Antenna Wavelength - double-slit interference

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Homework Help Overview

The problem involves determining the optimal placement of an antenna for receiving radio signals, considering the interference effects caused by boulders acting as slits in a double-slit interference setup. The radio station's wavelength is given as 2 meters, with specific distances to the boulders and their separation noted.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, questioning the number of slits represented by the boulders and the implications for interference patterns. There is an exploration of the conditions for constructive interference and the calculations involved in determining the optimal antenna height.

Discussion Status

The discussion is ongoing, with participants providing guidance and clarifications regarding the setup and calculations. Some participants express confusion about the number of boulders and their role in the interference pattern, while others attempt to derive the necessary equations for finding the next maximum in the interference pattern.

Contextual Notes

There is a mention of a drawing that accompanies the problem, which may clarify the configuration of the boulders and their relationship to the antenna placement. Participants are also considering the implications of the distances involved in relation to the wavelength and slit separation.

Koro21
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Homework Statement


The question is " Two large boulders lie between the house and Farmer Joe's favorite radio station. Farmer Joe cannot put the antenna on his house because the howl of the wind through it keeps Mrs Joe up all night. Where is the next best place Farmer Joe can put his antenna to get great reception?
Radio station wavelength = 2m
distance to boulder = 100m (the horizontal component)
y= antenna placement along the vertical
the slits around the boulders are 3m apart

Homework Equations



λ= c/f
sin theta = m( λ/d) ?


The Attempt at a Solution


this problem has really stumped me for a while
in essence this seems to be double-slit interference problem so would the first order fringe produce the strongest single?
 
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How can you have 2 slits with just 2 boulders? Did anyone provide a drawing?
 
Yeah the boulders confuse me too because this is the drawing taht came with the problem
 

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Koro21 said:
Yeah the boulders confuse me too because this is the drawing taht came with the problem

Aha. THREE boulders!
OK so this is a 2-slit setup.

You need to determine the next maximum located away from the house, marked by the asterisk on your drawing. What should be the difference in path lengths between each "slit" to the asterisk to get the next maximum? Then it's just trig to get the distance between the house and the asterisk.
 
thanks for the guiding hand
So for constructive interference m = an integer
λ=2m
L= 100m (distance to slits)
d= 3m (distance between slits)
m=1 ( the next constructive maxima)
so the final equation i get is y= (mλL)/d = 66.66 meters which sounds really big, did i mess up somewhere?
 
Koro21 said:
thanks for the guiding hand
So for constructive interference m = an integer
λ=2m
L= 100m (distance to slits)
d= 3m (distance between slits)
m=1 ( the next constructive maxima)
so the final equation i get is y= (mλL)/d = 66.66 meters which sounds really big, did i mess up somewhere?

That's what I got. The distance is large because, in relation to the wavelength, the slit separation is very small.
 

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