Minimun lens size (in meters) that will permit just resolving 2 stars

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Homework Help Overview

The discussion revolves around determining the minimum lens size required to resolve two stars that are 1.1 arcseconds apart, using light of a specific wavelength (510 nm). The problem involves concepts related to diffraction and resolution in optics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between slit width and wavelength in the context of resolving power. There are inquiries about calculating the angle theta and its conversion from arcseconds to radians. Some participants express uncertainty about the problem setup and seek clarification on the definitions involved.

Discussion Status

Several participants are actively engaging with the problem, offering hints and clarifications regarding the calculations needed. There is a recognition of the importance of converting units and understanding the implications of the given parameters. However, no consensus has been reached on a complete method or solution.

Contextual Notes

Participants note the specific angle given in arcseconds and the potential confusion surrounding the terminology used in the problem. There is an emphasis on the small angle approximation and its relevance to the calculations.

cheez
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What is the minimun lens size (in meters) that will permit just resolving 2 stars 1.1 seconds apart using light of 510 nm?
For a slit of width a, light of wavelength L, the first minimun occurs at an angle t from the center, assume a slit shaped lens! E=E max L* sin(pi a sin(t)/L)/(pi a sin(t) )
Hint: you don't really need all of this. Resolution implies that the maximun of one image falls on the minimun of the other.
I have really no clue on how to this:cry: please help!
 
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cheez said:
I have really no clue on how to this:cry: please help!
The first minimum occurs when [itex]a sin (\theta) = \lambda[/itex].



In these problems, often the angles are so small that you may replace sin and tan by the angle itself at the condition of working in radians.

Then [itex]a \theta \approx \lambda[/itex].

They give you lambda and theta. Just convert theta in radians and solve for "a" which will be the slit width "a"
 
how to calculate theta? thanks you so much!
 
cheez said:
how to calculate theta? thanks you so much!
If one is given the distance between the sources of light (perpendicular to the line of sight) and the distance between the sources and the lens, there is a formula in terms of a tangent. But here you are in luck, they give you the value of theta in the problem!
They are 1.1 arcsecond apart (if they said second, they really meant arcsecond). Do you know what an arc second is? It is one degree divided by 3600 (this comes from the fact that one degree is 60 arcminutes and one arcminute is 60 arcsenonds). So your angle is 1.1/3600 degrees. Now convert that in radians, plug that in the equation and you are done.

Patrick
 
thx for your answer!
 

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