Single slit diffraction pattern

The central bright fringe in a single slit diffraction pattern from light of wavelength 499 nm is 2.4 cm wide on a screen that is 1.05 m from the slit. (Define the width of a bright fringe as the linear distance from minimum to minimum.)

How wide is the slit?
How wide is the first bright fringe on either side of the central bright fringe?

the condition for a minimum in the single slit diffraction is

a sin(theta) = m L

a=slit width
theta = angle of minimum with respect to center line
m=order
L=lambda = wavelength

if the distance to the screen is D, the distance between the center line and minimum, y, is

y = D tan(theta)

since theta is small, tan(theta) = theta(radians) = sin(theta)

if the central bright is 2.4 cm, the value of y is 1.2 cm = 0.012m, and we find the slit size from

a sin(theta) = m L

but sin(theta) = tan(theta) = y/D, so we have

a y/D = m L

a = D m L/ y

a = 1.05m x 1 x 4.99x10^-7m/0.012m = 4.37x10^-5m

we can now find the y value of the minimum corresponding to m =2:

y = D m L/a = 1.05m x 2 x 4.99x10^-7m/4.37x10^-5m = 0.024m

the width of this fringe is the distance between the m=1 and m=2 minima, or 0.024m - 0.012m = 0.012m (half the width of the central bright)