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

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 $n_{2}^{t h}$ 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} \ldots \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 $n _{1} \lambda_{1}=\left( n _{1}+1\right) \lambda_{2}$ $\Rightarrow\left( n _{1}\right)\left(650 \times 10^{-9}\right)=\left( n _{1}+1\right.$ $\Rightarrow 65 n _{1}=52 n _{1}+52$ $\Rightarrow 13 n _{1}=52$ $\Rightarrow n _{1}=4$ Thus, $y = n _{1} \beta_{1}$ $= n _{1} \frac{\lambda_{1} D }{ d }$ $=4\left[\frac{\left(6.5 \times 10^{-7}\right)(1.5)}{3 \times 10^{-3}}\right]$ $=1.3 \times 10^{-3} m$ $=1.3 mm =\frac{2.6}{2} mm$