Answer
four
Work Step by Step
We will be finding our wavelengths, $\lambda$, using the conditions for constructive interference (equation 35-36):
$2L=(m+1/2) \lambda /n_2$ (for $m=0,1,2,3$)
Solving for $\lambda$, we obtain:
$\lambda=2*n_2*L(m+1/2) $
$\lambda=2*1.40*600nm(m+1/2) $
$\lambda_i=2*1.40*600nm(m_i+1/2) $
Therefore;
$\lambda_0=2*1.40*600nm(0+1/2) = 3360nm$
$\lambda_1=2*1.40*600nm(1+1/2) = 1120nm$
$\lambda_2=2*1.40*600nm(2+1/2) = 672nm$
$\lambda_3=2*1.40*600nm(3+1/2) = 480nm$
$\lambda_4=2*1.40*600nm(4+1/2) = 373nm$
$\lambda_5=2*1.40*600nm(5+1/2) = 305nm$
We see that we have four different wavelengths that are between 300 and 700nm.
