Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 mm wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 W/${\mathbf{m}}^{\mathbf{2}}$. (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

(a) Destructive interference (dark fringes) occurring at an angle

As the sine values bounded between -1 and 1, therefore m is bounded by;

Hence, this means there are 39 completely dark fringes on both sides of the central maximum, for a total of 78 completely dark fringes.

(b) For a maximum integer (m) smaller than $a/\lambda$, the dark fringe that is farthest from the central maximum appears;

Hence, the angle is$80.8°$

(c) The pattern begins with a dark fringe and ends with a light fringe. As a result, the 38^th light fringe is the light fringe that comes before the 39^th dark fringe. for large m, the maximum intensity of the mth light fringe, the equation is;

The most distant 39^th dark fringe occurs at an angle${\theta }_{d39}={\theta }_{dmax\text{'}}$ , therefore the 38^th dark fringe is expressed by the equation;

The midpoint between the angles of these two dark fringes is where the 38th light fringe reaches its maximum intensity.