Question

A beam of light consisting of two wavelengths, $650\text{nm}$ and $520nm$ is used to obtain interference fringes in 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 value of $x$ . The distance between the slits is $3mm$ and the distance between the plane of the slits and the 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 $n_2^{th}$ number of bright fringe with wavelength ${\lambda }_2$.
$\begin{array}{l}\therefore n_1{\beta }_1=n_2{\beta }_2\\ \Rightarrow n_1\frac{{\lambda }_{1}D}{d}=n_2\frac{{\lambda }_{2}D}{d}\\ \Rightarrow n_1{\lambda }_1=n_2{\lambda }_2\dots \text{ (i) }\end{array}$
Also at first position, the $n^{\text{th }}$ bright fringe of one will coincide with $(n+1{)}^{\text{th }}$ bright of other.
$\Rightarrow \left(n_1\right)\left(650\times 10^{-9}\right)=\left(n_1+1\right)\left(520\times 10^{-9}\right)$
If ${\lambda }_2<{\lambda }_1$ So then $n_2>n_1$ $\Rightarrow n_2=n_1+1$...(ii) Using equation (ii) and equation $n_1{\lambda }_1=\left(n_1+1\right){\lambda }_2$ $\Rightarrow \left(n_1\right)\left(650\times 10^{-9}\right)=(n_1+1$ $\Rightarrow 65n_1=52n_1+52$ $\Rightarrow 13n_1=52$ $\Rightarrow n_1=4$ Thus, $y=n_1{\beta }_1$ $=n_1\frac{{\lambda }_{1}D}{d}$ $=4\left[\frac{\left(6.5\times 10^{-7}\right)(1.5)}{3\times 10^{-3}}\right]$ $=1.3\times 10^{-3}m$ $=1.3mm=\frac{2.6}{2}mm$