Physics Question

Question

Two point sources of a wave are 1.0mm apart. At a distance of 1.0m away, the distance between the third nodal lines either side of the central antinodal line is 3.2mm. Describe the brightness of the fringe that is 1.3mm from the central fringe. (Carry out calculations)

Answer 1

Hi Vans!  It looks like this problems is set up as a Young's double-slit experiment (substituting the two point sources for the slits).  The clue is that the detection point is much, much farther (1000 times farther) from the sources than the sources are from each other. 

 

NOTE:  The following assumes some knowledge of the fundamentals of the Young double-slit setup (for far-field sources).  If that is not something you have, we can set it up in more detail.

 

In this configuration, and looking close to the central axis (which they call the central antinodal line), we can use the small-angle approximation version of the double slit formula:

 

dy/D = mλ

 

where d = distance between the sources

y = distance along the detection line (which is a screen if light is involved) from the central antinodal line (a maximum) the the point of observation

D = distance from the sources to the detection line

λ = wavelength of the wave

m = a real number.  If it is an integer, we have an antinode.  If it is a half-integer (1/2, 3/2, 5/2, and so on), we have a node.  If it is neither, we are somewhere in between.  We will solve for this value to characterize the solution.

 

With the information given, we can first solve for the wavelength λ.  We are told that the third nodes (which occur the third time away from zero that n is a half-integer on either side of the central antinodal line) are 3.2 mm apart.  Since the two nodes are symmetric around the center, This tells me that the distance from the central antinodal line to the third node on either side is half of 3.2 mm, or 1.6 mm.  Thus:

 

y = 1.6 mm

m = 5/2  (since this value represents the third time m is a half-integer, starting from m = 0)

 

We are also told:

 

d = 1.0 mm

D = 1.0 m

 

Using this, we can solve for the wavelength of light:

 

λ = (dy)/(Dm) = (1.0 mm)(1.6 mm)/(1000.0 mm * 5/2) = 0.00064 meters or (6.4 x 10^-4 meters)

 

With this information, we can now look at what the value of n is when y = 1.3 mm.  If it is an integer, we have an antinode.  If it is a half-integer, we have a node.  If it is neither, we are somewhere in between, and can describe the brightness in terms of being between bright and dark.

 

m = (dy)/(Dλ) = (1.0 mm)(1.3 mm)/(1000.0 mm * 6.4 x 10^-4 meters) = 2.03

 

This is *almost* m = 2, so we are very near an antinode, a bright fringe (specifically, the second antinode away from the central one).

 

Do double-check my calculations.  And if you have any questions, or want to look at this more closely, please let me know.  I hope this helps!