In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 Ã…$ was…

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 Ã…$ was $1.2 \mathrm{~mm}$ for parallel narrow slits placed $2 \mathrm{~mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is
  1. $0.9 \mathrm{~mm}$
  2. $0.8 \mathrm{~mm}$
  3. $1.8 \mathrm{~mm}$
  4. $1.6 \mathrm{~mm}$

Solution

By Young's double slit interference experiment $ \beta=\frac{\lambda D}{d} $ The given $ \begin{aligned} & \beta_1=1.2 \mathrm{~mm} \\ & \frac{d_2}{d_1}=1 \frac{1}{2}=1.5 \end{aligned} $ So, $\quad \beta \propto \frac{1}{d}$ $ \begin{aligned} & \frac{\beta_1}{\beta_2}=\frac{1 / d_1}{1 / d_2} \\ & \frac{\beta_1}{\beta_2}=\frac{d_2}{d_1}=1.5 \Rightarrow \frac{1.2}{\beta_2}=1.5 \\ \Rightarrow \quad & \beta_2=\frac{1.2}{1.5}=\frac{4}{5}=0.8 \mathrm{~mm} \end{aligned} $

Asked in: AP EAMCET 2014

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