Question 207: Waves Optics

Question

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 $$n 	imes 10^{-5} \mathrm{~m}$$. The value of $$n$$ is __________.

Accepted Answer

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:
$$\lambda_1 = 7 	imes 10^{-7} \mathrm{~m}$$
$$\lambda_2 = 5.5 	imes 10^{-7} \mathrm{~m}$$
$$d = 2.5 	imes 10^{-3} \mathrm{~m}$$
$$L = 1.5 \mathrm{~m}$$
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 $$n^{	ext{th}}$$ bright fringe of $$\lambda_1$$ match with the $$m^{	ext{th}}$$ 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:
$$y = n \beta_1 = n \frac{\lambda_1 L}{d} = \frac{11 	imes 7 	imes 10^{-7} \mathrm{~m} 	imes 1.5 \mathrm{~m}}{2.5 	imes 10^{-3} \mathrm{~m}}$$
$$y = k 	imes 10^{-5} \mathrm{~m}$$
Calculating the value of $$k$$:
$$k = \frac{11 	imes 7 	imes 10^{-7} \mathrm{~m} 	imes 1.5 \mathrm{~m}}{2.5 	imes 10^{-3} \mathrm{~m} 	imes 10^{-5} \mathrm{~m}} = 462$$
So, the least distance from the central fringe where the bright fringes due to both wavelengths coincide is $$462 	imes 10^{-5} \mathrm{~m}$$.

Waves Optics

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