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 :

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NEET 2020 Official Paper (Held On: 13 September, 2020)
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  1. four times
  2. one−fourth
  3. double
  4. half

Answer (Detailed Solution Below)

Option 1 : four times
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Concept:

In Young’s double slit experiment two coherent sources of light are placed at a very small distance away from each other. It is used to understand wave theory of light.

Constructive and destructive interference of light is shown.

Calculation:

Fringe width  \(β = \frac{{\lambda D}}{d}\)

Now, \(d' = \frac{d}{2}\) and D' = 2D

So, \(β ' = \frac{{\lambda (2D)}}{{d/2}} = \frac{{4\lambda D}}{d}\)

β' = 4β

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