A beam of light consisting of two wavelengths $7000~\mathop A\limits^o$ and $5500~\mathop A\limits^o$ is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is $2.5 \mathrm{~mm}$ and the distance between the plane of slits and the screen is $150 \mathrm{~cm}$. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is $n \times 10^{-5} \mathrm{~m}$. The value of $n$ is __________.

<p>In Young&#39;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.</p> <p>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.</p> <p>First, let&#39;s convert the wavelengths and distances to meters:</p> <p>$\lambda_1 = 7 \times 10^{-7} \mathrm{~m}$ $\lambda_2 = 5.5 \times 10^{-7} \mathrm{~m}$ $d = 2.5 \times 10^{-3} \mathrm{~m}$ $L = 1.5 \mathrm{~m}$</p> <p>The fringe width $\beta$ for a single wavelength is given by:</p> <p>$\beta = \frac{\lambda D}{d}$</p> <p>Now, let&#39;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:</p> <p>$n \beta_1 = m \beta_2$</p> <p>Substituting the expression for fringe width, we get:</p> <p>$n \frac{\lambda_1 L}{d} = m \frac{\lambda_2 L}{d}$</p> <p>Simplifying, we get:</p> <p>$\frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{11}{14}$</p> <p>The least values of $n$ and $m$ that satisfy this condition are $n = 11$ and $m = 14$.</p> <p>Now, let&#39;s find the position $y$ of the coincident bright fringe on the screen:</p> <p>$$y = n \beta_1 = n \frac{\lambda_1 L}{d} = \frac{11 \times 7 \times 10^{-7} \mathrm{~m} \times 1.5 \mathrm{~m}}{2.5 \times 10^{-3} \mathrm{~m}}$$</p> <p>$y = k \times 10^{-5} \mathrm{~m}$</p> <p>Calculating the value of $k$:</p> <p>$$k = \frac{11 \times 7 \times 10^{-7} \mathrm{~m} \times 1.5 \mathrm{~m}}{2.5 \times 10^{-3} \mathrm{~m} \times 10^{-5} \mathrm{~m}} = 462$$</p> <p>So, the least distance from the central fringe where the bright fringes due to both wavelengths coincide is $462 \times 10^{-5} \mathrm{~m}$.</p>