light sources has to do with something known as the Rayleigh criterium. This criterium says that if we have two light sources that are being observed, then if we consider the angle subtended by our vision that includes the two sources as well as the wavelength of the light coming from them, then the Rayleigh criterium says that that angle ๐œƒ in order for the sources to be resolvable as two separate sources rather than one combined signal must equal at least 1.22 times the wavelength ๐œ† divided by ๐‘‘ sub ๐‘œ, where ๐‘‘ sub ๐‘œ is the diameter of the observing optic.

When we apply this relationship to our scenario, we see that ๐œƒ, the angle subtended by the lines between our eye and the two headlamps, is equal to 1.22 times ๐œ† divided by ๐‘‘ sub ๐‘, the diameter of our eyeโ€™s pupil. Since weโ€™re given ๐œ† and ๐‘‘ sub ๐‘, we can plug those in now. When we do that, weโ€™re careful to make sure that ๐œ† and ๐‘‘ sub ๐‘ are in units of meters.

Now before we go ahead and calculate ๐œƒ, letโ€™s draw another sketch of our situation to see how ๐œƒ can be useful to us in arriving at capital ๐ท. If we take the triangle created by these dotted lines and divide that angle from our eye in half, then focusing on the top of the two triangles weโ€™ve created, we can see that, for one thing, weโ€™ve created a right triangle. And for another, we now have an angle from our eye of ๐œƒ over two.

We know the length of the opposite leg of this triangle. Itโ€™s 1.3 meters divided by two. And the adjacent leg of the triangle has a length of capital ๐ท, the distance between the eye and the headlamps. Bringing this all together through a trigonometric relationship, we can see that the tangent of the angle ๐œƒ divided by two is equal to 1.3 meters divided by two times ๐‘‘. Or rearranging for ๐‘‘, ๐‘‘ equals 1.3 meters divided by two times the tangent of ๐œƒ over two.

When we plug in for ๐œƒ in the expression we arrived at earlier, when we enter these values on our calculator, we find that ๐‘‘ is equal to two significant figures to 7.7 kilometers. Thatโ€™s the maximum distance we can be from the headlights and still resolve them as two separate sources of light.

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