Question

White light can be assumed as combination of three wavelength $\lambda$, $\frac{3\lambda}{2}$ and $2\lambda$ (where $\lambda = 3800 \overset{\circ}{A}$). Distance of 2nd white fringe from central white fringe in modified YDSE is N mm, then 5N is ($d = 3.8$ mm, $D = 1$m)

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (C)

6

Answer 2

In Young's Double Slit Experiment (YDSE), white light consists of various wavelengths. The central fringe is always white because all wavelengths interfere constructively at the center (path difference = 0). For subsequent white fringes, we need to find positions where constructive interference occurs for all the specified wavelengths simultaneously.

Let the three wavelengths be $\lambda_1 = \lambda$, $\lambda_2 = \frac{3\lambda}{2}$, and $\lambda_3 = 2\lambda$. The position of the $n$-th order fringe for a wavelength $\lambda_i$ is given by $y_i = \frac{n_i \lambda_i D}{d}$, where $n_i$ is an integer. For a white fringe to form at a position $y$, the path difference $\Delta x = y \frac{d}{D}$ must be an integer multiple of each of the given wavelengths. So, $\Delta x = n_1 \lambda_1 = n_2 \lambda_2 = n_3 \lambda_3$, where $n_1, n_2, n_3$ are integers.

Substituting the given wavelengths: $\Delta x = n_1 \lambda = n_2 \left(\frac{3\lambda}{2}\right) = n_3 (2\lambda)$

This implies that $\Delta x$ must be a common multiple of $\lambda$, $\frac{3\lambda}{2}$, and $2\lambda$. The smallest non-zero common multiple corresponds to the first white fringe after the central one. We find the Least Common Multiple (LCM) of these wavelengths: LCM$(\lambda, \frac{3\lambda}{2}, 2\lambda)$ To find the LCM of fractions, we can write them with a common denominator: $\lambda = \frac{2\lambda}{2}$ $\frac{3\lambda}{2}$ $2\lambda = \frac{4\lambda}{2}$ LCM$(\frac{2\lambda}{2}, \frac{3\lambda}{2}, \frac{4\lambda}{2}) = \frac{\text{LCM}(2\lambda, 3\lambda, 4\lambda)}{\text{GCD}(2, 2, 2)}$ LCM$(2\lambda, 3\lambda, 4\lambda) = \lambda \times \text{LCM}(2, 3, 4) = \lambda \times 12 = 12\lambda$. GCD$(2, 2, 2) = 2$. So, the smallest non-zero path difference for a white fringe is $\frac{12\lambda}{2} = 6\lambda$.

The path differences for consecutive white fringes are $0, 6\lambda, 12\lambda, 18\lambda, \ldots$. The positions of these white fringes are given by $y_m = \frac{m \cdot 6\lambda D}{d}$, where $m = 0, 1, 2, \ldots$.

• Central white fringe ($m=0$): $y_0 = 0$.
• 1st white fringe ($m=1$): $y_1 = \frac{6\lambda D}{d}$.
• 2nd white fringe ($m=2$): $y_2 = \frac{12\lambda D}{d}$.

The question asks for the distance of the 2nd white fringe from the central white fringe, which is $y_2$. $y_2 = \frac{12\lambda D}{d}$

Given values: $\lambda = 3800 \overset{\circ}{A} = 3800 \times 10^{-10}$ m $d = 3.8$ mm $= 3.8 \times 10^{-3}$ m $D = 1$ m

Substitute these values into the expression for $y_2$: $y_2 = \frac{12 \times (3800 \times 10^{-10} \text{ m}) \times (1 \text{ m})}{3.8 \times 10^{-3} \text{ m}}$ $y_2 = \frac{12 \times (3.8 \times 10^3 \times 10^{-10}) \times 1}{3.8 \times 10^{-3}} \text{ m}$ $y_2 = \frac{12 \times 3.8 \times 10^{-7}}{3.8 \times 10^{-3}} \text{ m}$ $y_2 = 12 \times 10^{-7} \times 10^3 \text{ m}$ $y_2 = 12 \times 10^{-4}$ m

The distance is given as $N$ mm. Convert $y_2$ to millimeters: $N \text{ mm} = 12 \times 10^{-4} \text{ m} \times \frac{1000 \text{ mm}}{1 \text{ m}}$ $N = 12 \times 10^{-1}$ mm $N = 1.2$ mm

The question asks for the value of $5N$: $5N = 5 \times 1.2 = 6.0$