22. 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) half
(b) one-fourth
(c) four times
(d) double
Answer : (c) four times
\( \beta = \dfrac{\lambda D}{d} \)
\( \beta_{2} = \dfrac{\lambda \, 2D}{d/2} \)
\( = \dfrac{\lambda D}{d} \cdot 4 \)
No comments:
Post a Comment
Please provide your valuable feedback. Students, Parents, Teachers.