Question

The source that illuminates the double slit in double slit interference experiment emits two distinct monochromatic waves of wavelength $500nm$ and $600nm$, each of them producing its own pattern on the screen. At the central point of the pattern when path difference is zero, maxima of both the patterns coincide and the resulting interference pattern is most distinct at the region of zero path difference. But as one moves out of this central region, the two fringe systems are gradually out of step such that maximum due to one wave to the other and the combined fringe system becomes completely indistinct. This may happen when path difference in $nm$ is:
A. 2000
B. 3000
C. 1000
D. 1500

Answer 1

Hint:To solve this question, we need to use the concept of Young’s double slit experiment. We will use the formulas for the path lengths constructive interference and destructive interference to solve this question. We will equate this both to find the value of the number of the wave and then put this value in the formula for the path difference for maxima to get our answer.

Complete step by step answer:
We know that to obtain constructive interference for a double slit, the path length difference must be an integral multiple of the wavelength which is given by:
$\mathrm{\Delta }x=n{\lambda }_1$
Also for destructive interference, the path length difference must be a half-integral multiple of the wavelength which is given by
$\mathrm{\Delta }x=\left(n+{\displaystyle \frac{1}{2}}\right){\lambda }_2$
For the given condition, we can equate both these,
$\Rightarrow n{\lambda }_1=\left(n+{\displaystyle \frac{1}{2}}\right){\lambda }_2$
We are given that ${\lambda }_1=500nm$ and ${\lambda }_2=600nm$ .
$$\begin{gathered} n\times 500=\left(n+{\displaystyle \frac{1}{2}}\right)\times 600 \\ \Rightarrow 500n=600n+300 \\ \Rightarrow n=-3 \end{gathered}$$
We know that the highest order for interference maxima is given by
$$\begin{gathered} n_{max}=\left|n\right| \\ \Rightarrow n_{max}=\left|-3\right| \\ \Rightarrow n_{max}=3 \end{gathered}$$
Now, for maxima, the path difference is given by:
$$\begin{gathered} \mathrm{\Delta }x=n_{max}{\lambda }_1 \\ \Rightarrow \mathrm{\Delta }x=3\times 500 \\ \therefore \mathrm{\Delta }x=1500nm \end{gathered}$$

Hence, option D is the right answer.

Note:Here, we have determined the value of the required path equation by putting the value of $n_{max}$ in the first equation. However, we can also use the othe equation to find the required answer. For example,
$$\begin{gathered} \mathrm{\Delta }x=\left(n_{max}+{\displaystyle \frac{1}{2}}\right){\lambda }_2 \\ \Rightarrow \mathrm{\Delta }x=\left(300+{\displaystyle \frac{1}{2}}\right)600 \\ \therefore \mathrm{\Delta }x=1500nm \end{gathered}$$