Fraunhofer Single Slit

 

Content . 301302303304 .

  1. Fraunhofer Diffraction Pdf
  2. Fraunhofer Diffraction At A Single Slit
  3. Fraunhofer Diffraction Pattern Single Slit
  4. Fraunhofer Diffraction Class 12
  5. Fraunhofer Diffraction Formula

S E C T I O N 3 8 . 2 • Diffraction Patterns from Narrow Slits

Fraunhofer DiffractionChapter 11. Fraunhofer Diffraction Last lecture. Numerical aperture of optical fiber. Allowed modes in fibers. Attenuation. Modal distortion, Material dispersion, Waveguide dispersion This lecture. Diffraction from a single slit. Diffraction from apertures: rectangular, circular. Resolution: diffraction limit. Fraunhofer Diffraction by single slit,diffraction of light waves and polarization,resolving power.

1209

Quick Quiz 38.1

Suppose the slit width in Figure 38.6 is made half as wide.

The central bright fringe (a) becomes wider (b) remains the same (c) becomes narrower.

Quick Quiz 38.2

If a classroom door is open slightly, you can hear sounds

coming from the hallway. Yet you cannot see what is happening in the hallway. Why is
there this difference? (a) Light waves do not diffract through the single slit of the open
doorway. (b) Sound waves can pass through the walls, but light waves cannot. (c) The
open door is a small slit for sound waves, but a large slit for light waves. (d) The open
door is a large slit for sound waves, but a small slit for light waves.

θ

sin

dark

= 2

/a

sin

dark

=

/a

sin

dark

= –

/a

sin

dark

= –2

/a

L

a

0

Naruto mugen online game. y

2

y

1

– y

1

– y

2

Viewing screen

θ

θ

θ

θ

λ

λ

λ

λ

Figure 38.6 Intensity distribution for a

Fraunhofer diffraction pattern from a

single slit of width a. The positions of two

minima on each side of the central

maximum are labeled. (Drawing not to

scale.)

Example 38.1 Where Are the Dark Fringes?

Light of wavelength 580 nm is incident on a slit having a
width of 0.300 mm. The viewing screen is 2.00 m from the
slit. Find the positions of the first dark fringes and the width
of the central bright fringe.

Solution The problem statement cues us to conceptualize
a single-slit diffraction pattern similar to that in Figure 38.6.
We categorize this as a straightforward application of our
discussion of single-slit diffraction patterns. To analyze the
problem, note that the two dark fringes that flank the
central bright fringe correspond to m ' # 1 in Equation
38.1. Hence, we find that

From the triangle in Figure 38.6, note that tan !

dark

'

y

1

/L.

Because !

dark

is very small, we can use the approximation

sin !

dark

Fraunhofer Diffraction Pdf

! tan !

dark

; thus, sin !

dark

! y

1

Fraunhofer Diffraction At A Single Slit

/L. Therefore, the

positions of the first minima measured from the central axis
are given by

The positive and negative signs correspond to the dark
fringes on either side of the central bright fringe. Hence,
the width of the central bright fringe is equal to 2

'y

1

' '

7.74 & 10

'

3

m '

To finalize this problem,

7.74 mm.

#

3.87 & 10

'

3

m

'

y

1

! L sin !

dark

'

(2.00 m)(#1.933 & 10

'

3

)

sin !

dark

' #

%
a

' #

5.80 & 10

'

7

m

0.300 & 10

'

3

m

' #

1.933 & 10

'

3

note that this value is much greater than the width of the
slit. We finalize further by exploring what happens if we
change the slit width.

What If?

What if the slit width is increased by an order

of magnitude to 3.00 mm? What happens to the diffraction
pattern?

Answer Based on Equation 38.1, we expect that the angles
at which the dark bands appear will decrease as a increases.
Thus, the diffraction pattern narrows. For a ' 3.00 mm, the
sines of the angles !

dark

for the m ' # 1 dark fringes are

The positions of the first minima measured from the central
axis are given by

and the width of the central bright fringe is equal to 2

'y

1

' '

7.74 & 10

'

4

m ' 0.774 mm. Notice that this is smaller than

the width of the slit.

In general, for large values of a, the various maxima and

minima are so closely spaced that only a large central bright
area resembling the geometric image of the slit is observed.
This is very important in the performance of optical instru-
ments such as telescopes.

' #3.87 & 10

'

4

m

y

Fraunhofer Diffraction Pattern Single Slit

1

! L sin !

dark

'

(2.00 m)(#1.933 & 10

'

4

)

sin !

dark

' #

%
a

Fraunhofer Single Slit

' #

5.80 & 10

'

7

m

3.00 & 10

'

3

m

' #

1.933 & 10

'

4

Fraunhofer Diffraction Class 12

Investigate the single-slit diffraction pattern at the Interactive Worked Example link at http://www.pse6.com.

Interactive

Fraunhofer Diffraction Formula