Answer
(a) $503.2nm$
(b) $670.9nm$ and $402.6nm$
Work Step by Step
(a) We know that
$\lambda_{vacuum}=\frac{2nt}{m}$
For $m=1$
$\lambda_1=\frac{2(1.480)(340nm)}{1}=1006nm$
For $m=2$
$\lambda_2=503.2nm$
For $m=3$
$\lambda_3=335.5nm$
Thus, the wavelength that is absent in the reflected beam is $503.2nm$.
(b) We know that in the given scenario
$\lambda_{vacuum}=\frac{4nt}{2m+1}$
For $m=0$
$\lambda_{0}=\frac{4(1.480)(340nm)}{1}=2013nm$
For $m=1$
$\lambda_1=670.9nm$
For $m=2$
$\lambda_2=402.6nm$
For $m=3$
$\lambda_3=287.5nm$
Thus, the wavelengths that are absent in the reflected beam are $670.9nm$ and $402.6nm$.
