Optics 6 Question 30
32. In Young’s double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is
$(1981,2 M)$
(a) unchanged
(b) halved
(c) doubled
(d) quadrupled
Match the Column
Show Answer
Solution:
- $\omega=\frac{\lambda D}{d}$
$d$ is halved and $D$ is doubled
$\therefore$ Fringe width $\omega$ will become four times.
$\therefore$ Correct option is $(d)$.
