Question:medium
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
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
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Physics Tip : In interference patterns, higher-order fringes or longer wavelengths result in greater distances from the central fringe.
Updated On: Apr 23, 2026
- $\frac{3\lambda_{1}}{4\lambda_{2}}$
- $\frac{3\lambda_{2}}{4\lambda_{1}}$
- $\frac{4\lambda_{1}}{3\lambda_{2}}$
- $\frac{4\lambda_{2}}{3\lambda_{1}}$
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The Correct Option is B
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