Question

A beam of light consisting two wavelengths, 650nm and 520nm is used to obtain the interference fringes in a Young’s double-slit experiment. What is the least distance from the central maximum where the bright fringes due to both the wavelength coincide? The width of the slit is 2mm and the distance of the screen is 1.2m
a) 1.17mm
b) 2.52mm
c) 1.56mm
d) 3.14mm

Answer 1

Hint: IN the above question it is given that in a Young’s double slit experiment a beam of light consisting two wavelengths, 650nm and 520nm is used to obtain the interference pattern. To obtain the least distance from the central maximum of the bright fringes we first need to obtain the ratio at which the bright fringes actually meet. Hence the least ratio will give us the position of the least distance from the central maximum of the bright fringes due to the respective wavelength of light used.
Formula used:
$y_n={\displaystyle \frac{n\lambda D}{d}}$

Complete answer:
Let us say we have a beam of light of wavelength $\lambda$. Suppose we illuminate this light on to the slits of width ‘d’ of Young’s double slit experiment the interference pattern will have the positions of the nth alternate bright fringes observed on to the screen at a distance ‘D’ from the slits is given by,
$y_n={\displaystyle \frac{n\lambda D}{d}}$
In the above question it is given that the light consisting of two wavelengths, ${\lambda }_1=650nm$ and ${\lambda }_2=520nm$ is used to obtain the interference fringes in a Young’s double-slit experiment. Let us say the $p^{th}$ bright fringe of wavelength ${\lambda }_1$ and the $q^{th}$ bright fringe of wavelength of light ${\lambda }_2$overlap each other at a point $y_n$ from the central fringe. Hence we can write,
$\begin{array}{rl}& y_n={\displaystyle \frac{p{\lambda }_{1}D}{d}}={\displaystyle \frac{q{\lambda }_{2}D}{d}}\\ & \Rightarrow p{\lambda }_1=q{\lambda }_2\\ & \because {\lambda }_1=650nm,{\lambda }_2=520nm\\ & \Rightarrow p\left(650nm\right)=q\left(520nm\right)\\ & \therefore {\displaystyle \frac{p}{q}}={\displaystyle \frac{520nm}{650nm}}={\displaystyle \frac{52}{65}}={\displaystyle \frac{4}{5}}\end{array}$
Hence the bright fringes of both the wavelengths will overlap each other in the ratio 4:5. Likewise the bright fringes will keep on overlapping such that the ratio between their $p^{th}$ bright fringe of wavelength ${\lambda }_1$ and the $q^{th}$ bright fringe of wavelength of light ${\lambda }_2$is always the same. The least distance is nothing but the above obtained ratio i.e. the fourth bright fringe of ${\lambda }_1$ and fifth bright fringe of ${\lambda }_2$will coincide with each other and so on . Hence the least distance ‘d’ from the central maximum is,
$\begin{array}{rl}& d={\displaystyle \frac{4\times 650nm\left(1.2m\right)}{2mm}}\\ & \Rightarrow d={\displaystyle \frac{4\times 650\times {10}^{-9}\left(1.2m\right)}{2\times {10}^{-3}m}}\\ & \Rightarrow d={\displaystyle \frac{4\times 650\times {10}^{-9}\left(1.2m\right)}{2\times {10}^{-3}m}}=2\times 650\times {10}^{-6}\left(1.2m\right)\\ & \therefore d=1.56\times {10}^{-3}m=1.56mm\end{array}$

Hence the correct answer of the question is option c.

Note:
The ratio of the fringes that overlap each other gives us the idea with what ratio the bright fringes of both the wavelength coincide from the central maximum. The ratio we obtained was 4:5. This is the minimum possible ratio as there are no possible natural numbers which can give this ratio. Hence we conclude that the 4th bright fringe of one coincides with the 5th bright fringe of the other.