Answer
a) (2) smaller angle of resolution
b) $\theta_{min}$ for slit of 0.55mm = $0.07^{\circ}$
$\theta_{min}$ for slit of 0.45mm = $0.087^{\circ}$
Work Step by Step
a) As per Rayleigh criterion,
$\theta_{min}=\frac{wavelength}{w}$
Hence, $\theta_{min}$ is inversely proportional to width of the slit w.
So, more the width, less is the value of $\theta_{min}$
(2) is the answer.
b) Red light wavelength = $680nm$
Width of first slit $w_{1}=0.55mm$
Width of second slit $w_{2}=0.45mm$
So, $\theta_{min}=\frac{wavelength}{w_{1}}=\frac{680\times10^{-9}}{0.55\times10^{-3}}radians=0.07^{\circ}$
For the second slit,
$\theta_{min}=\frac{wavelength}{w_{2}}=\frac{680\times10^{-9}}{0.45\times10^{-3}}radians=0.087^{\circ}$
