In a Young's double slit experiment, a laser light of wave length $560 \mathrm{~nm}$ produces an…

In a Young's double slit experiment, a laser light of wave length $560 \mathrm{~nm}$ produces an interference pattern with consecutive bright fringe's separation of $7.2 \mathrm{~mm}$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of $8.1 \mathrm{~mm}$. The wavelength of second light is
  1. $680 \mathrm{~nm}$
  2. $630 \mathrm{~nm}$
  3. $650 \mathrm{~nm}$
  4. $540 \mathrm{~nm}$

Solution

Wave length of light, $\lambda=560 \mathrm{~nm}=560 \times 10^{-9} \mathrm{~m}$ Fringes separation, $ \begin{aligned} & \beta_1=7.2 \times 10^{-3} \mathrm{~m} \\ & \beta_2=8.1 \times 10^{-3} \mathrm{~m} \end{aligned} $ As we know $ \begin{aligned} & \because \beta=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2} \Rightarrow \lambda_2=\frac{\lambda_1 \beta_2}{\beta_1} \\ & =\frac{560 \times 10^{-9} \times 8.1 \times 10^{-3}}{7.2 \times 10^{-3}}=630 \mathrm{~nm} \end{aligned} $

Asked in: AP EAMCET 2023 (19 May Shift 1)

Practice more Wave Optics questions on Aicharya