Question

In Young's double slit experiment, the $10^{\text{th}}$ maximum of wavelength '$\lambda_1$' is at a distance of '$Y_1$' from the central maximum. When the wavelength of the source is changed to '$\lambda_2$', the $5^{\text{th}}$ maximum is at a distance '$Y_2$' from the central maximum. The ratio $\frac{Y_1}{Y_2}$ is

• A. $\frac{2\lambda_1}{\lambda_2}$
• B. $\frac{\lambda_2}{2\lambda_1}$
• C. $\frac{2\lambda_2}{\lambda_1}$
• D. $\frac{\lambda_1}{2\lambda_2}$

Hint:

You can solve this very quickly by setting up a simple scale balance: $Y_1 / Y_2 = (n_1 / n_2) \times (\lambda_1 / \lambda_2)$. Since the first order is 10 and the second order is 5, the order ratio is exactly $10 / 5 = 2$. Multiplying this by the wavelength ratio gives $\frac{2\lambda_1}{\lambda_2}$ in a single thought step!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question relates to interference fringe positions in Young's Double Slit Experiment (YDSE). We are given the linear positions of two different bright fringes produced by two distinct wavelengths of light, and we need to determine the algebraic ratio of these positions, $\frac{Y_1}{Y_2}$.

Step 2: Key Formula or Approach:
The linear distance ($Y_n$) of the $n^{\text{th}}$ bright fringe (maximum) from the central principal maximum in a standard YDSE setup is given by the formula: $$Y_n = \frac{n \cdot \lambda \cdot D}{d}$$ Where $D$ is the distance between the slits and the screen, and $d$ is the separation distance between the two slits. Since the physical dimensions of the apparatus ($D$ and $d$) remain unchanged, the position is directly proportional to the product of the fringe order and the wavelength: $$Y \propto n \cdot \lambda$$

Step 3: Detailed Explanation:
Let's express the linear positions for both cases using our proportional relationship: 1. For the first configuration: fringe order $n_1 = 10$ and wavelength $\lambda = \lambda_1$. $$Y_1 = \frac{10 \cdot \lambda_1 \cdot D}{d}$$ 2. For the second configuration: fringe order $n_2 = 5$ and wavelength $\lambda = \lambda_2$. $$Y_2 = \frac{5 \cdot \lambda_2 \cdot D}{d}$$ Now, divide the first position expression by the second to find the required ratio: $$\frac{Y_1}{Y_2} = \frac{\frac{10 \cdot \lambda_1 \cdot D}{d}}{\frac{5 \cdot \lambda_2 \cdot D}{d}}$$ The geometric constants $\frac{D}{d}$ cancel out perfectly: $$\frac{Y_1}{Y_2} = \frac{10\lambda_1}{5\lambda_2}$$ Simplifying the numerical fraction $\frac{10}{5} = 2$ gives: $$\frac{Y_1}{Y_2} = \frac{2\lambda_1}{\lambda_2}$$

Step 4: Final Answer:
The ratio of the distances is $\frac{2\lambda_1}{\lambda_2}$, which corresponds to option (A).