In Young's double slit experiment, the slits are separated by 0.6 mm and screen is placed at a distance of 1…

In Young's double slit experiment, the slits are separated by 0.6 mm and screen is placed at a distance of 1.2 m from slit. It is observed that the tenth bright fringe is at a distance of 8.85 mm from the third dark fringe on the same side. The wavelength of light used is
  1. $5440 Å$
  2. $5890 Å$
  3. $5900 Å$
  4. $6630 Å$

Solution

For $\mathrm{n}^{\text {th }}$ bright fringe, $\mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}$ $\therefore \quad \mathrm{y}_{10}=\frac{10 \times \lambda \times 1.2}{0.6 \times 10^{-3}}=\left(20 \times 10^3\right) \lambda...(i)$
For $\mathrm{n}^{\text {th }}$ dark fringe, $y_n^{\prime}=\frac{(2 n-1) \lambda D}{2 d}$ $\therefore \quad y_3^{\prime}=\frac{5 \lambda \times 1.2}{2 \times\left(0.6 \times 10^{-3}\right)}=\left(5 \times 10^3\right) \lambda...(ii)$
Given, $\begin{aligned} & y_{10}-y_3^{\prime}=8.85 \times 10^{-3} \\ & \left(20 \times 10^3\right) \lambda-\left(5 \times 10^3\right) \lambda=8.85 \times 10^{-3} \end{aligned}$ ...[From(i) and (ii)] $\begin{aligned} & \lambda=\frac{8.85 \times 10^{-3}}{15 \times 10^3}=5.9 \times 10^{-7} \mathrm{~m} \\ & \lambda=5900 Å \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

Practice more Wave Optics questions on Aicharya