Video Transcript
Light with a wavelength of 563
nanometers passes through a sheet in which there are two parallel narrow slits 8.38
micrometers apart. The light from the slits is
incident on a screen parallel to the sheet, where a pattern of light and dark
fringes is observed. A line 𝐿 runs perpendicular to the
surface of the sheet and the direction of the slits. The line 𝐿 intersects the central
bright fringe of the pattern on the screen. How many bright fringes will be
present on a screen that can extend, without limit, on either side of 𝐿?
Clearing some space on screen, we
are told that light with a 563-nanometer wavelength passes through a sheet in which
there are two parallel narrow slits. The slits are 8.38 micrometers
apart. Once the light passes through the
slits, it is incident on a screen that lies parallel to the sheet. A pattern of light and dark fringes
are observed on the screen. We are also told that there is a
line, 𝐿, that runs perpendicular to the surface of the sheet in the direction of
the slits and intersects the central bright fringe on the screen. We are asked to determine how many
bright fringes will be present on that screen if it can extend without limit to
either side of a line 𝐿.
In order to know this, we will need
to remember what happens when light passes through two parallel narrow slits and
creates an interference pattern on the screen. When light passes through two
parallel narrow slits in a sheet, two wavefronts of the light will be produced at
these slits on the opposite side of the sheet to the side that the waves were
incident on. The waves that propagate from these
wavefronts will interfere with each other where they overlap. The interference of the overlapping
waves will produce a resultant wave amplitude at each point at which the waves
overlap.
Where the resultant amplitude has
maximum magnitude, this is called constructive interference of the waves. Where the resultant amplitude
equals zero, this is called destructive interference of the waves. We can see examples of constructive
and destructive interference and the resultant wave amplitudes here. When a screen is placed behind the
sheet that contains the slits, the resultant amplitudes of the waves can be seen at
different points on the screen. This is called an interference
pattern. The pattern consists of a set of
bright and dark fringes, where the bright fringes will appear at positions on the
screen as seen here. A dark fringe will appear between
any neighboring bright fringes.
Constructive interference produces
bright fringes, and destructive interference produces dark fringes. We can define the angle at which
the bright fringes occur. Because the question is asking us
to find the number of bright fringes that occur, we will use the equation that tells
us the angles required for constructive interference. The separation of the two slits,
𝑑, multiplied by the sin of the angle 𝜃, where 𝜃 is the angle between the line 𝐿
and a line from the center of the slits to one of the bright spots, is equal to the
number of that maximum from the central bright fringe — in general we label this as
the integer 𝑛 — multiplied by the wavelength of light 𝜆.
We are looking for the number of
fringes produced. So let’s modify this equation to
make 𝑛 the subject. We do this by dividing both sides
by the wavelength 𝜆. This cancels out the 𝜆 on the
right-hand side of the equation and brings one into the denominator of the left-hand
side. So 𝑛 is equal to 𝑑 multiplied by
the sin of 𝜃 divided by 𝜆.
Recall that the sine function is a
periodic function, and it has a maximum and minimum value. Because we are looking for the
maximum number of fringes, we can use the maximum value for the sine function to
find the maximum value of 𝑛. The maximum of the sine of an angle
occurs at an angle of 𝜋 over two radians, or 90 degrees, and has the value one.
So plugging this in, we see that
the maximum value for 𝑛 is equal to the distance between the slits divided by the
wavelength of light. But before we plug the values for
these in, we need to convert both these lengths into meters. A nanometer is equal to 10 to the
negative nine meters. And a micrometer is equal to 10 to
the negative six meters.
Plugging everything in, 𝑛 is equal
to 8.38 times 10 to the negative six meters divided by 563 times 10 to the negative
nine meters. This gives us a maximum 𝑛-value of
14.88. The greatest whole number of
fringes is then 14. There are then 14 bright fringes on
either side of the central bright fringe, which gives 28 fringes. We must then add one to include the
central bright fringe. So the maximum number of fringes
that can be present on the extended screen is 29.
