Q144 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Physics

If in a Young's double slit experiment the slit separation is doubled and the distance of the screen from the slits is reduced to half, then the fringe widths become how many times their original value?

  • A
    $\frac{1}{2}$
  • B
    2
  • Correct answer
    C
    $\frac{1}{4}$
  • D
    4

Solution

Fringe width in Young's double slit experiment is given as $\beta=\frac{D \lambda}{d}$ where, $D=$ distance between source and screen and $\quad d=$ slit width. $\begin{array}{ll} \therefore & \frac{\beta_2}{\beta_1}=\frac{d_1}{d_2} \times \frac{D_2}{D_1} \\ \Rightarrow & \frac{\beta_2}{\beta_1}=\frac{d_1}{2 d_1} \times \frac{D_1 / 2}{D_1} \quad\left[\because d_2=2 d_1 \text { and } D_2=\frac{D_1}{2}\right] \\ \Rightarrow & \frac{\beta_2}{\beta_1}=\frac{1}{4} \Rightarrow B_2=\frac{\beta_1}{4} \end{array}$

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

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