Answer
(a) $415nm $
(b) increase
(c) $587nm $
Work Step by Step
(a) We know that
$\lambda=\frac{dsin\theta}{m}$
We plug in the known values to obtain:
$\lambda=\frac{(48.0\times 10^{-5}m)sin(0.99^{\circ})}{2}$
$\lambda=414.6\times 10^{-9}m=415nm $
(b) We know that $ dsin\theta=m\lambda $. This equation shows that if the slit separation increases then the wavelength also increases.
(c) As $\lambda=\frac{dsin\theta}{m}$
We plug in the known values to obtain:
$\lambda=\frac{(68.0\times 10^{-5}m)sin(0.0990^{\circ})}{2}$
$\lambda=587nm $
