Problem 27.3

27.3. a)
We want the maximum when m = 2. So when the car is located where it is, we have
$\tan \theta = \frac{400m}{1000m} ~=~ .4$
$\theta = 21.8{}^{\circ}$
So, the wavelength of the signal is given by
$\lambda = \frac{d \sin \theta}{m}$
$\lambda = \frac{(300 m)(\sin 21.8{}^{\circ})}{2}$
$\lambda = 55.7 m$

b)
The next minimum encountered will occur when
$d \sin \theta = (m + \frac{1}{2}) \lambda$
$d \sin \theta = \frac{5\lambda}{2}$
$\sin \theta = \frac{5 \lambda}{2d}$
$\sin \theta = .464$
$\theta = 27.7{}^{\circ}$
so, to find the y value corresponding to this angle we take
$y = (1000 m) \tan 27.7{}^{\circ}$
y = 524 m
So, the car must travel an additional 124 m.