Video Transcript
Light of two different wavelengths
passes through a sheet in which there are two parallel narrow slits. 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. The distance on the screen from 𝐿
to the center of the bright fringe of the shorter wavelength, nearest to the central
bright fringe, is 5.55 centimeters. The distance on the screen from 𝐿
to the center of the bright fringe of the longer wavelength, nearest to the central
bright fringe, is 7.25 centimeters. What is the ratio of the
longer-wavelength light to that of the shorter-wavelength light? Give your answer to two decimal
places.
In this question, we are told that
light of two different wavelengths passes through a sheet with two parallel narrow
slits. After the light passes through, it
is incident on a screen that is parallel to the sheet with the slits. A pattern of light and dark fringes
is observed on the screen. There is a line 𝐿 that runs
perpendicular to the sheet and screen. It intersects the central bright
fringe of the pattern on the screen.
We are told that the distance on
the screen from 𝐿 to the center of the bright fringe of the shorter wavelength that
is closest to the central bright fringe is 5.55 centimeters. We are also told that the distance
on the screen from 𝐿 to the center of the bright fringe of the longer wavelength
that is closest to the central bright fringe is 7.25 centimeters. We are asked to find the ratio of
the longer wavelength of light to the shorter wavelength of light.
Before we calculate an answer,
let’s remember some information about what happens when light passes through narrow
slits.
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 one another 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.
Here are some examples of
constructive and destructive interference and the resultant wave amplitudes. 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 fringes 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.
Now that we know how light behaves
after traveling through two narrow slits and how an interference pattern is
produced, let’s look back at the question and answer it.
We are asked to find the ratio of
the longer wavelength of light to the shorter wavelength of light. Recall that we can find the
distance from the center fringe to the other fringes with the equation 𝑦 sub 𝑛 is
equal to the integer 𝑛 multiplied by the wavelength 𝜆 multiplied by the distance
from the sheet to the screen 𝐿 divided by the distance between the two slits
𝑑. Notice that the distance from a
bright fringe to the central bright fringe is directly proportional to the
wavelength of the light. This means that if we find the
ratio of these two lengths, we will find the ratio of the wavelengths as well. We can see this by setting up this
equation for the two different wavelengths.
Notice that the variables 𝑛, 𝐿,
and 𝑑 will be the same for both wavelengths. This is because we’re looking at
the first bright fringe for both. So 𝑛 is equal to one, and the
distance between the screen and the sheet as well as between the slits does not
change. For the longer wavelength, we have
7.25 centimeters is equal to 𝑛 multiplied by 𝜆 sub two multiplied by 𝐿 divided by
𝑑. For the shorter wavelength, we have
5.55 centimeters is equal to 𝑛 multiplied by 𝜆 sub one multiplied by 𝐿 divided by
𝑑.
Clearing some space, if we create a
ratio for the longer wavelength over the shorter wavelength, we find that 7.25
centimeters over 5.55 centimeters is equal to the quantity 𝑛, which is one,
multiplied by 𝜆 sub two multiplied by 𝐿 divided by 𝑑 divided by the quantity 𝑛,
again one, multiplied by 𝜆 sub one multiplied by 𝐿 divided by 𝑑. Notice that on the right-hand side
all the variables except the two 𝜆’s will cancel out. This leaves us with 𝜆 sub two over
𝜆 sub one is equal to 7.25 centimeters divided by 5.55 centimeters. The centimeters will cancel
out. And solving this, we find that the
ratio for the longer wavelength to the shorter wavelength is approximately equal to
1.306. We are asked to round to the
nearest two decimal places, which gives us an answer of 1.31 for the ratio.
Therefore, 1.31 is the ratio of the
longer-wavelength light to the shorter-wavelength light.
