Answer
$m_b = 6$
Work Step by Step
We start the same way we did in problem 5b. 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 = m_b *\lambda_b $
For $\lambda_a$, we have:
$\alpha \sin \theta = m_a *\lambda_a $
The slit width, $\alpha$ and the angle, $\theta$ are the same in each case, so we combine the equations to obtain:
$m_b *\lambda_b= m_a *\lambda_a $
We are given that they coincide at $m_a=1$ and $m_b=2$, so we obtain :
$2 *\lambda_b= \lambda_a $
Substituting in the previous equation, we obtain:
$m_b *\lambda_b= m_a *2 *\lambda_b $
so
$m_b = m_a *2 $
We were given that $m_a=2$, so we have:
$m_b = 3*2 $
$m_b = 6$
