Question

In Young's double slit experiment, the '$\text{n}^{\text{th}}$' maximum of wavelength '$\lambda_1$' is at a distance '$y_1$' from the central maximum. When the wavelength of the source is changed to '$\lambda_2$', the $(2n)^{\text{th}}$ maximum is at a distance of '$y_2$' from its central maximum. The ratio $\frac{y_1}{y_2}$ is

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

Hint:

Fringe position is proportional to the product of the fringe order and the wavelength ($y \propto m\lambda$). For the first case, the product factor is $n\lambda_1$. For the second case, the product factor is $2n\lambda_2$. Taking their ratio yields $\frac{n\lambda_1}{2n\lambda_2} = \frac{\lambda_1}{2\lambda_2}$ instantly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem presents a Young's Double Slit Experiment (YDSE) setup where two different monochromatic light wavelengths are used. We need to find the ratio of the spatial positions of their respective bright fringe maxima from the central zero-order reference fringe.

Step 2: Key Formula or Approach:
The linear distance $y$ of the $m^{\text{th}}$ bright fringe maximum from the central maximum in a standard YDSE layout is given by: $$ y = \frac{m \cdot \lambda \cdot D}{d} $$ where $m$ is the order of the maximum, $\lambda$ is the wavelength of light, $D$ is the distance from the slits to the screen, and $d$ is the distance between the two slits.

Step 3: Detailed Explanation:
Let's express the positions for both given conditions using our fringe formula:

• For the first condition (order $m = n$, wavelength $\lambda = \lambda_1$): $$ y_1 = \frac{n \cdot \lambda_1 \cdot D}{d} $$

• For the second condition (order $m = 2n$, wavelength $\lambda = \lambda_2$): $$ y_2 = \frac{2n \cdot \lambda_2 \cdot D}{d} $$
Now, let's divide $y_1$ by $y_2$ to evaluate their structural ratio: $$ \frac{y_1}{y_2} = \frac{\frac{n \cdot \lambda_1 \cdot D}{d}}{\frac{2n \cdot \lambda_2 \cdot D}{d}} $$ The geometric setup factors $\frac{D}{d}$ and the order integer $n$ cancel out perfectly from both the numerator and denominator: $$ \frac{y_1}{y_2} = \frac{\lambda_1}{2\lambda_2} $$ *(Note: If evaluating via a typo alternative layout where the order is halved to $\frac{n}{2}$, the factor flips to inverse values, but following the direct text parameters yields the canonical form $\frac{\lambda_1}{2\lambda_2}$.)*

Step 4: Final Answer:
The ratio of the distances $\frac{y_1}{y_2}$ matches $\frac{\lambda_1}{2\lambda_2}$, which corresponds to option (D).