Question

A light consisting of two wavelengths $650nm$ and $520nm$ 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 $3mm$ and the distance between the plane of the slits and screen is $150cm.$

Answer 1

Answer: 2

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 ${\lambda }_2$
$\therefore n_1{\beta }_1=n_2{\beta }_2$
$\Rightarrow n_1\frac{{\lambda }_{1}D}{d}=n_2\frac{{\lambda }_{2}D}{d}$
$\Rightarrow n_1{\left(\lambda \right)}_1=n_2{\left(\lambda \right)}_2\dots .(i)$
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 $n_2>n_1$
$\Rightarrow n_2=n_1+\mathrm{1...}\left(ii\right)$
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, $y=n_1{\beta }_1$
$=n_1\frac{{\lambda }_{1}D}{d}$
$=4\left[\frac{\left(6.5\times 10^{-7}\right)\left(1.5\right)}{3\times {\left(10\right)}^{-3}}\right]$
$=1.3\times 10^{-3}m$
$=1.3mm$