Q149 / 160 · AP EAMCET 2021 (25 Aug Shift 2) · Physics

In a double slit experiment, when the distance between slits is increased 10 times, while their distance from the screen is halved, then the fringe width

  • Correct answer
    A
    becomes $\left(\frac{1}{20}\right)$ times the original
  • B
    becomes $\left(\frac{1}{90}\right)$ times the original
  • C
    remains the same
  • D
    becomes $\left(\frac{1}{10}\right)$ times the original

Solution

Case 1 Fringe width, $\beta_1=\frac{\lambda D_1}{d_1}$ Case 2 Fringe width, $\beta_2=\frac{\lambda D_2}{d_2}$ $\begin{array}{ll}\text { Given, } d_2= & 10 d_1, D_2=\frac{D_1}{2} \\ \Rightarrow & \frac{\beta_1}{\beta_2}=\frac{\lambda D_1}{d_1} \times \frac{d_2}{\lambda D_2} \\ \Rightarrow & \frac{\beta_1}{\beta_2}=\frac{D_1}{D_2} \times \frac{d_2}{d_1}=2 \times 10 \\ \Rightarrow & \frac{\beta_1}{\beta_2}=20 \\ \Rightarrow & \beta_2=\frac{1}{20} \beta_1\end{array}$

Hence, the correct answer is option: 'A'.

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