MHT CET · Physics · Wave Optics
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
- A halved.
- B unchanged.
- C reduced to \(\left(\frac{1}{3}\right)^{r d}\)
- D doubled.
Answer & Solution
Correct Answer
(B) unchanged.
Step-by-step Solution
Detailed explanation
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 \(2 D\) :
\(\begin{aligned} & X^{\prime}=\frac{\lambda(2 D)}{(2 d)} \\ & \therefore X=X^{\prime}\end{aligned}\)
\(X=\frac{\lambda D}{d}\)
The fringe width under new situation, slit width \(2 d\) and the distance between the slit and the eyepiece is \(2 D\) :
\(\begin{aligned} & X^{\prime}=\frac{\lambda(2 D)}{(2 d)} \\ & \therefore X=X^{\prime}\end{aligned}\)
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