QJEEM23Phc9ab: JEE Main 2023 Physics - Class 12 JEEM23Phc9ab Solution: A beam of light consisting of two wavelengths and is used to…

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 $\lambda_1 = 7000~\mathop A\limits^o$ and $\lambda_2 = 5500~\mathop A\limits^o$ . 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 $462 \times 10^{-5} \mathrm{~m}$ .