Question

In Young's double slit experiment, the $8^{\text{th}}$ maximum with wavelength $\lambda_1$ is at a distance $d_1$ from the central maximum and the $6^{\text{th}}$ maximum with wavelength $\lambda_2$ is at a distance $d_2$. Then $\frac{d_2}{d_1}$ is

• A. $\frac{3\lambda_{1}}{4\lambda_{2}}$
• B. $\frac{3\lambda_{2}}{4\lambda_{1}}$
• C. $\frac{4\lambda_{1}}{3\lambda_{2}}$
• D. $\frac{4\lambda_{2}}{3\lambda_{1}}$

Hint:

Physics Tip : In interference patterns, higher-order fringes or longer wavelengths result in greater distances from the central fringe.

Answer 1

The Correct Option is B

Concept: Physics (Optics) - Young's Double Slit Experiment (YDSE).

Step 1: State the formula for the distance of the $n^{th}$ maximum. The distance of the $n^{th}$ bright fringe (maximum) from the central maximum is given by: $$d = \frac{n \lambda D}{d_{slit}}$$ where $n$ is the order, $\lambda$ is the wavelength, $D$ is the screen distance, and $d_{slit}$ is the slit separation. Since $D$ and $d_{slit}$ are constant for the same setup, $d \propto n\lambda$.

Step 2: Set up the ratio for the two wavelengths. For the first case ($8^{th}$ maximum, $\lambda_1$): $d_1 \propto 8\lambda_1$. For the second case ($6^{th}$ maximum, $\lambda_2$): $d_2 \propto 6\lambda_2$.

Step 3: Calculate the final ratio. $$\frac{d_{2}}{d_{1}} = \frac{n_{2}\lambda_{2}}{n_{1}\lambda_{1}} = \frac{6\lambda_{2}}{8\lambda_{1}}$$ Simplifying the fraction: $$\frac{d_{2}}{d_{1}} = \frac{3\lambda_{2}}{4\lambda_{1}}$$ $$ \therefore \text{The ratio } \frac{d_{2}}{d_{1}} \text{ is } \frac{3\lambda_{2}}{4\lambda_{1}}. \text{} $$