Answer
$\lambda_a =700nm$
Work Step by Step
First we will set up an equation using equation (36-3) for each wavelength.
$\alpha \sin \theta = m \lambda $ where $m=1,2,3,....$ (minima) (36-3)
For $\lambda_b$, we have:
$\alpha \sin \theta = 1 *350nm $
For $\lambda_a$, we have:
$\alpha \sin \theta = 2 *\lambda_a $
The slit width, $\alpha$ and the angle, $\theta$ are the same in each case, so we combine the equations to obtain:
$1 *350nm= 2 *\lambda_a $
$\lambda_a =2* 350nm$
$\lambda_a =700nm$
