Question

A light consisting of two wavelengths $650\,nm$ and $520\,nm$ is used to obtain interference fringes in a Young's double slit experiment. If the least distance (in $mm$ ) from a central maximum where the bright fringes due to both the wavelengths coincide is $\frac{2 . 6}{x}$ then find $x$ . It is given that the distance between the slits is $3\,mm$ and the distance between the plane of the slits and screen is $150\,cm.$

Answer 1

Let "y' be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let
$n_{1}^{\text{th }}$ number of bright fringe with wavelength $\lambda _{1}$
coincides with $n_{2}^{\text{th }}$ number of bright fringe with
wavelength $\lambda _{2}$
$\therefore n_{1}\beta _{1}=n_{2}\beta _{2}$
$\Rightarrow n_{1}\frac{\lambda _{1} D}{d}=n_{2}\frac{\lambda _{2} D}{d}$
$\Rightarrow n_{1}\left(\lambda \right)_{1}=n_{2}\left(\lambda \right)_{2}\ldots .\left(\right.i\left.\right)$
Also at first position, the $n^{\text{th }}$ bright fringe of one will coincide with $\left(n+1\right)^{\text{th }}$ bright of other.
If $\lambda _{2} < \lambda _{1}$
So then $n_{2}>n_{1}$
$\Rightarrow n_{2}=n_{1}+1...\left(i i\right)$
Using equation (ii) and equation (i), $n_{1}\left(\lambda \right)_{1}=\left(n_{1} + 1\right)\left(\lambda \right)_{2}$
$\Rightarrow \left(n_{1}\right)\left(650 \times 10^{- 9}\right)=\left(n_{1} + 1\right)\left(520 \times 10^{- 9}\right)$
$\Rightarrow 65n_{1}=52n_{1}+52$
$\Rightarrow 13n_{1}=52$
$\Rightarrow n_{1}=4$
Thus, $y=n_{1}\beta _{1}$
$=n_{1}\frac{\lambda _{1} D}{d}$
$=4\left[\frac{\left(6 . 5 \times 10^{- 7}\right) \left( 1 . 5 \right)}{3 \times \left(10\right)^{- 3}}\right]$
$=1.3\times 10^{- 3}m$
$=1.3\,mm$