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Last updated date: 06th Dec 2023
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MVSAT Dec 2023

The light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another light which produces interference fringes by 8.1 mm using the same pair of slits.

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Hint: In the problem a pair of slits are used to produce the interference pattern. First, the light of a certain wavelength is used and next time the light is changed without changing the pair of slits.
Also, keeping the same configuration means the distance from slits and order of fringe is unchanged as given in the question.

Complete answer:
In proceeding with a solution we first take the given things from the question.
Given, the first incident light has wavelength 630 nm
The separation between fringes initially = 7.2mm
The separation between fringes finally = 8.1 mm
We have the formula
 $ y = \dfrac{{mD\lambda }}{d}$
y is the distance between fringes
m is the order of the fringe which remains the same in our case. For both the incident wavelengths of laser light.
D is the distance between light and the slits.
D is slit width
For first incident light,
  y = \dfrac{{mD\lambda }}{d} \\
   \Rightarrow 7.2 = \dfrac{{mD(630)}}{d}...(i) \\
Let the second wavelength be $ \lambda $
For second incident light,
  y = \dfrac{{mD\lambda }}{d} \\
   \Rightarrow 8.1 = \dfrac{{mD\lambda }}{d}...(ii) \\
Now dividing the equation (ii) by (i) we get
  \dfrac{{8.1}}{{7.2}} = \dfrac{\lambda }{{630}} \\
   \Rightarrow \lambda = \dfrac{{630 \times 9}}{8} \\
   \Rightarrow \lambda = 78.75 \times 9 \\
So, after solving we get the wavelength of second light as
$ \lambda = 708.75\,nm$

Note: Double slits produce coherent sources of waves as we know from the experiment.
In one region light spreads out which is also said to be diffracted from each slit in the setup as the slits are narrow.
These then overlap constructively and destructively to form bright and dark bands respectively.