Question

In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:
A double
B half
(c) four times
D one-fourth

Answer 1

Solution:
Fringe width, \(\beta=\frac{\lambda D}{d}\)
\(\beta^{\prime}=\frac{\lambda D}{d^{\prime}}\)
Now, \(d^{\prime}=\frac{d}{2}\) and
\(D^{\prime}=2 D\)
So, \(\beta^{\prime}=\frac{\lambda \times 2 D}{d / 2}\)
\(=\frac{4 \lambda D}{d}\)
\(\beta^{\prime}=4 \beta\)
Fringe width becomes 4 times

Answer 2

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Answer 3

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