In Young's double slit experiment, the $6^{\text {th }}$ maximum with wavelength ${ }^{\prime} \lambda_{1}{…

In Young's double slit experiment, the $6^{\text {th }}$ maximum with wavelength ${ }^{\prime} \lambda_{1}{ }^{\prime}$ is at a distance ${ }^{\prime} \mathrm{d}_{1}{ }^{\prime}$ from the central maximum and the $4^{\text {th }}$ maximum with wavelength $\lambda_{2}$ is at distance $\mathrm{d}_{2}$. Then $\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}$ is
  1. $\frac{2}{3} \frac{\lambda_{1}}{\lambda_{2}}$
  2. $\frac{3}{2} \frac{\lambda_{1}}{\lambda_{2}}$
  3. $\frac{2}{3} \frac{\lambda_{2}}{\lambda_{1}}$
  4. $\frac{3}{2} \frac{\lambda_{2}}{\lambda_{1}}$

Solution

$\mathrm{d}_{1}=\frac{6 \lambda_{1} \mathrm{D}}{\mathrm{d}} \quad$ and $\quad \mathrm{d}_{2}=\frac{4 \lambda_{2} \mathrm{D}}{\mathrm{d}}$ $\therefore \frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}=\frac{3}{2} \frac{\lambda_{1}}{\lambda_{2}}$ ^

Asked in: MHT CET 2020 (13 Oct Shift 1)

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