Answer
(a) $25\mu m$
(b) less than
Work Step by Step
(a) We know that
$W=\frac{2\lambda}{sin(tan^{-1}(\frac{y}{L}))}$
We plug in the known values to obtain:
$W=\frac{2(632.8\times 10^{-9}m)}{sin(tan^{-1}(\frac{0.076m}{1.50}))}$
$W=25011\times 10^{-9}=25\mu m$
(b) We know that
$\theta=sin^{-1}(\frac{m\lambda}{W})$
We plug in the known values to obtain:
$\theta=sin^{-1}(\frac{2(591\times 10^{-9}m)}{25\times 10^{-6}})$
$\theta=2.708^{\circ}$
Now $y=Ltan\theta$
We plug in the known values to obtain:
$y=(1.50)tan(2.708^{\circ})$
$y=0.0709m=7.09cm$
Thus, the distance between the two bright bands is less than $15.2cm$.
