Car Headlight Resolution: Solving for Distance and Angle Using Physics

[wayneo]

the headlights of a car are separated by a distance of 1.4m. at what distance would these be resolved as 2 separate sources by a lens of diameter 5cm if a wavelength of 500nm is being used ?

using theta=1.22 * wavelenth / D I am stuck as I don't know what angle theta is or how to work it out.

 

[Hootenanny]

You need to find theta using the formula you stated above.

~H

 

[wayneo]

i have tried that but it doesn't work.

 

[Hootenanny]

Well, can you show me your working and then perhaps I can point out where you have gone wrong?

~H

 

[Gokul43201]

Did you draw a picture ? Theta is the angle subtended by the object (ie: the 2 headlights) at the center of the lens.

 

[wayneo]

using theta= 1.22 wavelength / b gives 1.22*(500*10^-9) / 0.05 = 1.22*10^-5
so 1.22*10^-6 = 1.22 wavelength / D

so D = 1.22*(500*10^-9) / (1.22*10^-5) = 0.05

the answer i have found in the book is 115km so I am stuck as how to get to that

 

[nrqed]

the headlights of a car are separated by a distance of 1.4m. at what distance would these be resolved as 2 separate sources by a lens of diameter 5cm if a wavelength of 500nm is being used ?

using theta=1.22 * wavelenth / D I am stuck as I don't know what angle theta is or how to work it out.

Usually, the angle is small enough that one uses \theta \approx tan(\theta) = separation between the sources /distance to the sources

 

[Hootenanny]

D is the thinkness of the lense. Once you have obtained theta, you need to use trig to find the distance. (As Gokul hinted at)

~H

 

[nrqed]

using theta= 1.22 wavelength / b gives 1.22*(500*10^-9) / 0.05 = 1.22*10^-5
so 1.22*10^-6 = 1.22 wavelength / D

so D = 1.22*(500*10^-9) / (1.22*10^-5) = 0.05

the answer i have found in the book is 115km so I am stuck as how to get to that

Yu are going in circle! Of course you will find D=0.05, this is what you used to start with! (I don't know why your exponent changed from -5 to -6 though)

You need another equation relating to the distance to the sources (which is NOT D) and I gave it in my other post.

 

[Gokul43201]

using theta= 1.22 wavelength / b gives 1.22*(500*10^-9) / 0.05 = 1.22*10^-5
so 1.22*10^-6 = 1.22 wavelength / D

so D = 1.22*(500*10^-9) / (1.22*10^-5) = 0.05

the answer i have found in the book is 115km so I am stuck as how to get to that

When you are given an equation, the first thing you must understand id what the various terms in the equation represent. You are using the same equation with two different meanings attached to the same term (D), and hence getting the same value for two different quantities.

If you merely attempt to juggle numbers in an equation so as to end up with the number that is the answer, you will gain no understanding of the physics involved.

In the equation you've written down, what do the different symbols represent ?

 

[nrqed]

D is the thinkness of the lense. Once you have obtained theta, you need to use trig to find the distance. (As Gokul hinted at)

~H

Just to avoid possible confusion for the OP: D is the diamter of the circular slit used (where slit may stand for the pupil of a person or a telecope lens or the slit of a pinhole camera)

 

[Hootenanny]

Just to avoid possible confusion for the OP: D is the diamter of the circular slit used (where slit may stand for the pupil of a person or a telecope lens or the slit of a pinhole camera)

Sorry, thinkess was the wrong word. I apologise.

~H

 

[wayneo]

When you are given an equation, the first thing you must understand id what the various terms in the equation represent. You are using the same equation with two different meanings attached to the same term (D), and hence getting the same value for two different quantities.

If you merely attempt to juggle numbers in an equation so as to end up with the number that is the answer, you will gain no understanding of the physics involved.

In the equation you've written down, what do the different symbols represent ?

D is the distance from the headlights to the lens and b is the diameter of the lens

 

[Hootenanny]

So we're all singing off the same hym sheet;

\theta = \frac{1.22\lambda}{D}

Where D is the diameter of the lense.

Your other equation D = 1.22*(500*10^-9) / (1.22*10^-5) is wrong.

Try drawing a diagram like Gokul suggests. HINT: Use triangles.

~H

 

[Gokul43201]

D is the distance from the headlights to the lens and b is the diameter of the lens

Do you not see the problem with that ? Okay, let's start over :

1. Write down the equation from the OP.

2. An equation is nothing but a statement relating various quantites, expressed in a concise mathematical fashion, using symbols to represent these quantities. In the equation above, what quantities do the different symbols represent ?

3. Now using these quantities, rewrite the statement of the equation in the form of a sentence.

You need to do this with every equation you come across, else you have not really understood what it is saying.