In: Physics

# Light from a He-Ne laser (wavelength 633nm) passes through a single slit of width 25μm. At...

Light from a He-Ne laser (wavelength 633nm) passes through a single slit of width 25μm. At the screen a distance away, the intensity at the center of the central maxima is 8.25 W/m^2.

a. Draw a clear diagram showing the slit and the intensity pattern seen on the screen. Label key quantities and key features.

b. Find the maximum number of totally dark fringes (minima) seen on the screen.

c. At what angle does the dark fringe (minima) that is most distant from the center occur? Note: The small angle approximation will NOT apply.

d. What is the intensity at the peak of the secondary maxima that occurs immediately before the dark fringe in part (b)? Hint: Approximate the angle at which this secondary maxima occurs by assuming it is midway between the angles to the dark fringes on either side of it. Note: The small angle approximation will NOT apply.

## Solutions

##### Expert Solution

(a)

_______________________________________

(b)

dsin = m

where d = 25e-6 m

we want to solve for m, therefore,

m = dsin /

To find maximum number of fringes, we need to set = 90

m = 25e-6 * sin 90 / 633e-9

m = 39.49

so,

total number of dark fringes on each side of central maximum is 39 ,

total dark fringes on screen

M = 39 + 39

M = 78

__________________________________

(c)

dsin = m

here, we need to solve for sin with m = 39 ( most distant fringe)

so,

sin = 39 * 633e-9 / 25e-6

sin = 0.98748

= 80.92 degree

so,

distant dark fringe occurs at +/ - 80.9 degree

________________________________________________

(d)

angle for next dark fringe ( m = 38)

sin = 38 * 633e-9 / 25e-6

sin = 0.96216

= 74.18 degree

the angle midway is

= 80.9 + 74.18 / 2

= 77.5 degree

now we find angle for maximum intensity,

/ 2 = dsin /

/ 2 = * 25e-6 * sin 77.5 / 633e-9

so,

I = Io ( ( sin ( / 2 ) / / 2 ) 2

I = 8.25 * (sin 121.13 / 121.13)2

I = 5.445e-4 W/m2

OR

I = 544.5 uW / m2

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