Two radio antennas separated by 300 m as shown in figure simultaneously broadcast indentical singals at the same wavelength. A radio in a car traveling due north receives the signals. (a) If the car is at the position of the second maximum, what is the wavelength of the signals? (b) How much farther must the car travel to encounter the next mininum in reception? (Note: Do not use the small- angle approximation in this problem.)

Explanation

Note, with the conditions given, then small angle appproxmination does not word well. That is, `sin theta, tan theta`, and `theta` are significantly different. The approach to be used is outlined below.
a. At `m = 2` maximum,
`tan theta = (400m)/(1000 m) = 0.400 = (2)/(5) implies sin theta = (2)/(sqrt 29)`
So, `lambda = (d sin theta)/(m) = ((300 m) (2)/(sqrt 29))/(2) = (300)/(sqrt 29) m = 55.7 m`
b. The next mininum encountered is the `m = 2` minimum.
At the point, `d sin theta' = (m + (1)/(2)) lambda`
which becomes ` d sin theta' = (5)/(2) lambda`
or `sin theta' = (5 lambda)/(2 d) = (5)/(2) (((300)/(sqrt 29)m)/(300 m)) = (5)/(2 sqrt 29)`
`implies tan theta' = (5)/(141)`
so, `y = (1000 m) (5)/(141) = 524 m`
Therefore, the car must travel an additional `124 m`.