I NEED HELP WITH PART B Learning Goal: To use the formulas for the locations of

Question

I NEED HELP WITH PART B

Learning Goal:

To use the formulas for the locations of the dark bands and understand Rayleigh's criterion of resolvability.

An important diffraction pattern in many situations is diffraction from a circular aperture. A circular aperture is relatively easy to make: all that you need is a pin and something opaque to poke the pin through. The figure shows a typical pattern. (Figure 1) It consists of a bright central disk, called the Airy disk, surrounded by concentric rings of dark and light.

While the mathematics required to derive the equations for circular-aperture diffraction is quite complex, the derived equations are relatively easy to use. One set of equations gives the angular radii of the dark rings, while the other gives the angular radii of the light rings. The equations are the following:

dark rings:  sin?=1.22?D  or  2.23?D  or  3.24?D,

bright rings:  sin?=1.63?D  or  2.68?D  or  3.70?D,

where ? is the wavelength of light striking the aperture, D is the diameter of the aperture, and ? is the angle between a line normal to the screen and a line from the center of the aperture to the point of observation. There are more alternating rings farther from the center, but they are so faint that they are not generally of practical interest.

Consider light from a helium-neon laser (?=632.8 nanometers) striking a pinhole with a diameter of 0.100mm .

Part A

At what angle ?1 to the normal would the first dark ring be observed?

Express your answer in degrees, to three significant figures.

0.442

?

Part B

Suppose that the light from the pinhole projects onto a screen 3.00 meters away. What is the radius r1 of the first dark ring on that screen? Notice that the angle from Part A is small enough that sin??tan?.

Express your answer in millimeters, to three significant figures.

0.442

?

Explanation / Answer

part A

sin(theta) = 1.22*lamda/D = 0.00772

theta = 0.442

part B

sin(theta) =(m*lamda) /d = tan(theta) = y/D

(m*lamda)/d = y/D

where m= 1.22 for first dark fringe,,,,,lamda = wavelength of incident light = 632.8 nanometers

d= diameter of aperature

y = dispalcement of fringe from centre

D = screen distance = 3m

d= (1.22*632.8*10^-9*3)/y =2.313*10^-6/y m

r1 =d/2 = 1.156*10^-6/ym = (1.156^10-3)/y mm

you need to provide the value of y to get the numerical value of r1

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