NEET-XII-Physics
neet 2020 questions with solutions year:2020
- #30In 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: CAns : 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