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Wave Optics question

2020 · Q145
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Wave Optics question

2020 · Q145

NEETPhysicsWave OpticsMCQ+4 / −1
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 :
  1. A
    half
  2. B
    four times
  3. C
    one-fourth
  4. D
    double
View written solutionFree

Correct answer: B

Given that the slit distance is made half and the screen distance made double than the original value, then,
Fringe width, β=λDd\beta = {{\lambda D} \over d}β=dλD​
Now, d′=d2d' = {d \over 2}d′=2d​ and D′=2DD' = 2DD′=2D
So, β′=λ(2D)d/2=4λDd\beta ' = {{\lambda (2D)} \over {d/2}} = {{4\lambda D} \over d}β′=d/2λ(2D)​=d4λD​
β\beta β = 4β\beta β

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