Antenna Wavelength - double-slit interference

[Koro21]

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?

 

[rude man]

How can you have 2 slits with just 2 boulders? Did anyone provide a drawing?

 

[Koro21]

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

 

[rude man]

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.

 

[Koro21]

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?

 

[rude man]

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.