Calculate Beam's Divergence Angle from Earth to Moon

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    Angle Divergence
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Homework Help Overview

The problem involves calculating the divergence angle of a laser beam directed from Earth to the Moon, given the beam's diameter at the Moon and the distance from Earth to the Moon.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the beam's diameter and the divergence angle, with one participant questioning the use of diameter as a variable in the angle calculation.

Discussion Status

Some participants have provided insights into the geometric interpretation of the problem, suggesting that the diameter can be considered in relation to the distance to the Moon. There is an ongoing exploration of the definitions and assumptions related to the variables involved.

Contextual Notes

Participants are navigating the definitions of variables and the geometric setup of the problem, with some expressing confusion about the terminology used in the calculations.

mars shaw
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Homework Statement


A tiny laser beam is directed from Earth to moon. If beam's diameter is 2.50 m at the moon, how small much the divergence angle be for the beam? The distance of moon from the Earth is 3.8x10^8 m




Homework Equations



The Attempt at a Solution

 
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Hi Mars shaw! :wink:

(try using the X2 tag just above the Reply box :wink:)

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
mars shaw said:

Homework Statement


A tiny laser beam is directed from Earth to moon. If beam's diameter is 2.50 m at the moon, how small much the divergence angle be for the beam? The distance of moon from the Earth is 3.8x10^8 m




Homework Equations


theta=s/r

The Attempt at a Solution


s=diameter=2.50 m
r=distance of moon from earth=3.8x10^8
theta=s/r
theta=2.50/3.8x10^8 = 6.6x10^-9 rad
I got the correct answer but I am confuse about diameter that how it can be taken as "s"?
Kindly explain.
 
Hi Mars shaw ! :smile:

Because you imagine the line from your eye to the bottom of the moon as being length r, and the diameter (s) of the moon as being at right-angles to that line, so the tangent of the angle is s/r.

(and if you're not happy about that, but you'd rather use the radius, and the line to the centre of the moon, you'll get the same result: the tangent of half the angle is (s/2)/r :wink:)
 

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