In a biprism experiment, the distance between the two sources is doubled and the distance between the slit…

In a biprism experiment, the distance between the two sources is doubled and the distance between the slit and the eyepiece is also doubled. Then the width of the fringe is
  1. halved.
  2. unchanged.
  3. reduced to $\left(\frac{1}{3}\right)^{r d}$
  4. doubled.

Solution

The fringe width under initial condition, slit width $d$ and the distance between the slit and the eyepiece is $D$ : $X=\frac{\lambda D}{d}$ The fringe width under new situation, slit width $2 d$ and the distance between the slit and the eyepiece is $2 D$ : $\begin{aligned} & X^{\prime}=\frac{\lambda(2 D)}{(2 d)} \\ & \therefore X=X^{\prime}\end{aligned}$ .

Asked in: MHT CET 2022 (10 Aug Shift 1)

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