JEE Main 2023 — Wave Optics Question with Solution

In Young's double slit experiment, we have two slits separated by a distance $$d$$, and a screen placed at a distance $$L$$ from the slits. When light with a single wavelength $$\lambda$$ passes through the slits, an interference pattern is formed on the screen with bright and dark fringes.

In this problem, we have a beam of light consisting of two wavelengths $${\lambda }_1=7000\stackrel{o}{A}$$ and $${\lambda }_2=5500\stackrel{o}{A}$$. We are given the distance between the slits $$d=2.5\mathrm{mm}$$ and the distance between the plane of the slits and the screen $$L=150\mathrm{cm}$$. Our goal is to find the least distance from the central fringe where the bright fringes due to both wavelengths coincide.

First, let's convert the wavelengths and distances to meters:

$${\lambda }_1=7\times 10^{-7}\mathrm{m}$$ $${\lambda }_2=5.5\times 10^{-7}\mathrm{m}$$ $$d=2.5\times 10^{-3}\mathrm{m}$$ $$L=1.5\mathrm{m}$$

The fringe width $$\beta$$ for a single wavelength is given by:

$$\beta =\frac{\lambda D}{d}$$

Now, let's consider the condition for the bright fringes due to both wavelengths to coincide. Let the $$n^{\text{th}}$$ bright fringe of $${\lambda }_1$$ match with the $$m^{\text{th}}$$ bright fringe of $${\lambda }_2$$. In this case, we have:

$$n{\beta }_1=m{\beta }_2$$

Substituting the expression for fringe width, we get:

$$n\frac{{\lambda }_{1}L}{d}=m\frac{{\lambda }_{2}L}{d}$$

Simplifying, we get:

$$\frac{n}{m}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{11}{14}$$

The least values of $$n$$ and $$m$$ that satisfy this condition are $$n=11$$ and $$m=14$$.

Now, let's find the position $$y$$ of the coincident bright fringe on the screen:

$$y=n{\beta }_1=n\frac{{\lambda }_{1}L}{d}=\frac{11\times 7\times 10^{-7}\mathrm{m}\times 1.5\mathrm{m}}{2.5\times 10^{-3}\mathrm{m}}$$

$$y=k\times 10^{-5}\mathrm{m}$$

Calculating the value of $$k$$:

$$k=\frac{11\times 7\times 10^{-7}\mathrm{m}\times 1.5\mathrm{m}}{2.5\times 10^{-3}\mathrm{m}\times 10^{-5}\mathrm{m}}=462$$

So, the least distance from the central fringe where the bright fringes due to both wavelengths coincide is $$462\times 10^{-5}\mathrm{m}$$.