#### Answer

a. Narrower.
b. Wider.

#### Work Step by Step

a. Equation 24-3b, $sin \theta = \frac{m \lambda}{D}$, predicts the location of the minima for nonzero integer values of m. For a D = 60 nm slit, the first minimum would occur at an angle satisfying $sin \theta = \frac{\lambda}{60 nm}\approx10$ which has no solution. Therefore, a slit width as small as 60 nm produces a very wide, white central maximum that covers the entire screen.
b. Equation 24-3b, $sin \theta = \frac{m \lambda}{D}$, predicts the location of the minima for nonzero integer values of m. For a D = 60000 nm slit, the first minimum would occur at an angle satisfying $sin \theta = \frac{\lambda}{60000 nm}$ which is exceedingly small. Most of the light intensity falls within that narrow central maximum, with indistinct fringes outside.