Young's Experiment - (problem with equation)

[FrogPad]

So my book has the following expression for Young's double slit experiment.

(maxima - bright fringes) d \sin \theta = m \lambda
for m = 0, \, 1 \, 2 \, \ldots.So what if you solve this for wavelength.

\lambda = \frac{d \sin \theta}{m}

How is this valid when m = 0

Is this because if m = 0, \theta HAS to equal 0?

by the way
d is the distance between the slits
theta is the angle from the central axis
lambda is the wavelength
m is the index of where the maxima occur

 

[Andrew Mason]

So my book has the following expression for Young's double slit experiment.

(maxima - bright fringes) d \sin \theta = m \lambda
for m = 0, \, 1 \, 2 \, \ldots.So what if you solve this for wavelength.

\lambda = \frac{d \sin \theta}{m}

How is this valid when m = 0

Is this because if m = 0, \theta HAS to equal 0?

The first maximum is always at 0 angle regardless of wavelength. So you can't determine the wavelength from the position of the central maximum.

AM