Question

A beam of light consisting of two wavelengths, $800nm$ and $600nm$ is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed $1.4m$ away. If the two slits are separated by $0.28mm$.Calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.

Answer 1

Hint:To find the least distance where the bright fringes by the two wavelengths coincide we equate the formulae for the distance of bright fringe for both wavelengths and find out the relation between the number of the fringes. Then the minimum number of the bright fringe is calculated, and then the distance of that fringe is calculated.

Formula used:
$x = \dfrac{{n\lambda D}}{d}$
where x is the distance of the bright fringe from the central bright fringe.

Complete step by step answer:
In Young's double slit experiment, if two wavelengths of light are used then both the light waves interfere with each other and produce a central bright maxima. But after that due to different fringe widths, the fringes will coincide only when the distance will be a common multiple of both the wavelengths.The distance of the nth bright fringe from the central maxima is given by this equation-
$x = \dfrac{{n\lambda D}}{d}$
Here n is the number of bright fringe, the central maxima is taken as $n = 0$
$\lambda $ is the wavelength of the light used
$D$ is the distance between slits and the screen
$d$ distance between two slits

In the given question, since the distance where both of the wavelengths interfere has to be same, therefore we equate both the distances-
$\dfrac{{{n_1}{\lambda _1}D}}{d} = \dfrac{{{n_2}{\lambda _2}D}}{d}$
$ \Rightarrow {n_1}{\lambda _1} = {n_2}{\lambda _2}$
On putting the values as- ${\lambda _1} = 800$ and ${\lambda _2} = 600$ we have,
$800{n_1} = 600{n_2}$
Reducing it to simple terms we have,
$4{n_1} = 3{n_2}$
$ \Rightarrow {n_1} = \dfrac{3}{4}{n_2}$

For the fringe numbers ${n_1}$and ${n_2}$ to be interfering, they must be a whole number and not fraction.On keeping ${n_2} = 4$, ${n_1}$ is $3$. This is the least value of n for both the wavelengths where their bright fringes will coincide. To calculate distance any one equation can be taken,
$x = \dfrac{{{n_1}{\lambda _1}D}}{d}$
Keeping the values ${n_1} = 3,{\lambda _1} = 800nm,D = 1.4m,d = 0.28mm$-
$ \Rightarrow x = \dfrac{{3 \times 800 \times {{10}^{ - 9}} \times 1.4}}{{0.28 \times {{10}^{ - 3}}}}$
$ \therefore x = 12 \times {10^{ - 3}}m$

They both coincide at $x = 12 \times {10^{ - 3}}m$ from the central bright fringe.

Note:The value of $n$for the central bright fringe is taken as 0, followed by dark and then bright fringes on its either side getting the value as $n = 1$. In this question, more than one wavelength of light is used, therefore the fringe width of the patterns formed by both of these wavelengths would be different. The distance calculated in the solution was the minimum distance where the fringes produced by both of these wavelengths will coincide, any distance where the relation$4{n_1} = 3{n_2}$is true, both wavelength fringes will coincide.