Answer
(a) $496nm$
(b) greater than
Work Step by Step
(a) We know that
$\theta=tan^{-1}(\frac{Y}{L})=tan^{-1}(\frac{0.076m}{0.855m})=5.07^{\circ}$
Now $\lambda=\frac{Wsin\theta}{m}$
We plug in the known values to obtain:
$\lambda=\frac{(11.2\times 10^{-6}m)sin(5.07^{\circ})}{2}$
$\implies \lambda=496nm$
(b) We know that $W=\frac{\lambda m}{sin(tan^{-1}(\frac{Y}{L}))}$. This equation shows that the width of the slit and distance between the fringes (y) are inversely proportional to each other. Thus, as the width of the slit decreases then the distance between the fringes increases -- that is, greater than $15.2m$.
