- #1
FrogPad
- 810
- 0
So my book has the following expression for Young's double slit experiment.
(maxima - bright fringes) [itex] d \sin \theta = m \lambda [/itex]
for [itex] m = 0, \, 1 \, 2 \, \ldots [/itex].So what if you solve this for wavelength.
[tex] \lambda = \frac{d \sin \theta}{m} [/tex]
How is this valid when [itex] m = 0 [/itex]
Is this because if [itex] m = 0 [/itex], [itex] \theta [/itex] 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
(maxima - bright fringes) [itex] d \sin \theta = m \lambda [/itex]
for [itex] m = 0, \, 1 \, 2 \, \ldots [/itex].So what if you solve this for wavelength.
[tex] \lambda = \frac{d \sin \theta}{m} [/tex]
How is this valid when [itex] m = 0 [/itex]
Is this because if [itex] m = 0 [/itex], [itex] \theta [/itex] 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