Q147 / 170 · Manipal (MET) 2016 · Physics

Two sources give interference pattern which is observed on a screen, placed at a distance $D$ from the sources. The fringe width is given as $2 W$. If the distance of sources from the screen is doubled, then the fringe width will

  • A
    become W/2
  • B
    remains the same
  • C
    become W
  • Correct answer
    D
    become 4 W

Solution

Given, $W_1=2 W$, Fringe width, $\quad W=\frac{\lambda D}{d}$ $\Rightarrow \quad 2 W=\frac{\lambda D}{d}$ When, $\quad D^{\prime}=2 D$ $W_2=\frac{\lambda D^{\prime}}{d}$ $\Rightarrow \quad W_2=\frac{\lambda 2 D}{d}$ $=2 \times 2 W=4 W$

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

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