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NEET-XII-Physics

neet 2020 questions with solutions year:2020

with Solutions - page 3
Qstn# 30 Prvs-QstnNext-Qstn
  • #30
    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
    digAnsr:   C
    Ans : Fringe width, `` \beta=\frac{\lambda\,D}{d} ``
    `` \beta' = \frac{\lambda\,D'}{d'} ``
    Now, `` d'=\frac{d}{2} `` and
    `` D'=2\,D ``
    So, `` \beta'=\frac{\lambda \times 2 D}{d / 2} ``
    `` =\frac{4 \,\lambda\,D}{d} ``
    `` \beta'=4\,\beta ``
    Fringe width becomes 4 times